% The following TeXfile comprises part of

%     MEASURE THEORY, by D.H.Fremlin
%          (University of Essex, Colchester, England;
%                   e-mail  fremdh@essex.ac.uk)

% Please contact me with any difficulties not covered by the remarks
% below.

% The file is set up for processing by plain TeX in the presence of
% files  amstex.tex ,  amssym.tex  and  amsppt.sty  suitably supported by the
% appropriate fonts.   If these are unavailable, a readable version
% can be obtained by removing the comment symbol  %  from the line

%        %   \amstexsupportedfalse

% below.   Some systems do not allow access to files  amstex.tex  and
% amsppt.sty , but instead employ a command "amstex <filename.tex>";
% for such systems it is sometimes right to delete or disable the
% commands

%        \input amstex
%        \documentstyle{amsppt}

% (I have little experience with this and should welcome comments.)

% Various formats of output are obtainable.   The principal variations
% are

%    1. "Standard" output uses the command
%               \smallprint
% and gives output adapted to A4 paper (21 cm x 29.7 cm).

%    2.  Using the command
%               \USsmallprint
% gives output adapted to 8.5" x 11" .

%    3.  Using the command
%               \noproofs
% produces an abridged version, in which proofs are omitted.

%    4.   Using the command
%               \resultsonly
% produces a super-abridged version, omitting comments and exercises
% as well as proofs.

% These commands are grouped together at the end of the introductory
% definitions;  they may be activated by deleting the initial comment
% symbols  % .

% Any diagrams included will be called for by inputting postscript files.
% (You can look for these by hunting for   \picture  in the code below.)
% To get these you will need the relevant  .ps  files with the expected
% names  ( \picture{mt133g}{...}  demands the file  mt133g.ps , for instance).


%(If you can read TeX fluently, you may wish to fast-forward to a line
%beginning   %file starts here   or   \frfilename
%if one of these is present.)

%TeXfiles with this heading are designed for processing with AmSTeX,
%accessed through TeXfiles  amstex.tex (with amssym.tex  and
% amsppt.sty  and possibly  amsppt.sti)  during compilation.   
%If these are unavailable, a readable
%version can (I hope) be produced by deleting the  %  before
% \amstexsupportedfalse  below.

%Some special difficulties arise with the importation of  .ps  files
%in diagrams.   Look for  \picture  below for notes on these.

%The following \count registers are used for special purposes:
%   \count1=\stylenumber  to specify style
%   \count2=\footnotenumber  to count footnotes

%AmSTeX declaration
\newif\ifamstexsupported
     \amstexsupportedtrue
%     \amstexsupportedfalse               %use when AMSTeX unavailable
\newif\ifDHFamstexloaded        %for detection of a non-standard AMSTeX
     \DHFamstexloadedfalse    
\newif\ifDHFamspptloaded        
     \DHFamspptloadedfalse    
\newif\iffourteensavailable
     \fourteensavailabletrue
    \fourteensavailablefalse
\newif\ifathome
     \athometrue     %to cope with a quirk     of my son's Ghostscript
     \athomefalse
\newif\iffirstpage
     \firstpagetrue

%Style assignments
\countdef\stylenumber=1
\countdef\footnotenumber=2
\def\fullsize{\magnification=\magstep1
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=1}
\def\smallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=2}
\def\printersize{\magnification=\magstep1
     \baselineskip24pt
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{8.5truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=3}
\def\refsize{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=4}
\def\rrversion{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=5}
\def\bigsize{\magnification=\magstep2
     \overfullrule0pt
     \def\pagewidth{325pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=6}
\def\bbigsize{\magnification=\magstep3
     \overfullrule0pt
     \def\pagewidth{271pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=7}
\def\USsmallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{482pt}
     \def\pageheight{8.7truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=8}
\def\halfmagsize{\magnification=\magstephalf
     \def\pagewidth{427pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=9}
           % register  stylenumber  is used to control the output format

%addresses, letterheads
\def\deptaddress{\hskip 3truein Department of Mathematical Sciences\par
   \noindent e-mail: {\tt fremdh}\hskip 2.9truein\today\par
   \medskip}
\long\def\univaddress#1{\hskip 3truein Department of Mathematical Sciences\par
   \hskip 3truein University of Essex\par
   \hskip 3truein Colchester CO4 3SQ\par
   \hskip 3truein England\par
   %\hskip 3truein van Vleck Hall\par
   %\hskip 3truein 480 Lincoln Drive\par
   %\hskip 3truein Madison\par
   %\hskip 3truein WI 53706-1388, U.S.A.\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
       %{\tt fremlin\@math.wisc.edu}
   \bigskip}
\long\def\homeaddress#1{\hskip 3truein 25 Ireton Road\par
   \hskip 3truein Colchester CO3 3AT\par
   \hskip 3truein England\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
   \bigskip}
\long\def\letterhead#1#2#3{\univaddress{#1}
   {\parindent=0pt
    #2
   }
   \medskip

   \noindent Dear #3,
     }
\def\shortaddress{\noindent Departmentof Mathematical Sciences, University of Essex,
   Colchester CO4 3SQ, England\par
   \noindent e-mail: david\@fremlin.org}
\def\valediction#1#2{\bigskip
   \hskip 3truein #1,
   \vskip 1truein
   \hskip 3.3truein #2
   \bigskip}

%picture importation
   %It is clear that there is no general standard.   The syntax I
   %use when calling for a  .ps  file is
   %    \def\Caption{<...>}
   %    \picture{<filename>}{<dim>}
   %which is supposed to import  <filename>.ps  at height  <dim> ,
   %centered, and adding a one-line centered caption  <...> .
   %The basic macro listed in  \def\picture  is that which
   %used to work at Essex.   I now generally use the macro in
   % \def\picturewithpsfig , which requires a previous input
   % psfig.tex  or  psfig.sty ;
   %if you interpolate  \atUEssex  early in your file it will
   %give you the latter.
   %   I find that  psfig  is erratic in its centering, especially
   %under magnification, so I have written a routine
   % \centerpicturewithpsfig  (which may be called from
   % \centerpicture ) with the syntax
   %    \centerpicture{<filename>}{<dim>}{<dim>}
   %the first dimension being width and the second height.

   % A second package is called for by  \usegraphicx .   This requires
   % miniltx.tex  and  graphicx.sty ;  miniltx.tex  is currently (April 2016)
   % available from
   % http://mirror.ox.ac.uk/sites/ctan.org/macros/plain/graphics/miniltx.tex
   % The advantage of this is that it seems to allow manipulation of the
   % dvi>pdf stage using  dvipdfmx .

\def\picture#1#2{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at height #2
         with dvitops}
  \vbox{\vskip #2
    \special{dvitops: import #1.ps \pagewidth #2}
    \Centerline{\Caption}
    }
   \medskip
   \def\Caption{}}
%an alternative version, without caption
\def\centerpicture#1#2#3{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with dvitops}
  \vbox{\vskip #3
    \special{dvitops: import #1.ps \pagewidth #3}
  }}
%to put pictures on same line
\def\sideshiftedpicture#1#2#3#4{\bigskip
   \bigskip
   \immediate\write0{importing diagram #1.ps at height #4
        and width #3 with dvitops}
   \vbox{\special{dvitops:  import #1.ps #3 #4}}
   \hskip #2 plus 0pt minus 0pt
   }
\def\startsideshiftedpicture{}
\newcount\scaleup
\newcount\scaledown
\scaleup=1
\scaledown=1
\def\scalesideshiftedpicture#1#2#3#4{
  \dimen4=#4
  \dimen3=#3
  \multiply\dimen4 by\scaleup
  \divide\dimen4 by \scaledown
  \multiply\dimen3 by\scaleup
  \divide\dimen3 by \scaledown
  \sideshiftedpicture{#1}{#2}{\dimen3}{\dimen4}}

\def\picturewithpsfig#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with psfig}
   \vbox{\hsize\pagewidth
   \centerline{\psfig{figure=#1.ps,height=#2}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithpsfig#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with psfig}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\psfig{file=#1.ps,width=\dimen3,height=\dimen4}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithpsfig#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with psfig}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\psfig{figure=#1.ps,height=\dimen4,width=\dimen3}}
   \psfig{figure=#1.ps,height=\dimen4,width=\dimen3}
   \hskip #2 plus 0pt minus 0pt}

\def\picturewithgraphicx#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with graphicx}
   \vbox{\hsize\pagewidth
   \centerline{\includegraphics[height=#2]{#1.ps}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithgraphicx#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with graphicx}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\includegraphics[width=\dimen3,height=\dimen4]{#1.ps}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithgraphicx#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with graphicx}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\includegraphics[height=\dimen4,width=\dimen3]{#1.ps}}
   \includegraphics[height=\dimen4,width=\dimen3]{#1.ps}
   \hskip #2 plus 0pt minus 0pt}

%alternative picture importation code;  I've been told about this
  %but have not used it successfully myself
%\def\Caption{}
%\input epsf
%\epsfverbosetrue
%\def\picture#1#2{\bigskip
%    \bigskip
%    \immediate\write0{importing diagram #1.ps at height #2 with epsf}
%    \epsfysize=#2
%    \centerline{\epsffile{#1.ps}}
%    \medskip
%    \Centerline{\Caption}
%    \medskip
%    }

%in addition it may be necessary to amend the imported  .ps  file
%by replacing the   %%BoundingBox   header by  %%BoundingBox:

\def\atUEssex{
   \input psfig.sty
   \def\picture##1##2{\picturewithpsfig{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithpsfig{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithpsfig{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
   }
\def\usegraphicx{
   \input miniltx
   \def\Gin@driver{dvips.def}
   \input graphicx.sty
   \def\picture##1##2{\picturewithgraphicx{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithgraphicx{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithgraphicx{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
}


%special fonts
\def\biggerscriptfonts{\scriptfont0=\eightrm
     \scriptfont1=\eighti
     \scriptfont2=\eightsy
     \scriptfont3=\eightex
     \scriptfont\bffam=\eightbf
     \scriptfont\msbfam=\eightmsb
     \scriptfont\eufmfam=\eighteufm
     }
\def\Loadcyrillic{\ifamstexsupported
      \input cyracc.def
      \font\tencyr=wncyr10
      \def\cyr{\tencyr\cyracc} \def\itcyr{\tenitcyr\cyracc}
    \else\def\cyr{\smc}\def\cdprime{u\i}\def\Cdprime{UI}\fi}
%http://wkweb3.cableinet.co.uk/linux/LDP/HOWTO/Cyrillic-HOWTO-7.html
%cyrillic.tex

\def\Loadfourteens{\iffourteensavailable
       \font\fourteenrm=cmr14
       \font\fourteenbf=cmbx14
       \font\fourteenit=cmsl14
   \else\font\fourteenrm=cmr10 scaled \magstep2
      \font\fourteenbf=cmbx10 scaled \magstep2
      \font\fourteenit=cmsl10 scaled \magstep2
   \fi}
\def\smallerfonts{\eightpoint
   \textfont\msbfam=\eightmsb
   \textfont\eufmfam=\eighteufm
   \textfont\bffam=\eightbf
   \textfont\eurmfam=\eighteurm
   \textfont\eusmfam=\eighteusm
   }
\font\twelveex=cmex10 scaled \magstep1


%Copyright declarations
\newcount\oldcopyrightdate
\newcount\newcopyrightdate
\oldcopyrightdate=0
\def\copyrightdate#1{\newcopyrightdate=#1
       \ifnum\newcopyrightdate=\oldcopyrightdate\else
       \oldcopyrightdate=#1
       \oldfootnote{}{\eightrm \copyright\ #1 D.~H.~Fremlin}
       \immediate\write0{Copyright #1}
       \fi}

%hyphenation
\hyphenation{equi-ver-id-ical}

%miscellaneous definitions
\def\Aut{\mathop{\text{Aut}}}
\def\backstep#1{\hskip -0.#1em plus 0 pt minus 0 pt}
\def\Bang{\vthsp\hbox{\bf X\hskip0.07em\llap X\hskip0.07em\llap X\rm}}
\def\BanG{\Bang{\hskip 1.111pt plus 3.333pt minus 0pt}}
%\def\BbbN{\mathchoice{\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb N$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb N$\hskip0.04em}}}
%\def\BbbQ{\mathchoice{\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Q$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Q$\hskip0.04em}}}
%\def\BbbR{\mathchoice{\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb R$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb R$\hskip0.04em}}}
%  %do not use with $\Bbb R^X$
%\def\BbbZ{\mathchoice{\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Z$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Z$\hskip0.04em}}}
\def\BbbN{\openBbb{N}}
\def\BbbQ{\openBbb{Q}}
\def\BbbR{\openBbb{R}}
  %do not use with $\Bbb R^X$  do not use in $_{\Bbb R^r}$
\def\BbbZ{\openBbb{Z}}
\def\Bbbone{\hbox{1\hskip0.13em\llap{1}}}
\def\biggerint{\textfont3=\twelveex\intop\nolimits}
  % you can get this size into a displayed formula with
  %  $$\hbox{$\biggerint_{\Cal S}\pmb{u}\,d\pmb{v}$}$$
  % but it may then seem cramped
  % or try \hbox{$\biggerint_{\Cal S}\,$}
\def\biggeriint{\textfont3=\twelveex\iint}
  % remarkable fact:  this seems to infect all \int within a given
  % $..$ ; but  $\biggerint$$\int$ seems to leave the $\int$ normal
  % really  \textfont3=\twelveex  is more reliable, since  \biggerint
  % as written gets turned off rather easily (e.g. by  \comment,
  % or by an extra  {...})
\def\bover#1#2{{{#1}\over{#2}}}
\def\Bover#1#2{\backstep4\hbox{
     \biggerscriptfonts${\shortstrut#1}\over{\shortstrut#2}$}}
  % when there are superscripts in the denominator, as in \Bover1{2^m}
  % or  \Bover1{\sqrt{2\pi}^r} , these can sometimes come out with better
  % horizontal alignment if you use  \Bover1{{\sshortstrut 2}^m}  or
  % \Bover1{{\sqrt{2\pi}}^r}
\def\bsover#1#2{\textstyle\bover{#1}{#2}}
\def\bsp{/\penalty-100}
  % \penalty seems to behave oddly in a \tt string in \fullsize mode
\def\Caption{}
\def\Center#1{

   \Centerline{#1}

   \noindent}
\def\Centerline#1{\vskip 5pt
     {\def\int{\text{}$$\textfont3=\twelveex\intop\nolimits}
      \def\dashint{\text{}$$\textfont3=\twelveex
         \mskip2mu\vrule height2.65pt width4.5pt depth-2.35pt\mskip-11mu
	 \intop\nolimits\mskip-2mu}
      \def\idashint{\text{}$$\int\mskip-7mu\dashint}	 
      \def\dashiint{\text{}$$\dashint\mskip-6mu\int}	 
      \def\dashidashint{\text{}$$\dashint\mskip-6mu\dashint}	 
      \def\underlineint{
           \text{}$$\textfont3=\twelveex\underline{\intop\nolimits}}
      \def\overlineint{
           \text{}$$\textfont3=\twelveex\overline{\intop\nolimits}}
      \def\iint{\text{}$$\textfont3=\twelveex
           \intop\nolimits\hskip-0.5em\intop\nolimits}
      \def\sum{\text{}$$\sumop}
      \def\prod{\text{}$$\prodop}
      \def\bigcap{\text{}$$\bigcapop}
      \def\bigcup{\text{}$$\bigcupop}
   \centerline{#1}}\vskip 5pt}
   \mathchardef\bigcapop="1354
   \mathchardef\bigcupop="1353
   \mathchardef\prodop="1351
   \mathchardef\sumop="1350
   %in  \Centerline{$\overline{}$, $\cmmnt{}$} must use \sumop etc
\def\cf{\mathop{\text{cf}}}
\def\circumflex#1{{\accent"5E #1}}
\def\closeplus{\hbox{+}}
\long\def\cmmnt#1{\ifresultsonly\else{#1}\fi}
\def\coint#1{\left[#1\right[}
\def\curl{\mathop{\text{curl}}}
\def\dashint{\mathchoice{-\mskip-20mu\intop\nolimits}
   {{\vrule height2.75pt width4.5pt depth-2.45pt\mskip-10mu\intop\nolimits}}
      {-\mskip-16.5mu\intop\nolimits} %not yet used
      {-\mskip-16.5mu\intop\nolimits}} %not yet used
\def\idashint{\mathchoice{\int\mskip-6mu\dashint}
      {\int\mskip-5mu\dashint}
      {\int\mskip-5mu\dashint} %not yet used
      {\int\mskip-5mu\dashint}} %not yet used
\def\dashiint{\mathchoice{\dashint\mskip-10mu\int}
      {\dashint\mskip-6mu\int}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashidashint{\mathchoice{\dashint\mskip-10mu\dashint}
      {\dashint\mskip-6mu\dashint}
      {\dashint\mskip-6mu\dashint} %not yet used
      {\dashint\mskip-6mu\dashint}} %not yet used
\def\dashidashiint{\mathchoice{\dashint\mskip-8mu\dashint\mskip-10mu\int}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashiidashint{\mathchoice{\dashint\mskip-10mu\int\mskip-8mu\dashint}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\idashidashint{\mathchoice{\int\mskip-8mu\dashint\mskip-8mu\dashint}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\diam{\mathop{\text{diam}}}
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
           \else\ifnum\stylenumber=3{\frnewpage}
           \else\ifnum\stylenumber=5{\frnewpage}
           \else\bigskip\fi\fi\fi}
\long\def\discrversionA#1#2{\ifnum\stylenumber=1#1\else#2\fi}
\long\def\discrversionB#1#2{\ifnum\stylenumber=2#1\else#2\fi}
\long\def\discrversionC#1#2{\ifnum\stylenumber=3#1\else#2\fi}
\long\def\discrversionD#1#2{\ifnum\stylenumber=4#1\else#2\fi}
\long\def\discrversionE#1#2{\ifnum\stylenumber=5#1\else#2\fi}
\long\def\discrversionF#1#2{\ifnum\stylenumber=6#1\else#2\fi}
\long\def\discrversionG#1#2{\ifnum\stylenumber=7#1\else#2\fi}
\long\def\discrversionH#1#2{\ifnum\stylenumber=8#1\else#2\fi}
           % facility for special interpolations in particular output
           % versions
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\diverg{\mathop{\text{div}}}
\def\dom{\mathop{\text{dom}}}
\def\eae{=_{\text{a.e.}}}

\def\esssup{\mathop{\text{ess sup}}}
\mathchardef\Exists="3239

\def\family#1#2#3{\langle#3\rangle_{#1\in#2}}
\def\familyi#1#2{\family{i}{#1}{#2}}
\def\familyiI#1{\familyi{I}{#1}}

\def\frfilename#1{\ifnum\stylenumber=1{\noindent filename #1}
               \else\ifnum\stylenumber=6{\noindent{\fiverm filename #1}}
               \else\ifnum\stylenumber=7{\noindent{\fiverm filename #1}}
               \else\fi\fi\fi}
\def\Footnote#1#2{\footnote"#1"{#2}}
\mathchardef\Forall="3238
\def\frakc{\frak c\phantom{.}}
\def\frnewpage{\vskip \pageheight plus 0pt minus\pageheight
           \eject
           \gdef\topparagraph{}
           \gdef\bottomparagraph{}
           }
\def\frsmallcirc{{\hskip 1pt plus 0pt minus 0pt}
     {\raise 1pt\hbox{$\smallcirc$}}{\hskip 1pt plus 0pt minus 0pt}}

\def\geae{\ge_{\text{a.e.}}}
\def\grad{\mathop{\text{grad}}}
\def\grheada{{\bf ($\pmb{\alpha}$)}}
\def\grheadb{{\bf ($\pmb{\beta}$)}}
\def\grheadc{{\bf ($\pmb{\gamma}$)}}
\def\grheadd{{\bf ($\pmb{\delta}$)}}
\def\grheade{{\bf ($\pmb{\epsilon}$)}}
\def\grheadz{{\bf ($\pmb{\zeta}$)}}
\def\grv#1{{\accent"12 #1}}

\def\hfillall{\hskip\pagewidth plus0pt minus\pagewidth}

\def\Imag{\mathop{\Cal I\text{m}}}
\long\def\inset#1{{\narrower\narrower{\vskip 1pt plus 1pt minus 0pt}
     #1
     {\vskip 1pt plus 1pt minus 0pt}}}
\def\Int#1#2{\displaystyle{\int_{#1}^{#2}}}
\def\interior{\mathop{\text{int}}}

\def\leae{\le_{\text{a.e.}}}
\long\def\leaveitout#1{}
\def\LLcorner{\mathchoice{\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height5pt width0.3pt depth0pt
       \vrule height0.3pt width4pt depth0pt}\,}
   {\,\hbox{\vrule height4pt width0.25pt depth0pt
       \vrule height0.25pt width3pt depth0pt}\,}
  }
\def\Ln{\mathop{\text{Ln}}}
\def\Loadfont#1{\font#1}
\def\loibr{\hbox{\hskip0.1em]\hskip0.05em}}
\def\Matrix#1{\pmatrix#1\endpmatrix}
\def\Mu{\text{M}}
\def\noamslogo{\def\amslogo{}}
\def\normalsubgroup{\vartriangleleft}
\def\Nu{\text{N}}
\def\ocint#1{\left]#1\right]}
\def\ofamily#1#2#3{\langle#3\rangle_{#1<#2}}
\let\oldcases=\cases
\let\oldfootnote=\footnote
\def\On{\mathop{\text{On}}}
\def\ooint#1{\left]#1\right[}
\def\openBbb#1{\Bbb#1\mskip1mu}
\def\pd#1#2{\bover{\partial#1}{\partial#2}}
\def\Pd#1#2{\Bover{\partial#1}{\partial#2}}
\def\pdd#1#2{{{\partial^2#1}\over{\partial#2^2}}}
\def\Pdd#1#2{\Bover{\partial^2#1}{\partial#2^2}}
\def\pde#1#2#3{{{\partial^2#1}\over{\partial#2\partial#3}}}
\def\Pde#1#2#3{\Bover{\partial^2#1}{\partial#2\partial#3}}
\def\Prf{\vthsp\hbox{\bf P\hskip0.07em\llap P\hskip0.07em\llap P\rm}}
\def\pushbottom#1{\hbox{\vrule height0pt depth#1 width0pt}}
\def\Qed{\vthsp\hbox{\bf Q\hskip0.07em\llap Q\hskip0.07em\llap Q\rm}}
\def\QeD{\Qed{\hskip 1.111pt plus 3.333pt minus 0pt}}
\def\Quer{\vthsp\hbox{\bf ?\hskip0.05em\llap ?\hskip0.05em\llap
   ?\thinspace\rm}}
\def\query{\discrversionA{\immediate\write0{query}
    \global\advance\footnotenumber by 1
    \oldfootnote{$^{\the\footnotenumber}$}{query}}{}}
\def\Real{\mathop{\Cal R\text{e}}}
\def\restr{\mathchoice
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\scriptstyle\restriction$}\mskip 1mu}
  {\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
\def\restrp{\mathchoice
  {\hbox{$\restriction$}\mskip 2mu}
  {\hbox{$\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptstyle\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
    %use in \restrp V \restrp W \restrp Y
    % \restrp\Cal P \restrp\Bbb R \restrp\frak
    % \restrp\Omega \restrp\Tau \restrp\Upsilon
    % \restrp[ \restrp\widehat{}
\def\Rho{\text{P}}
\def\roibr{\hbox{\hskip0.05em[\hskip0.1em}}
\def\RoverC{\hbox{\biggerscriptfonts${{\Bbb R}\atop{\Bbb C}}$}}
\def\sequence#1#2{\family{#1}{\Bbb N}{#2}}
\def\sequencen#1{\sequence{n}{#1}}
\def\setover#1#2{\overset{#1}\to{#2}}
\def\shortstrut{\hbox{\vrule height7.5pt depth2.5pt width0pt}}
\def\sshortstrut{\hbox{\vrule height5.8pt depth2.5pt width0pt}}
\def\smallcirc{{\scriptstyle\circ}}
\def\ssbullet{{\scriptscriptstyle\bullet}}
\def\symmdiff{\triangle}
\def\thicken#1{#1\hskip0.03em\llap #1\hskip0.03em\llap
   #1\hskip0.03em\llap #1}
\def\Tau{{\text{T}}}
\def\tbf#1{\text{\bf #1}}
\def\terror#1#2#3{#1:  `#2' for `#3'}
\def\today{\number\day.\number\month.\number\year}
\def\UEMDRR{University of Essex Mathematics Department Research Report}
\def\UEMDWP{University of Essex Mathematics Department Working Paper}
\def\versiondate#1{
    \immediate\write0{Version of #1}
    \def\vsdate{#1}
               \ifnum\stylenumber=1\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=2\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=6\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=7\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=8\hfill Version of #1
                              \medskip
               \else\fi\fi\fi\fi\fi}
\def\versionhead{\ifnum\stylenumber=3 \hfill PRINTER'S VERSION

                        \bigskip
                   \else\ifnum\stylenumber=4 \hfill REFEREE'S VERSION

                        \bigskip
                   \else \fi\fi}
\def\versionref{\ifnum\stylenumber=1\workingref
           \else\ifnum\stylenumber=2\workingref
           \else\ifnum\stylenumber=8\workingref
           \else\printerref\fi\fi\fi}
\def\vthsp{\hskip0.001em}

%Missing AmSTeX can be coped with by using the following lines:

\ifamstexsupported\else
   \immediate\write0{AmSTeX-unsupported version of  fremtex.tex }
   \fourteensavailablefalse
   \def\@{@}
   \def\\{\cr}
   \def\AmSTeX{AmS\TeX}
   \def\Bbb#1{\underline{\hbox{\bf #1}}}
   \def\biggerscriptfonts{}
   \def\bumpeq{=_{\text{approx}}}
   \def\Cal#1{{\cal#1}}
   \def\curlyeqprec{\preceq}
   \def\eightit{\it}
   \def\eightrm{\rm}
   \def\eightsmc{\smc}
   \def\eurm#1{\underline{\cal#1}}
   \def\Footnote#1#2{\footnote#1#2}
   \def\frac#1#2{{#1}\over{#2}}
   \def\frak#1{{\hbox{\bf #1\hskip0.1em\llap #1}}}
   \def\frnewpage{\vfill\eject}
   \def\iint{\int\!\!\!\int}
   \def\iiint{\int\!\!\!\int\!\!\!\int}
   \def\iiiint{\int\!\!\!\int\!\!\!\int\!\!\!\int}
   \def\llcorner{\lfloor}
   \def\loadeurm{}
   \def\loadeusm{}
%   \def\loadeusmx#1{\def\eusm#1{\underline{\cal#1}}}
   \def\ltimes{|\hskip-0.5em\times}
   \def\eusm#1{\underline{\cal#1}}
   \def\ltimes{|\times}
   \def\Loadfont#1{}
   \def\Matrix#1{\pmatrix{#1}}
   \def\noamslogo{}
   \def\nrightarrow{\not\rightarrow}
   \def\preccurlyeq{\preceq}
   \def\query{\discrversionA{\immediate\write0{query}
                             \footnote{$^*$}{query}}{}}
   \def\restriction{|}
   \def\rtimes{\times|}
   \def\setover#1#2{#2^{#1}}
   \def\shortstrut{\strut}
   \def\smallerfonts{\eightrm}
   \def\smallfrown{\wedge}
   \def\smc{}
   \def\spcheck{\check{\phantom{x}}}
   \def\sphat{\hat{\phantom{x}}}
   \def\square{\mathchoice\sqr64\sqr64\sqr43\sqr33}
   \def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt
          \hbox{\vrule width.#2pt height#1pt \kern#1pt
                  \vrule width.#2pt}
          \hrule height.#2pt}}}}
   \def\pmb#1{\underline{#1}}
   \def\text#1{\hbox{#1}}
   \def\therefore{\ssbullet\raise.9ex\hbox{$\ssbullet$}\ssbullet}
   \def\twelvebf{\bf}
   \def\vartriangleleft{\triangleleft}
   \def\Vdash{|\vdash}
   \def\vDash{|=}
   \fi

%End of lines for missing AmSTeX

\ifamstexsupported\input amstex
%              \documentstyle{amsppt}
               \input amsppt.sty
               \PSAMSFonts
	       \tenpoint
               %\font\eightmsb=msym8
               \font\eightmsb=msbm10
               \font\eighteufm=eufm10 at 8 pt
               \font\eighteurm=eurm8
               \font\eighteusm=eusm8
               \font\varseveneufm=eufm10 at 7 pt
               \scriptfont\eufmfam=\varseveneufm
   \def\.#1{{\accent"5F #1}}
   \def\=#1{{\accent"16 #1}}
   \gdef\nopagenumbers{\NoPageNumbers}
   \else\fi

\newcount\footmarkcount
\ifDHFamstexloaded
    \loadmsBm
    \def\eightBbb#1{{\fam\msBfam#1}}
    \loadmsFm
    \def\eightfrak#1{{\fam\msFfam#1}}
  \else
    \def\eightBbb#1{{\scriptstyle\Bbb#1}}
    \def\eightfrak#1{{\scriptstyle\frak#1}}
  \fi
\ifathome\athome\fi

%interword glue (\tenrm standard):
%  (space) 3.333 + 1.667 - 1.111
%  (,)     3.333 + 2.083 - 0.889
%  (.)     4.444 + 5.    - 0.370

%very odd fact:  \discrversionA{2.}{}  doesn't work,
%   \discrversionA{ 2.}{} does

%The false line feed introduced by my Fortran programs, leading to
%a surplus  ae  at the top of the next page, seems to be associated
%with  Fk(\032)3081  or something similar in the PostScript file.


\immediate\write0{Extract from MEASURE THEORY,
   by D.H.Fremlin, University of Essex, Colchester.
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   http://dsl.org/copyleft/dsl.txt.   This is a development version and the
   source files are not permanently archived.
   For further information contact david\@fremlin.org.}

%hyphenations
\hyphenation{Archi-me-dean}

%new style
\def\smallprintersize{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \nologo
     \def\copyrightdate##1{}
     \def\folio{\ifnum\pageno<0{}\else\number\pageno\fi}
     \def\mtfootline{}
     \stylenumber=10
     \pageno=-1000}

%With-proofs, results-only versions
\newif\ifwithproofs
\withproofstrue
\newif\ifresultsonly
\resultsonlyfalse
\def\noproofs{\withproofsfalse}
\def\resultsonly{\resultsonlytrue\withproofsfalse}
\long\def\proof#1{\ifwithproofs{\medskip

     \noindent{\bf proof} #1}\else\fi}
\long\def\prooflet#1{\ifwithproofs{#1}\else\fi}
%\long\def\Prooflet#1{\prooflet{\Prf\ #1 \Qed\ }}
\long\def\endnotes#1{\ifresultsonly\else{#1}\fi}
\long\def\exercises#1{\ifresultsonly\else{#1}\fi}
\long\def\references#1{\ifresultsonly\else{#1}\fi}
\long\def\dvro#1#2{\ifresultsonly{#1}\else{#2}\fi}

%Headline assignments

\def\topparagraph{}
\def\bottomparagraph{}
\def\newparagraph{}
\newif\ifpageturned

\newdimen\rhoffset
\newdimen\lhoffset
\lhoffset=0cm
\rhoffset=0cm

\newcount\volumeno
\volumeno=5
\newcount\luluvolumeno
\luluvolumeno=0

\let\plainpagebody=\pagebody
\def\pagebody{\ifodd\pageno\global\hoffset=\rhoffset
         \else\global\hoffset=\lhoffset\fi
     \plainpagebody
     \turnpage}
\def\turnpage{
     \iffirstpage\else{\baselineskip=24pt\line{\mtfootline}}\fi
%     \immediate\write0{turning page \folio,
%           top paragraph \topparagraph,   bottom \bottomparagraph,
%           new \newparagraph, pagetotal \the\pagetotal}
     \global\pageturnedtrue
     \global\firstpagefalse
     \global\let\topparagraph=\bottomparagraph
     \global\let\bottomparagraph=\newparagraph
     }

\newdimen\pagetotalplus
\newdimen\pagetotalminus

%wheader parameters:  #1 new paragraph name
 % #2#3#4  \vskip parameters if page break not forced (no. of points)
 % #5  spare height demanded if page break not to be forced
\long\def\wheader#1#2#3#4#5{\gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #5
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
%     \immediate\write0{forcing page break}
     \eject
     \else\vskip #2pt plus #3pt minus #4pt\fi
     \else\vskip #2pt plus #3pt minus #4pt\fi
  \ifpageturned\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
     \fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
%   \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  }
\def\header#1{
     \wheader{#1}{10}{4}{4}{24pt}}
\def\leader#1#2{\immediate\write0{#1 #2}
     \wheader{#1}{14}{7}{5}{24pt}
     {\bf #1 #2}}
\def\vleader#1#2#3{\immediate\write0{#2 #3}
     \wheader{#2}{14}{7}{5}{#1}
     {\bf #2 #3}}
\def\Notesheader#1{\immediate\write0{#1 Notes and comments}
     \wheader{#1 Notes}{16}{8}{5}{48pt}
     \noindent{\bf #1 Notes and comments}}
\def\vNotesheader#1#2{\immediate\write0{#2 Notes and comments}
     \wheader{#2 Notes}{16}{8}{5}{#1}
     \noindent{\bf #2 Notes and comments}}
\def\spheader#1#2#3#4#5{\header{#1#2#3#4#5}{\bf (#5)}}
\def\spstheader#1#2#3#4#5{\header{*#1#2#3#4#5}{\bf (#5)}}
\def\vspheader#1#2#3#4#5#6{
     \wheader{#2#3#4#5#6}{10}{4}{4}{#1}{\bf (#6)}}
\def\sqheader#1#2#3#4#5{\header{#1#2#3#4#5}{$\pmb{>}${\bf (#5)}}}
\def\errheader#1#2#3{
     \wheader{#3}{6}{4}{2}{24pt}
     {\bf p #1 l #2}}
\long\def\woddheader#1#2#3#4#5{
   \ifodd\pageno\wheader{#1}{#2}{#3}{#4}{#5}
   \else\vskip #2pt plus #3pt minus #4pt\fi}

\def\indexmedskip{
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by 18 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
      \else\medskip\fi\else\medskip\fi}

\def\indexheader#1{\immediate\write0{#1}\vindexheader{#1}{24}}
\def\vindexheader#1#2{
  \gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #2 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
%  \ifdim\pagetotal>\pagegoal
%      \immediate\write0{warning: pagetotal greater than pagegoal}
%      \immediate\write0{...   last paragraph reached is #1}
%      \immediate\write0{...   check for underfull vbox here}
%      \else\fi
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal{\eject}
      \fi\fi
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
  \ifpageturned{\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
      }\else\fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%    \bottomparagraph}
  }

\def\leftheadline{\rlap{\folio}\hfill{\eightit\chaptername}\hfill
       \llap{\eightbf\topparagraph}}
\def\rightheadline{\rlap{\eightbf\bottomparagraph}\hfill
       {\eightit\headlinesectionname}\hfill\llap{\folio}}

%Footline assignments
\def\mtfootline{\ifodd\pageno\rightfootline\else\leftfootline\fi}
\def\rightfootline{\hfill\eightsmc D.H.Fremlin}
\def\leftfootline{\ifwithproofs{\eightsmc Measure Theory\hfill}
    \else{\eightsmc Measure Theory \eightit (abridged version)\hfill}
    \fi}

%Style assignments
\def\newchapter#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Chap.\ #1 {\it intro.}}
   \gdef\newparagraph{Chap.\ #1 {\it intro.}}
   \footmarkcount=0
   \gdef\sectionname{Introduction}
   \gdef\headlinesectionname{Introduction}
   \centerline{\bf Chapter #1}
   \medskip
   \centerline{\bf \chaptername}
   \medskip
   }
\def\newsection#1{\gdef\bottomparagraph{\S#1 {\it intro.}}
   \noindent{\bf #1 \sectionname}
   \smallskip
   \let\headlinesectionname=\sectionname}
\def\newvolume#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Vol.\ #1 {\it intro.}}
   \Loadfont{\twelvebf=cmbx12}
   \def\chaptername{Volume #1}
   \def\sectionname{\volumename}
   \def\headlinesectionname{\volumename}
   \centerline{\twelvebf Volume #1:  \volumename}
   \bigskip
   }

%font variations
\newif\ifvolwp
\volwpfalse
\def\tenrmstretch#1{\fontdimen3\tenrm=#1}
\font\elevenex=cmex10 at 11pt

%contents and introductions
\newif\iflargelogo
\def\vtmpb#1{}
\def\newvolume#1{}
\def\pagereference#1#2{\hfill\hbox{\ifresultsonly#2\qquad \hskip 1truein
                                                 \else#1\qquad \fi}}
\long\def\section#1#2#3#4#5#6{\ifresultsonly
    \qquad #1 #2\pagereference{#4}{#5}
  \else\vskip 2pt plus 1pt minus 0pt
    \vbox{\qquad #1 #2\pagereference{#4}{#5}\par
    \abstr{#6}
    }\vskip -1pt plus 1pt minus 0pt
  \fi}
\long\def\chapintrosection#1#2#3{\ifresultsonly
    \qquad Introduction\pagereference{#2}{#3}
  \else\vskip 1pt plus 1pt minus 0pt
    \qquad Introduction\pagereference{#2}{#3}
    \vskip 1pt plus 1pt minus 0pt
  \fi}
\def\abstr#1{\ifresultsonly{}
   \else{\advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \vskip1pt plus 0pt minus0pt
     \parindent=0pt
     \smallerfonts #1%
     \hbox{\vrule height0pt depth2pt width0pt}
     \vskip1pt plus 0pt minus0pt}\fi}
     
%language variations
\newif\ifUSEnglish
\USEnglishfalse
%\USEnglishtrue

%"Later editions only"
\newcount\latereditionpage
\newcount\latereditionmark
\latereditionpage=-1
\latereditionmark=0
\newcount\newpageno

%removing the \write0 instruction altogether disorganizes it!
\def\Latereditions{\newpageno=\pageno%
  \ifdim\pagetotal>\pagegoal\advance\newpageno by 1\fi%
     \divide\newpageno by 2%
  \immediate\write0{%
%     "Later editions only";
%      pagetotal \the\pagetotal, pagegoal \the\pagegoal,
%      pageno \number\pageno,\par newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage%
  }\ifnum\latereditionpage=\newpageno%
     \ifDHFamspptloaded$^{\number\latereditionmark}$%
     \else\footnote{Later editions only.}\fi%
  \else\footnote{Later editions only.}%
  \global\latereditionpage=\newpageno%
  \global\latereditionmark=\footmarkcount\fi}%
%  \immediate\write0{newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage}

   
%general definitions
\def\action{\mathchoice{\lower-.3ex\hbox{$\ssbullet$}}
                       {\lower-.3ex\hbox{$\ssbullet$}}
                       {\ssbullet}
                       {\ssbullet}}  %see also \varaction
\def\add{\mathop{\text{add}}\nolimits}
\long\def\allowmorestretch#1#2{
  \ifdim\pagewidth<#1pt\else
    \fontdimen3\tenrm=2pt\fontdimen4\tenrm=1.67pt\fi
  #2
    \fontdimen3\tenrm=1.67pt\fontdimen4\tenrm=1.11pt}
\def\am{atom\discretionary{-}{}{}lessly-measurable}
\def\AmuA{\Aut_{\bar\mu}\mskip-2mu\frak A}
\def\ASint{\mathchoice{\Bbb A\mskip-5mu}
                  {\Bbb A\mskip-3mu}
                  {\Bbb A\mskip-2mu}
                  {\Bbb A\mskip-2mu}\Sint}
\def\Asisfour{$(\frak A,\bar\mu,T,\family{t}{T}{\frak A_t})$}
\def\Asissix{$(\frak A,\bar\mu,T,
   \penalty-100\family{t}{T}{\frak A_t},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Asisfourt{$(\frak A,\bar\mu,\coint{0,\infty},
   \langle\frak A_t\rangle_{t\ge 0})$}
\def\Asissixt{$(\frak A,\bar\mu,\coint{0,\infty},
   \penalty-100\langle\frak A_t\rangle_{t\ge 0},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Atop#1#2{{\scriptstyle #1}\atop{\scriptstyle #2}}
\def\Aut{\mathop{\text{Aut}}}

\def\Bcap{\Booleanop{\cap}}
\def\Bcup{\Booleanop{\cup}}
\def\Booleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
 %   {{\mskip2mu\scriptstyle#1\mskip2mu}}  %until 15.10.17
    {{\scriptscriptstyle#1}}  %from 15.10.17
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminus{\Booleanop{\setminus}}
\def\Bsubseteq{\hskip.1em\Booleanop{\subseteq}\hskip.1em}
\def\Bsubset{\Booleanop{\subset}}
\def\Bsupseteq{\Booleanop{\supseteq}}
\def\Bsupset{\Booleanop{\supset}}
\def\Bsymmdiff{\Booleanop{\symmdiff}}
\def\Bcapshort{\Booleanopshort{\cap}}
\def\Bcupshort{\Booleanopshort{\cup}}
\def\Booleanopshort#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hskip-1pt\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminusshort{\Booleanopshort{\setminus}}
\def\BSint{\mathchoice{\Bbb B\mskip-5mu}
                  {\Bbb B\mskip-3mu}
                  {\Bbb B\mskip-2mu}
                  {\Bbb B\mskip-2mu}\Sint}
%\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}\hskip.1em}
\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}}
\def\Bsubsetshort{\Booleanopshort{\subset}}
\def\Bsupseteqshort{\Booleanopshort{\supseteq}}
\def\Bsymmdiffshort{\Booleanopshort{\symmdiff}}

\def\closeBooleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\closeBcup{\closeBooleanop{\cup}}
\def\closeBcap{\closeBooleanop{\cap}}
\def\closeBsetminus{\closeBooleanop{\setminus}}
\def\closeBsymmdiff{\closeBooleanop{\symmdiff}}

\def\biggerHeint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 2.6pt{\sevenrm H}\hss}
   \intop\nolimits}
\def\biggerPfint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}
   \intop\nolimits\hskip 1.2pt}
\def\biggerSint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 3.6pt{\sevenrm S}\hss}
   \intop\nolimits}
\def\bu{\mathop{\text{bu}}}
\def\Bvalue#1{\mathchoice
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$\scriptstyle[\![#1]\!]$}}
   {\hbox{$\scriptscriptstyle[\![#1]\!]$}}}
\def\cadlag{c\`adl\`ag}
\def\CalBa{\Cal{B}\eurm{a}}
   %note: neither  \cal  nor  \eusm  has lower case
\def\CalFr{\Cal F_{\text{Fr}}}
\def\callal{c\`all\`al}
\def\CalNwd{\Cal{N}\hskip-1.2pt\eurm{w}\hskip-.8pt\eurm{d}}

\def\CalRbg{\Cal{R}\mskip-1mu\eurm{b}\mskip-1mu\eurm{g}}
\def\CalSmz{\Cal{S}\mskip-2mu\eurm{mz}}
\def\CalUn{\Cal{U}\eurm{n}}
\def\capacity{\mathop{\text{cap}}\nolimits}
\def\Caratheodory{Cara\-th\'eo\-dory}
\def\Cdlg{C_{\text{dlg}}}
\def\cff{\cf\mskip1mu}
\def\checkpmb#1{\check{\pmb #1}}
\def\ci{\mathop{\text{ci}}}
\def\cl{\mathop{\text{cl}}}
\def\Clll{\hbox{$C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\clMsimp{\overline{M}_{\text{simp}}}
\def\closeuparrow{\mathchar"0222}
\def\closedownarrow{\mathchar"0223}
\def\clstar{\text{cl*}}
\def\cov{\mathop{\text{cov}}}
\def\covar#1#2{\mathchoice
  {[#1{\lower4pt\hbox{${\scriptstyle *}\atop{\vrule height 18pt depth -10pt}$}}#2]}
  {[#1{*\atop{\vrule height 8pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}}
\def\covSh{\mathop{\text{cov}}\nolimits_{\text{Sh}}}
\def\cdownarrow{c\mskip2mu^{\downarrow}}
\def\cycle#1{(\overleftarrow{#1})}
\def\ddaggerloc{$(\ddagger)_{\text{loc}}$}
\def\ddownarrow{d\mskip2mu^{\downarrow}}
%\def\DiniD{\setover{
%  \hbox{\hskip.1em\vrule height-1pt width.5em depth1.4pt\hfil}}D}
%\def\Dinid{\hskip.1em
%  \hbox{\hskip.05em\vrule height-1.3pt width.5em depth1.7pt}\hskip-.65em D}
\def\Dinid{\hbox{$D$\hskip-8.3pt\vrule height-1pt width7pt depth1.85pt}
   \hskip1pt}
\def\DiniD{\hbox{$D$\hskip-8.3pt\vrule height9.2pt width7ptdepth-8.35pt}
   \hskip1pt}

\def\discrcenter#1#2{\ifdim\pagewidth>#1{#2}\else{

  \Centerline{#2}

  \noindent}\fi}
    %note:  in \discrcenter, make sure that #2 ends in a space
    % (if you want one to appear) and that the continuation
    % follows without
    % a space after the final } of #2, so that it will be
    % truly unindented
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
        \else\ifnum\stylenumber=3{\frnewpage}
        \else\ifnum\stylenumber=5{\frnewpage}
        \else\wheader{}{18}{14}{2}{72pt}\fi\fi\fi}
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\Displaycause#1{\cr\displaycause{#1}&}
\def\dotproduct{\hskip.1em\tbf{.}\hskip.1em}
\def\doubleheadrightarrow{\rightarrow\mskip-12mu\rightarrow}
\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\duparrow{d\mskip2mu^{\uparrow}}
\def\dvAformerly#1{\discrversionA{\footnote{Formerly #1.}}{}}
\def\dvAnew#1{\discrversionA{\footnote{New #1.}}{}}
\def\dvArevised#1{\discrversionA{\footnote{Revised #1.}}{}}
\def\dvrocolon#1{\dvro{#1: }{#1}}
\def\eightVVdash{\mskip5mu\vrule height 6pt depth 2pt width 0.4pt
  \mskip2mu\vrule height 6pt depth 2pt width 0.4pt
  \vrule height 2.25pt depth -1.8pt width 3.2pt\mskip2mu}
\def\embedsinto{\mskip5mu\raise1.5pt\hbox{$\subset$}\mskip-15mu
   \lower3.5pt\hbox{$\rightarrow$}\mskip5mu}
\def\energy{\mathop{\text{energy}}\nolimits}
%\eqalign:  see \S478 for breaks during \eqalign
\def\equivGT{\equiv_{\text{GT}}}
\def\equivT{\equiv_{\text{T}}}
\def\Eta{\text{H}}
\def\Expn{\Bbb E}
\long\def\Extract{\oldfootnote{}{
   \noindent{\eightit
       \ifresultsonly{Extract from {\eightsmc Measure Theory,}
          results-only version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else\ifwithproofs{Extract from {\eightsmc Measure Theory,}
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else{Extract from {\eightsmc Measure Theory,} abridged version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \fi\fi
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   {\eighttt http://dsl.org/copyleft/dsl.txt}.
   This is a development version and the
   source files are not permanently archived, but current versions are
normally
   accessible through
   {\eighttt https://www1.essex.ac.uk/maths/people/fremlin/mt.htm}.
   For further information contact {\eighttt david\@fremlin.org}.}
   }}
\def\Fn{\mathop{\text{Fn}}\nolimits}
\def\FN{\mathop{\text{FN}}\nolimits}
\def\formerly#1{\footnote{Formerly #1.}}
  %use only once in a chapter, never in same volume as reference
\def\fraction#1{\mathchoice{\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$\scriptstyle<$}#1\hbox{$\scriptstyle>$}}
     {\hbox{$\scriptscriptstyle<$}#1\hbox{$\scriptscriptstyle>$}}}
\def\frakmctbl{\frak m_{\text{countable}}}
\def\frmedcirc{{\raise 2.5pt\hbox{$_{\bigcirc}$}}}
\def\grad{\mathop{\text{grad}}\nolimits}
\def\hatpmb#1{\hat{\pmb{#1}}}
\def\hattilde#1{\skew2\hat{\tilde#1}}
\def\hattildepmb#1{\skew1\hat{\tildepmb#1}}
\def\hcov{\mathop{\text{hcov}}\nolimits}
\def\Heint{\mathchoice{\hbox to 0pt{\hskip 3.1pt H\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     \intop\nolimits}
\def\high#1{\lceil#1\rceil}
\def\highoverline#1{\overline{\vrule height6pt width0pt depth 0pt#1}}
\def\Hint#1{({\it Hint\/}: #1)}
\def\hL{\mathop{\text{hL}}}
\def\Hmp{\text{Hom}_{\bar\mu,\pi}}
\def\hp{\mathop{\text{hp}}\nolimits}
\def\hwtr{hereditarily weakly $\theta$-{\vthsp}refinable}
\def\iiv{ii_{\pmb{v}}}
\def\iivu{\iiv(\pmb{u})}
\def\imp{inverse-{\vthsp}measure-{\vthsp}preserving}
\def\innerprod#1#2{(#1|#2)}
\def\intstar{\text{int*}}
\def\CalIS{\Cal I(\Cal S)}
\def\jump{\mathop{\text{jump}}\nolimits}
\def\Jump{\mathop{\text{Jump}}\nolimits}
\def\Krein{Kre\v{\i}n}
\def\Lambdapv{\Lambda_{\text{pv}}}
\def\leRB{\le_{\text{RB}}}
\def\leRK{\le_{\text{RK}}}
\def\link{\mathord{\text{link}}}
\def\locmafp{\widehat{\otimes}_{\text{loc}}}
\def\Locmafp{\widehat{\bigotimes}^{\text{loc}}}
\def\low#1{\lfloor#1\rfloor}
\def\lti{left-\vthsp translation-\vthsp invariant}
\def\magnitude{\mathop{\text{mag}}}
\def\Mah{\mathop{\text{Mah}}}
\def\Mahcr{\Mah_{\text{cr}}}
\def\MahcrR{\Mah_{\text{crR}}}
\def\MahqR{\Mah_{\text{qR}}}
\def\MahR{\Mah_{\text{R}}}
\def\MahspcrR{\Mah_{\text{spcrR}}}
\def\Malm{M_{\text{alm}}}
\def\Marik{Ma\v r\'\i k} 
\def\Mbv{M_{\text{bv}}}
\def\Mcdlg{M_{\text{cdlg}}}
\def\Mdm{M_{\text{D-M}}}
\def\med{\mathop{\text{med}}\nolimits}
\def\Mfa{M_{\text{fa}}}
\def\Mhsr{\mathop{\text{Mhsr}}\nolimits}
\def\Mjf{\mathop{M_{\text{j-f}}}\nolimits}
\def\Mlbv{M_{\text{lbv}}}
\def\Mlob{M_{\text{lo-b}}}
\def\Mlns{M_{\text{ln-s}}}
\def\Mmo{M_{\text{mo}}}
\def\Mlmo{M_{\text{lmo}}}
\def\Mns{M_{\text{n-s}}}
\def\Mnsu{M_{\text{n-s}}^{\uparrow}}
%\def\Mnsu{M_{\text{n-s}\uparrow}}
\def\MMah{M_{\text{Mah}}}
\def\Mob{M_{\text{o-b}}}
\def\Mpob{M_{\text{po-b}}}
\def\Mpv{M_{\text{pv}}}
\def\MSi{M_{\text{S-i}}}
\def\Msimp{M_{\text{simp}}}
\def\Mth{Ma\-ha\-ram-{\vthsp}type-{\vthsp}homogeneous}
\def\Mvlm{M_{\text{vlm}}}
\def\negquad{\hskip-1em\relax}
\def\NN{\BbbN^{\Bbb N}}
\def\non{\mathop{\text{non}}}
\def\notBooleanop#1{
   \ifamstexsupported{\mathchoice
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
    {{\not\scriptstyle#1}}
    {{\not\scriptscriptstyle#1}}}
   \else{\mathchoice{{\not\scriptstyle#1}}{{\not\scriptstyle#1}}
    {{\not\scriptstyle#1}}{{\not\scriptscriptstyle#1}}}
    \fi}
\def\notBsubseteq{\notBooleanop{\subseteq}}
\def\notBsupseteq{\notBooleanop{\supseteq}}
\def\notVVdash{\not\mskip4mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip-2mu}
\def\nuprime{\nu^{\mskip0.7mu\prime}} %better  \nu'  in superscripts
\def\nw{\mathop{\text{nw}}}
\def\Osclln{\mathop{\text{Osclln}}\nolimits}
\def\otp{\mathop{\text{otp}}}
\def\partstar{\mathchoice{\partial\hbox{*}}
     {\partial\hbox{*}}
     {\partial\hbox{\sevenrm *}}
     {\partial\hbox{\fiverm *}}}
\def\per{\mathop{\text{per}}}
\def\Pfint{\mathchoice{\hbox to 0pt{\hskip 3.1pt Pf\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     \intop\nolimits
     \mathchoice{\hskip 2pt}{\hskip 1.2pt}{\hskip 1pt}{\hskip 1pt}}
\def\Pou{\mathop{\text{Pou}}\nolimits}
\def\prforc{\preccurlyeq_{\text{forcing}}}
\def\prGT{\preccurlyeq_{\text{GT}}}
\def\prT{\preccurlyeq_{\text{T}}}
\def\pssqa{power set $\sigma$-quotient algebra}
\def\qm{quasi-{\vthsp}measurable}
\def\Reverse#1{\setover
   {\lower0.4ex\hbox{$\scriptscriptstyle\leftrightarrow$}}{#1}}
\def\rhoKR{\rho_{\text{KR}}} 
\def\rhotv{\rho_{\text{tv}}} 
\def\Rint{\mathchoice{\hbox to 0pt{\hskip 3.1pt R\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     \intop\nolimits}
\def\RO{\text{RO}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}
\def\rvm{real-{\vthsp}valued-{\vthsp}measurable}
\def\sat{\mathop{\text{sat}}}
\def\sfamilysS#1{\family{\sigma}{\Cal S}{#1_{\sigma}}}
\def\sfcc{$\sigma$-finite-cc}
\def\sgn{\mathop{\text{sgn}}}
\def\shr{\mathop{\text{shr}}\nolimits}
\def\Sii{\text{S}ii}
\def\Siiv{\Sii_{\pmb v}}
\def\Sint{\mathchoice{\hbox to 0pt{\hskip 4.1pt\ninerm S\hss}} 
     {\raise 0.6pt\hbox to 0pt{\hskip 3.0pt{\sevenrm S}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     \intop\nolimits}
\def\Sinteightpoint{\raise 0.6pt\hbox to 0pt{\hskip 2.3pt{\fiverm S}\hss}
     \intop\nolimits}
\def\SL{\text{SL}}
\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind
\def\spread{\mathop{\text{spr}}}
\def\ssplrarrow{^{\scriptscriptstyle\leftrightarrow}}
\def\ssptilde{{\raise1.2ex\hbox{$\scriptscriptstyle\sim$}}}
%\def\sspvec{{\raise1.2ex\hbox{$\scriptscriptstyle\rightarrow$}}}
\def\sspvec{\mskip3mu\vec{\null}\mskip5mu}
\def\Sti{\mathop{\text{Sti}}\nolimits}
\def\strbar{^{\scriptscriptstyle\pmb{-}}}
\def\strplus{^{\scriptscriptstyle\pmb{+}}}
\def\strprime{^{\mathchoice{\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1mu\pmb{\prime}}}}
%\def\strprime{\,\pmb{\prime}}
\def\supp{\mathop{\text{supp}}}
\def\tausm{\tau_{\sigma\text{-m}}}
\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_<$}
   \phantom{\tbf 1^{(#1)}}}
%\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_-$}
%   \phantom{\tbf 1^{(#1)}}}
\def\tensorhat{\widehat{\otimes}}
\def\Tensorhat{\widehat{\bigotimes}}
\def\tensorhatsigma{\widehat{\otimes}_{\sigma}}
\def\tildeClll{\hbox{$\tilde C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\tildehat#1{\skew2\tilde{\hat#1}}
\def\tildehatpmb#1{\tilde{\hatpmb#1}}
\def\tildepmb#1{\tilde{\pmb{#1}}}
\def\Tr{\mathop{\text{Tr}}}
\def\trs{^{\top}\hskip-1pt}
\def\triplepc#1#2#3{(#1,#2,\hbox{$<$}#3)}
\def\TSi{\frak T_{\text{S-i}}}
\def\undphi{\underline{\phi}{}}
\def\undpsi{\underline{\psi}{}}
\def\undtheta{\underline{\theta}{}}
\def\updown{\mathchoice
  {\hbox{\hskip-1.5pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-3.3pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %OK for subscript, fails for superscript
    \hskip-1.7pt\vrule height6.5pt width.7pt depth0pt
    \hskip.9pt\vrule height6.5pt width.7pt depth0pt
    \hskip-2.5pt\lower5.4pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %not tested
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}}
\def\updownarrows{\uparrow\mskip-4mu\downarrow}
\def\updowncgt{$\updown$-convergent}
\def\upr{\mathop{\text{upr}}\nolimits}
\def\ureg{\mathop{\text{ureg}}}

\def\Var{\mathop{\text{Var}}\nolimits}

\def\varaction{\mathchoice{\lower-.2ex\hbox{$\smallcirc$}}
                       {\lower-.2ex\hbox{$\smallcirc$}}
                       {\ssbullet}
                       {\ssbullet}}
\def\varcheck#1{{\setover
       {\lower 0.4ex\hbox{$\scriptscriptstyle\vee$}}{#1}}}
\def\varcheckf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\vee$}}{f}}}
\def\varchecki{\mskip-3mu\text{\it{\accent"14{\hbox{\hskip-3.5pt\i}}}}\mskip2mu}
\def\varspcheck{^{\scriptscriptstyle\vee}}
%\def\varspcheck{{\setover{\scriptscriptstyle\vee}{\phantom{g}}}}
%\def\varspcheckf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\vee$}}{\phantom{f}}}}
\def\varhat#1{{\setover{\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{#1}}}
\def\varhatf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{f}}}
\def\varinnerprod#1#2{#1\dotproduct#2}
\def\varsphat{^{\scriptscriptstyle\wedge}}
% or use  \hbox{\^{\null}}
%\def\varsphat{{\setover{\scriptscriptstyle\wedge}{\phantom{g}}}}
%\def\varsphatf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\wedge$}}{\phantom{f}}}}
\def\varBspcheck{\mathchoice
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}}
\def\varCspcheck{\mathchoice
  {\hskip-1pt\check{\phantom x}}   %not tested
  {\hskip-1pt\check{\phantom x}}
  {\hskip-1.2pt\check{\phantom x}}
  {\hskip-1.5pt\check{\phantom x}}}  %not tested
\def\varCsphat{\mathchoice
  {\hskip-1pt\hat{\phantom x}}   %not tested
  {\hskip-1pt\hat{\phantom x}}
  {\hskip-1.2pt\hat{\phantom x}}  %not tested
  {\hskip-1.5pt\hat{\phantom x}}}  %not tested
\def\VeqL{$V\hbox{$=$}L$}
\def\VVdash{\mskip5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip2mu}
\def\VVdP{\VVdash_{\Bbb P}}
\def\VVdPk{\VVdash_{\Bbb P_{\kappa}}}
\def\wdistr{\mathop{\text{wdistr}}}
\def\weight{\mathop{\text{wt}}}
\def\w1lim{\mathop{\text{w1lim}}}
\def\wsid{weakly $(\sigma,\infty)$-{\vthsp}distributive}
\def\1lim{\mathop{1\mskip-1.5mu\text{lim}}}
\def\2vm{two-{\vthsp}valued-{\vthsp}measurable}

%Cara\-th\'e\-o\-dory

\def\chaptername{}
\def\sectionname{}
\def\headlinesectionname{}

\loadeusm\loadeurm
\Loadfourteens
\Loadfont{\ninerm=cmr9}

%note:  $..\cmmnt{=..}..$  is better as  \cmmnt{\mskip5mu
 
  
%----------------------------------------------------------%  
  
\smallprint  
%\USsmallprint  
%\noproofs  
%\resultsonly  
%\atUEssex  
\usegraphicx
  
\Extract  
  
%----------------------------------------------------------%  

\frfilename{mt5.tex}
\versiondate{9.1.15}

\noindent{\bf Introduction to Volume 5}

\medskip

For the final volume of this treatise, I have collected results which
demand more sophisticated set theory than elsewhere.   The line is not
sharp, but typically we are much closer to questions which are undecidable
in ZFC.   Only in Chapter 55 are these brought to the forefront of the
discussion, but elsewhere much of the work depends on formulations
carefully chosen to express, as arguments in ZFC, ideas which arose in
contexts in which some special axiom -- Martin's axiom, for instance -- was
being assumed.   This has forced the development of concepts -- e.g.,
cardinal functions of structures -- which have taken on vigorous lives of
their own, and
which stand outside the territory marked by the techniques of earlier
volumes.

In terms of the classification I have used elsewhere, this volume has
one preparatory chapter and five working chapters.   There is practically
no measure theory in Chapter 51, which is an introduction to some of the
methods which have been devised to make sense of abstract analysis in
the vast range of alternative mathematical
worlds which have become open to us in the last fifty years.
It is centered on a study of partially ordered sets, which provide a
language in which many of the most important principles can be expressed.
Chapter 52 looks at manifestations of these ideas in measure theory.   In
Chapter 53 I continue the work of Volumes 3 and 4, examining questions
which arise more or less naturally if we approach the topics of those
volumes with the new techniques.

The Banach-Ulam problem got a mention in Volume 2, a paragraph in Volume 3
and a section in Volume 4;
%232Hc 363S 438
at last, in Chapter 54 of the present volume, I try to give a proper
account of the extraordinary ideas to which it has led.   I have
regretfully abandoned the idea of describing even a representative sample
of the forcing models which have been devised to show that
measure-theoretic propositions are consistent, but in Chapter 55 I set out
some of the basic properties of random real forcing.   Finally, in Chapter
56, I look at what measure theory becomes in ZF alone, with countable
or dependent choice, and with the axiom of determinacy.

While I should like to believe that most of the material of this volume
will be accessible to those who have learnt measure theory from other
sources, it has obviously been written with earlier volumes constantly in
mind, and I have to advise you to make sure that Volumes 3 and 4, at
least, will be available in case of need.   Apart from these, I do of
course assume that readers will be at ease with modern set theory.   It is
not so much that I demand a vast amount of knowledge -- \S\S5A1-5A2 have a
good many proofs to help cover any gaps -- as that I present arguments
without much consideration for the inexperienced, and some of them may be
indigestible at first if you have not cut your teeth on
{\smc Just \& Weese 96} or {\smc Jech 78}.   What you may not need is any
prior knowledge of forcing.   But of course for Chapter 55 you will have to
take a proper introduction to forcing, e.g., {\smc Kunen 80}, in parallel
with \S5A3, since nothing here will make sense without an
acquaintance with forcing languages and the fundamental theorem of forcing.

\bigskip

\noindent{\bf Note on second printing}

There has been the usual crop of errors (most, but not all, minor) to be
corrected, and I have added a few new results.   The most important is
P.Larson's proof that it is relatively consistent with ZFC to suppose that
there is no medial limit.   %538Sb
In the process of preparing new editions of Volumes 1-4, I have I hope
covered all the items listed in the old \S5{}A6
(`Later editions only'), which I have therefore dropped, even though
there are one or two further entries under this heading.   As before,
these can be found on the Web edition at
{\tt http://www1.essex.ac.uk/maths{\bsp}people/fremlin/mtcont.htm}.
% 421Cf 367Rd
\frfilename{mt51.tex} 
\versiondate{3.1.15} 
\copyrightdate{2003} 
      
\def\chaptername{Cardinal functions} 
\newchapter{51} 
      
The primary object of this volume is to explore those topics in measure 
theory in which questions arise which are undecided by the ordinary 
axioms of set theory.   We immediately face a new kind of interaction 
between the propositions we consider.   If two statements are 
undecidable, we can ask whether either implies the other.   Almost at 
once we find ourselves trying to make sense of a bewildering tangle of 
uncoordinated patterns.   The most successful method so far found of 
listing the multiple connexions present is to reduce as many arguments 
as possible to investigations of the relationships between specially 
defined cardinal numbers.   In any particular model of set theory (so 
long as we are using the axiom of choice) these numbers must be in a 
linear order, so we can at least estimate the number of potential 
configurations, and focus our attention on the possibilities which seem 
most accessible or most interesting.   At the very beginning of the 
theory, for instance, we can ask whether $\frak c=2^{\omega}$ is equal 
to $\omega_1$, or $\omega_2$, or $\omega_{\omega_1}$, or $2^{\omega_1}$. 
For Lebesgue measure, perhaps the first question to ask is:  if 
$\ofamily{\xi}{\omega_1}{E_{\xi}}$ is a family of measurable sets, is 
$\bigcup_{\xi<\omega_1}E_{\xi}$ necessarily measurable?   If the 
continuum hypothesis is true, certainly not;  but if $\frak c>\omega_1$, 
either `yes' or `no' becomes possible.   The way in which it is now 
customary to express this is to say that 
`$\omega_1\le\add\Cal N\le\frak c$, and $\omega_1\le\add\Cal N<\frak c$, 
$\omega_1<\add\Cal N\le\frak c$ and $\omega_1<\add\Cal N<\frak c$ are 
all possible', where $\add\Cal N$ is defined as the least cardinal of 
any family $\Cal E$ of Lebesgue measurable sets such that 
$\bigcup\Cal E$ is not measurable.   (Actually it is not usually defined 
in quite this way, but that is what it comes to.) 
      
At this point I suggest that you turn to 522B, where you will find a 
classic picture (`Cicho\'n's diagram') of the relationships between ten 
cardinals intermediate between $\omega_1$ and $\frak c$, with 
$\add\Cal N$ immediately above $\omega_1$.   As this diagram already 
makes clear, one can define rather a lot of cardinal numbers. 
Furthermore, the relationships between them are not entirely expressible 
in terms of the partial order in which we say that 
$\kappa_{\frak a}\preceq\kappa_{\frak b}$ if we can prove in ZFC that 
$\kappa_{\frak a}\le\kappa_{\frak b}$.   Even in Cicho\'n's diagram we 
have results of the type $\add\Cal M=\min(\frak b,\cov\Cal M)$ in which 
three cardinals are involved.   It is clear that the framework which has 
been developed over the last thirty-five years is only a beginning. 
Nevertheless, I am confident that it will maintain a leading role as the 
theory evolves.   The point is that at least some of the cardinals 
($\add\Cal N$, $\frak b$ and $\cov\Cal M=\frakmctbl$, for instance) 
describe such 
important features of such important structures that they appear 
repeatedly in arguments relating to diverse topics, and give us a chance 
to notice unexpected connexions. 
      
The first step is to list and classify the relevant cardinals.   This is 
the purpose of the present chapter.   In fact the definitions here are 
mostly of a general type.   Associated with any ideal of sets, for 
instance, we have four cardinals (`additivity', `cofinality', 
`unformity' and `covering number';  see 511F).   Most of the cardinals 
examined in this volume can be defined by one of a limited number of 
processes from some more or less naturally arising structure;  thus 
$\add\Cal N$, already mentioned, is normally defined as the additivity 
of the ideal of Lebesgue negligible subsets of $\Bbb R$, and 
$\cov\Cal M$ is the covering number of the ideal of meager subsets of  
$\Bbb R$.   Another 
important type of definition is in terms of whole classes of structure: 
thus Martin's cardinal $\frak m$ can be regarded as the least Martin 
number (definition:  511Dg) of any ccc Boolean algebra. 
      
\S511 lists some of the cardinals associated with partially ordered 
sets, Boolean algebras, topological spaces and ideals of sets.   Which 
structures count as `naturally arising' is a matter of taste and 
experience, but it turns out that many important ideas can be expressed 
in terms of cardinals associated with relations, and some of these are 
investigated in \S512.   The core ideas of the chapter are most clearly  
manifest in their application to partially ordered sets,  
which I look at in \S513.   In \S514 I run through the elementary results 
connecting the cardinal functions of topological spaces and associated 
Boolean algebras and partially ordered sets.   \S515 is a brief 
excursion into abstract Boolean algebra. 
\S516 is a discussion of `precalibers'.   \S517 is an introduction to 
the theory of `Martin numbers', which (following the principles I have 
just tried to explain) I will use as vehicles for the arguments which 
have been used to make deductions from Martin's axiom.    
\S518 gives results on Freese-Nation numbers and tight filtrations of  
Boolean algebras which can be expressed in general terms  
and are relevant to questions in measure theory. 
      
\discrpage 
      
\frfilename{mt511.tex}
\versiondate{10.10.13}
\copyrightdate{2003}

\def\nw{\mathop{\text{nw}}}
\def\triplepc#1#2#3{(#1,#2,\hbox{$<$}#3)}
\def\bu{\mathop{\text{bu}}}

\def\chaptername{Cardinal functions}
\def\sectionname{Definitions}

\newsection{511}

A large proportion of the ideas of this volume will be expressed in
terms of cardinal numbers associated with the structures of measure
theory.   For any measure space $(X,\Sigma,\mu)$ we have, at least, the
structures $(X,\Sigma)$, $(X,\Sigma,\Cal N(\mu))$ (where $\Cal N(\mu)$
is the null ideal of $\mu$) and the measure algebra
$\frak A=\Sigma/\Sigma\cap\Cal N(\mu)$;  each of these types of
structure has
a family of cardinal functions associated with it, starting from the
obvious ones $\#(X)$, $\#(\Sigma)$ and $\#(\frak A)$.   For the measure
algebra $\frak A$, we quickly find that we have cardinals naturally
associated with its Boolean structure and others naturally associated
with the topological structure of its Stone space;  of course the most
important ones are those which can be described in both languages.   The
actual measure $\mu:\Sigma\to[0,\infty]$, and its daughter
$\bar\mu:\frak A\to[0,\infty]$, will be less conspicuous here;   for most
of the questions addressed in this volume, replacing a measure by
another with the same measurable sets and the same negligible sets will
make no difference.

In this section I list the definitions on which the rest of the chapter
depends, with a handful of elementary results to give you practice with
the definitions.

\leader{511A}{Pre-ordered sets}\cmmnt{ When we come to the theory of
forcing in Chapter 55, there will be technical advantages in using a
generalization of the concept of `partial order'.}   A {\bf pre-ordered
set} is a set $P$ together with a relation $\le$ on $P$ such that

\inset{if $p\le q$ and $q\le r$ then $p\le r$,

$p\le p$ for every $p\in P$\dvro{.}{;}}

\noindent\cmmnt{that is, $\le$ is transitive and reflexive but need not
be antisymmetric.  }As with partial orders,
I will write $p\ge q$ to mean $q\le p$;
$[p,q]=\{r:p\le r$ and $r\le q\}$;  $\coint{p,\infty}=\{q:p\le q\}$,
$\ocint{-\infty,p}=\{q:q\le p\}$.   An {\bf upper} (resp.\ {\bf lower})
{\bf bound} for a set $A\subseteq P$ will be a $p\in P$ such that $q\le p$
(resp.\ $p\le q$) for every $q\in A$.   If $(Q,\le)$ is another pre-ordered
set, I will say that $f:P\to Q$ is {\bf order-preserving} if
$f(p)\le f(p')$ whenever $p\le p'$ in $P$.   If $\familyiI{(P_i,\le_i)}$ is
a family of pre-ordered sets, their {\bf product} is the pre-ordered set
$(P,\le)$ where $P=\prod_{i\in I}P_i$ and, for $p$, $q\in P$,
$p\le q$ iff $p(i)\le_iq(i)$ for every $i\in I$\cmmnt{ (cf.\ 315C)}.

If $(P,\le)$ is a pre-ordered set, we have an equivalence relation $\sim$
on $P$ defined by saying that $p\sim q$ if $p\le q$ and $q\le p$.   Now we
have a canonical partial order on the set $\tilde P$ of equivalence
classes defined by saying that $p^{\ssbullet}\le q^{\ssbullet}$ iff
$p\le q$.   \cmmnt{For all ordinary purposes, $(P,\le)$ and
$(\tilde P,\le)$ carry the same structural information, and the move to the
true partial order is natural and convenient.   It occasionally happens
(see 512Ee below, for instance, and also the theory of iterated forcing
in {\smc Kunen 80}, chap.\ VIII) that it is
helpful to have a language which enables us to
dispense with this step, thereby simplifying some basic definitions.
However the extra generality leads to no new ideas, and I expect that
most readers will prefer to do nearly all their thinking in
the context of partially ordered sets.}

\leader{511B}{Definitions} Let $(P,\le)$ be any pre-ordered set.

\spheader 511Ba A subset $Q$ of $P$ is {\bf cofinal} with $P$ if for
every $p\in P$ there is a $q\in Q$ such that $p\le q$.   The
{\bf cofinality} of $P$, $\cf P$, is the least cardinal of any cofinal
subset of $P$.

\spheader 511Bb The {\bf additivity} of $P$, $\add P$, is the least
cardinal of any subset of $P$ with no upper bound in $P$.   If there is
no such set, write $\add P=\infty$.

\spheader 511Bc A subset $Q$ of $P$ is {\bf coinitial} with $P$ if for
every $p\in P$ there is a $q\in Q$ such that $q\le p$.   The
{\bf coinitiality} of $P$, $\ci P$, is the least cardinal of any
coinitial subset of $P$.

\spheader 511Bd Two elements $p$, $p'$ of $P$ are
{\bf compatible upwards} if
$\coint{p,\infty}\cap\coint{p',\infty}\ne\emptyset$\cmmnt{, that is, if
$\{p,p'\}$ has an upper bound in $P$};  otherwise they are
{\bf incompatible upwards}.   A subset $A$ of $P$ is an
{\bf up-antichain} if no two
distinct elements of $A$ are compatible upwards.   The {\bf
upwards cellularity} of $P$ is
$c^{\uparrow}(P)=\sup\{\#(A):A\subseteq P$ is an up-antichain in $P\}$;
the {\bf upwards saturation} of $P$, $\sat^{\uparrow}(P)$, is the least
cardinal $\kappa$ such that there is no up-antichain in $P$ with cardinal
$\kappa$.   $P$ is called {\bf upwards-ccc} if it has no uncountable
up-antichain\cmmnt{, that is, $c^{\uparrow}(P)\le\omega$, that is,
$\sat^{\uparrow}(P)\le\omega_1$}.

%Note:  Juhasz (2001) uses  $\hat c$  for $\sat$

\spheader 511Be Two elements $p$, $p'$ of $P$ are
{\bf compatible downwards} if
$\ocint{-\infty,p}\cap\ocint{-\infty,p'}\ne\emptyset$\cmmnt{, that is, if
$\{p,p'\}$ has a lower bound in $P$};  otherwise they are
{\bf incompatible downwards}.   A subset $A$ of $P$ is a
{\bf down-antichain} if no two
distinct elements of $A$ are compatible downwards.
The {\bf downwards cellularity} of $P$ is
$\cdownarrow(P)=\sup\{\#(A):A\subseteq P$ is a down-antichain in
$P\}$;  the {\bf downwards saturation} of $P$, $\sat^{\downarrow}(P)$,
is the least $\kappa$ such that there is no down-antichain in
$P$ with cardinal $\kappa$.   $P$ is called {\bf downwards-ccc} if it has
no uncountable down-antichain\cmmnt{, that is,
$\cdownarrow(P)\le\omega$, that is,
$\sat^{\downarrow}(P)\le\omega_1$}.

\spheader 511Bf If $\kappa$ is a cardinal, a subset $A$ of $P$ is
{\bf upwards-$\hbox{$<$}\kappa$-linked} in $P$ if every subset of $A$ of
cardinal less than $\kappa$ is bounded above in $P$.   The {\bf
upwards $\hbox{$<$}\kappa$-linking number} of $P$,
$\link_{<\kappa}^{\uparrow}(P)$, is the smallest
cardinal of any cover of $P$ by upwards-$\hbox{$<$}\kappa$-linked sets.

A subset $A$ of $P$ is {\bf upwards-$\kappa$-linked} in $P$ if it is
upwards-$\hbox{$<$}\kappa^+$-linked\cmmnt{, that is, every member of
$[A]^{\le\kappa}$ is bounded above in $P$}.   The {\bf
upwards $\kappa$-linking number} of $P$,
$\link_{\kappa}^{\uparrow}(P)=\link_{<\kappa^+}^{\uparrow}(P)$, is the
smallest cardinal of any cover of $P$ by upwards-$\kappa$-linked sets.

Similarly, a subset $A$ of $P$ is
{\bf downwards-$\hbox{$<$}\kappa$-linked} if every member of
$[A]^{<\kappa}$ has a lower bound in $P$, and
{\bf downwards-$\kappa$-linked} if it is
downwards-$\hbox{$<$}\kappa^+$-linked;  the
{\bf downwards $\hbox{$<$}\kappa$-linking number} of $P$,
$\link_{<\kappa}^{\downarrow}(P)$, is the smallest
cardinal of any cover of $P$ by
downwards-$\hbox{$<$}\kappa$-linked sets,
and $\link_{\kappa}^{\downarrow}(P)=\link_{<\kappa^+}^{\downarrow}(P)$.

\spheader 511Bg The most important cases of (f) above are $\kappa=2$ and
$\kappa=\omega$.   A subset $A$ of $P$ is {\bf upwards-linked} if any
two members of $A$ are compatible upwards in $P$, and
{\bf upwards-centered} if it is upwards-$\hbox{$<$}\omega$-linked\cmmnt{,
that is, any finite subset of $A$ has
an upper bound in $P$}.   The {\bf upwards linking number} of $P$,
$\link^{\uparrow}(P)=\link_2^{\uparrow}(P)$, is the least cardinal of any
cover of $P$ by upwards-linked sets, and the {\bf upwards centering
number} of $P$, $\duparrow(P)=\link_{<\omega}^{\uparrow}(P)$,
is the least cardinal of any cover of $P$ by upwards-centered sets.

\allowmorestretch{500}{\cmmnt{Similarly, }$A\subseteq P$ is
{\bf downwards-linked} if any two members
of $A$ are compatible downwards in $P$, and {\bf down\-wards-centered} if
any finite subset of $A$ has a lower bound in $P$;  the
{\bf downwards linking number} of $P$ is
$\link^{\downarrow}(P)=\link_2^{\downarrow}(P)$, and the
{\bf downwards centering number} of
$P$ is $\ddownarrow(P)=\link_{<\omega}^{\downarrow}(P)$.
}

If $\link^{\uparrow}(P)\le\omega$ (resp.\
$\link^{\downarrow}(P)\le\omega$) we say that $P$ is
{\bf $\sigma$-linked upwards} (resp.\ {\bf downwards}).
If $\duparrow(P)\le\omega$ (resp.\
$\ddownarrow(P)\le\omega$) we say that $P$ is
{\bf $\sigma$-centered upwards} (resp.\ {\bf downwards}).

\spheader 511Bh The {\bf upwards Martin number} $\frak m^{\uparrow}(P)$
of $P$ is the smallest cardinal of any family $\Cal Q$ of cofinal
subsets of $P$ such that there is some $p\in P$ such that no
upwards-linked subset of $P$ containing $p$ meets
every member of $\Cal Q$;  if there is no such family $\Cal Q$, write
$\frak m^{\uparrow}(P)=\infty$.

Similarly, the {\bf downwards Martin number} $\frak m^{\downarrow}(P)$
of $P$ is the smallest cardinal of any family $\Cal Q$ of coinitial
subsets of $P$ such that there is some $p\in P$ such that no
downwards-linked subset of $P$ containing $p$ meets every member
of $\Cal Q$, or $\infty$ if there is no such $\Cal Q$.

\spheader 511Bi A {\bf Freese-Nation function} on $P$ is a function
$f:P\to\Cal PP$ such that whenever $p\le q$ in $P$ then
$[p,q]\cap f(p)\cap f(q)$ is non-empty.   The {\bf Freese-Nation number}
of $P$, $\FN(P)$, is the least $\kappa$ such that there is a
Freese-Nation function $f:P\to[P]^{<\kappa}$.
The {\bf regular Freese-Nation number}
of $P$, $\FN^*(P)$, is the least regular infinite $\kappa$
such that there is a
Freese-Nation function $f:P\to[P]^{<\kappa}$.
%$P$ has the
%{\bf Freese-Nation property} if $\FN(P)\le\omega$, and the
%{\bf weak Freese-Nation property} if $\FN(P)\le\omega_1$.
If $Q$ is a subset of $P$, the {\bf Freese-Nation index} of $Q$ in $P$
is the least cardinal $\kappa$ such that
$\cf(Q\cap\ocint{-\infty,p})<\kappa$
and $\ci(Q\cap\coint{p,\infty})<\kappa$ for every $p\in P$.

\spheader 511Bj The {\bf (principal) bursting number} $\bu P$ of $P$ is
the least cardinal $\kappa$ such that there is a cofinal subset $Q$ of
$P$ such that

\Centerline{$\#(\{q:q\in Q$, $q\le p$, $p\not\le q\})<\kappa$}

\noindent for every $p\in P$.
%we want  bu \le cf  when  P  has top element

\spheader 511Bk\cmmnt{ It will be convenient to have a phrase for the
following phenomenon.}   I will say that $P$ is {\bf separative upwards} if
whenever $p$, $q\in P$ and $p\not\le q$ there is a $q'\ge q$ which is
incompatible upwards with $p$.   \cmmnt{Similarly, of course,} $P$ is
{\bf separative downwards} if whenever $p$,
$q\in P$ and $p\not\ge q$ there is a $q'\le q$ which is incompatible
downwards with $p$.

\cmmnt{
\leader{511C}{On the symbol $\infty$} I note that in the
definitions above I have introduced expressions of the form
`$\add P=\infty$'.   The `$\infty$' here must be rigorously
distinguished from the `$\infty$' of ordinary measure theory, which can
be regarded as a top point added to the set of real numbers.   The
`$\infty$' of 511B is rather a top point added to the class of ordinals.
But it is convenient, and fairly safe, to use formulae like
`$\add P\le\add Q$' on the understanding that
$\add P\le\infty$ for every pre-ordered set $P$, while
$\infty\le\add Q$ only when $\add Q=\infty$.   Of course we have to be
careful to distinguish between `$\add P<\infty$' (meaning that
there is a subset of $P$ with no upper bound in $P$) and `$\add P$ is
finite' (meaning that $\add P<\omega$).
}%end of comment

\leader{511D}{Definitions} Let $\frak A$ be a Boolean algebra.   I write
$\frak A^+$ for the set $\frak A\setminus\{0\}$ of non-zero elements of
$\frak A$ and
$\frak A^-$ for $\frak A\setminus\{1\}$, so that the partially ordered
sets $(\frak A^-,\Bsubseteqshort)$ and $(\frak A^+,\Bsupseteqshort)$ are
isomorphic.

\spheader 511Da The {\bf Maharam type} $\tau(\frak A)$ of $\frak A$ is
the smallest cardinal of any subset\cmmnt{ $B$} of $\frak A$ which
$\tau$-generates $\frak A$\cmmnt{ in the sense that the order-closed
subalgebra of $\frak A$ including $B$ is $\frak A$ itself}.
\cmmnt{(See Chapter 33.)}

\spheader 511Db The {\bf cellularity} of $\frak A$ is

\Centerline{$c(\frak A)
=c^{\uparrow}(\frak A^-)
=\cdownarrow(\frak A^+)
=\sup\{\#(C):C\subseteq\frak A^+$ is disjoint$\}$.}

\noindent The {\bf saturation} of $\frak A$ is

\Centerline{$\sat(\frak A)
=\sat^{\uparrow}(\frak A^-)
=\sat^{\downarrow}(\frak A^+)
=\sup\{\#(C)^+:C\subseteq\frak A^+$ is disjoint$\}$\dvro{.}{,}}

\cmmnt{\noindent that is, the smallest cardinal $\kappa$ such that
there is no disjoint family in $\frak A^+$ with cardinal $\kappa$.}

%following Koppelberg

\spheader 511Dc The {\bf $\pi$-weight} or {\bf density}
%Koppelberg 2.4.8
$\pi(\frak A)$ of $\frak A$ is
$\cf\frak A^-=\ci\frak A^+$\cmmnt{, that is, the smallest cardinal of
any order-dense subset of $\frak A$}.

\spheader 511Dd Let $\kappa$ be a cardinal.   A subset $A$ of
$\frak A^+$ is {\bf $\hbox{$<$}\kappa$-linked} if it is
downwards-$\hbox{$<$}\kappa$-linked in $\frak A^+$\cmmnt{, that is, no
$B\in[A]^{<\kappa}$ has infimum $0$}, and {\bf $\kappa$-linked} if it is
$\hbox{$<$}\kappa^+$-linked\cmmnt{, that is, every
$B\in[A]^{\le\kappa}$ has a non-zero lower bound}.
The {\bf $\hbox{$<$}\kappa$-linking number}
$\link_{<\kappa}(\frak A)$ of $\frak A$ is
$\link_{<\kappa}^{\downarrow}(\frak A^+)$,
the least cardinal of any family of
$\hbox{$<$}\kappa$-linked sets covering $\frak A^+$;  and the
{\bf $\kappa$-linking number} $\link_{\kappa}(\frak A)$ of $\frak A$ is
$\link_{<\kappa^+}(\frak A)$\cmmnt{, that is, the least cardinal of
any cover of $\frak A^+$ by $\kappa$-linked sets}.

\spheader 511De\cmmnt{ As in 511Bg, I say that} $A\subseteq\frak A^+$ is
{\bf linked} if no two members of $A$ are disjoint;  the
{\bf linking number} of $\frak A$ is $\link(\frak A)=\link_2(\frak A)$,
the least cardinal of any cover of $\frak A^+$ by linked sets.
\cmmnt{Similarly, }$A\subseteq\frak A^+$ is {\bf centered} if
$\inf I\ne 0$ for any finite $I\subseteq A$\cmmnt{;  that
is, if $A$ is downwards-centered in $\frak A^+$}.   The {\bf centering
number}
%my invention
$d(\frak A)$ of $\frak A$ is
$\duparrow(\frak A^-)=\ddownarrow(\frak A^+)$\cmmnt{, that is, the
smallest cardinal of any cover of $\frak A^+$ by centered sets}.
$\frak A$ is {\bf $\sigma$-$m$-linked} if
$\link_m(\frak A)\le\omega$;  \cmmnt{in particular,} it is
{\bf $\sigma$-linked} iff $\link(\frak A)\le\omega$.
$\frak A$ is {\bf $\sigma$-centered} if $d(\frak A)\le\omega$.

\spheader 511Df If $\kappa$ is any cardinal, $\frak A$ is {\bf weakly
$(\kappa,\infty)$-distributive} if whenever
$\ofamily{\xi}{\kappa}{A_{\xi}}$ is a family of partitions of unity in
$\frak A$, there is a partition $B$ of unity such that
$\{a:a\in A_{\xi}$, $a\Bcap b\ne 0\}$ is finite for every $b\in B$ and
$\xi<\kappa$.    Now the {\bf weak distributivity} $\wdistr(\frak A)$ of
$\frak A$ is the least cardinal $\kappa$ such that $\frak A$ is not
weakly $(\kappa,\infty)$-distributive.   (If there is no such cardinal,
write $\wdistr(\frak A)=\infty$.)

\spheader 511Dg The {\bf Martin number} %my invention
  %maybe `Baire number' or `Nov\'ak number'
$\frak m(\frak A)$ of $\frak A$ is the downwards Martin number of
$\frak A^+$\cmmnt{, that is, the smallest cardinal of any
family $\Cal B$ of coinitial subsets of $\frak A^+$ for
which there is some $a\in\frak A^+$ such that no
linked subset of $\frak A$ containing $a$ meets every member of
$\Cal B$;   or $\infty$ if there is no such $\Cal B$}.

\spheader 511Dh The {\bf Freese-Nation number} of $\frak A$,
$\FN(\frak A)$, is the Freese-Nation number of the partially ordered set
$(\frak A,\Bsubseteqshort)$.
%\cmmnt{;  we say that $\frak A$
%has the
%(weak) Freese-Nation property if it has the property for this partial
%ordering}.
The {\bf regular Freese-Nation number} $\FN^*(\frak A)$
of $\frak A$ is the regular Freese-Nation number of
$(\frak A,\Bsubseteqshort)$\cmmnt{, that is, the smallest regular infinite
cardinal greater than or equal to $\FN(\frak A)$}.

\spheader 511Di\dvArevised{2014} If
$\kappa$ is a cardinal, a {\bf tight $\kappa$-filtration} of $\frak A$
is a family $\ofamily{\xi}{\zeta}{a_{\xi}}$ in $\frak A$, where $\zeta$
is an ordinal, such that, writing
$\frak A_{\alpha}$ for the subalgebra of $\frak A$ generated by
$\{a_{\xi}:\xi<\alpha\}$,
($\alpha$) $\frak A_{\zeta}=\frak A$ ($\beta$) for every $\alpha<\zeta$,
the Freese-Nation index of $\frak A_{\alpha}$ in $\frak A$ is
at most $\kappa$.
If $\frak A$ has a tight $\kappa$-filtration, I will say that it is {\bf
tightly $\kappa$-filtered}.

\leader{511E}{Precalibers (a)} Let $(P,\le)$ be a pre-ordered set.

\medskip

\quad{\bf (i)} I will say that $\triplepc{\kappa}{\lambda}{\theta}$ is
an {\bf upwards precaliber triple} of $P$ if $\kappa$, $\lambda$ and
$\theta$
are cardinals, and whenever $\ofamily{\xi}{\kappa}{p_{\xi}}$ is a family
in $P$ then there is a set
$\Gamma\in[\kappa]^{\lambda}$ such that $\{p_{\xi}:\xi\in I\}$ has an
upper bound in $P$ for every $I\in[\Gamma]^{<\theta}$.

\cmmnt{Similarly, }$\triplepc{\kappa}{\lambda}{\theta}$ is a
{\bf downwards precaliber
triple} of $P$ if $\kappa$, $\lambda$ and $\theta$ are cardinals and
whenever $\ofamily{\xi}{\kappa}{p_{\xi}}$ is a family in $P$ then there
is a set
$\Gamma\in[\kappa]^{\lambda}$ such that $\{p_{\xi}:\xi\in I\}$ has a
lower bound in $P$ for every $I\in[\Gamma]^{<\theta}$.

\medskip

\quad{\bf (ii)} An {\bf upwards precaliber pair} of $P$ is a pair
$(\kappa,\lambda)$ of cardinals such that
$\triplepc{\kappa}{\lambda}{\omega}$ is
an upwards precaliber triple of $P$\cmmnt{, that is, whenever
$\ofamily{\xi}{\kappa}{p_{\xi}}$ is a family in $P$ there is a
$\Gamma\in[\kappa]^{\lambda}$ such that $\{p_{\xi}:\xi\in\Gamma\}$ is
upwards-centered in $P$}.

A {\bf downwards precaliber pair} of $P$ is a pair $(\kappa,\lambda)$ of
cardinals such that $\triplepc{\kappa}{\lambda}{\omega}$ is a downwards
precaliber triple of $P$.

\medskip

\quad{\bf (iii)} An {\bf up-} (resp.\ {\bf down-}) {\bf precaliber} of
$P$ is a cardinal $\kappa$ such that $(\kappa,\kappa)$ is an upwards
(resp.\ downwards) precaliber pair of $P$.

\spheader 511Eb Let $(X,\frak T)$ be a topological space.   Then
$\triplepc{\kappa}{\lambda}{\theta}$ is a
{\bf precaliber triple} of $X$ if it is
a downwards precaliber triple of $\frak T\setminus\{\emptyset\}$;
$(\kappa,\lambda)$ is a {\bf precaliber pair} of $X$ if it is a
downwards precaliber pair of
$\frak T\setminus\{\emptyset\}$;  and $\kappa$ is a {\bf precaliber} of
$X$ if it is a down-precaliber of $\frak T\setminus\{\emptyset\}$.

\spheader 511Ec Let $\frak A$ be a Boolean algebra.   Then
$\triplepc{\kappa}{\lambda}{\theta}$ is a
{\bf precaliber triple} of $\frak A$ if
it is a downwards precaliber triple of $\frak A^+$;
$(\kappa,\lambda)$ is a {\bf precaliber pair} of $\frak A$ if it is a
downwards precaliber pair of $\frak A^+$;  and $\kappa$ is a
{\bf precaliber} of $\frak A$ if it is a down-precaliber of
$\frak A^+$.

\spheader 511Ed If $(\frak A,\bar\mu)$ is a measure algebra, then
$\triplepc{\kappa}{\lambda}{\theta}$ is a {\bf measure-precaliber
triple} of $(\frak A,\bar\mu)$ if whenever
$\ofamily{\xi}{\kappa}{a_{\xi}}$ is a family in
$\frak A$ such that $\inf_{\xi<\kappa}\bar\mu a_{\xi}>0$, then there is
a $\Gamma\in[\kappa]^{\lambda}$ such that $\{a_{\xi}:\xi\in I\}$ has a
non-zero lower bound for every $I\in[\Gamma]^{<\theta}$.
Now $(\kappa,\lambda)$ is a {\bf measure-precaliber pair} of
$(\frak A,\bar\mu)$ if $\triplepc{\kappa}{\lambda}{\omega}$ is a
measure-precaliber triple, and $\kappa$ is a {\bf measure-precaliber} of
$(\frak A,\bar\mu)$ if $(\kappa,\kappa)$ is a measure-precaliber pair.

\spheader 511Ee In this context, I will say that
$(\kappa,\lambda,\theta)$ is a precaliber triple (in any sense) if
$\triplepc{\kappa}{\lambda}{\theta^+}$ is a precaliber triple as defined
above;  and similarly for measure-precaliber triples.

\spheader 511Ef I will say that one of the structures here satisfies
{\bf Knaster's condition} if it has $(\omega_1,\omega_1,2)$ as a
precaliber triple\cmmnt{, that is, if every uncountable set has an
uncountable linked subset}.  (For pre-ordered sets I will speak of
`Knaster's condition upwards' or `Knaster's condition downwards'.)
A structure satisfying Knaster's condition must be ccc\prooflet{, because
an uncountable set of mutually incompatible elements surely cannot have an
uncountable linked subset}.

\leader{511F}{Definitions} Let $X$ be a set and $\Cal I$ an ideal of
subsets of $X$.

\spheader 511Fa Taking $\Cal I$ to be partially ordered by
$\subseteq$, we can speak of $\add\Cal I$ and $\cf\Cal I$\cmmnt{ in
the sense of 511B}.   $\Cal I$ is called {\bf $\kappa$-additive} or
{\bf $\kappa$-complete} if $\kappa\le\add\Cal I$\cmmnt{, that is, if
$\bigcup\Cal E\in\Cal I$ for every $\Cal E\in[\Cal I]^{<\kappa}$}.

\cmmnt{In addition we have three other cardinals which will be
important to us.}

\spheader 511Fb The {\bf uniformity} of $\Cal I$ is

\Centerline{$\non\Cal I=\min\{\#(A):A\subseteq X$, $A\notin\Cal I\}$,}

\noindent or $\infty$ if there is no such set $A$.   \cmmnt{(Note the
hidden variable $X$ in this notation;  if any confusion seems possible,
I will write $\non(X,\Cal I)$.   Many authors prefer
unif $\Cal I$.)}

\spheader 511Fc The {\bf shrinking number} of
$\Cal I$, $\shr\Cal I$, is the smallest cardinal $\kappa$ such that
whenever $A\in\Cal PX\setminus\Cal I$ there is a
$B\in[A]^{\le\kappa}\setminus\Cal I$.
\cmmnt{(Again, we need to know $X$ as well as $\Cal I$ to determine
$\shr\Cal I$, and if necessary I will write $\shr(X,\Cal I)$.)}
The {\bf augmented shrinking number} $\shr^+(\Cal I)$ is the smallest
$\kappa$ such that
whenever $A\in\Cal PX\setminus\Cal I$ there is a
$B\in[A]^{<\kappa}\setminus\Cal I$.

\spheader 511Fd The {\bf covering number} of $\Cal I$ is

\Centerline{$\cov\Cal I=\min\{\#(\Cal E):\Cal E\subseteq\Cal I$,
$\bigcup\Cal E=X\}$,}

\noindent or $\infty$ if there is no such set $\Cal E$.
\cmmnt{(Once more, $X$ is a hidden variable here, and I may write
$\cov(X,\Cal I)$.)}

\leader{511G}{Definition} Let $(X,\Sigma,\mu)$ be a measure space.

\spheader 511Ga If $\kappa$ is a cardinal,
$\mu$ is {\bf $\kappa$-additive} if
$\bigcup\Cal E\in\Sigma$ and $\mu(\bigcup\Cal E)=\sum_{E\in\Cal E}\mu E$
for every disjoint family $\Cal E\in[\Sigma]^{<\kappa}$.   The {\bf
additivity} $\add\mu$ of $\mu$ is the largest cardinal $\kappa$ such
that $\mu$ is
$\kappa$-additive, or $\infty$ if $\mu$ is $\kappa$-additive for every
$\kappa$.

\spheader 511Gb The {\bf $\pi$-weight} $\pi(\mu)$ of $\mu$
is the coinitiality of $\Sigma\setminus\Cal N(\mu)$, where
$\Cal N(\mu)$ is the null ideal of $\mu$.

\cmmnt{
\spheader 511Gc Recall that the Maharam type $\tau(\mu)$ of $\mu$ is the
Maharam type of the measure algebra of $\mu$ (331Fc).
}

\leader{511H}{Elementary facts:  pre-ordered sets} Let $P$ be a
pre-ordered set.

\spheader 511Ha If $\tilde P$ is the partially ordered set of equivalence
classes in $P$, as described in 511A, all the cardinal functions defined in
511B have the same values for $P$ and
$\tilde P$.    \cmmnt{(The point is that $p$ is an upper bound for
$A\subseteq P$ iff $p^{\ssbullet}$ is an upper bound for
$\{q^{\ssbullet}:q\in A\}\subseteq\tilde P$.)}
\cmmnt{Similarly, }$P$ and $\tilde P$ will have the same triple
precalibers, precaliber pairs and precalibers.

\spheader 511Hb \cmmnt{Obviously,
}$c^{\uparrow}(P)\le\sat^{\uparrow}(P)$.   \cmmnt{(In fact
$c^{\uparrow}(P)$ is determined by $\sat^{\uparrow}(P)$;  see 513Bc
below.)}   If $\kappa\le\lambda$ are cardinals then

\Centerline{$\link_{<\kappa}^{\uparrow}(P)
\le\link_{<\lambda}^{\uparrow}(P)\le\cf P$\dvro{.}{,}}

\noindent\cmmnt{because every upwards-$\hbox{$<$}\lambda$-linked set is
upwards-$\hbox{$<$}\kappa$-linked and every set $\ocint{-\infty,p}$ is
upwards-$\hbox{$<$}\lambda$-linked.
}$c^{\uparrow}(P)\le\link^{\uparrow}(P)$\prooflet{, because if
$A\subseteq P$ is an up-antichain then no upwards-linked set can contain
more than one point of $A$}.   \cmmnt{It follows that}

\Centerline{$\link^{\uparrow}(P)
=\link_{<3}^{\uparrow}(P)\le\link_{<\omega}^{\uparrow}(P)
=\duparrow(P)\le\cf P$.}

\noindent\cmmnt{Of course }$\cf P\le\#(P)$.   \cmmnt{Similarly,}

\Centerline{$\link^{\downarrow}_{<\kappa}(P)
\le\link^{\downarrow}_{<\lambda}(P)\le\ci P$}

\noindent whenever $\kappa\le\lambda$, and

\Centerline{$\cdownarrow(P)\le\link^{\downarrow}(P)
\le\ddownarrow(P)\le\ci P\le\#(P)$,
\quad$\cdownarrow(P)\le\sat^{\downarrow}P$.}

\spheader 511Hc $P$ is empty iff $\cf P=0$ iff
$\ci P=0$ iff $\add P=0$ iff $\duparrow(P)=0$ iff
$\ddownarrow(P)=0$ iff $\link^{\uparrow}(P)=0$ iff
$\link^{\downarrow}(P)=0$ iff $c^{\uparrow}(P)=0$ iff
$\cdownarrow(P)=0$ iff
$\sat^{\uparrow}(P)=1$ iff $\sat^{\downarrow}(P)=1$ iff $\FN(P)=0$.

\spheader 511Hd $P$ is upwards-directed iff
$c^{\uparrow}(P)\le 1$ iff $\sat^{\uparrow}(P)\le 2$ iff
$\link^{\uparrow}(P)\le 1$ iff $\duparrow(P)\le 1$.   \cmmnt{Similarly,}
$P$ is downwards-directed iff $\cdownarrow(P)\le 1$ iff
$\sat^{\downarrow}(P)\le 2$ iff $\link^{\downarrow}(P)\le 1$ iff
$\ddownarrow(P)\le 1$.

If $P$ is not empty, it is upwards-directed iff $\add P>2$ iff
$\add P\ge\omega$.

\spheader 511He If $P$ is partially ordered, it has a greatest element
iff $\cf P=1$ iff
$\add P=\infty$.   Otherwise, $\add P\le\cf P$\prooflet{, since no
cofinal subset of $P$ can have an upper bound in $P$}.

\spheader 511Hf If $P$ is totally ordered, then $\cf P\le\add P$.
\prooflet{\Prf\ If $A\subseteq P$ has no upper bound in $P$ it must be
cofinal with $P$.\ \Qed}

\spheader 511Hg If $\familyiI{P_i}$ is a non-empty family of non-empty
pre-ordered sets with product $P$,
then $\add P=\min_{i\in I}\add P_i$.
\prooflet{\Prf\ A set $A\subseteq P$ lacks an upper bound in $P$ iff there
is an $i\in I$ such that $\{p(i):p\in A\}$ is unbounded above in $P_i$.\
\Qed}

\spheader 511Hh\dvAnew{2023} If $P$ is partially ordered and $f:P\to\Cal PP$ 
is a
Freese-Nation function then $p\in f(p)$ for every $p\in P$\prooflet{, because
$[p,p]$ meets $f(p)\cap f(p)$.}

\leader{511I}{Elementary facts:  Boolean algebras}\dvAformerly{5{}11J} Let
$\frak A$ be a Boolean algebra.

\spheader 511Ia

\Centerline{$\link_{<\kappa}(\frak A)\le\link_{<\lambda}(\frak A)
\le\pi(\frak A)$}

\noindent whenever $\kappa\le\lambda$,

\Centerline{$c(\frak A)\le\link(\frak A)\le d(\frak A)
\le\pi(\frak A)\le\#(\frak A)$,
\quad$c(\frak A)\le\sat(\frak A)$.}

\noindent\cmmnt{In addition, }$\tau(\frak A)
\le\pi(\frak A)$\prooflet{ because any order-dense
subset of $\frak A\,\,\tau$-generates $\frak A$}.

\spheader 511Ib{\bf (i)} $\frak A=\{0\}$ iff $\pi(\frak A)=0$ iff
$\link(\frak A)=0$ iff $d(\frak A)=0$ iff $c(\frak A)=0$ iff
$\sat(\frak A)=1$.

\medskip

\quad{\bf (ii)} $\tau(\frak A)=0$ iff $\frak A$ is either $\{0\}$ or
$\{0,1\}$.

\spheader 511Ic If $\frak A$ is finite, then
$c(\frak A)=\link(\frak A)=d(\frak A)=\pi(\frak A)$ is the number of
atoms of $\frak A$, $\sat(\frak A)=c(\frak A)+1$ and
$\#(\frak A)=2^{c(\frak A)}$, while
$\tau(\frak A)=\lceil\log_2c(\frak A)\rceil$, unless $\frak A=\{0\}$, in
which case $\tau(\frak A)=0$.   If
$\frak A$ is infinite then $c(\frak A)$, $\link(\frak A)$, $d(\frak A)$,
$\pi(\frak A)$, $\sat(\frak A)$
and $\tau(\frak A)$ are all infinite.

\spheader 511Id\cmmnt{ Note that} $\frak A$ is `ccc' just when
$c(\frak A)\le\omega$, that is, $\sat(\frak A)\le\omega_1$.   $\frak A$
is \wsid\cmmnt{, in the sense of
316G,} iff $\wdistr(\frak A)\ge\omega_1$.

\spheader 511Ie{\bf (i)} If $\frak A$ is purely atomic,
$\wdistr(\frak A)=\infty$.    \prooflet{\Prf\ Suppose that
$\ofamily{\xi}{\kappa}{A_{\xi}}$ is any family of partitions of unity in
$\frak A$.   Then the set $B$ of atoms of $\frak A$ is a partition of
unity, and $\{a:a\in A_{\xi}$, $a\Bcap b\ne 0\}$ has just one member for
every $b\in B$ and $\xi<\kappa$.   As
$\ofamily{\xi}{\kappa}{A_{\xi}}$ is arbitrary,
$\wdistr(\frak A)=\infty$.\ \Qed}

\medskip

\quad{\bf (ii)} If $\frak A$ is not purely atomic,
$\wdistr(\frak A)\le\pi(\frak A)$.
\prooflet{\Prf\ Let $c\in\frak A^+$ be disjoint
from every atom of $\frak A$, and $D\subseteq\frak A$ an order-dense set of
size $\pi(\frak A)$;  let $D'$ be $\{d:d\in D$, $d\Bsubseteq c\}$.
For $d\in D'$, there is a disjoint sequence of non-zero elements included
in $d$;  let $A_d$ be a partition of unity in $\frak A$ including
such a sequence.   If $B$ is any partition of unity in $\frak A$, there are
a $b\in B$ such that $b\Bcap c\ne 0$, and a $d\in D'$ such that
$d\Bsubseteq b\Bcap c$;  now $\{a:a\in A_d$, $b\Bcap a\ne 0\}$ is infinite.
So $\family{d}{D'}{A_d}$ witnesses that
$\wdistr(\frak A)\le\#(D')\le\pi(\frak A)$.\ \Qed}

\spheader 511If $\frak m(\frak A)=\infty$ iff $\frak A$ is purely
atomic.   \prooflet{\Prf\ Write $\Cal B$ for the family of all coinitial
subsets of $\frak A^+$.   (i) If $\frak A$ is purely atomic and
$a\in\frak A^+$, then there is an atom $d\Bsubseteq a$;  now $d\in B$
for every $B\in\Cal B$, so $\{d,a\}$ is a linked subset of $\frak A$
meeting every member of $\Cal B$.   Accordingly
$\frak m(\frak A)=\infty$.   (ii) If $\frak A$ is not purely atomic, let
$a\in\frak A^+$ be such that no atom of $\frak A$ is included in $a$.
\Quer\ If $A$ is a linked subset of $\frak A$ containing $a$ and meeting
every member of $\Cal B$, set
$B=\frak A^+\setminus A$.   If $b\in\frak A^+$, then either $b\Bcap a=0$
and $b\in B$, or there are non-zero disjoint $b'$,
$b''\Bsubseteq b\Bcap a$ and one of $b'$, $b''$ must belong to $B$.   So
$B\in\Cal B$, which is impossible.\ \BanG\
So $\frak m(\frak A)\le\#(\Cal B)<\infty$.\ \Qed}

\leader{511J}{Elementary facts:  ideals of sets}\dvAformerly{5{}11K}  Let
$X$ be a set and $\Cal I$ an ideal of subsets of $X$.

\spheader 511Ja $\add\Cal I\ge\omega$\prooflet{, by the definition of
`ideal of sets'}.

\spheader 511Jb
$\shr\Cal I=\sup\{\non(A,\Cal I\cap\Cal PA):A\in\Cal PX\setminus\Cal I\}$,
counting $\sup\emptyset$ as $0$;
$\shr\Cal I\le\#(X)$;   $\shr\Cal I\le\shr^+\Cal I\le(\shr\Cal I)^+$;
if $\shr\Cal I$ is a successor cardinal, $\shr^+\Cal I=(\shr\Cal I)^+$.

\spheader 511Jc Suppose that $\Cal I$ covers $X$ but does not contain
$X$.   Then $\add\Cal I\le\cov\Cal I\le\cf\Cal I$ and
$\add\Cal I\le\non\Cal I\le\shr\Cal I\le\cf\Cal I$.   \prooflet{\Prf\
Let $\Cal J$ be a subset of $\Cal I$ with cardinal $\cov\Cal I$
covering $X$;  let $\Cal K$ be a cofinal
subset of $\Cal I$ with cardinal $\cf\Cal I$;  let
$A\in\Cal PX\setminus\Cal I$ be such that $\#(A)=\non\Cal I$.
(i) $\Cal J$ cannot have an upper bound in $\Cal I$, so
$\add\Cal I\le\#(\Cal J)=\cov\Cal I$.   (ii)
$\bigcup\Cal K=\bigcup\Cal I=X$, so $\cov\Cal I\le\#(\Cal K)=\cf\Cal I$.
(iii) For each $x\in A$ we can
find an $I_x\in\Cal I$ containing $x$;  now $\{I_x:x\in A\}$ cannot have
an upper bound in $\Cal I$, so $\add\Cal I\le\#(A)=\non\Cal I$.   (iv)
By (b), $\shr\Cal I\ge\non\Cal I$.   (v) Take any $B\subseteq X$ such that
$B\notin\Cal I$.   Then for each $K\in\Cal K$ we can find an
$x_K\in B\setminus K$;  now $B'=\{x_K:K\in\Cal K\}$ is not included in
any member of $\Cal K$, so cannot belong to $\Cal I$, while
$B'\subseteq B$ and $\#(B')\le\#(\Cal K)=\cf\Cal I$.   As $B$ is
arbitrary, $\shr\Cal I\le\cf\Cal I$.\ \Qed}

\spheader 511Jd Suppose that $X\in\Cal I$.   Then
$\add\Cal I=\non\Cal I=\infty$,
$\cov\Cal I\le 1$ (with $\cov\Cal I=0$ iff $X=\emptyset$) and
$\shr\Cal I=0$.

\spheader 511Je Suppose that $\Cal I$ has a greatest member which is not
$X$.   Then $\add\Cal I=\cov\Cal I=\infty$ and
$\non\Cal I=\shr\Cal I=\cf\Cal I=1$.

\spheader 511Jf Suppose that $\Cal I$ has no greatest member and does
not cover $X$.   Then $\add\Cal I\le\cf\Cal I$\prooflet{ (511He)},
$\non\Cal I=\shr\Cal I=1$ and
$\cov\Cal I=\infty$.

\spheader 511Jg Suppose that $Y\subseteq X$, and set
$\Cal I_Y=\Cal I\cap\Cal PY$, regarded as an ideal of subsets of $Y$.
Then $\add\Cal I_Y\ge\add\Cal I$, $\non\Cal I_Y\ge\non\Cal I$,
$\shr\Cal I_Y\le\shr\Cal I$, $\shr^+\Cal I_Y\le\shr^+\Cal I$,
$\cov\Cal I_Y\le\cov\Cal I$ and $\cf\Cal I_Y\le\cf\Cal I$.


\exercises{\leader{511X}{Basic exercises (a)}
%\spheader 511Xa
Let $P$ be a partially ordered set and $\kappa\ge 3$ a  cardinal.   Show
that $\add P\ge\kappa$ iff $(\lambda,\lambda,\lambda)$ is an
upwards precaliber triple of $P$ for every $\lambda<\kappa$.
%511E

\sqheader 511Xb Let $X$ be a compact Hausdorff space.   Show that a pair
$(\kappa,\lambda)$ of cardinals is a precaliber pair of $X$ iff whenever
$\ofamily{\xi}{\kappa}{G_{\xi}}$ is a family of non-empty open subsets
of $X$ there is an $x\in X$ such that $\{\xi:x\in G_{\xi}\}$ has
cardinal at least $\lambda$.
%511E out of order query

\spheader 511Xc Let $(X,\Sigma,\mu)$ be a measure space.   For
$A\subseteq X$ write $\mu_A$ for the subspace measure on $A$, and
$\Cal N(\mu)$, $\Cal N(\mu_A)$ for the corresponding null ideals.   Show
that $\shr(X,\Cal N(\mu))
=\sup\{\non(A,\Cal N(\mu_A)):A\in\Cal PX\setminus\Cal N(\mu)\}$.
%511F

%\query gap

\spheader 511Xd Let $(X,\Sigma,\mu)$ be a measure space, and let
$\hat\mu$, $\tilde\mu$ be the completion and c.l.d.\ version of $\mu$.
(i) Let $\Cal N(\mu)=\Cal N(\hat\mu)$ and
$\Cal N(\tilde\mu)$ be the corresponding null ideals.   Show that
$\add\Cal N(\mu)\le\add\Cal N(\tilde\mu)$,
$\cov\Cal N(\mu)\ge\cov\Cal N(\tilde\mu)$,
$\non\Cal N(\mu)\le\non\Cal N(\tilde\mu)$,
$\shr\Cal N(\mu)\ge\shr\Cal N(\tilde\mu)$
and $\shr^+\Cal N(\mu)\ge\shr^+\Cal N(\tilde\mu)$.
(ii) Show that $\add\mu\le\add\hat\mu\le\add\tilde\mu$,
$\pi(\mu)=\pi(\hat\mu)\le\pi(\tilde\mu)$\dvAnew{2013} and
$\tau(\mu)=\tau(\hat\mu)\ge\tau(\tilde\mu)$\dvAnew{2014}.
%511G

%\query gap

\spheader 511Xe Show that if $P$ is a partially ordered set and
$c^{\uparrow}(P)<\omega$ then
$c^{\uparrow}(P)=\link^{\uparrow}(P)=\duparrow(P)$ and
$\frak m^{\uparrow}(P)=\infty$.
%511H

\spheader 511Xf Let $P$ be a partially ordered set.   Show that $\omega$
is an up-precaliber of $P$ iff $c^{\uparrow}(P)<\omega$.
%511H

\sqheader 511Xg(i) Show that if $P$ is a partially ordered set and
$\kappa$ is an up-precaliber of $P$, then $\cf\kappa$ is
also an up-precaliber of $P$.   (ii) Show that if $\kappa$ is a cardinal
and $\cf\kappa>\cf P$ then $\kappa$ is an up-precaliber of $P$.
%511H

\sqheader 511Xh Give $\Bbb R$, $\Bbb N$ and $\Bbb Q$ their usual total
orders.   Show that $\FN(\Bbb N)=\FN(\Bbb Q)=\omega$ and that
$\FN(\Bbb R)=\omega_1$.
%511H

\spheader 511Xi Show that if $P$ is a partially ordered set and
$\#(P)\ge 3$ then $\FN(P)\le\#(P)$.   \Hint{consider separately the
cases $P$ infinite, $P$ finite with no greatest member, and $P$ finite
with greatest and least members.}
%511H

\spheader 511Xj Let $P$ be a partially ordered set and $\Cal Q$ a
family of subsets of $P$ with $\#(\Cal Q)<\add P$.   Show that if
$\bigcup\Cal Q$ is cofinal with $P$ then one of the members of $\Cal Q$
is cofinal with $P$.
%511H

\spheader 511Xk Let $U$ be a Riesz space and $\kappa$ a cardinal.   Then
$U$ is {\bf weakly $(\kappa,\infty)$-distributive} if
whenever $\ofamily{\xi}{\kappa}{A_{\xi}}$ is a family of non-empty
downwards-directed subsets of $U^+$, each with infimum $0$, and
$\bigcup_{\xi<\kappa}A_{\xi}$ has an upper bound in $U$, then

\Centerline{$\{u:u\in U$, for every $\xi<\kappa$ there is a
$v\in A_{\xi}$ such that $v\le u\}$}

\noindent has infimum $0$ in $U$.   Show that an Archimedean Riesz space
is weakly $(\kappa,\infty)$-distributive iff its band algebra is.
\Hint{368R.}
%511I

\spheader 511Xl\dvAnew{2014} Let $X$ be a set and $\Cal I$ an ideal of
subsets of $X$.   Show that the coinitiality $\ci(\Cal PX\setminus\Cal I)$
is at most $\#(X)^{\shr\Cal I}$.
%511J

\leader{511Y}{Further exercises (a)}(i)
%\spheader 511Ya (i)
Show that $\duparrow(P)\le 2^{\link^{\uparrow}(P)}$ for every
partially ordered set $P$.
% if \ofamily}\xi}{\kappa}{A_{\xi}} maximal up-linked subsets covering  P
% for  p\in P  say  C_p=\{\x:p\in A_{\xi}\}
% if  C_p=C_q  and  p\le p'  then  p'\in A_{\xi} , \xi\in C_p=C_q  and
% q , p'  are linked;  hence  \{p:C_p=C\}  up-centered
(ii) Show that there is a partially ordered set $P$ such that
$\duparrow(P)=\omega$ but $P$ cannot be covered by countably many
upwards-directed sets.
%511H mt51bits n06328

\spheader 511Yb Let $\kappa$ be an infinite cardinal, with its usual
well-ordering.  Show that $\FN(\kappa)=\kappa$.
%511H

\spheader 511Yc(i) Find a semi-finite measure space $(X,\Sigma,\mu)$
such that $\cf\Cal N(\mu)<\cf\Cal N(\tilde\mu)$, where $\Cal N(\mu)$ and
$\Cal N(\tilde\mu)$
are the null ideals of $\mu$ and its c.l.d.\ version.
(ii) Find a semi-finite measure space $(X,\Sigma,\mu)$ such that
$\add\Cal N(\tilde\mu)>\add\Cal N(\mu)$ and
$\cf\Cal N(\tilde\mu)<\cf\Cal N(\mu)$.
%511J mt51bits

\spheader 511Yd\dvAnew{2017} 
Show that, for a set $I$, $(\omega_1,\omega,\omega)$ is a
precaliber triple of $\BbbN^I$ iff $I$ is countable.
}%end of exercises

\endnotes{
\Notesheader{511} Because $(P,\ge)$ is a pre-ordered set whenever
$(P,\le)$ is, any cardinal function on pre-ordered sets is bound
to appear in two mirror-image forms.   It does not quite follow that we
have to set up a language with a complete set of mirror pairs of
definitions, and indeed I have omitted the reflections of `additivity'
and `bursting number';   but the naturally arising pre-ordered
sets to which we shall want to apply these ideas may appear in either
orientation.   The most natural conversions to topological spaces and
Boolean algebras use the families of non-empty open sets and non-zero
elements, which are `active downwards', so that we have such formulae as
$\pi(\frak A)=\ci\frak A^+$ and
$c(X)=\cdownarrow(\frak T\setminus\{\emptyset\})$;  but we could
equally well
say that $\pi(\frak A)=\cf\frak A^-$ or that $c(X)$ is the
upwards-cellularity of the partially ordered set of proper closed
subsets of $X$.

Most readers, especially those acquainted with Volumes 3 and 4 of this
treatise, will be more familiar with topological spaces and Boolean
algebras than with general pre-ordered sets, and will prefer to
approach the concepts here through the formulations in 5A4A and 511D.
But even in the present chapter we shall be looking at questions which
demand substantial fragments of the theory of general partially ordered
sets, and I think it is useful to grapple with these immediately.
The list of definitions above
is a long one, and the functions here vary widely in importance;  but I
hope you will come to agree that all are associated with interesting
questions.

I apologise for introducing two cardinal functions to represent the
`breadth' of a pre-ordered set (or topological space or Boolean
algebra), its `cellularity'
and `saturation'.   It turns out that the saturation of a space
determines its cellularity (513Bc), which seems to render the concept of
`cellularity' unnecessary;  but it is well-established and makes some
formulae simpler.
%332E 332F 332R
This is an example of
a standard problem:  whenever we give a name to
a supremum, we find ourselves
asking whether the supremum is attained.   The question of whether
cellularity is attained turns out to be rather interesting (513B again).
In the case of shrinking numbers, the ordinary shrinking number
$\shr\Cal I$ is the one which has been most studied, but I shall have some
results which are more elegantly expressed in terms of the augmented
shrinking number $\shr^+\Cal I$.

I give very little space here to the functions $\frak m()$ and
$\wdistr()$ and to precalibers;   these are bound to be a bit
mysterious.   Later in the chapter I will explore their relations
with each other and with other cardinal functions.   You may recognise
them as belonging to the general area associated with Martin's axiom
({\smc Fremlin 84a}, or \S517 below).   `Precaliber pairs' have a
slightly more direct description in the context of compact Hausdorff
spaces (511Xb).   `Freese-Nation numbers' relate to quite different
aspects of the structure of ordered sets.   As will be made
clear in the next two sections, all the other cardinal functions defined
in 511B refer to the cofinal (or coinitial) structure of a partially
ordered set;  the Freese-Nation number, by contrast, tells us something
about the nature of intervals inside it.   We see a difference already
in the formula for the
Freese-Nation number of a Boolean algebra, which refers to the whole
algebra $\frak A$ rather than to $\frak A^+$.   Another
signal is the fact that it is not a trivial matter to calculate the
Freese-Nation number of a finite partially ordered set.

The only cardinal functions I have explicitly defined for measure spaces
are the additivity and $\pi$-weight of a measure (511G), and even these
are, in the most
important cases, reducible to the additivity of the null ideal (521A) and
the $\pi$-weight of the measure algebra (521Da).
I give a pair of warming-up exercises (511Xc-511Xd),
but we shall hardly see `measure' again until Chapter 52.
For the questions studied in this volume, the important
cardinals associated with a measure $\mu$ are those defined from its
measure algebra together with the four cardinals $\add\Cal N(\mu)$,
$\cov\Cal N(\mu)$, $\non\Cal N(\mu)$ and $\cf\Cal N(\mu)$.   In
particular, the
additivity of Lebesgue measure will have a special position.   In the
case of a topological measure space, of course, we can investigate
relationships between the cardinal functions of the topology and the
cardinal functions of the measure.   I will come to such questions in
Chapter 53.
}%end of notes

\discrpage

\frfilename{mt512.tex}
\versiondate{27.11.13}
\copyrightdate{2002}

\def\chaptername{Cardinal functions}
\def\sectionname{Galois-Tukey connections}

\def\Vojtas{Vojt\'a\v{s} 93}

\newsection{512}

One of the most powerful methods of relating the cardinals associated
with two partially ordered sets $P$ and $Q$ is to identify a `Tukey
function' from one to the other (513D).   It turns out that the idea can
be usefully generalized to other relational structures through the
concept of `Galois-Tukey connection' (512A).   In this section I give
the elementary theory of these connections and their effect on simple
cardinal functions.

\leader{512A}{Definitions (a)} A {\bf supported relation} is a triple
$(A,R,B)$ where $A$ and $B$ are sets and $R$ is a subset of $A\times B$.

\cmmnt{It will be convenient, and I think not dangerous, to abuse
notation by writing $(A,\in,B)$ or $(A,\subseteq,B)$ to mean $(A,R,B)$
where $R$ is $\{(a,b):a\in A,\,b\in B,\,a\in b\}$ or
$\{(a,b):a\in A,\,b\in B,\,a\subseteq b\}$.}

\spheader 512Ab If $(A,R,B)$ is a supported relation, its {\bf dual} is
the supported relation $(A,R,B)^{\perp}=(B,S,A)$ where

\Centerline{$S=(B\times A)\setminus R^{-1}
=\{(b,a):a\in A,\,b\in B,\,(a,b)\notin R\}$.}

\spheader 512Ac If $(A,R,B)$ and $(C,S,D)$ are supported relations, a
{\bf Galois-Tukey connection} from $(A,R,B)$ to $(C,S,D)$ is a pair
$(\phi,\psi)$ such that $\phi:A\to C$ and $\psi:D\to B$ are functions
and $(a,\psi(d))\in R$ whenever $(\phi(a),d)\in S$.

\leaveitout{alternatively:
$S\frsmallcirc\phi\subseteq\psi^{-1}\frsmallcirc R$}

\spheader 512Ad \cmmnt{({\smc Vojt\'a\v{s} 93})} If $(A,R,B)$ and
$(C,S,D)$ are supported relations, I write $(A,R,B)\prGT(C,S,D)$ if
there is a Galois-Tukey connection from $(A,R,B)$ to $(C,S,D)$, and
$(A,R,B)\equivGT(C,S,D)$ if $(A,R,B)\prGT(C,S,D)$ and
$(C,S,D)\prGT(A,R,B)$.

\leader{512B}{Definitions (a)} If $(A,R,B)$ is a
supported relation, its {\bf covering number}
$\cov(A,R,B)$\cmmnt{ (sometimes called {\bf norm} $\|(A,R,B)\|$)} is
the least cardinal of any set $C\subseteq B$ such
that $A\subseteq R^{-1}[C]$;  or $\infty$ if $A\not\subseteq R^{-1}[B]$.
Its {\bf additivity} is $\add(A,R,B)=\cov(A,R,B)^{\perp}$, that is, the
smallest cardinal of any subset $C\subseteq A$ such that
$C\not\subseteq R^{-1}[\{b\}]$ for any $b\in B$;  or $\infty$ if there
is no such $C$.

\cmmnt{Note that $\add(A,R,B)=0$ iff $B=\emptyset$, and that
$\add(A,R,B)=1$ iff $B\ne\emptyset$ and $\cov(A,R,B)=\infty$.}

\spheader 512Bb If $(A,R,B)$ is a supported relation, its {\bf
saturation} $\sat(A,R,B)$ is the least cardinal $\kappa$ such that
whenever $\ofamily{\xi}{\kappa}{a_{\xi}}$ is a family in $A$ then there
are distinct $\xi$, $\eta<\kappa$ and a $b\in B$ such that $(a_{\xi},b)$
and $(a_{\eta},b)$ both belong to $R$;  if there is no such
$\kappa$\cmmnt{ (that is, if $\cov(A,R,B)=\infty$)} I write
$\sat(A,R,B)=\infty$.

\spheader 512Bc If $(A,R,B)$ is a supported relation and $\kappa$ is a
cardinal, say that a subset $A'$ of $A$ is
{\bf $\hbox{$<$}\kappa$-linked} if for
every $I\in[A']^{<\kappa}$ there is a $b\in B$ such that
$I\subseteq R^{-1}[\{b\}]$, and {\bf $\kappa$-linked} if it is
$\hbox{$<$}\kappa^+$-linked\cmmnt{, that is, for every
$I\in[A']^{\le\kappa}$
there is a $b\in B$ such that $I\subseteq R^{-1}[\{b\}]$}.   Now the
{\bf $\hbox{$<$}\kappa$-linking number} $\link_{<\kappa}(A,R,B)$ of
$(A,R,B)$ is the
least cardinal of any cover of $A$ by $\hbox{$<$}\kappa$-linked sets, if
there is such a cover, and otherwise is $\infty$;  and the
{\bf $\kappa$-linking number} $\link_{\kappa}(A,R,B)$ of $(A,R,B)$ is
$\link_{<\kappa^+}(A,R,B)$\cmmnt{, that is, the least cardinal of any
cover of $A$ by $\kappa$-linked sets}.

If $\kappa\le\lambda$, then\cmmnt{ every $\hbox{$<$}\lambda$-linked
set is $\hbox{$<$}\kappa$-linked,
so} $\link_{<\kappa}(A,R,B)\le\link_{<\lambda}(A,R,B)$.
\cmmnt{Note also that} $\link_{\kappa}(A,R,B)$ is equal to
$\cov(A,R,B)$ for every $\kappa\ge\#(A)$, so that
$\link_{<\theta}(A,R,B)\le\cov(A,R,B)$ for every $\theta$.

\leader{512C}{}\cmmnt{ There are two things which should be done at
once:  to plainly state enough of the elementary theory to show at least
that the definitions here lead to a coherent structure;  and to give
examples.   I begin with the theory, which really is elementary.

\medskip

}\noindent{\bf Theorem} Let $(A,R,B)$, $(C,S,D)$ and $(E,T,F)$ be
supported relations.

(a) $(A,R,B)^{\perp\perp}=(A,R,B)$.

(b) If $(\phi,\psi)$ is a Galois-Tukey connection from $(A,R,B)$ to
$(C,S,D)$ and $(\phi',\psi')$ is a Galois-Tukey connection from
$(C,S,D)$ to $(E,T,F)$, then $(\phi'\phi,\psi\psi')$ is a Galois-Tukey
connection from $(A,R,B)$ to $(E,T,F)$.

(c) If $(\phi,\psi)$ is a Galois-Tukey connection from $(A,R,B)$ to
$(C,S,D)$, then $(\psi,\phi)$ is a Galois-Tukey connection from
$(C,S,D)^{\perp}$ to $(A,R,B)^{\perp}$.

(d) If $R'\subseteq R$ then $(A,R,B)\prGT(A,R',B)$.

(e) If $(A,R,B)\prGT(C,S,D)$ and $(C,S,D)\prGT(E,T,F)$ then
$(A,R,B)\prGT(E,T,F)$.

(f) $\equivGT$ is an equivalence relation on the class of supported
relations.

(g) If $(A,R,B)\prGT(C,S,D)$ then $(C,S,D)^{\perp}\prGT(A,R,B)^{\perp}$.
So if $(A,R,B)\equivGT(C,S,D)$ then
$(A,R,B)^{\perp}\equivGT(C,S,D)^{\perp}$.

\proof{(a)-(c) are immediate from the definitions.   (d) is trivial
because the identity functions from $A$ and $B$ to themselves form a
Galois-Tukey connection from $(A,R,B)$ to $(A,R',B)$.   (e) follows from
(b), and (g) from (c).   (f) is immediate from (d) and (e) and the
symmetry of the definition of $\equivGT$.
}%end of proof of 512C

\leader{512D}{Theorem} Let $(A,R,B)$ and $(C,S,D)$ be supported
relations such that $(A,R,B)\prGT(C,S,D)$.   Then

(a) $\cov(A,R,B)\le\cov(C,S,D)$;

(b) $\add(C,S,D)\le\add(A,R,B)$;
%formerly 512Da

(c) $\sat(A,R,B)\le\sat(C,S,D)$;

(d) $\link_{<\kappa}(A,R,B)\le\link_{<\kappa}(C,S,D)$
for every cardinal $\kappa$.
%formerly 512Dc

\proof{ Let $(\phi,\psi)$ be a Galois-Tukey connection from $(A,R,B)$ to
$(C,S,D)$.   If $D_0\subseteq D$
is such that $C=S^{-1}[D_0]$, then $A=R^{-1}[\psi[D_0]]$;  this shows
that $\cov(A,R,B)\le\cov(C,S,D)$.
If $\kappa=\sat(C,S,D)$ and $\ofamily{\xi}{\kappa}{a_{\xi}}$ is any
family in $A$, then there are a $d\in D$ and distinct $\xi$,
$\eta<\kappa$ such that $(\phi(a_{\xi}),d)\in S$ and
$(\phi(a_{\eta}),d)\in S$, in which case $(a_{\xi},\psi(d))$ and
$(a_{\eta},\psi(d))$ both belong to $R$;  so $\sat(A,R,B)\le\kappa$.
If $\Cal C$ is a cover of $C$ by
$\hbox{$<$}\kappa$-linked sets, then $\{\phi^{-1}[C']:C'\in\Cal C\}$ is
a cover of $A$ by $\hbox{$<$}\kappa$-linked sets;  this shows that
$\link_{<\kappa}(A,R,B)\le\link_{<\kappa}(C,S,D)$.

Finally,
$(C,S,D)^{\perp}\prGT(A,R,B)^{\perp}$, by 512Cc, so

\Centerline{$\add(C,S,D)=\cov(C,S,D)^{\perp}\le\cov(A,R,B)^{\perp}
=\add(A,R,B)$.}
}%end of proof of 512D

\leader{512E}{Examples} \cmmnt{ Of course `supported relations' appear
everywhere in mathematics.   They are important to us here because
covering numbers, saturation and linking numbers, as defined above,
correspond to
important cardinal functions as defined in \S511, and because surprising
Galois-Tukey connections exist, as we shall see in Chapter 52.   The
simplest examples are the following.

\medskip

}{\bf (a)} Let $(P,\le)$ be a pre-ordered set.   Then $(P,\le,P)$
and $(P,\ge,P)$ are supported relations, with duals $(P,\not\ge,P)$ and
$(P,\not\le,P)$.   $\cov(P,\le,P)=\cf P$, $\cov(P,\ge,P)=\ci P$,
$\add(P,\le,P)=\add P$ and $\sat(P,\le,P)=\sat^{\uparrow}(P)$.
\cmmnt{For any cardinal $\kappa$, a subset of $P$ is
upwards-$\hbox{$<$}\kappa$-linked
in the sense of 511Bf iff it is $\hbox{$<$}\kappa$-linked in $(P,\le,P)$
in the sense of
512Bc.   So} $\link_{<\kappa}^{\uparrow}(P)=\link_{<\kappa}(P,\le,P)$.
In particular, $\duparrow(P)=\link_{<\omega}(P,\le,P)$ (511Bg).

\spheader 512Eb\dvArevised{2011}
Let $(X,\frak T)$ be a topological space.   Then

\Centerline{$\pi(X)=\cov(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$,}

\Centerline{$d(X)=\cov(\frak T\setminus\{\emptyset\},\ni,X)
=\add(X,\notin,\frak T\setminus\{\emptyset\})$,}

\Centerline{$\sat(X)
=\sat(\frak T\setminus\{\emptyset\},\supseteq,
  \frak T\setminus\{\emptyset\})
=\sat(\frak T\setminus\{\emptyset\},\ni,X)$,}

\Centerline{$n(X)=\cov(X,\in,\CalNwd(X))
=\cov(X,\CalNwd(X))$}

\noindent where $\CalNwd(X)$ is the ideal of nowhere dense subsets of $X$.
Note that if $\Cal M(X)$ is the ideal of meager subsets of $X$, then
$\cov(X,\Cal M(X))=n(X)$ unless $n(X)=\omega$, in which case
$\cov(X,\Cal M(X))=1$.

\spheader 512Ec Let $\frak A$ be a Boolean algebra.   Write
$\frak A^+$ for $\frak A\setminus\{0\}$ and $\frak A^-$ for
$\frak A\setminus\{1\}$.   Then

\Centerline{$\pi(\frak A)=\cov(\frak A^+,\Bsupseteqshort,\frak A^+)
=\cov(\frak A^-,\Bsubseteqshort,\frak A^-)$,}

\Centerline{$\sat(\frak A)=\sat(\frak A^+,\Bsupseteqshort,\frak A^+)
=\sat(\frak A^-,\Bsubseteqshort,\frak A^-)$,}

\Centerline{$d(\frak A)
=\link_{<\omega}(\frak A^+,\Bsupseteqshort,\frak A^+)
=\link_{<\omega}(\frak A^-,\Bsubseteqshort,\frak A^-)$,}

\Centerline{$\link(\frak A)=\link_2(\frak A^+,\Bsupseteqshort,\frak A^+)
=\link_2(\frak A^-,\Bsubseteqshort,\frak A^-)$}

\noindent and generally

\Centerline{$\link_{<\kappa}(\frak A)
=\link_{<\kappa}(\frak A^+,\Bsupseteqshort,\frak A^+)
=\link_{<\kappa}(\frak A^-,\Bsubseteqshort,\frak A^-)$,}

\Centerline{$\link_{\kappa}(\frak A)
=\link_{\kappa}(\frak A^+,\Bsupseteqshort,\frak A^+)
=\link_{\kappa}(\frak A^-,\Bsubseteqshort,\frak A^-)$}

\noindent for every cardinal $\kappa$.

\spheader 512Ed Let $X$ be a set and $\Cal I$ an ideal of subsets of
$X$.   Then the dual of $(X,\in,\Cal I)$ is $(\Cal I,\not\ni,X)$;
$\cov(X,\in,\Cal I)=\cov\Cal I$ and $\add(X,\in,\Cal I)=\non\Cal I$.

\spheader 512Ee For a Boolean algebra $\frak A$, write
$\Pou(\frak A)$ for the set of partitions of unity in $\frak A$.
For $C$, $D\in\Pou(\frak A)$, say that $C\sqsubseteq^* D$
if every element of $D$ meets only finitely many members of $C$.
\cmmnt{Then} $\sqsubseteq^*$ is a pre-order on $\Pou(\frak A)$.
\cmmnt{Translating the definition 511Df into this language, we see that}
$\wdistr(\frak A)=\add\Pou(\frak A)$.

\leader{512F}{}\cmmnt{ I now turn to some constructions
involving supported relations and Galois-Tukey connections which will be
useful later.

\medskip

\noindent}{\bf Dominating sets} For any supported relation $(A,R,B)$ and
any cardinal $\kappa$, we can form a corresponding supported relation
$(A,R\strprime,[B]^{<\kappa})$, where

\Centerline{$R\strprime=\{(a,I):I\in[B]^{<\kappa}$,
$a\in R^{-1}[I]\}$.}

%call $R\strprime$ the `right extension' of $R$, or something?
%a name would be useful \query

\noindent\cmmnt{The most important cases to us will be $\kappa=\omega$
and $\kappa=\omega_1$.}   When $\kappa$ is a successor cardinal
I will normally write $(A,R\strprime,[B]^{\le\lambda})$ rather than
$(A,R\strprime,[B]^{<\lambda^+})$.

\leader{512G}{Proposition} Let $(A,R,B)$ and $(C,S,D)$ be supported
relations and $\kappa$, $\lambda$ cardinals.

(a) $(A,R,B)$ is isomorphic to $(A,R\strprime,[B]^1)$.

(b) If $(A,R,B)\prGT(C,S,D)$ and $\lambda\le\kappa$ then
$(A,R\strprime,[B]^{<\kappa})\prGT(C,S\strprime,[D]^{<\lambda})$.

(c) In particular, $(A,R\strprime,[B]^{<\kappa})\prGT(A,R,B)$ if
$\kappa\ge 2$.

(d) If $\cf\kappa\ge\lambda$ and
$(A,R\strprime,[B]^{<\kappa})\prGT(C,S,D)$ then
$(A,R\strprime,[B]^{<\kappa})\prGT(C,S\strprime,[D]^{<\lambda})$.

(e)(i) If $\cov(A,R,B)=\infty$ then
$\add(A,R\strprime,[B]^{<\kappa})\le 1$.

\quad(ii) If $\cov(A,R,B)<\infty$ then
$\add(A,R\strprime,[B]^{<\kappa})\ge\kappa$.

(f) $\cov(A,R,B)
\le\max(\omega,\kappa,\cov(A,R\strprime,[B]^{\le\kappa}))$;  if
$\kappa\ge 1$ and $\cov(A,R,B)>\max(\kappa,\omega)$ then
$\cov(A,R,B)\penalty-100=\cov(A,R\strprime,[B]^{\le\kappa})$.

\proof{{\bf (a)} is trivial.

\medskip

{\bf (b)} If $(\phi,\psi)$ is a Galois-Tukey connection from $(A,R,B)$
to $(C,S,D)$, then $(\phi,\psi')$ is a Galois-Tukey connection from
$(A,R\strprime,[B]^{<\kappa})$ to $(C,S\strprime,[D]^{<\lambda})$,
where $\psi'(J)=\psi[J]$ for every $J\in[D]^{<\lambda}$.

\medskip

{\bf (c)} Setting $\phi(a)=a$ for $a\in A$ and $\psi(b)=\{b\}$ for
$b\in B$, $(\psi,\phi)$ is a Galois-Tukey connection from
$(A,R\strprime,[B]^{<\kappa})$ to $(A,R,B)$.

\medskip

{\bf (d)} Let $(\phi,\psi)$ be a Galois-Tukey connection from
$(A,R\strprime,[B]^{<\kappa})$
to $(C,S,D)$.   Set $\psi'(I)=\bigcup_{d\in I}\psi(d)$ for
$I\in[D]^{<\lambda}$;  then $(\phi,\psi')$ is a
Galois-Tukey connection from $(A,R\strprime,[B]^{<\kappa})$ to
$(C,S\strprime,[D]^{<\lambda})$.

\medskip

{\bf (e)(i)} There is an $a\in A\setminus R^{-1}[B]$;  now
$(a,I)\notin R\strprime$ for any $I\in[B]^{<\kappa}$, so
$\add(A,R\strprime,[B]^{<\kappa})\le 1$.

\medskip

\quad{\bf (ii)} For every $a\in A$ there is a $b_a\in B$ such that
$(a,b_a)\in R$.   If $A'\subseteq A$ and $\#(A')<\kappa$, then
$I=\{b_a:a\in A'\}$ belongs to $[B]^{<\kappa}$, and
$(a,I)\in R\strprime$ for every $a\in A'$;  as $A'$ is arbitrary,
$\add(A,R\strprime,[B]^{<\kappa})\ge\kappa$.

\medskip

{\bf (f)} If $\lambda=\cov(A,R\strprime,[B]^{\le\kappa})$ is not
$\infty$, let
$\Cal D\subseteq[B]^{\le\kappa}$ be a set with cardinal $\lambda$ such that
$A=(R\strprime)^{-1}[\Cal D]$, and set $D=\bigcup\Cal D$;  then
$A\subseteq R^{-1}[D]$, so
$\cov(A,R,B)\le\#(D)\le\max(\omega,\kappa,\lambda)$.

If $\kappa\ge 1$, then
$\cov(A,R\strprime,[B]^{\le\kappa})\le\cov(A,R,B)$, by (c) and 512Da,
so if the latter is greater than $\max(\kappa,\omega)$ they are equal.
}%end of proof of 512G

\leader{512H}{Simple products (a)} If $\familyiI{(A_i,R_i,B_i)}$ is any
family of supported relations, its
{\bf simple product} is $(\prod_{i\in I}A_i,T,\prod_{i\in I}B_i)$ where
$T=\{(a,b):(a(i),b(i))\in R_i$ for every $i\in I\}$.

\spheader 512Hb Let $\familyiI{(A_i,R_i,B_i)}$ and
$\familyiI{(C_i,S_i,D_i)}$ be two families of supported relations, with
simple products $(A,R,B)$ and $(C,S,D)$.   If
$(A_i,R_i,B_i)\prGT(C_i,S_i,D_i)$ for every $i$, then
$(A,R,B)\prGT(C,S,D)$.   \prooflet{\Prf\ For each $i$, let
$(\phi_i,\psi_i)$ be a Galois-Tukey connection from $(A_i,R_i,B_i)$ to
$(C_i,S_i,D_i)$.   Define $\phi:A\to C$ and $\psi:D\to B$ by setting
$\phi(\familyiI{a_i})=\familyiI{\phi_i(a_i)}$,
$\psi(\familyiI{d_i})=\familyiI{\psi_i(d_i)}$ for $\familyiI{a_i}\in A$,
$\familyiI{d_i}\in D$;  then

$$\eqalign{(\phi(\familyiI{a_i}),\familyiI{d_i})\in S
&\Longrightarrow (\phi_i(a_i),d_i)\in S_i\text{ for every }i\in I\cr
&\Longrightarrow (a_i,\psi_i(d_i))\in R_i\text{ for every }i\in I\cr
&\Longrightarrow (\familyiI{a_i},\psi(\familyiI{d_i}))\in R.\cr}$$

\noindent So $(\phi,\psi)$ is a Galois-Tukey connection and
$(A,R,B)\prGT(C,S,D)$.\ \Qed}%end of prooflet

\spheader 512Hc Let $\familyiI{(A_i,R_i,B_i)}$ be a family of supported
relations with simple product $(A,R,B)$.   Suppose that no $A_i$ is
empty.   Then $\add(A,R,B)=\min_{i\in I}\add(A_i,R_i,B_i)$, interpreting
$\min\emptyset$ as $\infty$ if $I=\emptyset$.   \prooflet{\Prf\ Set
$\kappa=\add(A,R,B)$ and
$\kappa'=\min_{i\in I}\add(A_i,R_i,B_i)$.   If $I=\emptyset$ then
$A=B=\{\emptyset\}$ and $R=\{(\emptyset,\emptyset)\}$ so
$\add(A,R,B)=\infty$.    Otherwise, if $C\subseteq A$ and
$\#(C)<\kappa'$, then, for each $i$,
$\#(\{c(i):c\in C\})<\add(A_i,R_i,B_i)$, so there is a $b_i\in B_i$ such
that $(c(i),b_i)\in R_i$ for every $c\in C$;  now
$(c,\familyiI{b_i})\in R$ for every $c\in C$;  as $C$ is arbitrary,
$\kappa\ge\kappa'$.   In the other direction, if $i\in I$ and
$C'\in[A_i]^{<\kappa}$, then (because no $A_j$ is empty) there is a
$C\in[A]^{<\kappa}$ such that $C'=\{c(i):c\in C\}$.   Now there is a
$b\in B$ such that $(c,b)\in R$ for every $c\in C$, so that
$(c',b(i))\in R_i$ for every $c'\in C'$.  As $i$ and $C'$ are arbitrary,
$\kappa'\le\kappa$.\ \Qed}

\spheader 512Hd Suppose that $(A,R,B)$ and $(C,S,D)$ are supported
relations with simple product $(A\times C,T,B\times D)$.   Let $\kappa$
be an infinite cardinal and define $(A,R\strprime,[B]^{<\kappa})$,
$(C,S\strprime,[D]^{<\kappa})$ and
$(A\times C,T\strprime,[B\times D]^{<\kappa})$ as in 512F.   Then

\Centerline{$(A,R\strprime,[B]^{<\kappa})
  \times(C,S\strprime,[D]^{<\kappa})
\equivGT(A\times C,T\strprime,[B\times D]^{<\kappa})$.}

\prooflet{\noindent\Prf\ Express
$(A,R\strprime,[B]^{<\kappa})\times(C,S\strprime,[D]^{<\kappa})$
as $(A\times C,\tilde T,[B]^{<\kappa}\times[D]^{<\kappa})$.

\medskip

\quad{\bf (i)} Set
$\phi(a,c)=(a,c)$ for all $a\in A$, $c\in C$, and for
$I\in[B\times D]^{<\kappa}$ set

\Centerline{$\psi(I)
=(\pi_1[I],\pi_2[I])\in[B]^{<\kappa}\times[D]^{<\kappa}$,}

\noindent where $\pi_1(b,d)=b$ and $\pi_2(b,d)=d$ for $b\in B$,
$d\in D$.   If $a\in A$, $c\in C$ and $I\in[B\times D]^{<\kappa}$ are
such that $(\phi(a,c),I)\in T\strprime$, then there must be a
$(b,d)\in I$ such that $((a,c),(b,d))\in T$, that is, $(a,b)\in R$ and
$(c,d)\in S$;  now $b\in\pi_1[I]$ and $d\in\pi_2[I]$, so
$(a,\pi_1[I])\in R\strprime$ and $(c,\pi_2[I])\in S\strprime$ and

\Centerline{$((a,c),\psi(I))
=((a,c),(\pi_1[I],\pi_2[I]))\in\tilde T$.}

\noindent As $a$, $c$ and $I$ are arbitrary, $(\phi,\psi)$ is a
Galois-Tukey connection and

\Centerline{$(A,R\strprime,[B]^{<\kappa})
  \times(C,S\strprime,[D]^{<\kappa})
\prGT(A\times C,T\strprime,[B\times D]^{<\kappa})$.}

\medskip

\quad{\bf (ii)} In the other direction, given
$(J,K)\in[B]^{<\kappa}\times[D]^{<\kappa}$ set
$\psi'(J,K)=J\times K\in[B\times D]^{<\kappa}$.   (This is where we need
to suppose that $\kappa$ is infinite.)   If now
$(\phi(a,c),(J,K))\in\tilde T$, that is,
$(a,J)\in R\strprime$ and $(c,K)\in S\strprime$, there are
$b\in J$ and $d\in K$ such that $(a,b)\in R$ and $(c,d)\in S$, so that
$((a,c),(b,d))\in T$ and $((a,c),\psi'(J,K))\in T\strprime$.
As $a$, $c$, $J$ and $K$ are arbitrary, $(\phi,\psi')$ is a
Galois-Tukey connection and

\Centerline{$(A\times C,T\strprime,[B\times D]^{<\kappa})
\prGT(A,R\strprime,[B]^{<\kappa})
\times(C,S\strprime,[D]^{<\kappa})$.\ \Qed}
}%end of prooflet

\spheader 512He If $\familyiI{(P_i,\le_i)}$ is a family of
pre-ordered sets, with product $(P,\le)$\cmmnt{ (511A)},
then $(P,\le\nobreak,P)$ is just
\penalty-100$\prod_{i\in I}(P_i,\le_i\nobreak,P_i)$ in the sense here.

\vleader{72pt}{512I}{Sequential compositions}
Let $(A,R,B)$ and $(C,S,D)$ be supported relations.   Their {\bf
sequential composition} $(A,R,B)\ltimes(C,S,D)$ is
$(A\times C^B,T,B\times D)$, where

\Centerline{$T
=\{((a,f),(b,d)):(a,b)\in R,\,f\in C^B,\,(f(b),d)\in S\}$.}

\noindent Their {\bf dual sequential composition}
$(A,R,B)\rtimes(C,S,D)$ is $(A\times C,\tilde T,B\times D^A)$ where

$$\eqalign{\tilde T
&=\{((a,c),(b,g)):a\in A,\,b\in B,\,c\in C,\,g\in D^A\cr
&\mskip150mu
  \text{and either }(a,b)\in R\text{ or }(c,g(a))\in S\}.\cr}$$

\leader{512J}{Proposition} Let $(A,R,B)$ and $(C,S,D)$ be supported
relations.

(a)
$(A,R,B)\rtimes(C,S,D)=((A,R,B)^{\perp}\ltimes(C,S,D)^{\perp})^{\perp}$.

(b) $\cov((A,R,B)\ltimes(C,S,D))$ is the cardinal product
$\cov(A,R,B)\cdot\cov(C,S,D)$ unless
$B=C=\emptyset\ne A$, if we use the interpretations

\Centerline{$0\cdot\infty=\infty\cdot 0=0$,
\quad$\kappa\cdot\infty=\infty\cdot\kappa=\infty\cdot\infty=\infty$ for
every cardinal $\kappa\ge 1$.}

(c) $\add((A,R,B)\ltimes(C,S,D))=\min(\add(A,R,B),\add(C,S,D))$ unless
$A\times C=\emptyset\ne B\times D$.

\proof{{\bf (a)} is just a matter of disentangling the
definitions.

\medskip

{\bf (b)} Define $T\subseteq (A\times C^B)\times(B\times D)$ as in 512I.

\medskip

\quad{\bf (i)}
Suppose first that neither $A$ nor $C$ is empty, that
$A\subseteq R^{-1}[B]$ and that $C\subseteq S^{-1}[D]$.   If
$B_0\subseteq B$ and $D_0\subseteq D$ are such that
$A\subseteq R^{-1}[B_0]$ and $C\subseteq S^{-1}[D_0]$, then for any
$a\in A$ and $f\in C^B$ there are $b\in B_0$ and $d\in D_0$ such that
$(a,b)\in R$ and $(f(b),d)\in S$, so that
$(a,f)\in T^{-1}[B_0\times D_0]$.   So
$\cov((A,R,B)\ltimes(C,S,D))\le\cov(A,R,B)\cdot\cov(C,S,D)$.

On the other hand, if $H\subseteq B\times D$ is such that
$A\times C^B\subseteq T^{-1}[H]$, set
$B_0=\{b:C\subseteq S^{-1}[H[\{b\}]]\}$.   Then
$\#(H[\{b\}])\ge\cov(C,S,D)$ for $b\in B_0$.   Also
$A\subseteq R^{-1}[B_0]$.   \Prf\Quer\ Otherwise, take
$a\in A\setminus R^{-1}[B_0]$.   For $b\in B\setminus B_0$, choose
$f(b)\in C\setminus S^{-1}[H[\{b\}]]$;  for $b\in B_0$,
take $f(b)$ to be any member of $C$.   There is supposed to be a
member
$(b,d)$ of $H$ such that $((a,f),(b,d))\in T$, that is, $(a,b)\in R$ and
$(f(b),d)\in S$.   But now $b\notin B_0$, by the choice of $a$, and
$(f(b),d)\notin S$, by the choice of $f$;  so we have a
contradiction.\
\Bang\Qed

So $\#(B_0)\ge\cov(A,R,B)$ and $\#(H)\ge\cov(A,R,B)\cdot\cov(C,S,D)$;
as $H$ is arbitrary,
$\cov((A,R,B)\ltimes(C,S,D))\ge\cov(A,R,B)\cdot\cov(C,S,D)$.

\medskip

\quad{\bf (ii)}
If $A=\emptyset$ then $A\times C^B=\emptyset$ so $\cov(A,R,B)$ and
$\cov((A,R,B)\ltimes(C,S,D))$ are both zero.   If $C=\emptyset$ and
$B\ne\emptyset$ then $\cov(C,S,D)=\cov((A,R,B)\ltimes(C,S,D))=0$.
If $A$ and $C$ are non-empty and $A\not\subseteq R^{-1}[B]$, then
$A\times C^B\not\subseteq T^{-1}[B\times D]$, so
$\cov(A,R,B)=\cov((A,R,B)\ltimes\cov(C,S,D))=\infty$, while
$\cov(C,S,D)\ge 1$.   If $A$ and $C$ are non-empty and
$A\subseteq R^{-1}[B]$ and $C\not\subseteq S^{-1}[D]$, then
$B\ne\emptyset$;  if we take $c\in C\setminus S^{-1}[D]$ and any member
$a$ of $A$, and set $f(b)=c$ for every $b\in B$, then
$(a,f)\notin T^{-1}[B\times D]$, so
$\cov((A,R,B)\ltimes(C,S,D))=\cov(C,S,D)=\infty$, while
$\cov(A,R,B)\ge 1$.   So with the single exception of
$B=C=\emptyset\ne A$ (in which case the empty function belongs to $C^B$,
so that $\cov((A,R,B)\ltimes(C,S,D))=\infty$, while $\cov(C,S,D)=0$) we
have $\cov((A,R,B)\ltimes(C,S,D))=\cov(A,R,B)\cdot\cov(C,S,D)$.

\medskip

{\bf (c)} Assume throughout that either $A\times C\ne\emptyset$ (so that
$A\times C^B\ne\emptyset$) or that $B\times D=\emptyset$.

\medskip

\quad{\bf (i)} $\add((A,R,B)\ltimes(C,S,D))\le\add(A,R,B)$.   \Prf\ If
$\add(A,R,B)=\infty$ the result is trivial.   If $B\times D=\emptyset$
then $\add((A,R,B)\ltimes(C,S,D))=0\le\add(A,R,B)$.   Otherwise, our
hypothesis ensures that $C$ is not empty;  take
$A_0\subseteq A$ such that $\#(A_0)=\add(A,R,B)$ and
$A_0\not\subseteq R^{-1}[\{b\}]$ for any $b\in B$, take any $b_0\in B$
and any $f_0\in C^B$;  then there is no $(b,d)\in B\times D$ such that
$((a,f_0(b_0)),(b,d))\in T$ for every $a\in A_0$, so
$\add((A,R,B)\ltimes(C,S,D))\le\#(A_0)=\add(A,R,B)$.\ \Qed

\medskip

\quad{\bf (ii)} $\add((A,R,B)\ltimes(C,S,D))\le\add(C,S,D)$.   \Prf\
Again, if $\add(C,S,D)=\infty$ or $B\times D=\emptyset$ the result is
immediate.   Otherwise, $A\ne\emptyset$.   Take
$C_0\subseteq C$ such that $\#(C_0)=\add(C,S,D)$ and there is no
$d\in D$ such that $C_0\subseteq S^{-1}[\{d\}]$, for
$c\in C_0$ set $f_c(b)=c$ for every $b\in B$, and fix any $a_0\in A$;
then there is no $(b,d)\in B\times D$ such that
$((a_0,f_c),(b,d))\in T$ for every $c\in C_0$, so
$\add((A,R,B)\ltimes(C,S,D))\le\#(C_0)=\add(C,S,D)$.\ \Qed

\medskip

\quad{\bf (iii)}
$\add((A,R,B)\ltimes(C,S,D))\ge\min(\add(A,R,B),\add(C,S,D))$.
\Prf\ If
$H\subseteq A\times C^B$ and $\#(H)$ is less than
$\min(\add(A,R,B),\add(C,S,D))$, set
$A_0=\{a:(a,f)\in H\}$ and $F=\{f:(a,f)\in H\}$.
Then there are a $b\in B$ such that $(a,b)\in R$ for any $a\in A_0$,
and a $d\in D$ such that $(f(b),d)\in S$ for any $f\in F$, so that
$((a,f),(b,d))\in T$ for any $(a,f)\in H$.   As $H$ is arbitrary,
$\add((A,R,B)\ltimes(C,S,D))\ge\min(\add(A,R,B),\add(C,S,D))$.\ \Qed
}%end of proof of 512J

\leader{512K}{}\cmmnt{ The following fact will be used in \S526.
%526F

\medskip

\noindent}{\bf Lemma} Suppose that $(A,R,B)$ and $(C,S,D)$ are supported
relations, and $P$ is a partially ordered set.   Suppose that
$\family{p}{P}{A_p}$ is a family of subsets of $A$ such that

\Centerline{$(A_p,R,B)\prGT(C,S,D)$ for every $p\in P$,}

\Centerline{$A_p\subseteq A_q$ whenever $p\le q$ in $P$,
\quad$\bigcup_{p\in P}A_p=A$.}

\noindent Then $(A,R,B)\prGT(P,\le,P)\ltimes(C,S,D)$.

\proof{ If $C=\emptyset$ the result is trivial, since every $A_p$ is
empty and $B$ can be empty only if $D$ is.   So we may suppose that
$C\ne\emptyset$.   For each $p\in P$, let $(\phi_p,\psi_p)$ be a
Galois-Tukey
connection from $(A_p,R,B)$ to $(C,S,D)$.   For $a\in A$, let
$r(a)\in P$ be such that $a\in A_{r(a)}$, and set $f_a(p)=\phi_p(a)$
whenever $p\in P$ and $a\in A_p$;  for other $p\in P$ take $f_a(p)$ to
be any member of $C$.   Set $\phi(a)=(r(a),f_a)$ for $a\in A$.   For
$q\in P$, $d\in D$ set $\psi(q,d)=\psi_q(d)\in B$.   Now $(\phi,\psi)$
is a Galois-Tukey connection from $(A,R,B)$ to
$(P,\le,P)\ltimes(C,S,D)$.
\Prf\ Suppose that $a\in A$ and $(q,d)\in P\times D$ are such that
$r(a)\le q$ and $(f_a(q),d)\in S$.   Then $a\in A_{r(a)}\subseteq A_q$
so $f_a(q)=\phi_q(a)$.   Because $(\phi_q,\psi_q)$ is a Galois-Tukey
connection, $(a,\psi(q,d))=(a,\psi_q(d))\in R$.\ \Qed

So we have the result.
}%end of proof of 512K

\exercises{\leader{512X}{Basic exercises (a)}%
%\spheader 512Xa
(i) Suppose that $A\subseteq A'$, $B'\subseteq B$ and
that $R$ is any relation.   Show that $(A,R,B)\prGT(A',R,B')$.
(ii) Show that $(\emptyset,\emptyset,\{\emptyset\})\prGT(A,R,B)
\prGT(\{\emptyset\},\emptyset,\emptyset)$ for every supported relation
$(A,R,B)$.
%512A

\spheader 512Xb Let $(A,R,B)$ be any supported relation.   Show that
$\sat(A,R,B)\le(\link_2(A,R,B))^+$.
%512B

\spheader 512Xc\dvAformerly{5{}14Xa}
Let $(X,\frak T)$ be a topological space and
$(Y,\frak T_Y)$ an open subspace.    Show that
$(\frak T_Y\setminus\{\emptyset\},\supseteq,
  \frak T_Y\setminus\{\emptyset\})
\prGT(\frak T\setminus\{\emptyset\},\supseteq,
  \frak T\setminus\{\emptyset\})$.
%5A4B 512E

\spheader 512Xd Let $(X,\frak T)$ and $(Y,\frak S)$ be topological spaces.
(i) Show that if $Y$ is a continuous image of $X$,
$(\frak S\setminus\{\emptyset\},\penalty-50\ni\nobreak,Y)\penalty-100
   \prGT(\frak T\setminus\{\emptyset\},\ni,X)$.
(ii) Show that if $X$ and $Y$ are compact and Hausdorff and
there is an irreducible continuous surjection from $X$ onto $Y$, then
$(\frak T\setminus\{\emptyset\},\ni,X)\equivGT
(\frak S\setminus\{\emptyset\},\ni,Y)$ and
$(\frak T\setminus\{\emptyset\},\supseteq,\frak T\setminus\{\emptyset\})
\equivGT
(\frak S\setminus\{\emptyset\},\supseteq,\frak S\setminus\{\emptyset\})$,
so $d(X)=d(Y)$ and $\pi(X)=\pi(Y)$.
%512E 512Xc

\spheader 512Xe Let $\familyiI{(A_i,R_i,B_i)}$ be a family of supported
relations with simple product $(A,R,B)$.   Show that $(A,R,B)^{\perp}$
can be naturally identified with the simple product of
$\familyiI{(A_i,R_i,B_i)^{\perp}}$.
%512H

\spheader 512Xf Let $(A,R,B)$ be a supported relation and $\kappa>0$ a
cardinal.   Show that
$(A,R\strprime,[B]^{\le\kappa})\prGT(A,R,B)^{\kappa}$,
where $(A,R,B)^{\kappa}$ is the simple product of $\kappa$ copies of
$(A,R,B)$ and $R\strprime=\{(a,J):a\in R^{-1}[J]\}$ as
usual.
%512H

\spheader 512Xg Let $(A,R,B)$ and $(C,S,D)$ be supported relations, and
$(A\times C,T,B\times D)$ their simple product.   (i) Show that if
$C\ne\emptyset$, then
$(A,R,B)\prGT(A\times C,T,B\times D)$.   (ii) Show that
$(A\times C,T,B\times D)\prGT(A,R,B)\ltimes(C,S,D)$.
(iii) Show that (using the conventions of 512Jb)
$\cov(A\times C,T,B\times D)=\cov(A,R,B)\cdot\cov(C,S,D)$.
%512I

\spheader 512Xh Let $(A_0,R_0,B_0)$, $(A_1,R_1,B_1)$, $(C_0,S_0,D_0)$
and $(C_1,S_1,D_1)$ be supported relations such
that \ifdim\pagewidth>467pt\break\fi
$(A_0,R_0,B_0)\prGT(A_1,R_1,B_1)$ and
$(C_0,S_0,D_0)\prGT(C_1,S_1,D_1)$.   Show that

\Centerline{$(A_0,R_0,B_0)\ltimes(C_0,S_0,D_0)
\prGT(A_1,R_1,B_1)\ltimes(C_1,S_1,D_1)$,}

\Centerline{$(A_0,R_0,B_0)\rtimes(C_0,S_0,D_0)
\prGT(A_1,R_1,B_1)\rtimes(C_1,S_1,D_1)$.}
%512J
}%end of exercises

\endnotes{
\Notesheader{512} Much of this section is cluttered by
the repeated names $(A,R,B)$ of `supported relations'.   In fact these
could probably be dispensed with.   While I am reluctant to alter the
general convention I use in this book, that a `relation' is neither more
nor less than a class of ordered pairs, it is clear that in all
significant cases our supported relation $(A,R,B)$ will be such that
$A=\{a:(a,b)\in R\}$ and $B=\{b:(a,b)\in R\}$, so that $A$ and $B$ can
be recovered from the set $R$.   But this would make impossible the very
useful convention that `$(X,\in,\Cal A)$' is to be interpreted as
`$(X,\{(x,A):A\in\Cal A,\,x\in X\cap A\},\Cal A)$', and since nearly
every mathematical argument in this context demands names for the
domains and codomains of the relations, it seems easier to write these
in each time.

An important feature of the theory here is that while it is very common
for our relations to be reasonably well-behaved by some
criterion (for instance, we may have Polish spaces $A$ and
$B$ and a coanalytic set $R\subseteq A\times B$), the functions in a
Galois-Tukey connection are not required to have any properties beyond
those declared in the definition.   Of course the most important
Galois-Tukey
connections are those which are `natural' in some sense, and are
constructed in a way which does not involve totally unscrupulous use
of the axiom of choice.
I will return to this question in the next section.
}%end of notes

\discrpage

\frfilename{mt513.tex}
\versiondate{23.2.14}
\copyrightdate{2004}

\def\chaptername{Cardinal functions}
\def\sectionname{Partially ordered sets}

\newsection{513}

In \S\S511-512 I have given long
lists of definitions.   It is time I filled in details of
the most elementary relationships between the various concepts
introduced.   Here I treat some of those which can be expressed in the
language of partially ordered sets.   I begin with notes on cofinality
and saturation, with the Erd\H{o}s-Tarski theorem (513B).
In this context, Galois-Tukey connections take on particularly direct
forms (513D-513E);  for directed sets, we have an alternative definition
of Tukey equivalence (513F).   The majority of the cardinal functions
defined so far on partially ordered sets are determined by their cofinal
structure (513G, 513If\cmmnt{;  see also 516Ga below}).

In the last third of the section
(513K-513O), %513K 513L 513M 513N 513O
I discuss Tukey functions between directed sets with a special kind of
topological structure, which I call `metrizably compactly based';  the
point is
that for Polish metrizably compactly based directed sets, if there is
any Tukey function between them, there must be one
which is measurable in an appropriate sense (513O).

\leader{513A}{}\cmmnt{ It will help to have an elementary lemma on
maximal antichains.

\medskip

\noindent}{\bf Lemma} Let $P$ be a partially ordered set.

(a) If $Q\subseteq P$ is cofinal and $A\subseteq Q$ is an up-antichain,
there is a maximal up-antichain $A'$ in $P$ such that
$A\subseteq A'\subseteq Q$.   In particular, $Q$ includes a maximal
up-antichain.

(b) If $A\subseteq P$ is a maximal up-antichain,
$Q=\bigcup_{q\in A}\coint{q,\infty}$ is cofinal with $P$.

\proof{{\bf (a)} Let $A'\subseteq Q$ be maximal subject to being an
up-antichain in $P$ including $A$.   Then for any $p\in P\setminus A'$,
there are a $q\in Q$ such that $p\le q$, and an $r\in A'$ such that

\Centerline{$\emptyset\ne\coint{r,\infty}\cap\coint{q,\infty}
\subseteq\coint{r,\infty}\cap\coint{p,\infty}$,}

\noindent so $A'\cup\{p\}$ is not an up-antichain.   But this means that
$A'$ is a maximal up-antichain in $P$.

Starting from $A=\emptyset$, we see that $Q$ includes a maximal
up-antichain.

\medskip

{\bf (b)} If $p\in P$, then either $p\in A\subseteq Q$, or $A\cup\{p\}$
is not an up-antichain, so that there is some $q\in A$ such that

\Centerline{$\emptyset\ne\coint{p,\infty}\cap\coint{q,\infty}
\subseteq Q\cap\coint{p,\infty}$.}
}%end of proof of 513A

\leader{513B}{Theorem} Let $P$ be a partially ordered set.

(a) $\bu P\le\cf P\le\#(P)$.

(b) $\sat^{\uparrow}(P)$ is either finite or a regular uncountable
cardinal.

(c) $c^{\uparrow}(P)$ is the predecessor of $\sat^{\uparrow}(P)$ if
$\sat^{\uparrow}(P)$ is a successor cardinal, and otherwise is equal to
$\sat^{\uparrow}(P)$.

\proof{{\bf (a)} To see that $\cf P\le\#(P)$ all we have to note is that
$P$ is a cofinal subset of itself.   To see that $\bu P\le\cf P$, set
$\kappa=\cf P$ and let $\ofamily{\xi}{\kappa}{p_{\xi}}$ enumerate a
cofinal subset of $P$.   Set

\Centerline{$A
=\{\xi:\xi<\kappa,\,p_{\xi}\not\le p_{\eta}$ for any $\eta<\xi\}$,
\quad$Q=\{p_{\xi}:\xi\in A\}$.}

\noindent If $p\in P$ there is a least $\xi<\kappa$ such that
$p\le p_{\xi}$, and now $\xi\in A$;  so $Q$ is cofinal with $P$.   If
$p\in P$, there is some $\xi\in A$ such that $p\le p_{\xi}$, and now
$\{q:q\in Q$, $q\le p$, $p\not\le q\}\subseteq\{p_{\eta}:\eta<\xi\}$
has cardinal less than $\kappa$, so that $Q$ witnesses that
$\bu P\le\kappa$.

\medskip

{\bf (b)(i)} Set $\kappa=\sat^{\uparrow}(P)$.
For $p\in P$, set $\theta(p)=\sat^{\uparrow}(\coint{p,\infty})$.   Note
that if $p\in P$ and $B\subseteq P$ is any up-antichain, then
$B_p=\{q:q\in B$, $\coint{q,\infty}\cap\coint{p,\infty}\ne\emptyset\}$
has less than $\theta(p)$ members.   \Prf\ For $q\in B_p$, choose
$q'\in\coint{q,\infty}\cap\coint{p,\infty}$.   Because $B$ is an
up-antichain, $\{q':q\in B_p\}$ is an up-antichain and $q\mapsto q'$ is
injective;  so $\#(B_p)=\#(\{q':q\in B_p\})<\theta(p)$.\ \Qed

If $p\le q$ in $P$, any up-antichain in $\coint{q,\infty}$ is also an
up-antichain in $\coint{p,\infty}$, so $\theta(q)\le\theta(p)$.   It
follows that $Q=\{p:p\in P,\,\theta(q)=\theta(p)$ for every $q\ge p\}$
is cofinal with $P$.   Let $A\subseteq Q$ be a maximal up-antichain
(513Aa);  then $\#(A)<\kappa$.

\medskip

\quad{\bf (ii)} \Quer\ Suppose, if possible, that $\kappa=\omega$.

\medskip

\qquad{\bf case 1} Suppose there is a $p\in A$ such that
$\theta(p)=\omega$.   Then we can choose $\sequencen{p_n}$,
$\sequencen{q_n}$ inductively, as follows.   $p_0=p$.   Given that
$p_n\in Q$ and $\theta(p_n)=\omega$, there must be $p_{n+1}$,
$q_n\in\coint{p_n,\infty}$ such that
$\coint{p_{n+1},\infty}\cap\coint{q_n,\infty}=\emptyset$;  now
$p_{n+1}\in Q$ and $\theta(p_{n+1})=\omega$.   Continue.   At the end of
the induction, $\{q_n:n\in\Bbb N\}$ is an infinite up-antichain in $P$,
which is impossible.

\medskip

\qquad{\bf case 2} Suppose that $\theta(p)<\omega$ for every $p\in A$.
Then $n=\sum_{p\in A}\theta(p)$ is finite.   Let $B\subseteq P$ be any
up-antichain.   For each $p\in A$, set
$B_p=\{q:q\in B,\,\coint{q,\infty}\cap\coint{p,\infty}\ne\emptyset\}$;
as noted in (i), $\#(B_p)<\theta(p)$ for every $p\in A$, so
$\#(\bigcup_{p\in A}B_p)\le n$.   But
$B=\bigcup_{p\in A}B_p$, because $A$ is a maximal up-antichain, so
$\#(B)\le n$.   As $B$ is arbitrary,
$\sat^{\uparrow}(P)\le n+1<\omega$.\ \Bang

Thus $\kappa\ne\omega$.

\medskip

\quad{\bf (iii)} \Quer\ Suppose, if possible, that $\kappa$
is a singular infinite cardinal.   Set $\lambda=\cf\kappa$ and let
$\ofamily{\xi}{\lambda}{\kappa_{\xi}}$ be a strictly increasing family
of cardinals with supremum $\kappa$.

\medskip

\qquad{\bf case 1} Suppose there is a $p\in Q$ such that
$\theta(p)=\kappa$.   Then, because $\lambda<\kappa$, there is an
up-antichain $B\subseteq\coint{p,\infty}$ with cardinal $\lambda$;
enumerate $B$ as $\ofamily{\xi}{\lambda}{p_{\xi}}$.   For each
$\xi<\lambda$, $\theta(p_{\xi})>\kappa_{\xi}$, so there is an
up-antichain $C_{\xi}\subseteq\coint{p_{\xi},\infty}$ with cardinal
$\kappa_{\xi}$.   Now $C=\bigcup_{\xi<\lambda}C_{\xi}$ is an
up-antichain in $P$ with cardinal $\kappa$, which is supposed to be
impossible.

\medskip

\qquad{\bf case 2} Suppose that $\theta(p)<\kappa$ for every $p\in Q$.

\medskip

\qquad\quad{\bf case 2a} Suppose that $\sup_{p\in A}\theta(p)<\kappa$.
Then there is an up-antichain $C\subseteq P$ such that
$\#(C)$ is greater than
$\max(\omega,\#(A),\sup_{p\in A}\theta(p))$.   For each $p\in A$ set
$C_p=\{q:q\in C,\,\coint{q,\infty}\cap\coint{p,\infty}\ne\emptyset\}$,
so that $\#(C_p)<\theta(p)$.   It follows that
$C\ne\bigcup_{p\in A}C_p$.   But if
$q\in C\setminus\bigcup_{p\in A}C_p$, there is a $q'\in Q$ such that
$q'\ge q$, and now $A\cup\{q'\}$ is an up-antichain in $Q$ strictly
including  $A$, which is impossible.

\medskip

\qquad\quad{\bf case 2b} Suppose that $\sup_{p\in A}\theta(p)=\kappa$.
Then we can choose inductively a family
$\ofamily{\xi}{\lambda}{p_{\xi}}$ in $A$ such that
$\theta(p_{\xi})>\max(\kappa_{\xi},\sup_{\eta<\xi}\theta(p_{\eta}))$ for
each $\xi$;  the point being that when we come to choose $p_{\xi}$,
$\ofamily{\eta}{\xi}{\theta(p_{\eta})}$ is a family of fewer than
$\cf\kappa$ cardinals less than $\kappa$, so has supremum less than
$\kappa$.   Now all the $p_{\xi}$ must be distinct.   For each $\xi$,
let $B_{\xi}\subseteq\coint{p_{\xi},\infty}$ be an up-antichain with cardinal
$\kappa_{\xi}$;  then $\bigcup_{\xi<\lambda}B_{\xi}$ is an up-antichain
in $P$ with cardinal $\kappa$, which is impossible.\ \Bang

Thus $\kappa$ cannot be a singular infinite cardinal.

\medskip

{\bf (c)} All we need to know is that
$c^{\uparrow}(P)=\sup\{\kappa:\kappa<\sat^{\uparrow}(P)\}$.
}%end of proof of 513B

\cmmnt{\medskip

\noindent{\bf Remark} (b) is sometimes called the {\bf Erd\H{o}s-Tarski
theorem}.}

\leader{513C}{Cofinalities of cardinal
\dvrocolon{functions}}\cmmnt{ We can say a little about the possible
cofinalities of the cardinals which have appeared so far.

\medskip

\noindent}{\bf Proposition} (a) Let $P$ be a partially ordered set with
no greatest member.

\quad(i) If $\add P$ is greater than $2$\cmmnt{ (that is, $P$ is
upwards-directed)}, it is a regular infinite cardinal, and there is a
family $\ofamily{\xi}{\add P}{p_{\xi}}$ in $P$ such that
$p_{\eta}<p_{\xi}$ whenever $\eta<\xi<\add P$, but
$\{p_{\xi}:\xi<\add P\}$ has no upper bound in $P$.

\quad(ii) If $\cf P$ is infinite, its cofinality is at least $\add P$.

(b) Let $\Cal I$ be an ideal of subsets of a set $X$ such that
$\bigcup\Cal I=X\notin\Cal I$.

\quad(i) $\cf(\add\Cal I)=\add\Cal I\le\cf(\cf\Cal I)$.

\quad(ii) $\cf(\non\Cal I)\ge\add\Cal I$.

\quad(iii) If $\cov\Cal I=\cf\Cal I$ then $\cf(\cf\Cal I)\ge\non\Cal I$.

\proof{{\bf (a)(i)} By 511Hd, $\add P\ge\omega$;  by 511He,
$\add P<\infty$, so $\add P$ is an infinite cardinal.   Let
$\ofamily{\xi}{\add P}{q_{\xi}}$ be a family in $P$ with no upper bound
in $P$.   Choose $\ofamily{\xi}{\add P}{p_{\xi}}$ inductively, as follows.   Given $p_{\xi}$, where $\xi<\add P$, there is a
$p'_{\xi}\in P$ such that $p'_{\xi}\not\le p_{\xi}$;  let $p_{\xi+1}$ be an upper bound of $\{p_{\xi},p'_{\xi},q_{\xi}\}$.
For a limit ordinal
$\xi<\add P$, let $p_{\xi}$ be an upper bound of $\{p_{\eta}:\eta<\xi\}$.   This will ensure that $p_{\eta}<p_{\xi}$ whenever $\xi<\eta<\add P$ and that $\{p_{\xi}:\xi<\add P\}$ has no upper bound, since such a bound would have to be a bound for
$\{q_{\xi}:\xi<\add P\}$.

\Quer\ If $\add P$ is singular, express it as
$\sup_{\xi<\lambda}\kappa_{\xi}$, where $\lambda<\add P$ and
$\kappa_{\xi}<\add P$ for each $\xi<\lambda$.   Then for
each $\xi<\lambda$, $\{p_{\eta}:\eta<\kappa_{\xi}\}$ has an upper bound
$r_{\xi}$ in $P$;  but now $\{r_{\xi}:\xi<\lambda\}$ has an upper bound
in $P$, which is also an upper bound of $\{p_{\eta}:\eta<\add P\}$.\
\BanG\   Thus $\add P$ is regular.

\medskip

\quad{\bf (ii)} \Quer\ If $\cf(\cf P)<\add P$, express $\cf P$ as
$\sup_{\xi<\lambda}\kappa_{\xi}$ where $\lambda<\add P$ and
$\kappa_{\xi}<\cf P$ for each $\xi<\lambda$.   Let
$\ofamily{\eta}{\cf P}{p_{\eta}}$ enumerate a cofinal subset of $P$.
Then $\{p_{\eta}:\eta<\kappa_{\xi}\}$ is never cofinal with $P$, so
there is for each $\xi<\lambda$ a $q_{\xi}\in P$ such that
$q_{\xi}\not\le p_{\eta}$ for every $\eta<\kappa_{\xi}$.   But now there
is a $q\in P$ which is an upper bound for $\{q_{\xi}:\xi<\lambda\}$, and
$q\not\le p_{\eta}$ for any $\eta<\cf P$.\ \Bang

\medskip

{\bf (b)(i)} Because $\bigcup\Cal I=X\notin\Cal I$, $\cf\Cal I$ and
$\add\Cal I$ are both infinite, so this is just a special case of (a).

\medskip

\quad{\bf (ii)} \Quer\ If $\cf(\non\Cal I)<\add\Cal I$, express
$\non\Cal I$ as $\sup_{\xi<\lambda}\kappa_{\xi}$ where
$\lambda<\add\Cal I$ and
$\kappa_{\xi}<\non\Cal I$ for each $\xi<\lambda$.    Let
$\ofamily{\eta}{\non\Cal I}{x_{\eta}}$ enumerate a subset of $X$ not
belonging to $\Cal I$.   Then $I_{\xi}=\{x_{\eta}:\eta<\kappa_{\xi}\}$
belongs to $\Cal I$ for each $\xi<\lambda$;  but this means that
$\bigcup_{\xi<\lambda}I_{\xi}=\{x_{\eta}:\eta<\non\Cal I\}$ belongs to
$\Cal I$.\ \Bang

\medskip

\quad{\bf (iii)} Set $\cf(\cf\Cal I)=\kappa$.   Let
$\ofamily{\xi}{\kappa}{\Cal A_{\xi}}$ be a family of subsets of
$\Cal I$, all with cardinal less than $\cf\Cal I=\cov\Cal I$, such that
$\bigcup_{\xi<\kappa}\Cal A_{\xi}$ is cofinal with $\Cal I$.   Because
$\#(\Cal A_{\xi})<\cov\Cal I$, there is an
$x_{\xi}\in X\setminus\bigcup\Cal A_{\xi}$ for each $\xi<\kappa$.   Now
$\{x_{\xi}:\xi<\kappa\}$  is not included in any member of
$\Cal A_{\xi}$ for any $\xi$, so cannot belong to $\Cal I$ and witnesses
that $\non\Cal I\le\kappa$.
}%end of proof of 513C

\leader{513D}{}\cmmnt{ Galois-Tukey connections between partial orders
have some distinctive features which make a special language
appropriate.

\medskip

\noindent}{\bf Definition} Let $P$ and $Q$ be pre-ordered sets.
A function $\phi:P\to Q$ is a {\bf Tukey function} if $\phi^{-1}[B]$ is
bounded above in $P$ whenever $B\subseteq Q$ is bounded above in $Q$.
A function $\psi:Q\to P$ is a {\bf dual Tukey function} (also called
`cofinal function', `convergent function') if $\psi[B]$ is
cofinal with $P$ whenever $B\subseteq Q$ is cofinal with $Q$.

If $P$ and $Q$ are pre-ordered sets, I
will write `$P\prT Q$' if $(P,\le,P)\prGT(Q,\le,Q)$, and `$P\equivT Q$'
if $(P,\le,P)\equivGT(Q,\le,Q)$;  in the latter case I say that $P$ and
$Q$ are {\bf Tukey equivalent}.   It follows immediately from 512C that
$\prT$ is reflexive and transitive, and of course $P\equivT Q$ iff
$P\prT Q$ and $Q\prT P$.

\leader{513E}{Theorem} Let $P$ and $Q$ be pre-ordered sets.

(a) If $(\phi,\psi)$ is a Galois-Tukey connection from $(P,\le,P)$ to
$(Q,\le,Q)$ then $\phi:P\to Q$ is a Tukey function and $\psi:Q\to P$ is
a dual Tukey function.

(b)(i) A function $\phi:P\to Q$ is a Tukey function iff there is a
function $\psi:Q\to P$ such that $(\phi,\psi)$ is a Galois-Tukey
connection from $(P,\le,P)$ to $(Q,\le,Q)$.

\quad(ii) A function $\psi:Q\to P$ is a dual Tukey function iff there is
a function $\phi:P\to Q$ such that $(\phi,\psi)$ is a Galois-Tukey
connection from $(P,\le,P)$ to $(Q,\le,Q)$.

\quad(iii) If $\psi:Q\to P$ is order-preserving and $\psi[Q]$ is cofinal
with $P$, then $\psi$ is a dual Tukey function.

(c) The following are equiveridical, that is, if one is true so are the
others:

\quad(i) $P\prT Q$;

\quad(ii) there is a Tukey function $\phi:P\to Q$;

\quad(iii) there is a dual Tukey function $\psi:Q\to P$.

\wheader{513E}{0}{0}{0}{48pt}
(d)(i) Let $f:P\to Q$ be such that $f[P]$ is cofinal with $Q$ and, for $p$,
$p'\in P$, $f(p)\le f(p')$ iff $p\le p'$.   Then $P\equivT Q$.

\quad(ii) Suppose that $A\subseteq P$ is cofinal with $P$.   Then
$A\equivT P$

\quad(iii) For $p$, $q\in P$ say that $p\equiv q$ if $p\le q$ and
$q\le p$;  let $\tilde P$ be the partially ordered set of equivalence
classes in $P$ under the equivalence relation $\equiv$\cmmnt{ (511A,
511Ha)}.   Then $P\equivT\tilde P$.

\wheader{513E}{0}{0}{0}{36pt}
(e) Suppose now that $P\prT Q$.   Then

\quad (i) $\cf P\le\cf Q$;

\quad (ii) $\add P\ge\add Q$;

\quad (iii) $\sat^{\uparrow}(P)\le\sat^{\uparrow}(Q)$,
$c^{\uparrow}(P)\le c^{\uparrow}(Q)$;

\quad (iv)
$\link_{<\kappa}^{\uparrow}(P)\le\link_{<\kappa}^{\uparrow}(Q)$ for any
cardinal $\kappa$;

\quad (v) $\link^{\uparrow}(P)\le\link^{\uparrow}(Q)$,
$\duparrow(P)\le\duparrow(Q)$.

(f) If $P$ and $Q$ are Tukey equivalent, then $\cf P=\cf Q$ and
$\add P=\add Q$.

(g) If $\familyiI{P_i}$ and $\familyiI{Q_i}$ are families of pre-ordered
ordered sets such that $P_i\prT Q_i$ for every $i$, then
$\prod_{i\in I}P_i\prT\prod_{i\in I}Q_i$.

(h) If $0<\kappa<\add P$ then
$P\equivT P^{\kappa}$.   In particular, if $P$ is upwards-directed then
$P\equivT P\times P$.

\proof{{\bf (a)} To say that $(\phi,\psi)$ is a Galois-Tukey connection
from $(P,\le,P)$ to $(Q,\le,Q)$ means just that $p\le\psi(q)$ whenever
$\phi(p)\le q$.   Now if $B\subseteq Q$ has an upper bound $q$,
$\psi(q)$ is an upper bound for $\phi^{-1}[B]$;  as $B$ is arbitrary,
$\phi$ is a Tukey function.   Similarly, if
$B\subseteq Q$ is cofinal, then for any $p\in P$ there is a $q\in B$
such that $\phi(p)\le q$ and $p\le\psi(q)$, so $\psi[B]$ is cofinal with
$P$.   As $B$ is arbitrary, $\psi$ is a dual Tukey function.

\medskip

{\bf (b)(i)} If $\phi:P\to Q$ is a Tukey function, then for each
$q\in Q$ set $A_q=\{p:p\in P,\,\phi(p)\le q\}$.
$A_q$ must be bounded above in $P$;  take $\psi(q)$ to be any upper
bound for $A_q$ in $P$.   Then we see that $p\le\psi(q)$ whenever
$\phi(p)\le q$, so that $(\phi,\psi)$ is a Galois-Tukey connection from
$(P,\le,P)$ to $(Q,\le,Q)$.   Together with (a), this proves (i).

\medskip

\quad{\bf (ii)} If $\psi:Q\to P$ is a dual Tukey function, then for each
$p\in P$ set $B_p=\{q:q\in Q,\,\psi(q)\not\ge p\}$.   Then $\psi[B_p]$
is not cofinal with $P$, so $B_p$ cannot be cofinal with $Q$, and there
must be a $\phi(p)\in P$ such that $\phi(p)\not\le q$ for any
$q\in B_p$.   Turning
this round, if $\phi(p)\le q$ then $q\notin B_p$ and $p\le\psi(q)$;  so
$(\phi,\psi)$ is a Galois-Tukey connection from $(P,\le,P)$ to
$(Q,\le,Q)$.   Together with (a), this proves (ii).

\medskip

\quad{\bf (iii)} Because $\psi[Q]$ is cofinal with $P$, we have a
function $\phi:P\to Q$ such that $p\le\psi\phi(p)$ for every $p\in P$.
Now, for any $p\in P$ and $q\in Q$,

\Centerline{$\phi(p)\le q\Longrightarrow p\le\psi\phi(p)\le\psi(q)$,}

\noindent so $(\phi,\psi)$ is a Galois-Tukey connection and $\psi$ is a
dual Tukey function.

\medskip

{\bf (c)} This follows immediately from (a) and (b).

\medskip

{\bf (d)(i)} $f$ is a Tukey function.   \Prf\ If $A\subseteq P$ and $f[A]$
is bounded above in $Q$, let $q$ be an upper bound for $f[A]$.   Because
$f[P]$ is cofinal with $Q$, there is a $p_0\in P$ such that $q\le f(p_0)$.
If now $p\in A$, we have $f(p)\le q\le f(p_0)$ so $p\le p_0$;  thus $A$ is
bounded above in $P$.   As $A$ is arbitrary, $f$ is a Tukey function.\
\QeD\   So $P\prT Q$.

$f$ is a dual Tukey function.   \Prf\ If $A\subseteq P$ is cofinal with
$P$, and $q\in Q$, there are a $p\in P$ such that $q\le f(p)$, and a
$p'\in A$ such that $p\le p'$;  in which case

\Centerline{$q\le f(p)\le f(p')\in f[A]$.}

\noindent Thus $f[A]$ is cofinal with $Q$;  as $A$ is arbitrary, $f$ is a
dual Tukey function.\ \QeD\  So $Q\prT P$ and $P\equivT Q$.

\medskip

\quad{\bf (ii)} Apply (i) to the identity map from $A$ to $P$.

\medskip

\quad{\bf (iii)} Apply (i) to the canonical map from $P$ to $\tilde P$.

\medskip

{\bf (e)} This is just a restatement of the results in 512D, using the
identifications listed in 512Ea.   The only omission concerns
cellularities, for which I have not set out a formal definition in the
context of supported relations;  but if $A\subseteq P$ is an
up-antichain and $\phi:P\to Q$ is a Tukey function, then
$\{\phi(a),\phi(a')\}$ can have no upper bound in $Q$ for any distinct
$a$, $a'\in A$, so $\phi[A]$ is an up-antichain in $Q$ with the same
cardinality as $A$, and $\#(A)\le c^{\uparrow}(Q)$.   As $A$ is
arbitrary, $c^{\uparrow}(P)\le c^{\uparrow}(Q)$ and
$\sat^{\uparrow}(P)\le\sat^{\uparrow}(Q)$.

\medskip

{\bf (f)} follows at once from (e).

\medskip

{\bf (g)} This is a special case of 512H.

\medskip

{\bf (h)} Let $Q\subseteq P^{\kappa}$ be the set of constant functions.
Because $\kappa\ge 1$, $Q$ is isomorphic to $P$;  because
$\kappa<\add P$, $Q$ is cofinal with $P^{\kappa}$;  so
$P\cong Q\equivGT P^{\kappa}$.
}%end of proof of 513E

\leader{513F}{Theorem}\cmmnt{ ({\smc Tukey 1940})} Suppose that $P$ and
$Q$ are upwards-directed partially ordered sets.   Then $P$ and $Q$ are
Tukey equivalent iff there is a partially ordered set $R$
such that $P$ and $Q$ are both isomorphic, as partially ordered sets, to
cofinal subsets of $R$.

\proof{{\bf (a)} Suppose that $P$ and $Q$ are Tukey equivalent.   Then
there are Tukey functions $\phi:P\to Q$ and $\psi:Q\to P$.   Set
$S=(P\times\{0\})\cup(Q\times\{1\})$, with a relation $\le$ defined by
saying that

\inset{$(p,0)\le(q,1)$ iff ($\alpha$) there is a $p'\ge p$ in $P$ such
that $\phi(p')\le q$ in $Q$ ($\beta$) $q'\le q$ in $Q$ whenever
$q'\in Q$ and $\psi(q')\le p$ in $P$,

$(q,1)\le(p,0)$ iff ($\alpha$) there is a $q'\ge q$ in $Q$ such that
$\psi(q')\le p$ in $P$ ($\beta$) $p'\le p$ in $P$ whenever $p'\in P$ and
$\phi(p')\le q$ in $Q$,

$(p',0)\le(p,0)$ iff $p'\le p$ in $P$,

$(q',1)\le (q,1)$ iff $q'\le q$ in $Q$.}

\noindent Of course $\le$ is reflexive.   To see that it is transitive,
observe that if $(p,0)\le(q,1)\le(\tilde p,0)$ then there is a
$p'\ge p$ such that $\phi(p')\le q$, and now $p'\le\tilde p$, so
$(p,0)\le(\tilde p,0)$.   Similarly, $(q,1)\le(\tilde q,1)$ whenever
$(q,1)\le(p,0)\le(\tilde q,1)$.   The other cases to check are equally
easy.   It is {\it not} necessarily the case that $\le$ is
antisymmetric, since it is possible to have $(p,0)\le(q,1)\le(p,0)$;
but we have an equivalence relation $\cong$ on $S$ defined by saying
that $s\cong t$ if $s\le t$ and $t\le s$, and a natural partial order on
the set $R$ of equivalence classes defined by saying that
$s^{\ssbullet}\le t^{\ssbullet}$ iff $s\le t$.

The map $p\mapsto(p,0)^{\ssbullet}:P\to R$ is an order-isomorphism
between $P$ and its image $\tilde P\subseteq R$.   Now for any $q\in Q$
there is a $p\in P$ such that $(q,1)\le (p,0)$.   \Prf\ Since $\phi$ is
a Tukey function, $A=\{p':\phi(p')\le q\}$ must be bounded above in $P$;
let $p_0$ be an upper bound for $A$.   Now because $P$ is
upwards-directed,
there is a $p\in P$ such that $p_0\le p$ and $\psi(q)\le p$, and in this
case $(q,1)\le (p,0)$.\ \QeD\   This is what we need to see that
$\tilde P$ is cofinal with $R$.   Similarly, $Q$ is order-isomorphic to
its canonical image in $R$, and this too is cofinal with $R$.
So both $P$ and $Q$ are isomorphic to cofinal subsets of $R$.

\medskip

{\bf (b)} Conversely, if $P$ and $Q$ are both isomorphic to cofinal
subsets of a partially ordered set $R$, then $P$, $R$ and $Q$ are all
Tukey equivalent, by 513Ed.
}%end of proof of 513F

\leader{513G}{}\cmmnt{ We shall repeatedly want to use some elementary
facts about cofinal subsets.

\medskip

\noindent}{\bf Proposition} Let $P$ be a pre-ordered set and $Q$ a
cofinal subset of $P$.   Then

(a) $\add Q=\add P$;

(b) $\cf Q=\cf P$;

(c) $\sat^{\uparrow}(Q)=\sat^{\uparrow}(P)$,
$c^{\uparrow}(Q)=c^{\uparrow}(P)$;

(d) $\link_{<\kappa}^{\uparrow}(Q)=\link_{<\kappa}^{\uparrow}(P)$ for
any cardinal $\kappa$;  in particular,
$\link^{\uparrow}(Q)=\link^{\uparrow}(P)$ and
$\duparrow(Q)=\duparrow(P)$;

(e) $\bu Q=\bu P$.

\proof{ All except (e) are consequences of 513Ed and 513Ee.
As for bursting numbers, every cofinal subset of $Q$ is also cofinal
with $P$, so $\bu P\le\bu Q$.   For the reverse inequality, let $Q_1$ be
a cofinal subset of $P$ such that
$\#(\{q:q\in Q_1$, $q\le p$, $p\not\le q\})<\bu P$
for every $p\in p$.   Let $\phi:Q_1\to Q$ be any function such that
$\phi(q)\ge q$ for every $q\in Q_1$, so that $\phi[Q_1]$ is cofinal with
$Q$.   If $q\in Q$, then

\Centerline{$\{q':q'\in\phi[Q_1]$, $q'\le q$, $q\not\le q'\}
\subseteq\{\phi(q''):q''\in Q_1$, $q''\le q$, $q\not\le q''\}$}

\noindent has cardinal less than $\bu P$, and $\phi[Q_1]$ witnesses
that $\bu Q\le\bu P$.
}%end of proof of 513G

\leader{513H}{Definition} Let $P$ be a partially ordered set.   Its
{\bf $\sigma$-additivity} $\add_{\omega}P$ is the smallest cardinal of
any subset $A$ of $P$ such
that $A\not\subseteq\bigcup_{q\in D}\ocint{-\infty,q}$ for any countable
set $D\subseteq P$.   If there is no such set\cmmnt{, that is, if
$\cf P\le\omega$}, I write $\add_{\omega}P=\infty$.

\leader{513I}{Proposition} Let $P$ be a partially ordered set.   As in
512F, write $p\le\strprime A$, for $p\in P$ and $A\subseteq P$, if
there is a $q\in A$ such that $p\le q$.

(a) $\add_{\omega}P=\add(P,\le\strprime,[P]^{\le\omega})$.

(b) $\max(\omega_1,\add P)\le\add_{\omega}(P)$.

(c) If $\add_{\omega}P$ is an infinite cardinal, it is regular.

(d) If $2\le\kappa\le\add P$,
then $(P,\le\strprime,[P]^{<\kappa})\equivGT(P,\le,P)$.   So if
$\add P>\omega$, $\add_{\omega}(P)=\add P$.

(e) If $Q$ is another partially ordered set and
$(P,\le\strprime,[P]^{\le\omega})
\prGT(Q,\le\strprime,[Q]^{\le\omega})$\cmmnt{ (in particular, if
$P\prT Q$)} then $\add_{\omega}P\ge\add_{\omega}Q$.

(f) If $Q\subseteq P$ is cofinal with $P$, then
$\add_{\omega}Q=\add_{\omega}P$.

(g) If $\kappa\le\cf P$ then
$\add(P,\le\strprime,[P]^{<\kappa})\le\cf P$.
So if $\cf P>\omega$ then $\add_{\omega}P\le\cf P$.

(h) If $\cf(\cf P)>\omega$ then $\cf(\cf P)\ge\add_{\omega}P$.

\proof{{\bf (a)} All we have to do is to disentangle the definitions in
512Ba, 512F and 513H.

\medskip

{\bf (b)} is immediate from the definition of $\add_{\omega}$.

\medskip

{\bf (c)} \Quer\ Suppose, if possible, that $\add_{\omega}P=\kappa$
where $\kappa>\max(\omega,\cf\kappa)$.   Express $\kappa$ as
$\sup_{\xi<\lambda}\kappa_{\xi}$ where $\kappa_{\xi}<\kappa$ for every
$\xi<\lambda=\cf\kappa$.   Let $A\subseteq P$ be a set with cardinal
$\kappa$ such that $A\not\subseteq\bigcup_{q\in D}\ocint{-\infty,q}$ for
any countable set $D\subseteq P$.   Express $A$ as
$\bigcup_{\xi<\lambda}A_{\xi}$ where $\#(A_{\xi})=\kappa_{\xi}$ for each
$\xi<\lambda$.   For each $\xi<\lambda$, there is a countable set
$D_{\xi}\subseteq P$ such that
$A_{\xi}\subseteq\bigcup_{q\in D_{\xi}}\ocint{-\infty,q}$.   Set
$B=\bigcup_{\xi<\lambda}D_{\xi}$;  then $\#(B)\le\lambda<\kappa$, so
there is a countable set $D\subseteq P$ such that
$B\subseteq\bigcup_{q\in D}\ocint{-\infty,q}$.   But now
$A\subseteq\bigcup_{q\in D}\ocint{-\infty,q}$.\ \Bang

\medskip

{\bf (d)} By 512Gc, $(P,\le\strprime,[P]^{<\kappa})\prGT(P,\le,P)$.
In the other direction, because $\kappa\le\add P$, we have a function
$\psi:[P]^{<\kappa}\to P$ such that $I\subseteq\ocint{-\infty,\psi(I)}$
for every $I\in[P]^{<\kappa}$;  so if we set $\phi(p)=p$ for $p\in P$,
$(\phi,\psi)$ will be a Galois-Tukey connection from $(P,\le,P)$ to
$(P,\le\strprime,[P]^{<\kappa})$, and
$(P,\le,P)\prGT(P,\le\strprime,[P]^{<\kappa})$.

Now if $\add P>\omega$,

\Centerline{$\add_{\omega}P=\add(P,\le\strprime,[P]^{\le\omega})
=\add(P,\le,P)=\add P$.}

\medskip

{\bf (e)} Use (a) with 512Db and 512Gb.

\medskip

{\bf (f)} Use (e) and 513Ed.

\medskip

{\bf (g)} Let $Q\subseteq P$ be a cofinal subset of $P$ with cardinal
$\cf P$.   If $A\subseteq P$ is such that every member of $Q$ is
dominated by a member of $A$, then $A$ also is cofinal, so
$\#(A)\ge\kappa$;  thus $Q$ witnesses that
$\add(P,\le\strprime,[P]^{<\kappa})\le\cf P$.   Putting
$\kappa=\omega_1$
we see that if $\cf P>\omega$ then $\add_{\omega}P\le\cf P$.

\medskip

{\bf (h)} \Quer\ If
$\omega<\cf(\cf P)=\lambda<\add_{\omega}P$ let
$Q\subseteq P$ be a cofinal set with cardinal $\cf P$ and express $Q$ as
$\bigcup_{\xi<\lambda}Q_{\xi}$ where $\#(Q_{\xi})<\cf P$ and
$Q_{\xi}\subseteq Q_{\eta}$ whenever $\xi\le\eta<\lambda$.   For each
$\xi<\lambda$, $Q_{\xi}$ cannot be cofinal with $P$, so there is a
$p_{\xi}\in P$ such that $p_{\xi}\not\le q$ for any $q\in Q_{\xi}$.
Now $A=\{p_{\xi}:\xi<\lambda\}$ has cardinal less than $\add_{\omega}P$,
so there is a countable set $D\subseteq P$ such that
$A\subseteq\bigcup_{r\in D}\ocint{-\infty,r}$.   For each $r\in D$ there
is a $q_r\in Q$ such that $r\le q_r$;  let $\xi_r<\lambda$ be such that
$q_r\in Q_{\xi_r}$.   Because $\lambda$ is uncountable and regular
(being the cofinality of a cardinal), $\zeta=\sup_{r\in D}\xi_r$ is less
than $\lambda$, and $q_r\in Q_{\zeta}$ for every $r\in D$.   But now
there is an $r\in D$ such that $p_{\zeta}\le r\le q_r\in Q_{\zeta}$,
contrary to the choice of $p_{\zeta}$.\ \Bang
}%end of proof of 513I

\cmmnt{\medskip

\noindent{\bf Remark} The point of (b) and (d) here is that there are
significant cases in which $\add P<\omega_1<\add_{\omega}P$.
}%end of comment

\leader{*513J}{Cofinalities
of \dvrocolon{products}}\dvAformerly{5{}11I}\cmmnt{ It is easy to find
the additivity of a product of partially ordered sets (511Hg).
Calculating the cofinality of a
product of partially ordered sets is surprisingly difficult, and there
are some extraordinary results in this area.   (See
{\smc Burke \& Magidor 90};  there is a taster in 542J below.)
Here I will give just one special fact which will be useful.

\medskip

\noindent}{\bf Proposition} Suppose that the generalized continuum
hypothesis is true.    Let $\familyiI{P_i}$ be a family of
non-empty partially ordered sets with product $P$.  Set

\Centerline{$\kappa=\#(\{i:i\in I$, $\cf P_i>1\})$,
\quad$\lambda=\sup_{i\in I}\cf P_i$.}

\noindent Then

(i) if $\kappa$ and $\lambda$ are both finite,
$\cf P$ is the cardinal product $\prod_{i\in I}\cf P_i$;

(ii) if $\lambda>\kappa$ and there is some $\gamma<\lambda$ such that
$\cf\lambda>\#(\{i:i\in I$, $\cf P_i>\gamma\})$, then
$\cf P=\lambda$;

(iii) otherwise, $\cf P=\max(\kappa^+,\lambda^+)$.

% all except (a) work for supported relations

\proof{{\bf (a)} For each $i\in I$, let $Q_i\subseteq P_i$ be a cofinal
set with cardinal $\cf P_i$.   Then $Q=\prod_{i\in I}Q_i$ is cofinal with
$P=\prod_{i\in I}P_i$, so $\cf P\le\#(Q)$.   If $\lambda<\omega$, then
every $Q_i$ must be just the set of maximal elements of $P_i$, so $Q$ is
the set of maximal elements of $P$, and $\cf P=\#(Q)$.   This deals with
case (i).

\medskip

{\bf (b)} $\cf P>\kappa$.   \Prf\ Set $J=\{i:i\in I$, $\cf P_i>1\}$.
If $\family{i}{J}{p_i}$ is any family in
$P$, then we can choose $q\in P$ such that $q(i)\not\le p_i(i)$ for
every $i\in J$;
accordingly $\{p_i:i\in J\}$ cannot be cofinal with $P$.\ \QeD\  So if
$\max(\omega,\lambda)\le\kappa$,

$$\eqalignno{\cf P
&\ge\kappa^+
=2^{\kappa}\cr
\displaycause{because we are assuming the generalized continuum
hypothesis}
&=2^{\max(\kappa,\lambda)}
\ge\#(\Cal P\{(i,q):i\in J,\,q\in Q_i\})
\ge\#(\prod_{i\in J}Q_i)
=\#(Q)\cr
\displaycause{because $\#(Q_i)=1$ for $i\in I\setminus J$}
&\ge\cf P,\cr}$$

\noindent and $\cf P=\kappa^+=\max(\kappa^+,\lambda^+)$, as required by
(iii).

\medskip

{\bf (c)} Note that $\cf P\ge\lambda$, because if
$R\subseteq P$ is cofinal with $P$ then $\{p(i):p\in R\}$ is cofinal
with $P_i$ for each $i$.   So if $\kappa$ is finite and $\lambda$ is
infinite,

\Centerline{$\lambda\le\cf P\le\#(Q)
\le\max(\omega,\sup_{i\in J}\#(Q_i))=\lambda$}

\noindent and $\cf P=\lambda$, as required by (ii).

\medskip

{\bf (d)} If $\lambda$ is infinite and $\kappa<\cf\lambda$ then every
function from $\kappa$ to
$\lambda$ is bounded above in $\lambda$.   So

$$\eqalignno{\lambda
&\le\cf P\le\#(Q)\le\#(\lambda^{\kappa})\cr
\displaycause{where $\lambda^{\kappa}$, for once, denotes the set of
functions from $\kappa$ to $\lambda$}
&=\#(\bigcup_{\zeta<\lambda}\zeta^{\kappa})
\le\max(\omega,\lambda,\sup_{\zeta<\lambda}\#(\zeta^{\kappa}))
\le\max(\omega,\lambda,\sup_{\zeta<\lambda}2^{\max(\zeta,\kappa)})
=\lambda,\cr}$$

\noindent again using GCH.   Thus in this case also we have
$\cf P=\lambda$, as required by (ii).

\medskip

{\bf (e)} So we are left with the case in which
$\cf\lambda=\theta\le\kappa<\lambda$.   Let
$\ofamily{\eta}{\theta}{\lambda_{\eta}}$ be a family of cardinals less
than $\lambda$ with supremum $\lambda$.

\medskip

\quad\grheada\ Suppose that we are in case (iii), that is,
$\#(\{i:i\in I$, $\cf P_i>\gamma\})\ge\theta$ for every
$\gamma<\lambda$.   Then $\cf P>\lambda$.   \Prf\ We can choose
$\ofamily{\eta}{\theta}{i(\eta)}$ inductively in $I$ so that
$\cf P_{i(\eta)}>\lambda_{\eta}$ and $i(\eta)\ne i(\xi)$ when
$\xi<\eta<\theta$.   If $\ofamily{\xi}{\lambda}{p_{\xi}}$ is any family
in $P$, we can find $q\in P$ such that
$q(i(\eta))\not\le p_{\xi}(i(\eta))$ for any $\eta<\theta$ and
$\xi<\lambda_{\eta}$, so that $q\not\le p_{\xi}$ for any $\xi<\lambda$.
As $\ofamily{\xi}{\lambda}{p_{\xi}}$ is arbitrary, $\cf P>\lambda$.\
\QeD\  Now

\Centerline{$\cf P\le\#(Q)\le\#(\lambda^{\kappa})
\le 2^{\max(\kappa,\lambda)}=\lambda^+\le\cf P$,}

\noindent so $\cf P=\lambda^+=\max(\kappa^+,\lambda^+)$, as required.

\medskip

\quad\grheadb\  Otherwise, we are in case (ii), and there is a cardinal
$\gamma<\lambda$ such that
$\#(K)<\theta$, where $K=\{i:i\in I$, $\cf P_i>\gamma\}$.   Then
$\sup_{i\in K}\cf P_i=\lambda$, so (d) tells us that
$\cf(\prod_{i\in K}P_i)=\lambda$.   On the other hand,

\Centerline{$\#(\prod_{i\in I\setminus K}\cf P_i)
\le 2^{\max(\gamma,\kappa)}\le\lambda$.}

\noindent Since we can identify $P$ with the product of
$\prod_{i\in K}P_i$ and $\prod_{i\in I\setminus K}P_i$,
$\cf P\le\#(\lambda\times\lambda)=\lambda$.   But we noted in
(c) that $\cf P\ge\lambda$, so $\cf P=\lambda$, as required.
This completes the proof.
}%end of proof of 513J

\leader{*513K}{}\dvAformerly{5{}13J}\cmmnt{ I remarked in the notes to
\S512 that
Galois-Tukey correspondences are not required to have any special
properties, and of course the same is true of Tukey functions.   But it
is also the case that the `natural' Tukey functions arising in Chapter
52 can in many cases be derived from Borel
measurable functions between Polish
spaces.   I now present some ideas taken from {\smc Solecki \&
Todor\v{c}evi\'c 04} which may be regarded as a partial explanation of
the phenomenon.

\medskip

\noindent}{\bf Definition} I will say that a {\bf metrizably compactly
based directed set} is a partially ordered set $P$ endowed with a
metrizable topology such that

(i) $p\vee q=\sup\{p,q\}$ is defined for all $p$, $q\in P$, and
$\vee:P\times P\to P$ is continuous;

(ii) $\{p:p\le q\}$ is compact for every $q\in P$;

(iii) every convergent sequence in $P$ has a subsequence which is bounded
above.

\noindent In this context, I will say that $P$ is `separable' or
`analytic' if it is separable, or analytic, in the topological sense.

\cmmnt{I leave it to you to check that many significant partially
ordered sets are compactly based in the sense defined here
(513Xj-513Xn, %513Xj 513Xk 513Xl 513Xm 513Xn
513Yg).
}%end of comment

\leader{*513L}{Proposition}\dvAformerly{5{}13K} Let $P$ be a metrizably
compactly based directed set.

(a) The ordering of $P$ is a closed subset of $P\times P$.

(b) $P$ is Dedekind complete.

(c)(i) A non-decreasing sequence in $P$ has an upper bound iff it is
topologically convergent, and in this case its supremum is its limit.

\quad(ii) A non-increasing sequence in $P$ converges topologically
to its infimum.

(d) If $p\in P$ and $\sequence{i}{p_i}$ is a sequence in $P$, then
$\sequence{i}{p_i}$ is topologically convergent to $p$ iff for every
$I\in[\Bbb N]^{\omega}$ there is a $J\in[I]^{\omega}$ such that
$p=\inf_{n\in\Bbb N}\sup_{i\in J\setminus n}p_i$.

(e) Suppose that $p\in P$ and a double sequence
$\langle p_{ni}\rangle_{n,i\in\Bbb N}$ in $P$ are such that
$\lim_{i\to\infty}p_{ni}=p_n$ is defined in $P$ and less than or equal
to $p$ for each $n$.   Then there is a $q\in P$ such that
$\{i:p_{ni}\le q\}$ is infinite for every $n\in\Bbb N$.

\proof{ Let $\rho$ be a metric on $P$ inducing its topology.

\medskip

{\bf (a)} We have only to observe that
$\{(p,q):p\le q\}=\{(p,q):p\vee q=q\}$.

\medskip

{\bf (b)} Suppose that $A\subseteq P$ is non-empty and bounded above.
Let $B$ be the set of upper bounds of $A$.   Then
$\Cal E=\{[p,q]:p\in A$, $q\in B\}$ is a non-empty family of compact
sets with the finite intersection property, because any non-empty finite
subset of $A$ has a least upper bound.   So there is a
$q_0\in\bigcap\Cal E$ and now $q_0$ must be the supremum of $A$.

\medskip

{\bf (c)(i)} Suppose that $\sequence{i}{p_i}$ is a non-decreasing
sequence in $P$.   ($\alpha$) If it has a topological limit $p$, then

\Centerline{$p\vee p_j=\lim_{i\to\infty}p_i\vee p_j
=\lim_{i\to\infty}p_i=p$}

\noindent for each $j$, so $p$ is an upper bound for
$\{p_i:i\in\Bbb N\}$;  while if $q$ is an upper bound for
$\{p_i:i\in\Bbb N\}$ then $p\le q$ by (a).   Thus
$p=\sup_{i\in\Bbb N}p_i$.   ($\beta$) If $\{p_i:i\in\Bbb N\}$ is bounded
above, then it has a least upper bound $p$, by (b).   Now
$\ocint{-\infty,p}$ is compact, therefore sequentially compact, and
every subsequence of $\sequence{i}{p_i}$ has a convergent
sub-subsequence;  by ($\alpha$), the limit of this sub-subsequence is
always its supremum, which must be $p$;  so $\sequence{i}{p_i}$ itself
converges to $p$.

\medskip

\quad{\bf (ii)} Suppose that $\sequence{i}{p_i}$ is a non-increasing
sequence in $P$.   Then it lies in the compact set $\ocint{-\infty,p_0}$
so has a convergent subsequence $\sequence{i}{p'_i}$ with limit $p$ say.
As in (i) just above,

\Centerline{$p\vee p'_j=\lim_{i\to\infty}p'_i\vee p'_j
=\lim_{i\to\infty}p'_j=p'_j$}

\noindent for each $j$, so $p$ is a lower bound for
$\{p'_i:i\in\Bbb N\}$;  while if $q$ is a lower bound for
$\{p'_i:i\in\Bbb N\}$ then $q\le p$ by (a).   Thus
$p=\inf_{i\in\Bbb N}p'_i=\inf_{i\in\Bbb N}p_i$.   What this shows is
that $\inf_{i\in\Bbb N}p_i$ is the only cluster point of
$\sequence{i}{p_i}$ and is therefore its topological limit.

\medskip

{\bf (d)(i)} Suppose that $p=\lim_{i\to\infty}p_i$.
Note first that if $q\in P$ then

\Centerline{$\limsup_{i\to\infty}\rho(q\vee p_i,p)
\le\lim_{i\to\infty}\rho(q\vee p_i,q\vee p)+\rho(q\vee p,p)
=\rho(q\vee p,p)$,}

\Centerline{$\limsup_{i\to\infty}\rho(p\vee q\vee p_i,p)
\le\rho((p\vee q)\vee p,p)=\rho(q\vee p,p)$}

\noindent because $\vee$ is continuous.   Now let $I\subseteq\Bbb N$ be
infinite.   By 513K(iii), there is an infinite $I'\subseteq I$ such that
$\{p_i:i\in I'\}$ is bounded above.
We can choose inductively a strictly increasing
sequence $\sequencen{i_n}$ in $I'$ such that

\Centerline{$\rho(\sup_{j\le n\le k}p_{i_n},p)<2^{-j}$,
\quad$\rho(p\vee\sup_{j\le n\le k}p_{i_n},p)<2^{-j}$}

\noindent whenever $j\le k$ in $\Bbb N$.   Set $J=\{i_n:n\in\Bbb N\}$;
then $\rho(\sup_{j\in J,m\le j\le k}p_j,p)<2^{-m}$ whenever
$m\le k\in\Bbb N$ and $[m,k]$ meets $J$.   For each $m$,
$q_m=\sup_{j\in J\setminus m}p_j$ is
defined in $P$, by (b) above;  moreover, (c-i) tells us that
$q_m=\lim_{k\to\infty}\sup_{j\in J,m\le j\le k}p_j$ so

\Centerline{$\rho(q_m,p)
=\lim_{k\to\infty}\rho(\sup_{j\in J,m\le j\le k}p_j,p)\le 2^{-m}$.}

\noindent But this means that $p$ is the topological limit of the
non-increasing sequence $\sequence{m}{q_m}$ and must be
$\inf_{m\in\Bbb N}q_m$.  Thus $\sequence{i}{p_i}$ satisfies the
condition proposed.

\medskip

\quad{\bf (ii)} Now suppose that for every $I\in[\Bbb N]^{\omega}$ there
is a $J\in[I]^{\omega}$ such that
$p=\inf_{n\in\Bbb N}\sup_{i\in J\setminus n}p_i$.
Then any convergent subsequence of $\sequence{i}{p_i}$ has limit $p$.
\Prf\ Suppose the subsequence is $\sequencen{p_{i_n}}$ where
$\sequencen{i_n}$ is strictly increasing.   Set $I=\{i_n:n\in\Bbb N\}$.
Then we must have an infinite $J\subseteq\Bbb N$ such that
$p=\inf_{m\in\Bbb N}\sup_{k\in J\setminus m}p_{i_k}$.   Now (i) tells us
that we also have an infinite $K\subseteq J$ such that the limit $p'$ of
$\sequence{n}{p_{i_n}}$ is
$\inf_{m\in\Bbb N}\sup_{k\in K\setminus m}p_{i_k}$.   Since
$\sup_{k\in K\setminus m}p_{i_k}\le\sup_{k\in J\setminus m}p_{i_k}$ for
every $m$, $p'\le p$.   On the other hand, we also have an infinite
$L\subseteq K$ such that
$p=\inf_{m\in\Bbb N}\sup_{k\in L\setminus m}p_{i_k}$;  so that
$p\le p'$ and $p=p'$.\ \Qed

Since the condition tells us also that every subsequence of
$\sequence{i}{p_i}$ has a sub-subsequence which is bounded above, and
therefore has a convergent sub-sub-subsequence, $p$ is
actually the limit of $\sequence{i}{p_i}$.

\medskip

{\bf (e)} Note first that if $\sequence{i}{q_i}$ is a sequence in $P$
converging to $q^*\in P$, and $\epsilon>0$, there is a $q'\in P$ such
that $\rho(q',q^*)\le\epsilon$ and $\{i:q_i\le q'\}$ is infinite.
\Prf\  By (d), there is an infinite $J\subseteq\Bbb N$ such that
$q^*=\inf_{n\in\Bbb N}\sup_{i\in J\setminus n}q_i$;  by (c-ii), we can
take $q'=\sup_{i\in J\setminus n}q_i$ for some $n$.\ \Qed

For $m$, $i\in\Bbb N$, set $q_{mi}=p\vee\sup_{n\le m}p_{ni}$.
Then $\lim_{i\to\infty}q_{mi}=p\vee\sup_{n\le m}p_n=p$ for each
$m$.   We can therefore find, for each $m\in\Bbb N$, a $q_m\in P$ such
that $\rho(q_m,p)\le 2^{-m}$ and $\{i:q_{mi}\le q_m\}$ is infinite.   As
$\sequence{m}{q_m}\to p$, there is a
$q\in P$ such that $\{m:q_m\le q\}$ is infinite.   Now, for any
$n\in\Bbb N$, there is an $m\ge n$ such that $q_m\le q$, so that

\Centerline{$\{i:p_{ni}\le q\}\supseteq\{i:q_{mi}\le q_m\}$}

\noindent is infinite.
}%end of proof of 513L

\leader{*513M}{Proposition}\dvAformerly{5{}13L} Let $P$ be a separable
metrizably compactly based directed set, and give the set
$\Cal C$ of closed subsets of $P$ its Vietoris topology.
Let $\Cal K_b\subseteq\Cal C$ be the family of non-empty compact subsets
of $P$ which are bounded above in $P$.   Then
$K\mapsto\sup K:\Cal K_b\to P$ is Borel measurable.

\proof{ Writing $\Cal K$ for the family of non-empty compact subsets of
$P$, we have a sequence $\sequencen{f_n}$ of Borel measurable functions
from $\Cal K$ to $P$ such that $K=\overline{\{f_n(K):n\in\Bbb N\}}$ for
every $K\in\Cal K$ (5A4Dc).
Set $g_n(K)=\sup_{i\le n}f_i(K)$ for each
$K\in\Cal K$ and $n\in\Bbb N$;   because $P$ is separable, every
$g_n$ is Borel measurable (put 418Bd and 418Ac together).   For
$K\in\Cal K_b$, $\sequencen{g_n(K)}$ is a non-decreasing bounded
sequence, so converges to $g(K)\in P$, by 513L(c-i);  now
$g:\Cal K_b\to P$ is Borel measurable (418Ba).   Since
$\{q:q\le g(K)\}$ is a closed set including $\{f_i(K):i\in\Bbb N\}$, it
includes $K$, and $g(K)$ is an upper bound for $K$;  because
$g(K)=\sup_{i\in\Bbb N}f_i(K)$, $g(K)=\sup K$.   So we have the result.
}%end of proof of 513M

\leader{*513N}{Lemma}\dvAformerly{5{}13M} Let $P$ and $Q$ be non-empty
metrizably compactly based directed sets
of which $P$ is separable, and $\phi:P\to Q$ a Tukey function.   Set

\Centerline{$R=\overline{\{(q,p):p\in P,\,q\in Q,\,\phi(p)\le q\}}$.}

\noindent Then

(a) $R[\,\ocint{-\infty,q}\,]$ is bounded above in $P$ for every
$q\in Q$;

(b) $R\subseteq Q\times P$ is usco-compact.

\proof{{\bf (a)} Because $P$ is non-empty, we need consider only the case
in which $R[\,\ocint{-\infty,q}\,]$ is non-empty.
Let $\sequencen{p_n}$ be a sequence running over a
dense subset of $R[\,\ocint{-\infty,q}\,]$.   For each $n\in\Bbb N$ we
have sequences $\sequence{i}{p_{ni}}$ in $P$ and $\sequencen{q_{ni}}$ in
$Q$ such that $\phi(p_{ni})\le q_{ni}$, $\lim_{i\to\infty}p_{ni}=p_n$
and $\lim_{i\to\infty}q_{ni}=q_n\le q$.   By 513Le, there is a $q'\in Q$
such that $I_n=\{i:q_{ni}\le q'\}$ is infinite for every $n\in\Bbb N$.
Because $\phi$ is a Tukey function, there is a $p'\in P$ such that
$p_{ni}\le p'$ whenever $n\in\Bbb N$ and $i\in I_n$.   But now
$\{p:p\le p'\}$ is closed, so it contains every $p_n$ and $p'$ is an upper
bound for $R[\,\ocint{-\infty,q}\,]$.

\medskip

{\bf (b)} In particular, for any $q\in Q$, $R[\{q\}]$ is bounded above
in $P$, therefore relatively compact;  since $R$ is closed, every
$R[\{q\}]$ is closed and therefore compact.   Now suppose that
$F\subseteq P$ is closed and that $\sequencen{q_n}$ is a sequence in
$R^{-1}[F]$ converging to $q\in Q$.   Then there is a $q^*\in Q$ such
that $J=\{n:q_n\le q^*\}$ is infinite.   For $n\in J$, let $p_n\in F$ be
such that $(q_n,p_n)\in R$.   Then $\{p_n:n\in J\}$ is included in the
order-bounded set $R^{-1}[\,\ocint{-\infty,q^*}\,]$, so is relatively
compact, and $\sequencen{p_n}$ has a cluster point $p$ say.   Of course
$p\in F$;  also $(q,p)$ is a cluster point of $\sequencen{(q_n,p_n)}$,
so belongs to $\overline{R}=R$, and $q\in R^{-1}[F]$.   As $q$ is
arbitrary, $R^{-1}[F]$ is closed;  as $F$ is arbitrary, $R$ is
usco-compact.
}%end of proof of 513N

\leader{*513O}{Theorem}\dvAformerly{5{}13N}\cmmnt{ ({\smc Solecki \&
Todor\v{c}evi\'c
04})} Let $P$ and $Q$ be metrizably compactly based directed sets such
that $P\prT Q$.   Let $\Sigma$ be the $\sigma$-algebra of subsets of $P$
generated by the Souslin-F sets.

(a) If $P$ is separable, there is a Borel measurable dual Tukey function
$\psi:Q\to P$.

(b) If $P$ is separable and $Q$ is analytic,
there is a $\Sigma$-measurable Tukey function $\phi:P\to Q$.

\proof{ If either $P$ or $Q$ is empty, so is the other, and the result
is trivial;  suppose that they are non-empty.

\medskip

{\bf (a)} Let $\phi_0:P\to Q$ be a Tukey function, and set
$R=\overline{\{(q,p):p\in P,\,q\in Q,\,\phi_0(p)\le q\}}$, so that $R$
is usco-compact (513N).
Let $\Cal C$ be the set of closed subsets of $P$ with its Vietoris
topology;  then $q\mapsto R[\{q\}]$ is Borel measurable (5A4Db).   Since
$\emptyset$ is an isolated point of $\Cal C$,
$Q_0=\{y:R[\{y\}]\ne\emptyset\}$ is a Borel set in $Q$.   If $q\in Q_0$,
then $R[\{q\}]$ is a non-empty compact subset of $P$ which is bounded
above (513Na), so 513M tells us that $q\mapsto\sup R[\{q\}]:Q_0\to P$ is
Borel measurable.   Fix any $p_0\in P$ and set $\psi(q)=\sup R[\{q\}]$
if $q\in Q_0$, $p_0$ if $q\in Q\setminus Q_0$.   Then $\psi$ is Borel
measurable.   If $p\in P$, $q\in Q$ and $\phi_0(p)\le q$, then
$(q,p)\in R$, $p\in R[\{q\}]$ and $p\le\psi(q)$;  thus $(\phi_0,\psi)$
is a Galois-Tukey connection and $\psi$ is a dual Tukey function.

\medskip

{\bf (b)} $R\subseteq Q\times P$ is a closed set (422Da) and $R[Q]=P$.
Because $P$ and $Q$ are separable and metrizable, $R$ can be obtained by
Souslin's operation from products of closed sets.
By the von Neumann-Jankow selection theorem (423N), there is a
$\Sigma$-measurable $\phi:P\to Q$ such that $(\phi(p),p)\in R$ for every
$p\in P$.   If $q\in Q$, then
$\{p:\phi(p)\le q\}\subseteq R[\,\ocint{-\infty,q}\,]$ is bounded above
in $P$, so $\phi$ is a Tukey function.
}%end of proof of 513O

\leader{513P}{}\dvAformerly{5{}13O}\cmmnt{ The last result in this
section is entirely
unconnected with the rest, and is a standard trick;  but it will be useful
later and contains an implicit challenge (513Yj).

\medskip

\noindent}{\bf Lemma} Let $P$ and $Q$ be
non-empty partially ordered sets, and
suppose that (i) every non-decreasing sequence in $P$ has an upper bound in
$P$ (ii) there is no strictly increasing family
$\ofamily{\xi}{\omega_1}{q_{\xi}}$ in $Q$.   Let $f:P\to Q$ be an
order-preserving function.   Then there is a $p\in P$ such that
$f(p')=f(p)$ whenever $p'\in P$ and $p'\ge p$.

\proof{ \Quer\ Otherwise, we can choose $\ofamily{\xi}{\omega_1}{p_{\xi}}$
inductively so that

\inset{$p_0\in P$,

$p_{\xi+1}\ge p_{\xi}$ and $f(p_{\xi+1})>f(p_{\xi})$ for every
$\xi<\omega_1$,

$p_{\xi}$ is an upper bound for $\{p_{\eta}:\eta<\xi\}$ for
every non-zero limit ordinal $\xi<\omega_1$.}

\noindent But now $\ofamily{\xi}{\omega_1}{f(p_{\xi})}$ is strictly
increasing, which is impossible.\ \Bang
}%end of proof of 513P

\exercises{\leader{513X}{Basic exercises (a)}
%\spheader 513Xa
Let $P$ be a partially ordered set, and $\Cal A$ the
family of subsets of $P$ which are not cofinal with $P$.   Show that
$(\Cal A,\not\ni,P)\prGT(P,\le,P)$.   Explain the relation of this fact to
511Xj, 513C(a-ii) and 513Xb.
%513C

\spheader 513Xb Let $P$ be a partially ordered set such that
$\bu P\ge\omega$.   Show that $\cf(\bu P)\ge\add P$.
%513C

\spheader 513Xc
Let $P$, $Q$ and $R$ be partially ordered sets.   (i)
Show that if $\phi_1:P\to Q$ and $\phi_2:Q\to R$ are Tukey functions,
then $\phi_2\phi_1:P\to R$ is a Tukey function.
(ii) Show that if $\psi_1:P\to Q$ and $\psi_2:Q\to R$ are dual Tukey
functions, then $\psi_2\psi_1:P\to R$ is a dual Tukey function.
%513D

\spheader 513Xd Let $P$ and $Q$ be partially ordered sets, and
$g:Q\to P$ a function.   Show that $g$ is a dual Tukey function iff for
every $p_0\in P$ there is a $q_0\in Q$ such that $g(q)\ge p_0$ for every
$q\ge q_0$.
%513D

\spheader 513Xe(i) Show that if $P$, $Q$ are partially ordered sets,
$P$ is Dedekind complete and $P\prT Q$, there is an order-preserving
dual Tukey function from $Q$ to $P$.
(ii) Set $P=[\{0,1,2\}]^{\le 2}$ and $Q=[\{0,1,2\}]^2$.   Show
that there is no order-preserving Tukey function from $P$ to $Q$.
%513E

\spheader 513Xr\dvAformerly{5{}27Xh}
Let $P$ be a partially ordered set.   Show that if
$\kappa\ge\cf P$ and $\lambda\le\add P$ then $P\prT[\kappa]^{<\lambda}$.
%513E  out of order query

\spheader 513Xf Suppose that $P$ is a partially ordered set and
$\add P=\cf P=\kappa\ge\omega$.    Show that $P\equivT\kappa$.
%513E

\spheader 513Xg Prove (a)-(d) of 513G directly, without mentioning Tukey
functions or Galois-Tukey connections.
%513G

\sqheader 513Xh(i) Show that if $P$ and $Q$ are two partially ordered
sets such that $\sat^{\uparrow}(P)=\#(P)^+=\#(Q)^+=\sat^{\uparrow}(Q)$
then $P$ and $Q$ are Tukey equivalent.   \Hint{if $B\subseteq Q$ is an
up-antichain, any injective function $\phi:P\to B$ is a Tukey function
from $P$ to $Q$.}   (ii) Give an example of such a pair $P$, $Q$ such
that $\frak m(P)\ne\frak m(Q)$ and $\bu P\ne\bu Q$.
%513G

\spheader 513Xi Let $P$, $Q_1$, $Q_2$ be partially ordered sets such
that $(P,\le,P)\prGT(Q_1,\le,Q_1)\ltimes(Q_2,\le\nobreak,Q_2)$ (definition:
512I).   Show that
$\add_{\omega}P\ge\min(\add_{\omega}Q_1,\add_{\omega}Q_2)$.
%513H

\spheader 513Xj Show that $\BbbN^{\Bbb N}$, with its usual ordering and
topology, is a metrizably compactly based directed set.
%513K

\spheader 513Xk Let $X$ be a set, $1\le p<\infty$ and $P$ the positive cone
$(\ell^p(X))^+$ of the Banach lattice $\ell^p(X)$, with the topology
induced by the norm of $\ell^p(X)$.   Show that $P$ is a metrizably
compactly based directed set.
%513K

\spheader 513Xl Let $\Cal Z$ be the ideal of subsets of $\Bbb N$ with
asymptotic density zero, with its natural ordering and the topology
induced by the metric
$(a,b)\mapsto\sup_{n\ge 1}\Bover1n\#((a\Bsymmdiff b)\cap n)$.   Show
that $\Cal Z$ is a metrizably compactly based directed set.
%513K

\spheader 513Xm Let $X$ be a metrizable space, and
$\Cal F$ the set of nowhere dense compact subsets of $X$.   Show that
$\Cal F$, with its natural ordering and its Vietoris topology, is a
metrizably compactly based directed set.   \Hint{use a Hausdorff metric.}
%513K

\spheader 513Xn Let $X$ be a metrizable space, $\Cal K$ the family
of compact subsets of $X$, and $\Cal I$ a $\sigma$-ideal of subsets of
$X$.   Show that $\Cal K\cap\Cal I$, with the natural partial order and
the Vietoris topology, is a metrizably compactly based directed set.
%513K

\spheader 513Xo Let $\familyiI{P_i}$ be a countable family of metrizably
compactly based directed sets, with product $P$.
Show that $P$ is metrizably compactly based.
%513K

\spheader 513Xp Let $P$ be a metrizably compactly based directed set.
(i) Show that $P$ is a lattice iff it has a least element.   (ii) Show
that if we adjoin a least element $-\infty$ to $P$ as an isolated point,
$P\cup\{-\infty\}$ is a metrizably compactly based directed set.
%513L

\spheader 513Xq\dvAnew{2014}
Let $\familyiI{P_i}$ be a family of partially ordered
sets, and $P$ their product.
(i) Show that $\cf P$ is at most the cardinal
product $\prod_{i\in I}\cf P_i$, with equality if $I$ is finite.
(ii) Show that if $P\ne\emptyset$ then $\sup_{i\in I}\cf P_i\le\cf P$.
(iii) Show that if $P\ne\emptyset$ and for every $i\in I$ there is a
$j\in I$ such that $\cf P_i<\cf P_j$, then $\sup_{i\in I}\cf P_i<\cf P$.
%513J out of order query

\leader{513Y}{Further exercises (a)}
%\spheader 513Ya
Show that for a cardinal $\kappa$, there is a partially ordered set $P$
such that $c^{\uparrow}(P)=\sat^{\uparrow}(P)=\kappa$ iff $\kappa$ is
weakly inaccessible.   \Hint{for such a $\kappa$, take $X$ to be a product of
discrete spaces, one of each cardinality less than $\kappa$, and
$P$ the family of proper closed subsets of $X$.}
%513B

\spheader 513Yb Show that for any cardinal $\kappa>0$ there is a
supported relation $(A,R,B)$ such that $\sat(A,R,B)=\kappa$.
%513B

\spheader 513Yc For a non-empty upwards-directed set $P$, a topological
space $X$ and $A\subseteq X$, write $\text{cl}_{P}(A)$ for the set of
points $x\in X$ for which there is a function $f:P\to A$ such that
$x\in\overline{f[C]}$ for every cofinal set $C\subseteq P$;
equivalently, $f[[\Cal F^{\uparrow}(P)]]\to x$, where
$\Cal F^{\uparrow}(P)$ is the filter on $P$ generated by sets of the
form $\coint{p,\infty}$ as $p$ runs over $P$.   Now let $P$ and $Q$ be
upwards-directed sets.   Show that $P\prT Q$ iff
$\text{cl}_P(A)\subseteq\text{cl}_Q(A)$ for any subset $A$ of any
topological space.
%513D

\spheader 513Yd For partially ordered sets $P$ and $Q$, say that
$P\approx Q$ if there is a partially ordered set $R$ into which both $P$
and $Q$ can be embedded as cofinal subsets.
(i) Show that $P\approx Q$ iff there is a Galois-Tukey connection
$(\phi,\psi)$ from $(P,\le,P)$ to $(Q,\le,Q)$ such that $(\psi,\phi)$
is a Galois-Tukey connection from $(Q,\le,Q)$ to $(P,\le,P)$.
(ii) Show that if $P$, $R$ and
$R'$ are partially ordered sets such that $P$ can be embedded as a cofinal
subset into both $R$ and $R'$, then $R\approx R'$.   (iii) Show that
$\approx$ is an equivalence relation on the class of partially ordered
sets.   (iv) Show that if $\Cal P$ is a set of partially ordered sets,
and $P\approx P'$ for all $P$, $P'\in\Cal P$, then there is a partially
ordered set $R$ such that every member of $\Cal P$ can be embedded into $R$
as a cofinal set.   (v) Give an example of partially ordered sets $P$
and $Q$ such that $P\equivT Q$ but $P\not\approx Q$.
%513F mt51bits "co-cofinal sets"

\spheader 513Ye For a cardinal $\kappa$ and a supported relation
$(A,R,B)$ set
$\add_{<\kappa}(A,R,B)=\add(A,R\strprime,[B]^{<\kappa})$.   Which of
the ideas of 513I can be extended to the general context?
%513H

\spheader 513Yf\dvAformerly{5{}12Yb}
Show that there are two families
$\familyiI{(A_i,R_i,B_i)}$ and
$\familyiI{(C_i,S_i,D_i)}$ of supported relations, with simple products
$(A,R,B)$ and $(C,S,D)$ respectively, such that
$\cov(A_i,R_i,B_i)=\cov(C_i,S_i,D_i)$ for each $i$, but
$\cov(A,R,B)\ne\cov(C,S,D)$.   \Hint{examine the proof of 513J.}
%512H 513J mt51bits query out of order

\spheader 513Yg Let $X$ be a set and $U$ a solid linear subspace of
$\BbbR^X$ with an order-continuous norm under which it is a Banach
lattice.   Show that its positive cone, with its norm topology, is a
metrizably compactly based directed set.
%513Xk 513K

\spheader 513Yh Explore possible definitions of `compactly based'
partially ordered set which do not require the topology to be
metrizable.
%513L

\spheader 513Yi Let $P$ be an analytic metrizably compactly based
directed set.
Show that $P$ is Polish.   \Hint{{\smc Solecki \& Todor\v{c}evi\'c
04}.} %Theorem 4.1 show G_{\delta} in completion
%513O

\spheader 513Yj For partially ordered sets $P$ and $Q$, say that
$Q\maltese P$ if for every order-preserving $f:P\to Q$ there is a
$p\in P$ such that $f(p')=f(p)$ for every $p'\ge p$.   Explore the
properties of the relation $\maltese$.
%513P

}%end of exercises

\endnotes{
\Notesheader{513} Most of the first part of this section consists of
elementary verifications;  an exception is the Erd\H{o}s-Tarski theorem
on the cellularity and saturation of a partially ordered set
(513Bb-513Bc), which can equally well be regarded as a theorem about
topological spaces or Boolean algebras (see 514Da and 514Nc).   As usual,
I have presented the ideas of the last two sections in an ahistorical
manner;  the original objective of {\smc Tukey 1940} was to classify
directed sets from the point of view of net-convergence (513Yc).

I have starred 513K-513O %513K 513L 513M 513N 513O
because I do not expect to rely on them in the rest of this work.
Nevertheless I think that they give a useful support to the ideas here,
particularly in the context of \S526, where these `compactly based'
partial orders appear naturally.   Note that 513Ld tells us that if a
directed set $P$ is metrizably compactly based, there is a unique
witnessing topology;  every topological property of $P$ must be a
reflection of a property of the ordering.
}%end of notes

\discrpage


\frfilename{mt514.tex}
\versiondate{16.5.14}
\copyrightdate{2003}

\def\nw{\mathop{\text{nw}}}
\def\triplepc#1#2#3{(#1,#2,\hbox{$<$}#3)}
\def\congforc{\cong_{\text{forcing}}}
\def\chaptername{Cardinal functions}
\def\sectionname{Boolean algebras}

\newsection{514}

The cardinal functions of Boolean algebras and topological spaces are
intimately entwined;  necessarily so, because we have a functorial
connexion between Boolean algebras and zero-dimensional compact
Hausdorff spaces\cmmnt{ (312Q)}.   In this section I run through the
elementary ideas.   In 514D-514E I list properties of cardinal
functions of
Boolean algebras, corresponding to the relatively familiar results in
5A4B for topological spaces;
Stone spaces (514B), regular open algebras (514H) and category algebras
(514I) provide links of
different kinds between the two theories.   It turns out that some of
the most important features of the cofinal structure of a
pre-ordered set can also be described in terms of its
`up-topology' (514L-514M) and the associated regular open algebra
(514N-514S).
I conclude with a brief note on finite-support products (514T-514U).

\leader{514A}{}\cmmnt{ I put a special property of locally compact
spaces into the language of this chapter.

\medskip

\noindent}{\bf Lemma} Let $(X,\frak T)$ be a topological space.
Then $d(X)\ge\ddownarrow(\frak T\setminus\{\emptyset\})$.
If $X$ is locally compact and Hausdorff, then
$d(X)=\ddownarrow(\frak T\setminus\{\emptyset\})$.

\proof{ If $x\in X$, then $\{G:x\in G\in\frak T\}$ is downwards-centered in
$\frak T\setminus\{\emptyset\}$.   So

\Centerline{$\ddownarrow(\frak T\setminus\{\emptyset\})
\le\cov(\frak T\setminus\{\emptyset\},\ni,X)=d(X)$.}

Now suppose that $X$ is locally compact and Hausdorff.   Set
$\kappa=\ddownarrow(\frak T\setminus\{\emptyset\})$,
and let $\ofamily{\xi}{\kappa}{\Cal H_{\xi}}$ be a cover of
$\frak T\setminus\{\emptyset\}$ by downwards-centered sets.   For
$\xi<\kappa$ set
$F_{\xi}=\bigcap\{\overline{H}:H\in\Cal H_{\xi}\}$, and let
$D\subseteq X$ be a set with cardinal at most $\kappa$ such that
$D\cap F_{\xi}\ne\emptyset$ whenever $\xi<\kappa$ and
$F_{\xi}\ne\emptyset$.    If $G\subseteq X$ is a non-empty open set,
then there is a non-empty relatively compact open set $H_0$ such that
$\overline{H}_0\subseteq G$ (recall that $X$, being locally compact and
Hausdorff, is certainly regular).   There is some $\xi<\kappa$ such that
$H_0\in\Cal H_{\xi}$;  because $\{\overline{H}:H\in\Cal H_{\xi}\}$ is a
family of closed sets with the finite intersection property containing
the compact set $\overline{H}_0$, its intersection $F_{\xi}$ is not
empty.   Also $F_{\xi}\subseteq\overline{H}_0\subseteq G$, so $D\cap
G\supseteq D\cap F_{\xi}$ is non-empty.   As $G$ is arbitrary, $D$ is
dense, and $d(X)\le\#(D)\le\kappa$.   We know already that
$\kappa\le d(X)$, so they are equal.
}%end of proof of 514A

\leader{514B}{Stone \dvrocolon{spaces}}\cmmnt{ Necessarily, any
cardinal function $\zeta$ of topological spaces corresponds to a
cardinal function $\tilde{\zeta}$ of Boolean algebras, taking
$\tilde{\zeta}(\frak A)=\zeta(Z)$ where $Z$ is the Stone space of
$\frak A$.   Working through the functions described in 5A4A and
511D, we have the following results.

\medskip

\noindent}{\bf Theorem}
Let $\frak A$ be any Boolean algebra and $Z$ its Stone space.   For
$a\in\frak A$ let $\widehat{a}$ be the corresponding open-and-closed
subset of $Z$.

(a) $\#(\frak A)$ is $2^{w(Z)}=2^{\#(Z)}$ if $\frak A$ is finite, $w(Z)$
otherwise.

(b) $\sat(\frak A)=\sat(Z)$, $c(\frak A)=c(Z)$.

(c) $\pi(\frak A)=\pi(Z)$.

(d) $d(\frak A)=d(Z)$.

(e) Let $\CalNwd(Z)$ be the ideal of nowhere dense subsets of $Z$.   Then
$\wdistr(\frak A)=\add\CalNwd(Z)$.

\proof{ Let $\Cal E$ be the algebra of open-and-closed subsets of $Z$,
so that $a\mapsto\widehat{a}$ is an isomorphism from $\frak A$ to
$\Cal E$.   The essential fact here is that
$\Cal E\setminus\{\emptyset\}$ is coinitial with
$\frak T\setminus\{\emptyset\}$, where $\frak T$ is the topology of $Z$,
so that (writing $\frak A^+$ for $\frak A\setminus\{0\}$, as usual)

\Centerline{$(\frak A^+,\Bsupseteqshort,\frak A^+)
\cong(\Cal E\setminus\{\emptyset\},\supseteq,
  \Cal E\setminus\{\emptyset\})
\equivGT(\frak T\setminus\{\emptyset\},\supseteq,
  \frak T\setminus\{\emptyset\})$}

\noindent by 513E(d-ii), inverted.

\medskip

{\bf (a)} If $\frak A$ is finite, so is $Z$, and in
this case $\frak A\cong\Cal PZ$ has cardinal $2^{\#(Z)}$.   If
$\frak A$ is infinite, so are $Z$ and $w(Z)$.   Because
$\Cal E$ is a base for the topology of $Z$,
$w(Z)\le\#(\Cal E)=\#(\frak A)$.   On the other hand, let $\Cal U$ be a
base for the topology of $Z$ with $\#(\Cal U)=w(Z)$.   Then every member
of $\Cal E$ is expressible as the union of a finite subset of $\Cal U$,
so

\Centerline{$\#(\frak A)=\#(\Cal E)
\le\#([\Cal U]^{<\omega})\le\max(\omega,\#(\Cal U))=w(Z)$.}

\medskip

{\bf (b)-(c)}

\Centerline{$c(\frak A)=c(\Cal E)
=\cdownarrow(\Cal E\setminus\{\emptyset\})
=\cdownarrow(\frak T\setminus\{\emptyset\})
=c(Z)$,}

\Centerline{$\sat(\frak A)=\sat(\Cal E)
=\sat^{\downarrow}(\Cal E\setminus\{\emptyset\})
=\sat^{\downarrow}(\frak T\setminus\{\emptyset\})
=\sat(Z)$,}

\Centerline{$\pi(\frak A)=\pi(\Cal E)
=\ci(\Cal E\setminus\{\emptyset\})
=\ci(\frak T\setminus\{\emptyset\})
=\pi(Z)$}

\noindent using 513Gb, inverted, to move between
$\Cal E\setminus\{\emptyset\}$ and $\frak T\setminus\{\emptyset\}$.

\medskip

{\bf (d)}

$$\eqalignno{d(\frak A)
&=\ddownarrow(\frak A^+)
=\ddownarrow(\Cal E\setminus\{\emptyset\})
=\ddownarrow(\frak T\setminus\{\emptyset\})\cr
\displaycause{513Gd, inverted}
&=d(Z)\cr}$$

\noindent because $Z$ is compact and Hausdorff (514A).

\medskip

{\bf (e)} Let $(\Pou(\frak A),\sqsubseteq^*)$
be the pre-ordered set of partitions of unity in $\frak A$ as described in
512Ee.   For $C\in\Pou(\frak A)$, set

\Centerline{$f(C)=Z\setminus\bigcup_{c\in C}\widehat{c}$.}

\noindent Then $f(C)\in\CalNwd(Z)$.   \Prf\Quer\ Otherwise, since $f(C)$ is
certainly closed, its interior is non-empty, and there is a non-zero
$a\in\frak A$ such that $\widehat{a}\subseteq f(C)$;  but in this case
$a\Bcap c=0$ for every $c\in C$ and $C$ is not a partition of unity.\
\Bang\Qed

If $C$, $D\in\Pou(\frak A)$ and $C\sqsubseteq^*D$ then
$f(C)\subseteq f(D)$.
\Prf\ If $d\in D$, $C_0=\{c:c\in C$, $c\Bcap d\ne 0\}$ is finite and
$d\Bsubseteq\sup C_0$;  so
$\widehat{d}\subseteq\bigcup_{c\in C_0}\widehat{c}$ is disjoint from
$f(C)$.   Thus $Z\setminus f(D)\subseteq Z\setminus f(C)$ and
$f(C)\subseteq f(D)$.\ \Qed

If $C$, $D\in\Pou(\frak A)$ and $f(C)\subseteq f(D)$ then
$C\sqsubseteq^*D$.   \Prf\ If $d\in D$ then the compact set $\widehat{d}$
is included in the open set $\bigcup_{c\in C}\widehat{c}$.   So there is a
finite set $C_0\subseteq C$ such that
$\widehat{d}\subseteq\bigcup_{c\in C_0}\widehat{c}$ and
$\{c:c\in C$, $d\Bcap c\ne 0\}\subseteq C_0$ is finite.\ \Qed

$f[\Pou(\frak A)]$ is cofinal with $\CalNwd(Z)$.
\Prf\ If $F\in\CalNwd(Z)$, let
$C\subseteq\frak A$ be a maximal disjoint set such that
$F\cap\widehat{c}=\emptyset$ for every $c\in C$.   \Quer\ If $C$ is not a
partition of unity in $\frak A$, let $a\in\frak A^+$ be such that
$a\Bcap c=0$ for every $c\in C$.   Then $\widehat{a}\setminus F$ is a
non-empty open set, so there is a non-zero $b\in\frak A$ such that
$\widehat{b}\subseteq\widehat{a}\setminus F$;  in which case we ought to
have added $b$ to $C$.\ \BanG\  So $C\in\Pou(\frak A)$ and
$F\subseteq f(C)$.\ \Qed

By 513E(d-i), $\Pou(\frak A)$ and $\CalNwd(Z)$ are Tukey equivalent, and

\Centerline{$\add\CalNwd(Z)=\add\Pou(\frak A)=\wdistr(\frak A)$}

\noindent as remarked in 512Ee.
}%end of proof of 514B

\leader{514C}{}\cmmnt{ I begin the detailed study of cardinal
functions of Boolean algebras with two elementary remarks.

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be a Boolean algebra.

(a) $d(\frak A)$ is the smallest cardinal $\kappa$ such that $\frak A$
is isomorphic, as Boolean algebra, to a subalgebra of $\Cal P\kappa$.

(b) $\link(\frak A)$ is the smallest cardinal $\kappa$ such that
$\frak A$ is isomorphic, as partially ordered set, to a subset of
$\Cal P\kappa$.

\proof{{\bf (a)(i)} If we have an isomorphism $\pi$ from $\frak A$ to a
subalgebra of $\Cal P\kappa$, then $A_{\xi}=\{a:\xi\in\pi a\}$ is
centered for each $\xi<\kappa$, and
$\bigcup_{\xi<\kappa}A_{\xi}=\frak A^+$;  so
$d(\frak A)\le\kappa$.

\medskip

\quad{\bf (ii)} Let $Z$ be the Stone space of $\frak A$, and for
$a\in\frak A$ let $\widehat{a}\subseteq Z$ be the corresponding
open-and-closed set.   There is a dense set $D\subseteq Z$ with cardinal
$d(\frak A)$ (514Bd), and $a\mapsto D\cap\widehat{a}:\frak A\to\Cal PD$
is an injective Boolean homomorphism;  so $\frak A$ is isomorphic to a
subalgebra of $\Cal PD\cong\Cal P(d(\frak A))$.

\medskip

{\bf (b)(i)} $\kappa\le\link(\frak A)$.   \Prf\ Let
$\ofamily{\xi}{\link(\frak A)}{A_{\xi}}$ be a family of linked subsets
of $\frak A^+$ covering $\frak A^+$.   Set
$A'_{\xi}=\{b:\,\Exists a\in A_{\xi},\,b\Bsupseteq a\}$;  then each
$A'_{\xi}$ is still linked in $\frak A$.   Define
$h:\frak A\to\Cal P\kappa$ by setting $h(a)=\{\xi:a\in A'_{\xi}\}$.
Then $h$ is order-preserving.   Now if $a$, $b\in\frak A$ and
$a\notBsubseteq b$, there is a $\xi<\kappa$ such that
$a\Bsetminus b\in A_{\xi}$, in which case $\xi\in h(a)\setminus h(b)$.
Thus $h$ is an embedding and $\kappa\le\link(\frak A)$.\ \Qed

\medskip

\quad{\bf (ii)} $\link(\frak A)\le\kappa$.   \Prf\ Let
$h:\frak A\to\Cal P\kappa$ be an order-isomorphism between $\frak A$ and
a subset of $\Cal P\kappa$.   For each $\xi$, set

\Centerline{$A_{\xi}
=\{a:a\in\frak A,\,\xi\in h(a)\setminus h(1\Bsetminus a)\}$.}

\noindent If $a$, $b\in A_{\xi}$ then
$\xi\in h(b)\setminus h(1\Bsetminus a)$ so
$b\notBsubseteq 1\Bsetminus a$ and $a\Bcap b\ne 0$;  thus $A_{\xi}$ is
linked.   If $a\in\frak A^+$ then $a\notBsubseteq 1\Bsetminus a$ so
$h(a)\not\subseteq h(1\Bsetminus a)$ and there is a $\xi<\kappa$ such
that $\xi\in h(a)\setminus h(1\Bsetminus a)$;  thus
$\frak A^+=\bigcup_{\xi<\kappa}A_{\xi}$ and
$\link(\frak A)\le\kappa$.\ \Qed
}%end of proof of 514C

\leader{514D}{Theorem} Let $\frak A$ be a Boolean algebra.

(a)

\Centerline{$c(\frak A)\le\link(\frak A)\le d(\frak A)
\le\pi(\frak A)\le\#(\frak A)\le 2^{\link(\frak A)}$,
\quad$\tau(\frak A)\le\pi(\frak A)$,}

\noindent $\sat(\frak A)=c(\frak A)^+$ unless $\sat(\frak A)$ is weakly
inaccessible, in which case $\sat(\frak A)=c(\frak A)$.

(b) If $A\subseteq\frak A$, there is a $B\in[A]^{<\sat(\frak A)}$ with
the same upper bounds as $A$;  \cmmnt{similarly,} there is a
$B\in[A]^{<\sat(\frak A)}$ with the same lower bounds as $A$.

(c) $\link_{c(\frak A)}(\frak A)=\link_{<\sat(\frak A)}(\frak A)
=\pi(\frak A)$.

(d) If $\frak A$ is not purely atomic,
$\wdistr(\frak A)\le\min(d(\frak A),2^{\tau(\frak A)})$ is a regular
infinite cardinal.

(e) $\#(\frak A)
\le\max(4,\sup_{\lambda<\sat(\frak A)}\tau(\frak A)^{\lambda})$\cmmnt{,
where $\tau(\frak A)^{\lambda}$ is the cardinal power}.

\proof{ Let $Z$ be the Stone space of $\frak A$;  for $a\in\frak A$, let
$\widehat{a}\subseteq Z$ be the corresponding open-and-closed set.

\medskip

{\bf (a)} This is mostly a repetition of 511Ia.
By 514Cb, $\#(\frak A)\le 2^{\link(\frak A)}$.
By 513Bc, inverted, and the definitions in 511Db,

\Centerline{$\sat(\frak A)=\sat^{\downarrow}(\frak A^+)
=c^{\downarrow}(\frak A^+)^+=c(\frak A)^+$}

\noindent unless $\sat(\frak A)$ is a regular uncountable limit cardinal,
that is, is weakly inaccessible, and otherwise $\sat(\frak A)=c(\frak A)$.
(See also 5A4Ba.)

\medskip

{\bf (b)} By 5A4Bd, applied to $\{\widehat{a}:a\in A\}$, there is a
$B\in[A]^{<\sat(\frak A)}$ such that
$\overline{\bigcup_{b\in B}\widehat{b}}
=\overline{\bigcup_{a\in A}\widehat{a}}$.   Now if $c$ is an upper bound
of $B$, then $\widehat{c}$ is a closed set including $\widehat{b}$ for
every $b\in B$, so it also includes $\widehat{a}$ for every $a\in A$,
and $c$ is an upper bound of $A$.

Applying this to $\{1\Bsetminus a:a\in A\}$ we see that there is a set
$B'\in[A]^{<\sat(\frak A)}$ with the same lower bounds as $A$.

\medskip

{\bf (c)} Set $\kappa=\link_{<\sat(\frak A)}(\frak A)$.   By 511Ia,
$\kappa\le\link_{c(\frak A)}(\frak A)\le\pi(\frak A)$.
On the other hand, if $A\subseteq\frak A^+$ is
$\hbox{$<$}\sat(\frak A)$-linked,
it has a lower bound in $\frak A^+$.   \Prf\ By (b), there is a set
$B\subseteq A$, with the same lower bounds as $A$, such that
$\#(B)<\sat(\frak A)$.   Now $B$ has a
non-zero lower bound because $A$ is $\hbox{$<$}\sat(\frak A)$-linked, so
$A$ also has a non-zero lower bound.\ \QeD\
We have a cover $\ofamily{\xi}{\kappa}{A_{\xi}}$ of $\frak A^+$ by
$\hbox{$<$}\sat(\frak A)$-linked sets;  each $A_{\xi}$ has a non-zero
lower bound $a_{\xi}$ say;  and $\{a_{\xi}:\xi<\kappa\}$ is a $\pi$-base
for
$\frak A$, so $\pi(\frak A)\le\kappa$.

\medskip

{\bf (d)(i)} Because $\wdistr(\frak A)=\add\CalNwd(Z)$, where
$\CalNwd(Z)$ is
the ideal of nowhere dense subsets of $Z$ (514Be), and is not $\infty$
(511Ie), it must be a regular infinite cardinal (513C(a-i)).   (Or argue
directly from 511Df.)

\medskip

\quad{\bf (ii)} As for the upper bound for $\wdistr(\frak A)$, suppose
that $a\in\frak A^+$ includes no atom and that
$D\in[\frak A]^{\tau(\frak A)}\,\,\tau$-generates $\frak A$.   Since
$\frak A$ and $\tau(\frak A)$ are surely infinite, the subalgebra
$\frak B$ of $\frak A$ generated by $D\cup\{a\}$ is still with cardinal
$\tau(\frak A)$ (331Gc).   For $B\subseteq\frak B$ set
$E_B=Z\cap\bigcap_{b\in B}\widehat{b}$, and set
$\Cal C=\{B:B\subseteq\frak B,\,E_B$ is nowhere dense$\}$.   Then
$\bigcup_{B\in\Cal C}E_B\supseteq\widehat{a}$.   \Prf\ Take any
$z\in\widehat{a}$.   Set $B=\{b:b\in\frak B,\,z\in\widehat{b}\}$.
\Quer\ If $E_B$ has non-empty interior, it includes $\widehat{c}$ for
some non-zero $c\Bsubseteq a$.   But now, for any $d\in D$, either
$d\in B$ and $c\Bsubseteq d$, or $1\Bsetminus d\in B$ and $c\Bcap d=0$.
So the order-closed subalgebra
$\{d:$ either $c\Bsubseteq d$ or $c\Bcap d=0\}$ includes $D$ and must be
the whole of $\frak A$, and $c\Bsubseteq a$ is an atom.\ \BanG\  So
$\interior E_B=\emptyset$, $B\in\Cal C$ and $z\in E_B$.   As $z$ is
arbitrary, $\widehat{a}\subseteq\bigcup_{B\in\Cal C}E_B$.\ \Qed

By 514Be, with 514Bd,

\Centerline{$\wdistr(\frak A)\le\#(\Cal C)\le 2^{\#(\frak B)}
=2^{\tau(\frak A)}$.}

At the same time, if $Y\subseteq Z$ is any dense set with cardinal $d(Z)$,
then $\{\{y\}:y\in Y\cap\widehat{a}\}$ is a family of nowhere dense sets
with no upper bound in the ideal of nowhere dense subsets of $Z$;  so
514Be also tells us that

\Centerline{$\wdistr(\frak A)\le\#(Y\cap\widehat{a})\le d(Z)
=d(\frak A)$.}

\medskip

{\bf (e)} (Compare 4A1O.)   Set
$\kappa=\sup_{\lambda<\sat(\frak A)}\tau(\frak A)^{\lambda}$.   If
$\#(\frak A)>4$ then $\tau(\frak A)\ge 2$ so
$\kappa\ge\sup_{\lambda<\sat(\frak A)}2^{\lambda}$, and the
result is immediate from 511Ic if $\frak A$ is finite.   If $\frak A$ is
infinite, so is $\sat(\frak A)$, while $\lambda<\kappa$ for every
$\lambda<\sat(\frak A)$, so $\sat(\frak A)\le\kappa$.   Let
$D\subseteq\frak A$ be a set with cardinal $\tau(\frak A)$ which
$\tau$-generates $\frak A$.   Define
$\ofamily{\xi}{\kappa}{D_{\xi}}$ inductively by setting

\Centerline{$D_0=D$,
\quad$D_{\xi}=\{1\Bsetminus a:a\in\frak A,\,a=\sup C$ for some
$C\subseteq\bigcup_{\eta<\xi}D_{\eta}\}$}

\noindent for $\xi<\kappa$.   Then
$\#(D_{\xi})\le\kappa$ for every $\xi<\kappa$.   \Prf\ The
point is that, by (b), every member of $D_{\xi}$ is expressible in the
form $1\Bsetminus\sup C$ for some
$C\in[\bigcup_{\eta<\xi}D_{\eta}]^{<\sat(\frak A)}$.   But the inductive
hypothesis tells us that $\bigcup_{\eta<\xi}D_{\eta}$ has cardinal at
most $\kappa$, so the number of its subsets with
cardinal less than $\sat(\frak A)$ is also $\kappa$ (5A1Ff, because
$\sat(\frak A)$ is regular), and
$\#(D_{\xi})\le\kappa$.\ \Qed

At the end of the induction, set
$\frak B=\bigcup_{\xi<\sat(\frak A)}D_{\xi}$.
Then $1\Bsetminus(b\Bcup b')\in\frak B$ for every $b$, $b'\in\frak B$,
so $\frak B$ is a subalgebra of $\frak A$.   Also it is order-closed.
\Prf\ If $B\subseteq\frak B$ has a supremum $a\in\frak A$, there is a
$C\subseteq B$ such that $\#(C)<\sat(\frak A)$ and $a=\sup C$.   Now
there must be some set $J\subseteq\sat(\frak A)$ such that
$\#(J)<\sat(\frak A)$ and $C\subseteq\bigcup_{\eta\in J}D_{\eta}$.
Since $\sat(\frak A)$ is
regular (513Bb), $\zeta=\sup J$ is less than $\sat(\frak A)$.
Now $1\Bsetminus a\in D_{\zeta+1}$ and $a\in\frak B$.\ \Qed

By the choice of $D$, $\frak B=\frak A$, so
$\#(\frak A)=\#(\frak B)\le\kappa$.
}%end of proof of 514D

\vleader{108pt}{514E}{Subalgebras, homomorphic images, products:
Theorem} Let $\frak A$ be a Boolean algebra.

(a) If $\frak B$ is a subalgebra of $\frak A$, then

\Centerline{$\sat(\frak B)\le\sat(\frak A)$,
\quad$c(\frak B)\le c(\frak A)$,}

\Centerline{$\link_{<\kappa}(\frak B)\le\link_{<\kappa}(\frak A)$}

\noindent for every $\kappa\le\omega$\cmmnt{, in particular},

\Centerline{$d(\frak B)\le d(\frak A)$,
\quad$\link(\frak B)\le\link(\frak A)$.}

(b) If $\frak B$ is a regularly embedded subalgebra of $\frak A$
then\cmmnt{, in addition,}
$\link_{<\kappa}(\frak B)\le\link_{<\kappa}(\frak A)$ for
$\kappa>\omega$, $\pi(\frak B)\le\pi(\frak A)$ and
$\wdistr(\frak A)\le\wdistr(\frak B)$.

(c) If $\frak B$ is a Boolean algebra and $\phi:\frak A\to\frak B$ is a
surjective order-continuous Boolean homomorphism, then

\Centerline{$\sat(\frak B)\le\sat(\frak A)$,
\quad$c(\frak B)\le c(\frak A)$,
\quad$\pi(\frak B)\le\pi(\frak A)$,}

\Centerline{$\link_{<\kappa}(\frak B)\le\link_{<\kappa}(\frak A)$
for every cardinal $\kappa$,}

\Centerline{$d(\frak B)\le d(\frak A)$,
\quad$\link(\frak B)\le\link(\frak A)$,}

\cmmnt{\noindent and also}

\Centerline{$\wdistr(\frak A)\le\wdistr(\frak B)$,
\quad$\tau(\frak B)\le\tau(\frak A)$.}

(d) If $\frak B$ is a principal ideal of $\frak A$, then

\Centerline{$\sat(\frak B)\le\sat(\frak A)$,
\quad$c(\frak B)\le c(\frak A)$,
\quad$\pi(\frak B)\le\pi(\frak A)$,}

\Centerline{$\link_{<\kappa}(\frak B)\le\link_{<\kappa}(\frak A)$
for every $\kappa$,}

\Centerline{$d(\frak B)\le d(\frak A)$,
\quad$\link(\frak B)\le\link(\frak A)$;}

\cmmnt{\noindent moreover,}

\Centerline{\quad$\wdistr(\frak A)\le\wdistr(\frak B)$,
\quad$\tau(\frak B)\le\tau(\frak A)$.}

(e) If $\frak B$ is an order-dense subalgebra of $\frak A$ then

\Centerline{$\sat(\frak B)=\sat(\frak A)$,
\quad$c(\frak B)=c(\frak A)$,
\quad$\pi(\frak B)=\pi(\frak A)$,}

\Centerline{$\link_{<\kappa}(\frak B)=\link_{<\kappa}(\frak A)$
for every $\kappa$,}

\Centerline{$d(\frak B)=d(\frak A)$,
\quad$\link(\frak B)=\link(\frak A)$,}

\cmmnt{\noindent and finally}

\Centerline{$\wdistr(\frak B)=\wdistr(\frak A)$,
\quad$\tau(\frak A)\le\tau(\frak B)$.}
%query:  is $\tau(\frak B)\le\max(c(\frak A),\tau(\frak A))$?

(f) If $\frak A$ is the simple product of a family
$\familyiI{\frak A_i}$ of Boolean algebras, then

\Centerline{$\tau(\frak A)
\le\max(\omega,\sup_{i\in I}\tau(\frak A_i),
   \min\{\lambda:\#(I)\le 2^{\lambda}\})$,}

\Centerline{$\sat(\frak A)
\le\max(\omega,\#(I)^+,\sup_{i\in I}\sat(\frak A_i))$,}

\Centerline{$c(\frak A)
\le\max(\omega,\#(I),\sup_{i\in I}c(\frak A_i))$,}

\Centerline{$\pi(\frak A)
\le\max(\omega,\#(I),\sup_{i\in I}\pi(\frak A_i))$,}

\Centerline{$\link_{<\kappa}(\frak A)
\le\max(\omega,\#(I),\sup_{i\in I}\link_{<\kappa}(\frak A_i))$
for every $\kappa$,}

\Centerline{$\link(\frak A)
\le\max(\omega,\#(I),\sup_{i\in I}\link(\frak A_i))$,}

\Centerline{$d(\frak A)
\le\max(\omega,\#(I),\sup_{i\in I}d(\frak A_i))$,}

\cmmnt{\noindent and}

\Centerline{$\wdistr(\frak A)=\min_{i\in I}\wdistr(\frak A_i)$.}

\proof{ Write $Z$ for the Stone space of $\frak A$.

\medskip

{\bf (a)} Any disjoint subset of $\frak B^+$ is a disjoint subset of
$\frak A^+$, so $\sat(\frak B)\le\sat(\frak A)$ and
$c(\frak B)\le c(\frak A)$.   If $\kappa\le\omega$ and $\Cal A$ is
a cover of $\frak A^+$ by sets which are downwards
$\hbox{$<$}\kappa$-linked in $\frak A^+$, then $A\cap\frak B$ is
downwards
$\hbox{$<$}\kappa$-linked in $\frak B^+$ for each $A\in\Cal A$, so
$\link_{<\kappa}(\frak B)\le\link_{<\kappa}(\frak A)$.

\medskip

{\bf (b)} For each non-zero $a\in\frak A$, the set
$B_a=\{b:b\in\frak B,\,a\Bsubseteq b\}$ does not have infimum $0$ in
$\frak A$ so cannot have infimum $0$ in $\frak B$;  let
$\psi(a)\in\frak B^+$ be a lower bound for $B_a$.   If now we set
$\phi(b)=b$ for $b\in\frak B$, $(\phi,\psi)$ is a
Galois-Tukey connection from
$(\frak B^+,\Bsupseteqshort,\frak B^+)$ to
$(\frak A^+,\Bsupseteqshort,\frak A^+)$.   It follows at once that

\Centerline{$\link_{<\kappa}(\frak B)
=\link_{<\kappa}(\frak B^+,\Bsupseteqshort,\frak B^+)
\le\link_{<\kappa}(\frak A^+,\Bsupseteqshort,\frak A^+)
=\link_{<\kappa}(\frak A)$}

\noindent for arbitrary $\kappa$ (512Dd), and that

\Centerline{$\pi(\frak B)
=\cov(\frak B^+,\Bsupseteqshort,\frak B^+)
\le\cov(\frak A^+,\Bsupseteqshort,\frak A^+)
=\pi(\frak A)$}

\noindent (512Da).

Now suppose that $\kappa<\wdistr(\frak A)$ and that
$\ofamily{\xi}{\kappa}{B_{\xi}}$ is a family of partitions of unity in
$\frak B$.   Then

\Centerline{$D=\{d:d\in\frak B$, $\{b:b\in B_{\xi}$, $b\Bcap d\ne 0\}$
is finite for every $\xi<\kappa\}$}

\noindent is order-dense in $\frak B$.   \Prf\ Take any non-zero
$d\in\frak B$.   $\sup B_{\xi}=1$ in $\frak A$, that is, $B_{\xi}$ is
still a partition of unity in $\frak A$, for each $\xi$.   So there
is a partition $A$ of unity in $\frak A$ such that
$\{b:b\in B_{\xi}$, $b\Bcap a\ne 0\}$ is finite for every $\xi<\kappa$
and $a\in A$.   Let $a\in A$ be such that $d\Bcap a\ne 0$, and set
$e_{\xi}=\sup\{b:b\in B_{\xi}$, $b\Bcap a\ne 0\}$ for each
$\xi<\kappa$.   Then $a\Bsubseteq e_{\xi}\in\frak B$ for each $\xi$.
This means that $\{d\}\cup\{e_{\xi}:\xi<\kappa\}$ has a
non-zero lower bound $d\Bcap a$ in $\frak A$;  as $\frak B$ is regularly
embedded in $\frak A$, there is a non-zero $d'\Bsubseteq d$ which is
also a lower bound for $\{e_{\xi}:\xi<\kappa\}$.   But this means that
$d'\in D$.   As $d$ is arbitrary, $D$ is order-dense in $\frak B$.\ \Qed

There is therefore a partition of unity included in $D$.   As
$\ofamily{\xi}{\kappa}{B_{\xi}}$ is arbitrary,
$\wdistr(\frak B)\ge\wdistr(\frak A)$.

\medskip

{\bf (c)(i)} For any $b\in\frak B^+$ there is a $\psi(b)\in\frak A^+$ such
that $\phi\psi(b)\Bsubseteq b$ and $a=0$ whenever
$a\Bsubseteq\psi(b)$ and $\phi a=0$.   \Prf\ Consider
$D=\{d:d\in\frak A$, $\phi d\supseteq b\}$.   This is a
non-empty downwards-directed subset of $\frak A$ and $b$ is a non-zero
lower bound of $\phi[D]$.   As $\phi$ is supposed to be order-continuous,
$D$ must have a non-zero lower bound in $\frak A$;  let $\psi(b)$ be such a
lower bound.   Since there is a $d\in\frak A$ such that $\phi d=b$, and now
$d\in D$, we must have $\phi\psi(b)\Bsubseteq\phi d=b$.   If
$a\Bsubseteq\psi(b)$ and $\phi a=0$, then
$\phi(1\Bsubseteq a)=1\Bsupseteq b$, $1\Bsetminus a\in D$ and
$a\Bsubseteq\psi(b)\Bsubseteq 1\Bsetminus a$, so $a=0$.\ \Qed

\medskip

\quad{\bf (ii)} If $\kappa=\sat(\frak A)$ and
$\ofamily{\xi}{\kappa}{b_{\xi}}$ is a family in $\frak B^+$, then
$\ofamily{\xi}{\kappa}{\psi(b_{\xi})}$ is a family in $\frak A^+$ and there
are distinct $\xi$, $\eta<\kappa$ such that
$a=\psi(b_{\xi})\Bcap\psi(b_{\eta})$ is non-zero.   Now

\Centerline{$0\ne\phi a
\Bsubseteq\phi\psi(b_{xi})\Bcap\phi\psi(b_{\eta})
\Bsubseteq b_{\xi}\Bcap b_{\eta}$.}

\noindent As $\ofamily{\xi}{\kappa}{b_{\xi}}$ is arbitrary,
$\sat(\frak B)\le\sat(\frak A)$.   By a similar argument, or using 514Da,
we see that $c(\frak B)\le c(\frak A)$.

\medskip

\quad{\bf (iii)} Let $A$ be a coinitial subset of $\frak A^+$ of
cardinal $\pi(\frak A)$.   Set $B=\phi[A]\setminus\{0\}$.   If
$b\in\frak B^+$, there is an $a\in A$ such that $a\Bsubseteq\psi(b)$, and
now $\phi a\in B$ and $\phi a\Bsubseteq\phi\psi(b)\Bsubseteq b$.   So $B$
is cofinal with $\frak B^+$ and

\Centerline{$\pi(\frak B)\le\#(B)\le\#(A)=\pi(\frak A)$.}

\medskip

\quad{\bf (iv)}
Let $W$ be the Stone space of $\frak B$.   Write $\CalNwd(Z)$,
$\CalNwd(W)$ for
the ideals of nowhere dense subsets of $Z$ and $W$, so that
$\add\CalNwd(Z)=\wdistr(\frak A)$ and $\add\CalNwd(W)=\wdistr(\frak B)$
(514Be).   Corresponding to
$\phi:\frak A\to\frak B$ we have an injective continuous function
$\theta:W\to Z$ such that $\theta^{-1}[E]\in\CalNwd(W)$ for every
$E\in\CalNwd(Z)$ (312Sb, 313R).   Also $\theta[F]\in\CalNwd(Z)$ for every
$F\in\CalNwd(W)$.   \Prf\Quer\ Otherwise, because $\theta[\overline{F}]$ is
compact, therefore closed, there is a non-empty open set
$G\subseteq\theta[\overline{F}]$.   Now $\theta^{-1}[G]$ is a non-empty
open set, and is included in $\overline{F}$, because $\theta$ is
injective;  but this is impossible.\ \Bang\QeD\  So if
$\Cal J_0\subseteq\CalNwd(W)$ and $\#(\Cal J_0)<\wdistr(\frak A)$,
$E=\bigcup\{\theta[F]:F\in\Cal J_0\}$ belongs to $\CalNwd(Z)$ and
$\bigcup\Cal J_0\subseteq\theta^{-1}[E]$ belongs to $\CalNwd(W)$.   This
shows that $\add\CalNwd(W)\ge\add\CalNwd(Z)$, so that
$\wdistr(\frak B)\ge\wdistr(\frak A)$.

\medskip

\quad{\bf (v)} As for $\tau(\frak B)$, we have only to recall that if
$D\subseteq\frak A$ is a $\tau$-generating set with cardinal $\tau(\frak A)$,
the order-closed subalgebra of $\frak B$ generated by $\phi[D]$ includes
$\phi[\frak A]=\frak B$ (313Mb), and

\Centerline{$\tau(\frak B)\le\#(\phi[D])\le\tau(\frak A)$.}

\medskip

{\bf (d)} If $\frak B$ is the principal ideal generated by $b$, then
$a\mapsto a\Bcap b:\frak A\to\frak B$ is an order-continuous surjection,
so we can repeat the list in (c).

\medskip

{\bf (e)(i)} Because $\frak B^+$ is coinitial with
$\frak A^+$ we can use 513Gc, inverted, to see that

\Centerline{$\sat(\frak B)=\sat^{\downarrow}(\frak B^+)
=\sat^{\downarrow}(\frak A^+)=\sat(\frak A)$,
\quad$c(\frak B)=c(\frak A)$,}

\Centerline{$\pi(\frak B)=\ci(\frak B^+)
=\ci(\frak A^+)=\pi(\frak A)$,}

\Centerline{$\link_{<\kappa}(\frak B)
=\link_{<\kappa}^{\downarrow}(\frak B^+)
=\link_{<\kappa}^{\downarrow}(\frak A^+)
=\link_{<\kappa}(\frak A)$.}

\medskip

\quad{\bf (ii)} From (b) we know that
$\wdistr(\frak B)\ge\wdistr(\frak A)$.   For the reverse inequality,
suppose that $\kappa<\wdistr(\frak B)$ and that
$\ofamily{\xi}{\kappa}{A_{\xi}}$ is any family of partitions of unity in
$\frak A$.    For each $\xi<\kappa$ set
$B_{\xi}=\{b:b\in\frak B,\,\Exists a\in A_{\xi},\,b\Bsubseteq a\}$.
Then $B_{\xi}$ is order-dense in $\frak B$ and includes a partition of
unity $B'_{\xi}$ (313K).
Now there is a partition $C$ of unity in $\frak B$
such that $D'_{\xi c}=\{b:b\in B'_{\xi}$, $b\Bcap c\ne 0\}$ is finite
for any $\xi<\kappa$ and $c\in C$.   $C$ is still a partition of unity
in $\frak A$, and $D_{\xi c}=\{a:a\in A_{\xi}$, $a\Bcap c\ne 0\}$ is
finite for every $c\in C$ and $\xi<\kappa$.   (For if $a$, $a'$ are
distinct elements of $D_{\xi c}$, then $\{b:b\in D'_{\xi c}$,
$b\Bsubseteq a\}$ and $\{b:b\in D'_{\xi c}$, $b\Bsubseteq a'\}$ are
disjoint and not empty.)   As
$\ofamily{\xi}{\kappa}{A_{\xi}}$ is arbitrary,
$\wdistr(\frak B)\le\wdistr(\frak A)$.

\medskip

\quad{\bf (iii)} If $D\subseteq\frak B\,\,\tau$-generates $\frak B$,
then $D$ also $\tau$-generates $\frak A$.   \Prf\ Applying 313Mb to the
identity map from $\frak B$ to $\frak A$, we see that the
order-closed subalgebra $\frak D$ of $\frak A$ generated by $D$ includes
$\frak B$;  but as any member of $\frak A$ is the supremum of a subset
of $\frak B$, $\frak D=\frak A$.\ \QeD\   So
$\tau(\frak A)\le\tau(\frak B)$.

\medskip

{\bf (f)} We can identify each $\frak A_i$ with the principal ideal of
$\frak A$ generated by an element $a_i$, where $\familyiI{a_i}$ is a
partition of unity in $\frak A$ (315E).   If
\discrcenter{390pt}{$\kappa=\max(\omega,\sup_{i\in I}\tau(\frak A_i),
   \min\{\lambda:\#(I)\le 2^{\lambda}\})$, }then for each $i\in I$
choose $\ofamily{\xi}{\kappa}{a_{i\xi}}$ in $\frak A_i$ such that
$\{a_{i\xi}:\xi<\kappa\}\,\,\tau$-generates $\frak A_i$, and let
$\phi:I\to\Cal P\kappa$ be injective.   For $\xi<\kappa$, set

\Centerline{$b_{\xi}=\sup_{i\in I}a_{i\xi}$,
\quad$c_{\xi}=\sup_{i\in\phi(\xi)}a_i$.}

\noindent Let $\frak B$ be the order-closed subalgebra of $\frak A$
generated by $\{b_{\xi}:\xi<\kappa\}\cup\{c_{\xi}:\xi<\kappa\}$.
Then

\Centerline{$a_i
=\inf\{c_{\xi}:\xi\in\phi(i)\}
  \Bsetminus\sup\{c_{\xi}:\xi\in\kappa\setminus\phi(i)\}
\in\frak B$}

\noindent for each $i$.   Because $\{b:b\in\frak B,\,b\Bsubseteq a_i\}$
is an order-closed subalgebra of $\frak A_i$ containing
$b_{\xi}\cap a_i=a_{i\xi}$ for every $\xi<\kappa$, it is the whole of
$\frak A_i$, so $\frak A_i\subseteq\frak B$ for every $i\in I$.   It
follows at once that $\frak B=\frak A$, so that
$\tau(\frak A)\le\kappa$.

The other parts are all elementary.
}%end of proof of 514E

\leader{514F}{}\cmmnt{ For measure algebras, Maharam type is not only
the cardinal function which gives most information, but is also,
as a rule, easy to
calculate.   For other Boolean algebras, it may not be obvious what the
Maharam type is.   The following result sometimes helps.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Dedekind complete Boolean
algebra, and $\langle a_{ij}\rangle_{i\in I,j\in J}$ a
$\tau$-generating family in $\frak A$ such that

\Centerline{$\family{j}{J}{a_{ij}}$ is disjoint for
every $i\in I$,
\quad$\sup_{i\in I}a_{ij}=1$ for every $j\in J$.}

\noindent Then $\tau(\frak A)\le\max(\omega,\#(I))$.
%528R

\proof{{\bf (a)} We may suppose that $J=\kappa$ is a cardinal.
For $i$, $j\in I$ set

\Centerline{$a^*_i=\sup_{\xi<\kappa}a_{i\xi}$,
\quad$b_{ij}=\sup_{\xi<\eta<\kappa}a_{i\eta}\Bcap a_{j\xi}$.}

Then

\Centerline{$\sup_{\eta\le\zeta}a_{i\eta}
=a^*_i\Bsetminus\sup_{j\in I}(b_{ij}\setminus\sup_{\xi<\zeta}a_{j\xi})$}

\noindent whenever $i\in I$ and $\zeta<\kappa$.
\Prf\ (i) If $\eta\le\zeta$ and $j\in I$, then $a_{i\eta}\Bsubseteq a^*_i$ and

$$\eqalignno{a_{i\eta}\Bcap b_{ij}
&=\sup_{\xi<\theta<\kappa}a_{i\eta}\Bcap a_{i\theta}\Bcap a_{j\xi}
=\sup_{\xi<\eta}a_{i\eta}\Bcap a_{j\xi}\cr
\displaycause{because $\ofamily{\theta}{\kappa}{a_{i\theta}}$ is disjoint}
&\Bsubseteq\sup_{\xi<\zeta}a_{j\xi},\cr}$$

\noindent so

\Centerline{$\sup_{\eta\le\zeta}a_{i\eta}
\Bsubseteq a^*_i
   \Bsetminus\sup_{j\in I}(b_{ij}\setminus\sup_{\xi<\zeta}a_{j\xi})$.}

\noindent(ii) If $0\ne c\Bsubseteq a_i^*$ and $c\Bcap a_{i\eta}=0$ for every
$\eta\le\zeta$, there are
an $\eta>\zeta$ such that $c'=c\Bcap a_{i\eta}$ is non-zero, and
a $j\in I$ such that $c''=c'\Bcap a_{j\zeta}$ is non-zero.   In this case
$c''\Bsubseteq b_{ij}\Bsetminus\sup_{\xi<\zeta}a_{j\xi}$, so
$c\Bcap b_{ij}\Bsetminus\sup_{\xi<\zeta}a_{j\xi}\ne 0$.   Accordingly

\Centerline{$\sup_{\eta\le\zeta}a_{i\eta}
\Bsupseteq a^*_i
  \Bsetminus\sup_{j\in I}(b_{ij}\setminus\sup_{\xi<\zeta}a_{j\xi})$.
\Qed}

\medskip

{\bf (b)} Let $\frak B$ be the order-closed subalgebra of $\frak A$
generated by $\{a^*_i:i\in I\}\cup\{b_{ij}:i$, $j\in I\}$.
Using (a) for the inductive step, we see that
$\sup_{\xi\le\zeta}a_{i\xi}\in\frak B$
for every $i\in I$ and $\zeta<\kappa$.   Consequently
$a_{i\zeta}=\sup_{\xi\le\zeta}a_{i\xi}\Bsetminus\sup_{\xi<\zeta}a_{i\xi}$
belongs
to $\frak B$ whenever $i\in I$ and $\zeta<\kappa$, and $\frak A=\frak B$ is
$\tau$-generated by $\{a^*_i:i\in I\}\cup\{b_{ij}:i$, $j\in I\}$,
so has Maharam type at most $\max(\omega,\#(I))$.
}%end of proof of 514F

\leader{514G}{Order-preserving functions of Boolean algebras (a)} Let
$F$ be an ordinal function of Boolean algebras, that is, a function
defined on the class of Boolean algebras, taking ordinal values, and
such that $F(\frak A)=F(\frak B)$ whenever $\frak A$ and $\frak B$ are
isomorphic.   We say that $F$ is {\bf order-preserving} if
$F(\frak B)\le F(\frak A)$ whenever $\frak B$ is a principal ideal of
$\frak A$.   \cmmnt{It is easy to check that all the
cardinal functions defined in 511D are
order-preserving;  see 514Ed.}   Now a Boolean algebra $\frak A$ is
{\bf $F$-homogeneous} if $F(\frak B)=F(\frak A)$ for every non-zero
principal ideal $\frak B$ of $\frak A$.   Of course any principal ideal
of an $F$-homogeneous Boolean algebra is again $F$-homogeneous.

\cmmnt{We have already seen `\Mth' algebras in
Chapter 33.   I mention {\bf cellularity-homogeneous} algebras
as a class which will be used later.   The proof of the Erd\H{o}s-Tarski
theorem in 513Bb is based on the idea of upwards-saturation-homogeneous
partially ordered set.   Of course all the most important ordinal
functions of Boolean algebras actually take cardinal values.}

\spheader 514Gb If $F$ is any order-preserving ordinal function of
Boolean algebras, and $\frak A$ is a Boolean algebra,
then\cmmnt{ (writing $\frak A_a$ for the principal ideal generated by
$a$)} $\{a:a\in\frak A,\,\frak A_a$ is $F$-homogeneous$\}$ is
order-dense in $\frak A$.   \prooflet{\Prf\ If $a\in\frak A^+$, set
$\xi=\min\{F(\frak A_b):0\ne b\Bsubseteq a\}$, and let $b$ be such that
$0\ne b\Bsubseteq a$ and $F(\frak A_b)=\xi$;  then $\frak A_b$ is
$F$-homogeneous.\ \QeD}   So if $\frak A$ is a Dedekind complete Boolean
algebra, it is isomorphic to a simple product of $F$-homogeneous Boolean
algebras.   \cmmnt{(Argue as in the proof of 332B.)}

\spheader 514Gc Similarly, if $F_0,\ldots,F_n$ are order-preserving
ordinal functions of Boolean algebras, and $\frak A$ is any Boolean
algebra, then $\{a:\frak A_a$ is $F_i$-homogeneous for every $i\le n\}$
is order-dense in $\frak A$;  and if $\frak A$ is Dedekind complete, it
is isomorphic to a simple product of Boolean algebras all of which are
$F_i$-homogeneous for every $i\le n$.

\cmmnt{\spheader 514Gd Of course any Boolean algebra which is
homogeneous in the full sense (316N) is $F$-homogeneous for every
function $F$ of Boolean algebras.   Maharam's theorem tells us that
a Maharam-type-homogeneous {\it measurable} algebra is homogeneous
(331N).}

\leader{514H}{Regular open algebras:
Proposition} Let $(X,\frak T)$ be a topological space and
$\RO(X)$ its regular open algebra\cmmnt{ (314O {\it et seq.})}.

(a)(i) $(\RO(X)^+,\supseteq,\RO(X)^+)
\prGT(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$.

\quad (ii) If $X$ is regular,
$(\RO(X)^+,\supseteq,\RO(X)^+)
\equivGT(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$.

(b)(i) $\sat(\RO(X))=\sat(X)$, $c(\RO(X))=c(X)$, $\pi(\RO(X))\le\pi(X)$ and
$d(\RO(X))\le d(X)$.

\quad(ii) If $X$ is regular, $\pi(\RO(X))=\pi(X)$.

\quad(iii) If $X$ is locally compact and Hausdorff, $d(\RO(X))=d(X)$.

(c) Let $\CalNwd(X)$ be the ideal of nowhere dense subsets of $X$.

\quad(i) If $X$ is regular, $\wdistr(\RO(X))\le\add\CalNwd(X)$.

\quad(ii) If $X$ is locally compact and Hausdorff,
$\wdistr(\RO(X))=\add\CalNwd(X)$.

(d) If $Y\subseteq X$ is dense, then $G\mapsto G\cap Y$ is a Boolean
isomorphism from $\RO(X)$ to $\RO(Y)$.

\proof{{\bf (a)} For $G\in\frak T\setminus\{\emptyset\}$, set
$\psi(G)=\interior\overline{G}$.   If we set $\phi(G)=G$ for
$G\in\RO(X)^+$, then $(\phi,\psi)$ is a Galois-Tukey connection from
$(\RO(X)^+,\supseteq,\RO(X)^+)$ to
$(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$.

If $X$ is regular, then $\RO(X)^+$ is coinitial with
$\frak T\setminus\{\emptyset\}$, so 513Ed, inverted, shows that they are
equivalent.

\medskip

{\bf (b)(i)} Any disjoint family in $\RO(X)^+$ is a disjoint family of
non-empty open subsets of $X$, so $c(\RO(X))\le c(X)$ and
$\sat(\RO(X))\le\sat(X)$.   On the other hand, if $\Cal G$ is a disjoint
family of non-empty open subsets of $X$, then
$\family{G}{\Cal G}{\interior\overline{G}}$ is a disjoint family in
$\RO(X)^+$, so $c(X)\le c(\RO(X))$ and $\sat(\RO(X))\le\sat(X)$.

By (a) and 513Ee, inverted,

\Centerline{$\pi(\RO(X))=\ci(\RO(X)^+)
\le\ci(\frak T\setminus\{\emptyset\})=\pi(X)$,}

\Centerline{$d(\RO(X))
=\ddownarrow(\RO(X)^+)
\le \ddownarrow(\frak T\setminus\{\emptyset\}
\le d(X)$}

\noindent by 514A.

\medskip

\quad{\bf (ii)} If $X$ is regular, $\RO(X)^+$ is coinitial
with $\frak T\setminus\{\emptyset\}$, so
$\ci(\RO(X)^+)=\ci(\frak T\setminus\{\emptyset\})$ and
$\pi(\RO(X))=\pi(X)$.

\medskip

\quad{\bf (iii)} If $X$
is locally compact and Hausdorff it is also regular, so $\RO(X)^+$ is
coinitial with $\frak T\setminus\{\emptyset\}$, and

$$\eqalignno{d(X)
&=\ddownarrow(\frak T\setminus\{\emptyset\})\cr
\displaycause{514A}
&=\ddownarrow(\RO(X)^+)
=d(\RO(X)).\cr}$$

\medskip

{\bf (c)(i)} Let $\ofamily{\xi}{\kappa}{E_{\xi}}$ be a family of nowhere
dense sets in $X$, where $\kappa<\wdistr(\RO(X))$.   For each
$\xi<\kappa$, set
$\Cal G_{\xi}=\{G:G\in\RO(X)$, $\overline{G}\cap E_{\xi}=\emptyset\}$.
Then $\Cal G_{\xi}$ is upwards-directed, and
$\bigcup\Cal G_{\xi}=X\setminus\overline{E}_{\xi}$, because any point of
$X\setminus\overline{E}_{\xi}$
belongs to a regular open set with closure disjoint from $E_{\xi}$.
But this means
that $\sup\Cal G_{\xi}=X$ in $\RO(X)$ (314P), and there is a partition
$\Cal G'_{\xi}$ of unity included in $\Cal G_{\xi}$.   Because
$\kappa<\wdistr(\RO(X))$, there is a partition $\Cal H$ of unity in
$\RO(X)$ such that $\{G:G\in\Cal G'_{\xi}$, $G\cap H\ne\emptyset\}$ is
finite for each $\xi$ and $H\in\Cal H$.   It follows that
$H\subseteq\bigcup\{\overline{G}:G\in\Cal G'_{\xi}\}$ is disjoint from
$E_{\xi}$ whenever $\xi<\kappa$ and $H\in\Cal H$.   Accordingly
$\bigcup_{\xi<\kappa}E_{\xi}$ is disjoint from $\bigcup\Cal H$ and is
nowhere dense.   As
$\ofamily{\xi}{\kappa}{E_{\xi}}$ is arbitrary,
$\add\CalNwd(X)\ge\wdistr(\RO(X))$.

\medskip

\quad{\bf (ii)} If $X$ is locally compact and Hausdorff,
suppose that $\kappa<\add\CalNwd(X)$ and that
$\ofamily{\xi}{\kappa}{\Cal G_{\xi}}$ is a family of partitions of unity
in $\RO(X)$.   Then $E_{\xi}=X\setminus\bigcup\Cal G_{\xi}$ is a
nowhere dense closed set for each $\xi$ (314P again).   So
$E=\bigcup_{\xi<\kappa}E_{\xi}$ is nowhere dense.   Set

\Centerline{$\Cal U=\{U:U\subseteq X$ is open,
$\overline{U}\subseteq X\setminus E$ is compact$\}$;}

\noindent then $\Cal U$ is an upwards-directed family with union
$X\setminus\overline{E}$, so includes a partition $\Cal G$ of unity.
But if $H\in\Cal G$ and $\xi<\kappa$, $\overline{H}$ is a
compact set disjoint from $E_{\xi}$, so must be included in the union of
some finite subfamily from
$\Cal G_{\xi}$, and $\{G:G\in\Cal G_{\xi}$, $G\cap H\ne\emptyset\}$ is
finite.   As
$\ofamily{\xi}{\kappa}{\Cal G_{\xi}}$ is arbitrary,
$\wdistr(\RO(X))\ge\add\CalNwd(X)$ and we have equality.

\medskip

{\bf (d)} If $Y\subseteq X$ is dense, and we write
$\interior_Y$, $\overline{\phantom{W}}^{(Y)}$ for interior and closure
in the subspace topology of $Y$, we have

\Centerline{$\interior_Y\overline{G\cap Y}^{(Y)}
=\interior_Y(Y\cap\overline{G\cap Y})
=\interior_Y(Y\cap\overline{G})
=Y\cap\interior\overline{G}$}

\noindent for every open set $G\subseteq X$.   Let $f:Y\to X$ be the
identity map.   Then $f$ is continuous and
$f^{-1}[M]=Y\cap M$ is nowhere dense in $Y$ whenever $M\subseteq X$ is
nowhere dense in $X$, so we have a corresponding Boolean homomorphism
$\pi:\RO(X)\to\RO(Y)$ defined by setting

\Centerline{$\pi G
=\interior_Y\overline{f^{-1}[G]}^{(Y)}
=\interior_Y\overline{G\cap Y}^{(Y)}
=Y\cap\interior\overline{G}
=G\cap Y$}

\noindent for every $G\in\RO(X)$ (314Ra).   Because $Y$ is dense,
$\pi G\ne\emptyset$ for every non-empty $G$, and $\pi$ is injective.
If $H\in\RO(Y)\setminus\{\emptyset\}$, then there is an open set
$G\subseteq X$ such that $H=G\cap Y$, so that

\Centerline{$\pi(\interior\overline{G})=Y\cap\interior\overline{G}
=\interior_Y\overline{G\cap Y}^{(Y)}=H$;}

\noindent thus $\pi$ is surjective and is an isomorphism.
}%end of proof of 514H

\leader{514I}{Category algebras} \cmmnt{For many topological spaces,
their regular open algebras can be understood better through their
expressions as quotients of Baire-property algebras.   It is time I
brought this approach into the main line of the argument.

\medskip

}{\bf (a)} Let $X$ be a topological space, and $\Cal M$ the
$\sigma$-ideal of meager subsets of $X$.   Recall that the
Baire-property algebra of $X$ is the $\sigma$-algebra
$\widehat{\Cal B}=\{G\symmdiff A:G\subseteq X$ is open,
$A\in\Cal M\}$, and that the category algebra of $X$ is
the quotient Boolean algebra $\frak G=\widehat{\Cal B}/\Cal M$
(4A3R\formerly{4{}A3Q}).
Note that if $G\subseteq X$ is any open set,
then\cmmnt{ $\overline{G}\setminus G$ and
$\overline{G}\setminus\interior\overline{G}$ are nowhere dense, so}

\Centerline{$G^{\ssbullet}=\overline{G}^{\ssbullet}
=(\interior\overline{G})^{\ssbullet}$}

\noindent in $\frak G$.

\spheader 514Ib For $G\in\RO(X)$, set $\pi G=G^{\ssbullet}\in\frak G$.
Then $\pi:\RO(X)\to\frak G$ is an order-continuous surjective Boolean
homomorphism.   \prooflet{\Prf\ (i) If $G$, $H\in\RO(X)$, then

\Centerline{$G\hskip.2em\Bcapshort_{\RO(X)}\hskip.2em H=G\cap H$,
\quad$X\hskip.2em\Bsetminusshort_{\RO(X)}\hskip.2em
G=X\setminus\overline{G}$,}

\noindent (314P), so

\Centerline{$(G\hskip.2em\Bcapshort_{\RO(X)}\hskip.2em H)^{\ssbullet}
=(G\cap H)^{\ssbullet}=G^{\ssbullet}\Bcap H^{\ssbullet}$,}

\Centerline{$(X\hskip.2em\Bsetminusshort_{\RO(X)}\hskip.2em
G)^{\ssbullet}
=(X\setminus\overline{G})^{\ssbullet}
=1\Bsetminus\overline{G}^{\ssbullet}
=1\Bsetminus G^{\ssbullet}$.}

\noindent By 312H(ii), this is enough to show that
$\pi$ is a Boolean homomorphism.   (ii) If $E\in\widehat{\Cal B}$, let
$G_0\subseteq X$ be an open set such that $G_0\symmdiff E\in\Cal M$;
then $G=\interior\overline{G}_0$ belongs to $\RO(X)$ and

\Centerline{$\pi G=G^{\ssbullet}=G_0^{\ssbullet}=E^{\ssbullet}$.}

\noindent Thus $\pi$ is surjective.   (iii) There is a regular open set
$W$ such that $X\setminus W$ is meager and every non-empty open subset
of $W$ is non-meager (4A3Sa\formerly{4{}A3Ra});  now the kernel of $\pi$ is just
$\{G:G\in\RO(X)$, $G\cap W=\emptyset\}$ which has a largest member
$\interior(X\setminus W)$.   This shows that the kernel of $\pi$ is
order-closed, so that $\pi$ is order-continuous (313P(a-ii)).\ \Qed}

\spheader 514Ic\cmmnt{ From the last part of the proof of (b), we see
that the kernel of $\pi$ is the principal ideal of $\RO(X)$ generated
by $X\setminus\overline{W}$, so that in fact} $\pi$ includes an isomorphism
between\dvro{ a
principal ideal of $\RO(X)$}{ the complementary principal ideal
generated by $W$} and $\frak G$.\cmmnt{

In particular, being isomorphic to a principal ideal in the
Dedekind complete Boolean algebra $\RO(X)$,} $\frak G$ is Dedekind
complete\cmmnt{ (314Xd, 314Ea)}.

\spheader 514Id It is useful to know that if $G\subseteq X$ is open,
then the category algebra of $G$ can be identified with the principal
ideal of $\frak G$ generated by $G^{\ssbullet}$\prooflet{;  this is
because a subset of $G$ is nowhere dense regarded as a subset of $G$ iff
it is nowhere dense regarded as a subset of $X$, so that
$\Cal M\cap\Cal PG$ is exactly the ideal of meager subsets of $G$ for
the subspace topology, while the Borel $\sigma$-algebra of $G$ is
$\{G\cap E:E\subseteq X$ is Borel$\}$ (4A3Ca)}.

\spheader 514Ie\cmmnt{ Recall from 431Fa that every $A\subseteq X$ has a
Baire-property envelope, that is,
a set $E\in\widehat{\Cal B}$ such that $A\subseteq E$ and
$E\setminus F$ is meager whenever $F\in\widehat{\Cal B}$ and
$A\subseteq F$.}
If $\sequencen{A_n}$ is any sequence of subsets of $X$, and $E_n$ is a
Baire-property envelope of $A_n$ for each $n$,
then $E=\bigcup_{n\in\Bbb N}E_n$ is a
Baire-property envelope of $A=\bigcup_{n\in\Bbb N}A_n$.
\prooflet{\Prf\ Of course $A\subseteq E\in\widehat{\Cal B}$.
If $A\subseteq F\in\widehat{\Cal B}$, then
$A_n\subseteq F$ for every $n$, so $E_n\setminus F$ is meager for every
$n$ and $E\setminus F$ is meager.\ \Qed}

If $A\subseteq X$, we can define $\psi(A)\in\frak G$ by setting
$\psi(A)=\inf\{F^{\ssbullet}:A\subseteq F\in\widehat{\Cal B}\}$\cmmnt{,
because $\frak G$ is Dedekind complete}.
\cmmnt{Note that} $\psi(A)=E^{\ssbullet}$ for any Baire-property envelope
$E$ of $A$.   It follows that
$\psi(\bigcup_{n\in\Bbb N}A_n)=\sup_{n\in\Bbb N}\psi(A_n)$ for any sequence
$\sequencen{A_n}$ of subsets of $X$;  also $\psi(A)=0$ in $\frak G$ iff $A$
is meager.

\spheader 514If\dvro{ When}{ The construction here is most useful
when} $X$ is a
Baire space,\cmmnt{ so that no non-empty open set is meager,}
$\pi$\cmmnt{ is injective and} is an isomorphism between $\RO(X)$ and
$\frak G$.

\spheader 514Ig If $X$ is a zero-dimensional
space, then the algebra $\Cal E$ of open-and-closed sets in $X$ is
an order-dense subalgebra of $\RO(X)$, so that $\RO(X)$ can be identified
with the Dedekind completion of $\Cal E$;  and if $X$ is a
zero-dimensional compact Hausdorff space, then the category algebra of $X$
can\cmmnt{ equally} be
identified with the Dedekind completion of $\Cal E$.

\spheader 514Ih\cmmnt{ Finally,} I note that if $X$ is an extremally
disconnected compact Hausdorff space, so that its algebra $\Cal E$ of
open-and-closed sets is already Dedekind complete\cmmnt{ (314S)},
then $\Cal E=\RO(X)$.   So if $X$ is the Stone space of a Dedekind complete
Boolean algebra $\frak A$, we have a Boolean isomorphism
$a\mapsto\widehat{a}^{\ssbullet}$ from $\frak A$ to $\frak G$, writing
$\widehat{a}$ for the open-and-closed subset of $X$ corresponding to
$a\in\frak A$.

\leader{514J}{}\cmmnt{ Now we have the following.

\medskip

\noindent}{\bf Proposition} Let $X$ be a topological space and $\frak C$
its category algebra.

(a) $\sat(\frak C)\le\sat(X)$, $c(\frak C)\le c(X)$,
$\pi(\frak C)\le\pi(X)$ and $d(\frak C)\le d(X)$.

(b) If $X$ is a Baire space, $\sat(\frak C)=\sat(X)$ and
$c(\frak C)=c(X)$.

(c) If $X$ is regular, $\wdistr(\frak C)\le\add\CalNwd(X)$, where
$\CalNwd(X)$ is the ideal of nowhere dense subsets of $X$.

\proof{ All we need to know is that $\frak C$ is isomorphic to a
principal ideal of $\RO(X)$, which is the whole of $\RO(X)$ if $X$ is a
Baire space (514Ic, 514If), and apply 514H and 514Ed.
}%end of proof of 514J

\leader{514K}{}\cmmnt{ Later in this volume, we shall see that the
Lebesgue measure algebra, in particular, can have weak distributivity
large compared with its cellularity and its Maharam type.   For such
algebras the following result gives us significant information.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Boolean algebra such that
$\sat(\frak A)\le\wdistr(\frak A)$.   Then whenever $A\subseteq\frak A$
and $\#(A)<\wdistr(\frak A)$ there is a set $C\subseteq\frak A$ such
that $\#(C)\le\max(c(\frak A),\tau(\frak A))$ and
$a=\sup\{c:c\in C,\,c\Bsubseteq a\}$ for every $a\in A$.

\proof{{\bf (a)} If $\frak A$ is finite we can take $C$ to be the set of
atoms of $\frak A$;  so let us henceforth suppose that $\frak A$ is
infinite.   Let $D\subseteq\frak A$ be a $\tau$-generating set of
cardinal $\tau(\frak A)$, and $\frak D$ the subalgebra of $\frak A$
generated by $D$, so that (because $\frak A$ is infinite)
$\#(\frak D)=\tau(\frak A)$.   For any $a\in\frak A$, write

\Centerline{$Q(a)
=\{b:b\in\frak A,\,\Exists\,d\in\frak D,\,(a\Bsymmdiff d)\Bcap b=0\}$,}

\Centerline{$\Cal E(a)=\{B:B$ is a maximal antichain, $\sup B'\in Q(a)$
for every finite $B'\subseteq B\}$.}

\noindent Now the first fact to establish is that
$\Cal E(a)\ne\emptyset$ for any $a\in\frak A$.

\Prf\ Set $E=\{a:\Cal E(a)\ne\emptyset\}$.   Then $\frak D\subseteq E$,
because $1\in Q(d)$ and $\{1\}\in\Cal E(d)$ for every $d\in\frak D$.
If $a\in E$, then $Q(1\Bsetminus a)=Q(a)$ (because
$1\Bsetminus d\in\frak D$ for every $d\in\frak D$), so
$\Cal E(1\Bsetminus a)=\Cal E(a)$ is non-empty, and $1\Bsetminus a\in E$.
If $F\subseteq E$ is
non-empty and has supremum $a\in\frak A$, then there is a non-empty set
$F_0\subseteq F$, still with supremum $a$, such that
$\#(F_0)<\sat(\frak A)$ (514Db).   For each $c\in F_0$ choose
$B_c\in\Cal E(c)$.   Because
$\#(F_0)<\wdistr(\frak A)$, there is a maximal antichain
$B\subseteq\frak A$ such that $\{e:e\in B_c$, $e\Bcap b\ne 0\}$ is
finite for every $c\in F_0$.   If $B'\subseteq B$ is finite and
$c\in F_0$, then $\sup B'\subseteq\sup B'_c$ where
$B'_c=\{e:e\in B_c$, $e\Bcap\sup B'\ne 0\}$, so $\sup B'\in Q(c)$.
Set

\Centerline{$\tilde D=\{b:$ there are $b'\in B$ and $c\in F_0$ such that
$b\Bsubseteq b'\Bsetminus(a\Bsetminus c)\}$.}

\noindent Because $\sup F_0=a$ and $\sup B=1$, $\sup\tilde D=1$ and
there is a maximal antichain $\tilde B\subseteq\tilde D$.
If $B'\subseteq\tilde B$
is finite, with supremum $b^*$, there are $c_0,\ldots,c_n\in F_0$ such
that $b^*$ is disjoint from $a\Bsetminus\sup_{i\le n}c_i$;  also
$b^*\in Q(c_i)$ for each $i$.   So we can find $d_i\in\frak D$ such that
$c_i\Bsymmdiff d_i$ is disjoint from $b^*$ for each $i\le n$;
accordingly $c\Bsymmdiff d$ is disjoint from $b^*$, where
$c=\sup_{i\le n}c_i$ and $d=\sup_{i\le n}d_i$, and

\Centerline{$a\Bsymmdiff d
\Bsubseteq(a\Bsymmdiff c)\Bcup(c\Bsymmdiff d)
\Bsubseteq(a\Bsetminus c)\Bcup(c\Bsymmdiff d)
\Bsubseteq 1\Bsetminus b^*$,}

\noindent while $d\in\frak D$.    This shows that
$b^*\in Q(a)$;  as $B'$ is arbitrary, $\tilde B\in\Cal E(a)$ and $a\in E$.

This shows that $E$ is closed under complements and arbitrary suprema.
It is therefore an order-closed subalgebra of $\frak A$ (312B(iii),
313E(a-i));  since it
includes $\frak D$, it is the whole of $\frak A$, which is what we need
to know.\ \Qed

\medskip

{\bf (b)} Now turn to the given set $A$.   For each $a\in A$ choose
$B_a\in\Cal E(a)$.   Then there is a maximal antichain $B$ such that
$\{e:e\in B_a$, $e\Bcap b\ne 0\}$ is finite for every $b\in B$ and
$a\in A$.   Of course $\#(B)<\sat(\frak A)$.   Set
$C=\{d\Bcap b:d\in\frak D$, $b\in B\}$.   Then

\Centerline{$\#(C)\le\max(\omega,\#(B),\#(\frak D))
\le\max(c(\frak A),\tau(\frak A))$.}

\noindent\Quer\ Suppose that $a\in A$ is not the supremum of
$C'=\{c:c\in C$, $c\Bsubseteq a\}$.   Let $a'\Bsubseteq a$ be non-zero
and disjoint from every member of $C'$.   Then there is a $b\in B$ such
that $b\Bcap a'\ne 0$.   As $b$ is covered by finitely many members of
$B_a$ it belongs to $Q(a)$, and there is a $d\in D$ such that
$(a\Bsymmdiff d)\Bcap b=0$;  which means that

\Centerline{$0\ne a'\Bcap b\Bsubseteq a\Bcap b=d\Bcap b$,}

\noindent while $d\Bcap b\in C$.   Thus $d\Bcap b\in C'$;  but $a'$ is
supposed to be disjoint from every member of $C'$.\ \Bang

Thus $C$ has the properties we need.
}%end of proof of 514K

\leader{514L}{The regular open algebra of a pre-ordered
\dvrocolon{set}}\cmmnt{ Many important features of pre-ordered
sets, at least in those aspects which are of concern to us here, can be
related to the regular open algebras of suitable topologies.

\medskip

\noindent}{\bf Definitions (a)} For any pre-ordered set $P$, a subset
$G$ of $P$ is {\bf up-open} if $\coint{p,\infty}\subseteq G$ whenever
$p\in G$.   The family of such sets is a topology on $P$, the
{\bf up-topology}.   Similarly, the {\bf down-topology} of $P$ is the
family of
{\bf down-open} sets $H$ such that $p\le q\in H\Rightarrow p\in H$.
\cmmnt{Note that} $G\subseteq P$ is up-open iff it is closed for the
down-topology\cmmnt{, and vice versa.   In particular, the intersection of
any non-empty family of up-open sets is again up-open.}.

\spheader 514Lb I will write $\RO^{\uparrow}(P)$ for the regular open
algebra of $P$ when $P$ is given its up-topology, and
$\RO^{\downarrow}(P)$ for the regular open
algebra of $P$ when $P$ is given its down-topology.

\leader{514M}{}\cmmnt{These up- and down-topologies, entirely
unrelated to the usual `order topology' on a totally ordered set (4A2A)
and the ideas of order-convergence considered in Volume 3, take a bit of
getting used to.   Their characteristic property is that every point $p$
has a smallest neighbourhood $\coint{p,\infty}$;  see 514Xj.   I begin
with an elementary lemma for practice.

\medskip

\noindent}{\bf Lemma} Let $P$ be a pre-ordered set endowed with
its up-topology.

(a)(i) For any $A\subseteq P$,
$\overline{A}=\{p:A\cap\coint{p,\infty}\ne\emptyset\}$.

\quad(ii) For any $p\in P$, $\overline{\coint{p,\infty}}$ is the set of
elements of $P$ which are compatible upwards with $p$.

\quad(iii) For any $p$, $q\in P$, the following are equiveridical:
($\alpha$) $q\in\interior\overline{\coint{p,\infty}}$;  ($\beta$) every
member of $\coint{q,\infty}$ is compatible upwards with $p$;  ($\gamma$)
$q$ is incompatible upwards with every $r\in P$ which is incompatible
upwards with $p$.

(b) A subset of $P$ is dense iff it is cofinal.

(c) If $Q$ is another pre-ordered set with its up-topology,
a function $f:P\to Q$ is continuous iff it is order-preserving.

(d)(i) A subset $G$ of $P$ is a regular open set iff

\Centerline{$G=\{p:G\cap\coint{q,\infty}\ne\emptyset$ for every
$q\ge p\}$.}

\quad(ii)\dvAnew{2011} If $\Cal G$ is a
non-empty family of regular open subsets of
$P$, then $\bigcap\Cal G$ is a regular open subset of $P$, and is
$\inf\Cal G$ in the regular open algebra $\RO^{\uparrow}(P)$.

(e) $P$ is separative upwards iff all the sets $\coint{p,\infty}$ are
regular open sets.

(f) If $P$ is separative upwards and $A\subseteq P$ has a supremum $p$ in
$P$, then $\coint{p,\infty}=\inf_{q\in A}\coint{q,\infty}$ in
$\RO^{\uparrow}(P)$.

\proof{{\bf (a)} For (i), we need only note that $\coint{p,\infty}$ is
the smallest open set containing $p$.   Now (ii) amounts to a
restatement of the definition of `compatible upwards'.   As for (iii),

$$\eqalignno{q\in\interior\overline{\coint{p,\infty}}
&\iff\coint{q,\infty}\subseteq\overline{\coint{p,\infty}}\cr
&\iff\coint{q',\infty}\cap\coint{p,\infty}\ne\emptyset
  \text{ for every }q'\ge q\cr
\displaycause{by (i)}
&\iff\coint{q,\infty}\cap\coint{r,\infty}=\emptyset
  \text{ whenever }\coint{r,\infty}\cap\coint{p,\infty}=\emptyset\cr}$$

\noindent because

\Centerline{$P\setminus\overline{\coint{p,\infty}}
=\bigcup\{\coint{r,\infty}:
   \coint{r,\infty}\cap\coint{p,\infty}=\emptyset\}$.}

\medskip

{\bf (b)} $\Cal U=\{\coint{p,\infty}:p\in P\}$ is a base for the
up-topology, so a subset of $P$ is dense iff it meets every member of
$\Cal U$;  but this is the same thing as saying that it is cofinal.

\medskip

{\bf (c)} If $f$ is order-preserving and $H\subseteq Q$ is up-open, then

\Centerline{$p'\ge p\in f^{-1}[H]\Longrightarrow f(p')\ge f(p)\in H
\Longrightarrow f(p')\in H$,}

\noindent so $f^{-1}[H]$ is up-open;  as $H$ is arbitrary, $f$ is
continuous.   If $f$ is continuous and $p\le p'$ in $P$, then
$H=\coint{f(p),\infty}$ is up-open, so $f^{-1}[H]$ is up-open and must
contain $p'$, that is, $f(p')\ge f(p)$;  as $p$ and $p'$ are arbitrary,
$f$ is order-preserving.

\medskip

{\bf (d)(i)} For any set $A\subseteq P$,

\Centerline{$\{p:A\cap\coint{q,\infty}\ne\emptyset
  \text{ for every }q\ge p\}
=\{p:\coint{p,\infty}\subseteq\overline A\}
=\interior\overline{A}$}

\noindent (using (a)).

\medskip

\quad{\bf (ii)} As noted in 514L, $\bigcap\Cal G$ is open, so is equal
to its interior;  but 314P tells us that $\interior\bigcap\Cal G$ is
$\inf\Cal G$ in $\RO(P)$.

\medskip

{\bf (e)}

$$\eqalignno{&P\text{ is separative upwards}
\cr&\mskip80mu
\Longleftrightarrow\,
   \Forall\,p,\,q\in P,\text{ either }p\le q
   \text{ or }\Exists\,r,\,r\ge q,
     \,\coint{r,\infty}\cap\coint{p,\infty}=\emptyset
\Displaycause{511Bk}
\mskip80mu
\Longleftrightarrow\,\Forall\,p,\,q\in P,\text{ either }q\in\coint{p,\infty}
  \text{ or }q\notin\interior\overline{\coint{p,\infty}}
\Displaycause{(a-iii) above}
\mskip80mu
\Longleftrightarrow\,\Forall\,p\in P,\,\interior\overline{\coint{p,\infty}}
   \subseteq\coint{p,\infty}
\cr&\mskip80mu
\Longleftrightarrow\,\Forall\,p\in P,\,\coint{p,\infty}
  \text{ is a regular open set}.
\cr}$$

\medskip

{\bf (f)} $\coint{p,\infty}$ is actually the intersection
$\bigcap_{q\in A}\coint{q,\infty}$.
}%end of proof of 514M

\leader{514N}{Proposition} Let $(P,\le)$ be a pre-ordered set, and
write $\frak T^{\uparrow}$ for the up-topology of $P$ and
$\RO^{\uparrow}(P)$ for the regular open algebra of
$(P,\frak T^{\uparrow})$.

(a) $(\RO^{\uparrow}(P)^+,\supseteq,\RO^{\uparrow}(P)^+)
\prGT(\frak T^{\uparrow}\setminus\{\emptyset\},\supseteq,
  \frak T^{\uparrow}\setminus\{\emptyset\})
\equivGT(P,\le,P)$.   If $P$ is separative upwards, then
$(\RO^{\uparrow}(P)^+,\supseteq,\penalty-100\RO^{\uparrow}(P)^+)
\equivGT(P,\le,P)$.

(b) $\pi(\RO^{\uparrow}(P))\le\pi(P,\frak T^{\uparrow})
=d(P,\frak T^{\uparrow})=\cf P$.   If $P$ is separative upwards,
then we have equality.

(c) $\sat^{\uparrow}(P,\le)
=\sat(P,\frak T^{\uparrow})=\sat(\RO^{\uparrow}(P))$ and
$c^{\uparrow}(P,\le)
=c(P,\frak T^{\uparrow})=c(\RO^{\uparrow}(P))$.

(d) For any cardinal $\kappa$,

\Centerline{$\link_{<\kappa}(\RO^{\uparrow}(P))
\le\link_{<\kappa}^{\uparrow}(P,\le)$,}

\noindent with equality if {\it either} $P$ is separative upwards {\it or}
$\kappa\le\omega$.   In particular, we always have

\Centerline{$\link^{\uparrow}(P,\le)=\link(\RO^{\uparrow}(P))$,
\quad$\duparrow(P,\le)=d(\RO^{\uparrow}(P))$.}

(e) If $Q\subseteq P$ is cofinal, then
$\RO^{\uparrow}(Q)\cong\RO^{\uparrow}(P)$.

(f) If $A\subseteq P$ is a maximal up-antichain, then
$\RO^{\uparrow}(P)\cong\prod_{a\in A}\RO^{\uparrow}(\coint{a,\infty})$.

(g) If $\tilde P$ is the partially ordered set of equivalence classes
associated with $P$, then $\RO^{\uparrow}(\tilde P)\cong\RO^{\uparrow}(P)$.

%pre-orders needed in 5A3O

\proof{{\bf (a)} By 514Ha,

\Centerline{$(\RO^{\uparrow}(P)^+,\supseteq,\RO^{\uparrow}(P)^+)
\prGT(\frak T^{\uparrow}\setminus\{\emptyset\},\supseteq,
  \frak T^{\uparrow}\setminus\{\emptyset\})$.}

\noindent Next, observe that $\Cal U=\{\coint{p,\infty}:p\in P\}$ is a
base for $\frak T^{\uparrow}$, so that

\Centerline{$(\frak T^{\uparrow}\setminus\{\emptyset\},\supseteq,
  \frak T^{\uparrow}\setminus\{\emptyset\})
\equivGT(\Cal U,\supseteq,\Cal U)$}

\noindent by 513Ed (inverted, as usual).
If we set $\phi(p)=\coint{p,\infty}$ for $p\in P$,
and choose $\psi(U)\in P$ such that $U=\coint{\psi(U),\infty}$ for
$U\in\Cal U$, then $(\phi,\psi)$ is a Galois-Tukey connection from
$(P,\le,P)$ to $(\Cal U,\supseteq,\Cal U)$, while $(\psi,\phi)$ is a
Galois-Tukey connection in the reverse direction;  so
$(P,\le)\equivGT(\Cal U,\supseteq)$.

If $P$ is separative upwards, then $\Cal U$ is included in
$\RO^{\uparrow}(P)$ (514Me) and is coinitial with $\RO^{\uparrow}(P)^+$, so

\Centerline{$(\RO^{\uparrow}(P)^+,\supseteq,\RO^{\uparrow}(P)^+)
\equivGT(\Cal U,\supseteq,\Cal U)\equivGT(P,\le,P)$.}

\medskip

{\bf (b)} Now

$$\eqalignno{\pi(\RO^{\uparrow}(P))
&\le\pi(P,\frak T^{\uparrow})\cr
\displaycause{514H(b-i)}
&=\ci(\frak T^{\uparrow}\setminus\{\emptyset\})
=\ci\Cal U
=\cf P,\cr}$$

\noindent  defining $\Cal U$ as in (a) above.
By 514Mb, $\cf P=d(P,\frak T^{\uparrow})$.   If $P$
is separative upwards, then $\pi(\RO^{\uparrow}(P))=\cf P$ because
$(\RO^{\uparrow}(P)^+,\supseteq,\RO^{\uparrow}(P)^+)\equivGT(P,\le,P)$.

\medskip

{\bf (c)} Similarly, again using 514H(b-i), and with 512Dc at the last
step,

\Centerline{$\sat(\RO^{\uparrow}(P))
=\sat(P,\frak T^{\uparrow})
=\sat^{\downarrow}(\frak T^{\uparrow}\setminus\{\emptyset\})
=\sat^{\downarrow}(\Cal U)=\sat^{\uparrow}(P)$.}

\noindent Now we saw in 514Da and 513Bc that cellularity is determined by
saturation both for partially ordered sets and for Boolean algebras, so
$c(\RO^{\uparrow}(P))=c^{\uparrow}(P)$.   (Of course this is easily shown
by a direct argument.)

\wheader{514N}{6}{2}{2}{48pt}

{\bf (d)} Using (a) and 512Dd, we see that

$$\eqalign{\link_{<\kappa}(\RO^{\uparrow}(P))
&=\link_{<\kappa}(\RO^{\uparrow}(P)^+,\supseteq,
  \RO^{\uparrow}(P)^+)\cr
&\le\link_{<\kappa}(P,\le,P)
=\link_{<\kappa}^{\uparrow}(P,\le),\cr}$$

\noindent with equality if $P$ is separative upwards.   For other $P$, if
$\kappa\le\omega$, set $\lambda=\link_{<\kappa}(\RO^{\uparrow}(P))$ and
let $\ofamily{\xi}{\lambda}{\Cal H_{\xi}}$ be a cover of
$\RO^{\uparrow}(P)^+$ by $\hbox{$<$}\kappa$-linked sets.   Set
$A_{\xi}=\{p:\interior\overline{\coint{p,\infty}}\in\Cal H_{\xi}\}$ for
each $\xi<\kappa$.   Then any $A_{\xi}$ is
upwards-$\hbox{$<$}\kappa$-linked in $P$.   \Prf\Quer\ Otherwise, there
is an $I\in[A_{\xi}]^{<\kappa}$ which has no upper bound in $P$, that
is, $\bigcap_{p\in I}\coint{p,\infty}=\emptyset$.   Now

\Centerline{$\bigcap_{p\in I}\interior\overline{\coint{p,\infty}}
\subseteq
\bigcup_{i\in I}(\overline{\coint{p,\infty}}
  \setminus\coint{p,\infty})$}

\noindent is an open set covered by finitely many nowhere dense sets and
is therefore empty, so we have a finite
subset of $\Cal H_{\xi}$ with empty intersection.\ \Bang\QeD\    So
$\ofamily{\xi}{\lambda}{A_{\xi}}$ witnesses that
$\link_{<\kappa}^{\uparrow}(P,\le)\le\lambda$ and again we have
equality.   In particular,

\Centerline{$\link(\RO^{\uparrow}(P))=\link_{<3}(\RO^{\uparrow}(P))
=\link_{<3}^{\uparrow}(P,\le)=\link^{\uparrow}(P,\le)$,}

\Centerline{$d(\RO^{\uparrow}(P))=\link_{<\omega}(\RO^{\uparrow}(P))
=\link_{<\omega}^{\uparrow}(P,\le)=\duparrow(P,\le)$.}

\medskip

{\bf (e)} Put 514Mb and 514Hd together.

\medskip

{\bf (f)} Because $A$ is an up-antichain,
$\family{a}{A}{\coint{a,\infty}}$ is a disjoint family of open sets in
$P$;  because $A$ is maximal, $\bigcup_{a\in A}\coint{a,\infty}$ is
cofinal, therefore dense.   So 315H gives the result.

\medskip

{\bf (g)} Let $Q\subseteq P$ be a set meeting each equivalence class in
just one point, so that $q\mapsto q^{\ssbullet}:Q\to\tilde P$ is a
bijection.   Then $Q$ is cofinal with $P$, while with its subspace ordering
$Q$ is isomorphic to $\tilde P$.  So

\Centerline{$\RO^{\uparrow}(\tilde P)
\cong\RO^{\uparrow}(Q)\cong\RO^{\uparrow}(P)$}

\noindent by (e).
}%end of proof of 514N

\leader{514O}{}\cmmnt{ Of course we very much want to be able to
recognise cases in which two partially ordered sets have isomorphic
regular open algebras;  and it is also important to know when one
$\RO^{\uparrow}(P)$ can be regularly embedded in another.   The next
four results give some of the known sufficient conditions for these.

\medskip

\wheader{514O}{0}{0}{0}{48pt}
\noindent}{\bf Proposition} Suppose that $P$ and $Q$ are pre-ordered sets
and $f:P\to Q$ is an order-preserving function such that
$f^{-1}[Q_0]$ is cofinal with $P$ for every up-open cofinal
$Q_0\subseteq Q$.   Then there is an order-continuous Boolean
homomorphism $\pi:\RO^{\uparrow}(Q)\to\RO^{\uparrow}(P)$ defined by setting
$\pi H=\interior\overline{f^{-1}[H]}$ (taking the closure and interior with
respect to the up-topology on $P$) for every
$H\in\RO^{\uparrow}(Q)$.   If
$f[P]$ is cofinal with $Q$ then $\pi$ is injective, so is a regular
embedding of $\RO^{\uparrow}(Q)$ in $\RO^{\uparrow}(P)$.

\proof{ By 514Mc, $f$ is continuous for the up-topologies.   Moreover,
$f^{-1}[M]$ is nowhere dense in $P$ whenever $M\subseteq Q$ is nowhere
dense in $Q$.   \Prf\ $Q_0=Q\setminus\overline{M}$ is up-open and dense,
therefore cofinal (514Mb), so $f^{-1}[Q_0]$ is up-open and dense, and
$f^{-1}[M]\subseteq P\setminus f^{-1}[Q_0]$ is nowhere dense.\ \Qed

By 314Ra again, there is an order-continuous Boolean
homomorphism $\pi:\RO^{\uparrow}(Q)\to\RO^{\uparrow}(P)$ defined by
setting $\pi H=\interior\overline{f^{-1}[H]}$ for every
$H\in\RO^{\uparrow}(Q)$.   Now

$$\eqalign{f[P]\text{ is cofinal}
&\iff f[P]\text{ is dense}\cr
&\mskip9mu\Longrightarrow\mskip5mu f[P]\cap H\ne\emptyset
  \text{ for every }H\in\RO^{\uparrow}(Q)\setminus\{\emptyset\}\cr
&\iff f^{-1}[H]\ne\emptyset
  \text{ for every }H\in\RO^{\uparrow}(Q)\setminus\{\emptyset\}\cr
&\iff\pi H\ne\emptyset
  \text{ for every }H\in\RO^{\uparrow}(Q)\setminus\{\emptyset\}
\iff\pi\text{ is injective}.\cr}$$

\noindent So in this case $\pi$ is a regular embedding of
$\RO^{\uparrow}(Q)$ in $\RO^{\uparrow}(P)$.
}%end of proof of 514O

\leader{514P}{Corollary} Suppose that $P$ and $Q$ are pre-ordered sets,
that $f:P\to Q$ is an order-preserving function
and whenever $p\in P$, $q\in Q$ and $f(p)\le q$, there is a $p'\ge p$ such
that $f(p')\ge q$.    If $f[P]$ is {\it either} cofinal with $Q$ {\it or}
coinitial with $Q$, then $\RO^{\uparrow}(Q)$ can be regularly
embedded in $\RO^{\uparrow}(P)$.

\proof{ If $Q_0\subseteq Q$ is up-open and cofinal, then $f^{-1}[Q_0]$ is
cofinal with $P$.   \Prf\ Take any $p\in P$.   Then there are a $q\in Q_0$
such that $q\ge f(p)$ and a $p'\ge p$ such that $f(p')\ge q$;  as $Q_0$ is
up-open, $p'\in f^{-1}[Q_0]$;  as $p$ is arbitrary, $f^{-1}[Q_0]$ is
cofinal.\ \QeD\  So if $f[P]$ is cofinal with $Q$, we can use 514O.   On
the other hand, if $f[P]$ is coinitial
with $Q$ it is also cofinal with $Q$.
\Prf\ For $q\in Q$ there is a $p\in P$ such that $f(p)\le q$;
now our main hypothesis tells us that there is a $p'\in P$ such that
$f(p')\ge q$.\ \QeD\  So we have the result in this case also.
}%end of proof of 514P

\vleader{48pt}{514Q}{Proposition} Let $P$ and $Q$ be pre-ordered sets,
endowed with their up-topologies, and $f:P\to Q$ a function such that

\inset{\noindent whenever $A\subseteq P$ is a maximal up-antichain then
$f\restr A$ is injective and $f[A]$ is a maximal up-antichain in $Q$.}

\noindent Then there is an injective order-continuous Boolean
homomorphism $\pi:\RO^{\uparrow}(P)\to\RO^{\uparrow}(Q)$ defined by
setting $\pi(\interior\overline{\coint{p,\infty}})
=\interior\overline{\coint{f(p),\infty}}$ for every $p\in P$.   In
particular, $\RO^{\uparrow}(P)$ can be regularly embedded in
$\RO^{\uparrow}(Q)$.   If $f[P]$ is cofinal with $Q$, then $\pi$ is an
isomorphism.

\proof{{\bf (a)} For $p\in P$, set
$H_p=\interior\overline{\coint{f(p),\infty}}\in\RO^{\uparrow}(Q)$.   If
$A\subseteq P$ is a maximal up-antichain,
$\family{p}{A}{\coint{f(p),\infty}}$ is a disjoint family of up-open
subsets of $Q$ with dense union, so $\family{p}{A}{H_p}$ is a partition
of unity in $\RO^{\uparrow}(Q)$.   It follows that $\family{p}{A}{H_p}$
must be disjoint for every up-antichain $A\subseteq P$.   Moreover, if
$p_0\in P$ and $p_1\in\interior\overline{\coint{p_0,\infty}}$ in $P$,
we have a maximal up-antichain $A$ containing $p_0$, and
$A'=(A\setminus\{p_0\})\cup\{p_1\}$ is an up-antichain;  as
$H_{p_1}\cap\bigcup_{p\in A,p\ne p_0}H_p=\emptyset$, $H_{p_1}$ must be
included in $H_{p_0}$.

\medskip

{\bf (b)} For $G\in\RO^{\uparrow}(P)$, set $\pi G=\sup\{H_p:p\in G\}$,
the supremum being taken in $\RO^{\uparrow}(Q)$.   If $G$,
$G'\in\RO^{\uparrow}(P)$ are disjoint, then $p$ and $p'$ are
incompatible upwards, so $H_p$ and $H_{p'}$ are disjoint, whenever
$p\in G$ and $p'\in G'$;  accordingly $\pi G$ and $\pi G'$ must be
disjoint.

\medskip

{\bf (c)} If $p\in P$, then of course
$H_p\subseteq\pi(\interior\overline{\coint{p,\infty}})$.
On the other hand, if
$p'\in\interior\overline{\coint{p,\infty}}$, then we saw in (a) that
$H_{p'}\subseteq H_p$, so that $\pi(\interior\overline{\coint{p,\infty}})$
must be exactly $H_p$.

\medskip

{\bf (d)} If $\Cal G\subseteq\RO^{\uparrow}(P)$ has supremum $G_0$ in
$\RO^{\uparrow}(P)$, $\pi G_0=\sup_{G\in\Cal G}\pi G$ in
$\RO^{\uparrow}(Q)$.   \Prf\ Of course $\pi G_0\supseteq\pi G$ for every
$G\in\Cal G$.   Let $A$ be maximal among the up-antichains included in
$\bigcup\Cal G$, and extend $A$ to a maximal up-antichain
$A'\subseteq P$.   Then $\family{p}{A'}{H_p}$ is a partition of unity in
$\RO^{\uparrow}(Q)$, so $H=\sup_{p\in A}H_p$ and
$H'=\sup_{p\in A'\setminus A}H_p$ are complementary elements of
$\RO^{\uparrow}(Q)$.   For every $p\in A$ there is a $G\in\Cal G$ with
$p\in G$, so that $H_p\subseteq\pi G$;  accordingly
$H\subseteq\sup_{G\in\Cal G}\pi G$.   On the other hand, take any
$p\in G_0$.   By the maximality of $A$,
$G\cap\coint{p',\infty}=\emptyset$ for every
$p'\in A'\setminus A$ and $G\in\Cal G$, so
$\coint{p,\infty}\cap\coint{p',\infty}
\subseteq G_0\cap\coint{p',\infty}=\emptyset$ for every
$p'\in A'\setminus A$ and
$H_p\cap H_{p'}=\emptyset$ for every $p'\in A'\setminus A$, that is,
$H_p\cap H'=\emptyset$ and $H_p\subseteq H$.   As $p$ is arbitrary,

\Centerline{$\pi G_0\subseteq H\subseteq\sup_{G\in\Cal G}\pi G
\subseteq\pi G_0$}

\noindent and we have equality.\ \Qed

\medskip

{\bf (e)} Now we see that

\Centerline{$\pi\emptyset=\emptyset$,}

\Centerline{$\pi P=Q$}

\noindent (because if we take any maximal up-antichain $A\subseteq P$,
$\pi P$ includes $\sup_{p\in A}H_p$),

\Centerline{$\pi G\cap\pi H=\emptyset$ whenever $G$,
$H\in\RO^{\uparrow}(P)$ and $G\cap H=\emptyset$,}

\Centerline{$\pi(\sup\Cal G)
=\sup\pi[\Cal G]$ for every $\Cal G\subseteq\RO^{\uparrow}(P)$.}

\noindent By 312H(iv), $\pi$ is a Boolean homomorphism, and by
313L(b-iv) it is order-continuous.   Finally, $\pi G\ne\emptyset$
whenever $G\in\RO^{\uparrow}(P)\setminus\{\emptyset\}$, so $\pi$ is
injective and is a regular embedding.

\medskip

{\bf (f)} If $f[P]$ is cofinal with $Q$, then $\pi[\RO^{\uparrow}(P)]$
is order-dense in $\RO^{\uparrow}(Q)$.   \Prf\ Let
$H\in\RO^{\uparrow}(Q)$ be non-empty.   As $f[P]$ is dense, there is a
$p\in P$ such that $f(p)\in H$.   Now

\Centerline{$\emptyset\ne\pi(\interior\overline{\coint{p,\infty}})
=\interior\overline{\coint{f(p),\infty}}
\subseteq\interior\overline{H}=H$;}

\noindent as $H$ is arbitrary, we have the result.\ \QeD\   By 314Ia,
$\pi$ is an isomorphism.   This completes the proof.
}%end of proof of 514Q

\leader{514R}{Corollary} Let $P$ and $Q$ be pre-ordered sets.
Suppose that there is a function $f:P\to Q$ such that $f[P]$ is cofinal
with $Q$ and, for $p$, $p'\in P$, $p$ and $p'$ are compatible upwards in
$P$ iff $f(p)$ and $f(p')$ are compatible upwards in $Q$.   Then
$\RO^{\uparrow}(P)\cong\RO^{\uparrow}(Q)$.

\proof{ The point is that $f$ satisfies the condition of 514Q.   \Prf\
Suppose that $A\subseteq P$ is a maximal up-antichain.   If $p$, $p'$
are distinct elements of $A$, then $p$ and $p'$ are incompatible upwards
in $P$, so $f(p)$ and $f(p')$ are incompatible upwards in $Q$.   This
shows simultaneously that $f\restr A$ is injective and that $f[A]$ is an
up-antichain in $Q$.   If $q$ is any element of $Q$, there is a $p\in P$
such that $f(p)\ge q$;  now there must be a $p'\in A$ such that $p'$ is
compatible upwards with $p$, in which case $f(p')$ is compatible upwards
with $f(p)$ and therefore with $q$.   So $f[A]$ is maximal;  as $A$ is
arbitrary, we have the result.\ \Qed

So 514Q gives the result.
}%end of proof of 514R

\vleader{72pt}{514S}{Proposition}
(a) Let $\frak A$ be a Dedekind complete Boolean algebra and $P$ a
pre-ordered set.   Suppose that we have a function $f:P\to\frak A^+$ such
that, for $p$, $q\in P$,

\Centerline{$f(p)\Bsubseteq f(q)$ whenever $p\le q$,}

\Centerline{$f(p)\Bcap f(q)=0$ whenever $p$ and $q$ are incompatible
downwards in $P$,}

\Centerline{$f[P]$ is order-dense in $\frak A$.}

\noindent Then $\RO^{\downarrow}(P)\cong\frak A$.

(b) Let $\frak A$ be a Dedekind complete Boolean algebra and
$D\subseteq\frak A$ an order-dense set not containing $0$.    Give $D$
the ordering $\Bsubseteq$, and write $\RO^{\downarrow}(D)$ for the
regular open algebra of $D$ with its
down-topology.   Then $\RO^{\downarrow}(D)\cong\frak A$.

(c) Let $(X,\frak T)$ be a topological space and $P$ a pre-ordered set.
Suppose we have a function $g:P\to\frak T\setminus\{\emptyset\}$ such that,
for $p$, $q\in P$,

\Centerline{$g(p)\subseteq g(q)$ whenever $p\le q$,}

\Centerline{$g(p)\cap g(q)=\emptyset$ whenever $p$ and $q$ are incompatible
downwards in $P$,}

\Centerline{$g[P]$ is a $\pi$-base for $\frak T$.}

\noindent Then $\RO^{\downarrow}(P)\cong\RO(X)$.

(d) Let $(X,\frak T)$ be a topological space and
$\Cal U$ a $\pi$-base for the topology of $X$ not containing
$\{\emptyset\}$.   Give $\Cal U$ the ordering $\subseteq$.
Then $\RO^{\downarrow}(\Cal U)\cong\RO(X)$.

\proof{{\bf (a)(i)}
The key is the following fact:  if $p\in P$, $a\in\frak A$ and
$a\Bcap f(p)\ne 0$, then there is a $q\le p$ such that $f(q)\Bsubseteq a$.
\Prf\ There is a $q_0\in P$ such that $f(q_0)\Bsubseteq a\Bcap f(p)$.
Now $q_0$ and $p$ cannot be incompatible downwards, so there is a
$q\in\ocint{-\infty,q_0}\cap\ocint{-\infty,p}$, and in this case
$f(q)\Bsubseteq f(q_0)\Bsubseteq a$.\ \Qed

\medskip

\quad{\bf (ii)} For $G\in\RO^{\downarrow}(P)$, set $\pi G=\sup f[G]$ in
$\frak A$.   Then $\pi:\RO^{\downarrow}(P)\to\frak A$ is order-preserving.
Of course $\pi(\emptyset)=0$.

$\pi(G\cap H)=\pi G\Bcap\pi H$ for all $G$, $H\in\RO^{\downarrow}(P)$.
\Prf\ Because $\pi$ is order-preserving,
$\pi(G\cap H)\Bsubseteq\pi G\Bcap\pi H$.   \Quer\ If
$a=\pi G\Bcap\pi H\Bsetminus\pi(G\cap H)\ne 0$, take $p\in G$ such that
$a\Bcap f(p)\ne 0$;  then there is a $p'\le p$ such that
$f(p')\Bsubseteq a$.   Next, there must be a $q\in H$ such that
$f(q)\Bcap f(p')\ne 0$, and a $q'\le q$ such that $f(q')\Bsubseteq f(p')$.
But now $q'\in\ocint{-\infty,p}\cap\ocint{-\infty,q}\subseteq G\cap H$,
so $f(q')\Bsubseteq\pi(G\cap H)$;  while at the same time
$f(q')\Bsubseteq a$.\ \BanG\
Thus $\pi(G\cap H)=\pi G\Bcap\pi H$.\ \Qed

$\pi(P\setminus\overline{G})=1\Bsetminus\pi G$ for every
$G\in\RO^{\downarrow}(P)$.   \Prf\ (Perhaps I should say that
$\overline{G}$ here is the closure of $G$ for the down-topology of $P$.)
Set $H=P\setminus\overline{G}$.   Then
$\pi G\Bcap\pi H=\pi(G\cap H)=0$
by what we have just seen.   \Quer\ If
$a=1\Bsetminus(\pi G\Bcup\pi H)$ is non-zero, let
$p_0\in P$ be such that $f(p_0)\Bsubseteq a$.   Then $\ocint{-\infty,p_0}$
is a non-empty open set so must meet one of $G$, $H$.   But if
$p\in G\cup H$ then $f(p_0)\Bcap f(p)=0$ so $p_0$ and $p$ are incompatible
downwards and, in particular, $p\not\le p_0$.\ \Bang\Qed

So $\pi$ is a Boolean homomorphism,
%313H(ii) again
and it is injective because
$\pi G\Bsupseteq f(p)\ne 0$ whenever $p\in G\in\RO^{\downarrow}(P)$.
Finally, $\pi$ is surjective.   \Prf\ If $a\in\frak A$, set
$G=\{p:f(p)\Bsubseteq a\}$.   Then $G$ is down-open.   If $q\notin G$,
$f(q)\Bsetminus a\ne 0$, so there is a $q_1\le q$ such that
$f(q_1)\Bcap a=0$ and $\ocint{-\infty,q_1}$ does not meet $G$;  accordingly
$\ocint{-\infty,q}\not\subseteq\overline{G}$
and $q\notin\interior\overline{G}$.   So $G\in\RO^{\downarrow}(P)$.
Because $f[P]$ is order-dense, $a=\sup f[G]=\pi G$ belongs to
$\pi[\RO^{\downarrow}(P)]$.\ \Qed

Thus we have an isomorphism between $\RO^{\downarrow}(P)$ and $\frak A$.

\medskip

{\bf (b)} Apply (a) to the identity map from $D$ to $\frak A$.

\medskip

{\bf (c)} Apply (a) to the map
$p\mapsto\interior\overline{g(p)}:P\to\RO(X)$.

\medskip

{\bf (d)} Apply (c) to the identity function from $\Cal U$ to $\frak T$.
}%end of proof of 514S

\leader{514T}{Finite-support \dvrocolon{products}}\cmmnt{ At many
points in this chapter we find ourselves seeking to relate partially
ordered sets to Boolean algebras and topological spaces.   In 511D
and 512Eb
I sought to describe the cardinal functions of topological spaces and
Boolean algebras in terms of naturally associated partially ordered
sets, and in 514L and 514N of this section I described constructions of
topologies and Boolean algebras from partial orders.   One of the
most important constructions of general topology is that of `product'.
The matching construction in Boolean algebra is that of `free product'
(315I).   I now come to the corresponding idea for partial orders.

\medskip

\wheader{514T}{0}{0}{0}{60pt}
\noindent}{\bf Definition} Let $\familyiI{P_i}$ be a family of non-empty
partially ordered sets.   The {\bf upwards finite-support product}
$\bigotimes^{\uparrow}_{i\in I}P_i$ of $\familyiI{P_i}$ is the set
$\bigcup\{\prod_{i\in J}P_i:J\in[I]^{<\omega}\}$, ordered by saying that
$p\le q$ iff $\dom p\subseteq\dom q$ and $p(i)\le q(i)$ for every
$i\in\dom p$.   Similarly, the {\bf downwards finite-support product}
$\bigotimes^{\downarrow}_{i\in I}P_i$ of $\familyiI{P_i}$ is\cmmnt{
the same set} $\bigcup\{\prod_{i\in J}P_i:J\in[I]^{<\omega}\}$\cmmnt{,
but} ordered by saying that $p\le q$ iff $\dom q\subseteq\dom p$ and
$p(i)\le q(i)$ for every $i\in\dom q$.

\vleader{48pt}{514U}{Proposition} Let $\familyiI{P_i}$ be a family of
non-empty partially ordered sets, with upwards finite-support product
$P=\bigotimes^{\uparrow}_{i\in I}P_i$.

(a) The regular open algebra $\RO^{\uparrow}(P)$ is isomorphic to the
regular open algebra of $P^*=\prod_{i\in I}P_i$ when every $P_i$ is
given its up-topology.

(b) If $I$ is finite, $P^*$ is a cofinal subset of $P$, and the ordering
of $P^*$, regarded as a subset of $P$, is the usual product partial
order on $P^*$.

(c) If $Q_i\subseteq P_i$ is cofinal for each $i\in I$, then
$\bigcup_{J\in[I]^{<\omega}}\prod_{i\in J}Q_i$ is cofinal with $P$.   So
\ifnum\stylenumber=12$\cf P\le\max(\omega,\#(I),\sup_{i\in I}\cf P_i)$.
\else$\cf P$ is at most $\max(\omega,\#(I),\sup_{i\in I}\cf P_i)$.\fi

(d) $c^{\uparrow}(P)
=\sup_{J\in[I]^{<\omega}}c^{\uparrow}(\prod_{i\in J}P_i)$.

\proof{{\bf (a)} For $p\in P$, set

\Centerline{$G_p=\{q:q\in P^*,\,q(i)\ge p(i)$ for every $i\in\dom p\}$.}

\noindent Then $G_p$ is a non-empty open set in $P^*$.   If $p\le p'$ in
$P$, then $G_p\supseteq G_{p'}$.   If $p$, $p'\in P$ are
incompatible upwards in $P$, there must be an $i\in\dom p\cap\dom p'$ such
that $p(i)$ and $p'(i)$ are incompatible upwards in $P_i$, in which case
$G_p\cap G_{p'}$ is empty.
If $V\subseteq P^*$ is a non-empty open set, take any $q\in V$.   There
is a finite set $J\subseteq I$ such that
$V\supseteq\{q':q'\in P^*,\,q'(i)\ge q(i)$ for every $i\in J\}$.
Set $p=q\restr J$;  then $G_p\subseteq V$.   So $p\mapsto G_p$ satisfies
the conditions of 514Sc, inverted, and $\RO^{\uparrow}(P)$ is isomorphic to
$\RO(P^*)$.

\medskip

{\bf (b)-(c)} These are immediate from the definition of the ordering of
$P$.   For the estimate of the cofinality of $P$, just take cofinal sets
$Q_i\subseteq P_i$ such that $\#(Q_i)=\cf P_i$ for each $i$, and
estimate $\#(\bigcup_{J\in[I]^{<\omega}}\prod_{i\in J}Q_i)$.

\medskip

{\bf (d)} We have

$$\eqalignno{c^{\uparrow}(P)
&=c(\RO^{\uparrow}(P))\cr
\displaycause{514Nc}
&=c(\RO^{\uparrow}(P^*))\cr
\displaycause{(a) above)}
&=c(P^*)\cr
\displaycause{514Hb}
&=\sup_{J\in[I]^{<\omega}}c(\prod_{i\in J}P_i)\cr
\displaycause{5A4Be, here taking the product topology on
$\prod_{i\in I}P_i$}
&=\sup_{J\in[I]^{<\omega}}c^{\uparrow}(\prod_{i\in J}P_i)\cr}$$

\noindent because if $J$ is finite then the up-topology
$\frak T^{\uparrow}_J$ on
$\prod_{i\in J}P_i$ is just the product of the up-topologies on the
$P_i$, so we can use the identification of
$c(\prod_{i\in J}P_i,\frak T^{\uparrow}_J)$ with
$c^{\uparrow}(\prod_{i\in J}P_i)$.
}%end of proof of 514U

\exercises{\leader{514X}{Basic exercises (a)}
%\spheader 514Xa
Let $\frak A$ be a Boolean algebra.   Show that
$\link_{<\kappa}(\frak A)=\pi(\frak A)$ for any
$\kappa\ge\sat(\frak A)$.
%514D

\spheader 514Xb Let $\familyiI{\frak A_i}$ be a family of Boolean
algebras and $\frak A$ their free product.   Show that

\Centerline{$d(\frak A)\le\max(\omega,\#(I),\sup_{i\in I}d(\frak A_i))$,
\quad$\pi(\frak A)\le\max(\omega,\#(I),\sup_{i\in I}\pi(\frak A_i))$,}

\Centerline{$c(\frak A)\le\max(\omega,\sup_{i\in I}2^{c(\frak A_i)})$.\query}

\noindent\Hint{4A1Db, 5A1Ha.}
%514D mt51bits

\spheader 514Xc Let $\frak A$ be a Boolean algebra and $\frak B$
a chargeable Boolean algebra (definition: 391Bb).
Suppose that $\frak A\setminus\{1\}\prT\frak B\setminus\{1\}$.
Show that $\frak A$ is chargeable.   \Hint{391J.}
%514D

\spheader 514Xd Let $\kappa$ be a cardinal and $\frak A$ a Boolean algebra
with cardinal at most $2^{\kappa}$.   (i) Show that $\frak A$ is a homomorphic
image of a $\kappa$-centered Boolean algebra.   \Hint{if $\kappa$ is
infinite, $\{0,1\}^{2^{\kappa}}$ has density $\kappa$.}   (ii) Show that if
$\frak A$ is Dedekind complete it is a homomorphic image of $\Cal P\kappa$.
\Hint{514Ca, 314K.}
%514C 514D

\spheader 514Xe Let $\frak A$ be a Boolean algebra and $\frak B$ {\it
either} a
regularly embedded subalgebra of $\frak A$ {\it or} a quotient $\frak A/I$
where $I$ is an order-closed ideal in $\frak A$.
Let $\Pou(\frak A)$, $\Pou(\frak B)$ be
the pre-ordered sets of partitions of unity in $\frak A$, $\frak B$
respectively (512Ee).   Show that $\Pou(\frak B)\prT\Pou(\frak A)$, and
hence that $\wdistr(\frak B)\ge\wdistr(\frak A)$.
%514E

\spheader 514Xf\dvAnew{2011}
Let $X$ be a set.   Show that $\tau(\Cal PX)$ is the
least cardinal $\lambda$ such that $\#(X)\le 2^{\lambda}$.
%514E

\spheader 514Xg Let $\Sigma$ be the countable-cocountable algebra of
$\omega_1$.   Show that $\Sigma$ is an order-dense subalgebra of
$\Cal P\omega_1$, that $\tau(\Sigma)=\omega_1$, and that
$\tau(\Cal P\omega_1)=\omega$.
%514Xf 514E

\def\hc{\mathop{\text{hc}}\nolimits}
\spheader 514Xh\dvAnew{2011}
For a Boolean algebra $\frak A$, write
$\hc(\frak A)
=\min\{c(\frak B):\frak B$ is a non-zero principal ideal of $\frak A\}$,
counting $\min\emptyset$ as $\infty$.   (i) Show that if $\frak B$ is a
regularly embedded subalgebra of $\frak A$, then
$\hc(\frak B)\le\hc(\frak A)$.
(ii) Show that if $\frak B$ is a Boolean algebra and there is a surjective
order-continuous Boolean homomorphism from $\frak A$ onto $\frak B$, then
$\hc(\frak B)\le\hc(\frak A)$.   (iii) Show that if $\frak B$ is a
principal ideal of $\frak A$ then $\hc(\frak B)\ge\hc(\frak A)$.
(iv) Show that if $\frak B$ is an
order-dense subalgebra of $\frak A$ then $\hc(\frak B)=\hc(\frak A)$.
(v) Show that if $\frak A$ is the simple product of a family
$\familyiI{\frak A_i}$ of Boolean algebras, then
$\hc(\frak A)=\min_{i\in I}\hc(\frak A_i)$.
%514E

\sqheader 514Xi(i) ({\smc Solovay 66})
Let $I$ be any set, with its discrete topology, and
$X=I^{\Bbb N}$ with the product topology.
Show that $\tau(\RO(X))=\omega$.   \Hint{take $I$ to be well-ordered;
set $G_{ij}=\{x:x(i)<x(j)\}$ for $i$, $j\in\Bbb N$;  show that the closed
subalgebra of $\RO(X)$ generated by the $G_{ij}$ contains $\{x:x(n)\ge\xi\}$
for every $n\in\Bbb N$ and $\xi\in I$.}
(ii) Show that the subalgebra $\frak B$ of
$\RO(X)$ generated by $\{\{x:x(n)=i\}:n\in\Bbb N,\,i\in I\}$ is
an order-dense subalgebra of $\RO(X)$ and that $\tau(\frak B)\ge\#(I)$
if $\#(I)>1$.
%514D 514F

\spheader 514Xj Let $(X,\frak T)$ be a T$_0$ topological space.   Show
that we have a partial order on $X$ defined by saying that $x\le y$ iff
$x\in\overline{\{y\}}$.   Show that $\frak T$ is the up-topology on $X$
iff the family of $\frak T$-closed sets is a topology.
%514L

\spheader 514Xk Let $P$ be a partially ordered set.   Show that a subset
of $P$ is a regular open set for the up-topology iff it is of the form
$\bigcap_{q\in A}
\{p:p\in P,\,\coint{p,\infty}\cap\coint{q,\infty}=\emptyset\}$ for some
set $A\subseteq P$.
%514N

\sqheader 514Xl Rewrite the statement and proof of the Erd\H{o}s-Tarski
theorem (513Bb) (i) in terms of topological spaces (ii) in terms of
Boolean algebras.
%514N

\spheader 514Xm Find partially ordered sets $P$ and $Q$ such that the
regular open algebras of $P$ and $Q$ for their up-topologies are
isomorphic, but $\add P\ne\add Q$ and $\cf P\ne\cf Q$.
%514N

\spheader 514Xn Let $P$ be a non-empty partially ordered set such that
its regular open algebra $\RO^{\uparrow}(P)$ for the up-topology is
atomless, and let $Q$ be a set of the same size as $P$ with the trivial
partial order in which $q\le q'$ iff $q=q'$.   Show that $Q$ and the
product partially ordered set $P\times Q$ are Tukey equivalent but
$\RO^{\uparrow}(P\times Q)$ is atomless, while $\RO^{\uparrow}(Q)$ is
purely atomic.
%514N

\spheader 514Xo Let $P$ be a partially ordered set and $\kappa$
an infinite cardinal.   Show that $\kappa<\wdistr(\RO^{\uparrow}(P))$
iff for every family $\ofamily{\xi}{\kappa}{Q_{\xi}}$ of cofinal subsets
of $P$ there is a cofinal $Q\subseteq P$ such that for every $q\in Q$ and
$\xi<\kappa$ there is an $I\in[Q_{\xi}]^{<\omega}$ such that for every
$p\ge q$ there is an $r\in I$ which is compatible upwards with $p$.
%514N

\spheader 514Xp Suppose that $P$ is a
partially ordered set and that $A\subseteq P$ is such that

\Centerline{$Q=\{q:q\in P$, $q=\sup\{a:a\in A$, $a\le q\}\}$}

\noindent is cofinal with $P$.   Show that if $P$ is separative upwards,
then $\tau(\RO^{\uparrow}(P))\le\#(A)$.
%514N
%query do we need "separative"?

\spheader 514Xq Let $\frak A$ be the measure algebra of Lebesgue
measure.   Show that the simple products $\{0,1\}\times\frak A$ and
$\Cal P\Bbb N\times\frak A$ are not isomorphic, but that each can be
regularly embedded in the other.
%514Q

\spheader 514Xr Let $(X,\frak T)$ and $(Y,\frak S)$ be topological
spaces.   Suppose we have a $\pi$-base $\Cal U$ for $\frak T$ and a
function $f:\Cal U\to\frak S$ such that $f[\Cal U]$ is a $\pi$-base for
$\frak S$ and, for $U$, $U'\in\Cal U$, $U\cap U'=\emptyset$ iff
$f(U)\cap f(U')=\emptyset$.   Show that $\RO(X)\cong\RO(Y)$.
%514R

\spheader 514Xs Let $\familyiI{P_i}$ be a family of non-empty partially
ordered sets and $\family{j}{J}{I_j}$ a partition (that is, disjoint
cover) of $I$.   Show that
the upwards finite-support product $\bigotimes^{\uparrow}_{i\in I}P_i$
can be naturally identified with
$\bigotimes^{\uparrow}_{j\in J}\bigotimes^{\uparrow}_{i\in I_j}P_i$.
%514T

\leader{514Y}{Further exercises (a)}
%\spheader 514Ya
For a partially ordered set $P$, its {\bf
order-dimension} is the smallest cardinal $\kappa$ such that $P$ is
isomorphic, as partially ordered set, to a subset of a product
$\prod_{\xi<\kappa}X_{\xi}$ where every $X_{\xi}$ is a totally ordered
set (and the product is given its product partial order, as in 315C).
Show that the order-dimension of a Boolean algebra $\frak A$ is
$\link(\frak A)$.
%514C mt51bits

\spheader 514Yb Show that $\Cal P\Bbb N$ has a subalgebra with
uncountable $\pi$-weight.   \Hint{515H.}
%514D

\spheader 514Yc Let $\frak A$ be a Boolean algebra such that
$c(\frak A)\ne 1$, and $A$ a subset of $\frak A$.
Show that there is a $B\in[A]^{\le c(\frak A)}$
with the same upper and lower bounds as $A$.
%514D

\spheader 514Yd Show that for any cardinal $\kappa$ there are a ccc
Boolean algebra $\frak A$ and an ideal $\Cal I$ of $\frak A$ such that
$c(\frak A/\Cal I)=\kappa$.
%514E 514Xi mt51bits

\spheader 514Ye Let $\frak A$ be a Dedekind complete Boolean algebra, not
$\{0\}$, $\Bbb P$ the forcing notion
$(\frak A^+,\Bsubseteqshort,1,\downarrow)$ (5A3M), and $\kappa$ a cardinal.
Show that the following are equiveridical:  (i) there is an
atomless order-closed subalgebra of $\frak A$ with Maharam type at most
$\kappa$;  (ii)
$\VVdP\,\Cal P\check\kappa\ne(\Cal P\kappa)\varBspcheck$.
%514F mt51bits

\spheader 514Yf Let $\frak A$ be a Boolean algebra and $\kappa$ a
cardinal.   I will say that $\frak A$ has the
{\bf\hbox{$<$}$\kappa$-interpolation property}
if whenever $A$, $B\subseteq\frak A$, $a\Bsubseteq b$ whenever $a\in A$ and
$b\in B$, and $\#(A\cup B)<\kappa$, then there is a $c\in\frak A$ such that
$a\Bsubseteq c\Bsubseteq b$ for every $a\in A$, $b\in B$.
(Thus the $\sigma$-interpolation property of 466G is the
{\bf\hbox{$<$}$\omega_1$}-interpolation property.)
(i) Suppose that $\frak A$ has the \hbox{$<$}$\kappa$-interpolation
property and $I$ is an ideal of $\frak A$ such that $\kappa\le(\add I)^+$.
Show that the quotient $\frak A/I$ has the
\hbox{$<$}$\kappa$-interpolation property.
(ii) Suppose that $\frak A$ has the \hbox{$<$}$\kappa$-interpolation
property,
$\frak B$ is a Boolean algebra with cardinal at most $\kappa$, $\frak C$ is a
subalgebra of $\frak B$ and $\phi:\frak C\to\frak A$ is a
Boolean homomorphism.
Show that $\phi$ has an extension to a Boolean homomorphism from $\frak B$ to
$\frak A$.   (Compare 314K.)
(iii)\dvAnew{2010} Show that if $\frak A$ has the
\hbox{$<$}$\sat(\frak A)$-interpolation property it is Dedekind complete.
%514+ 466G 515K

\spheader 514Yg Let $\frak A$ be a ccc Dedekind complete Boolean algebra
with Maharam type $\kappa$.   Show that there is a $\sigma$-ideal
$\Cal J$ of the Baire $\sigma$-algebra $\CalBa(\{0,1\}^{\kappa})$ such that
$\frak A\cong\CalBa(\{0,1\}^{\kappa})/\Cal J$.
%4A3Na 314M
}%end of exercises

\endnotes{
\Notesheader{514}
With any mathematical object, the set-theorist's first concern is simply
to establish its cardinality.   There is therefore a natural distinction
to make between cardinal invariants which control the cardinality of a
space, as linking number, centering number and $\pi$-weight do for
Boolean algebras
(514Da), and others, like weak distributivity, which are measures of
some kind of complexity not directly linked with cardinality.
Observe that for general Boolean algebras $\frak A$ not even
the cellularity is controlled by the Maharam type (514Xi);  in fact, of
the cardinals here, only $\wdistr(\frak A)$ is controlled by
$\tau(\frak A)$ alone (514Dd).   Maharam
type and cellularity together control the size of the algebra (514De),
and for measurable algebras, of course,
Maharam type almost completely determines the algebra and even the
measure (see Chapter 33).

I use the language of Galois-Tukey connections in many of the proofs of
this section.   This is not because there is any real need for it (there
is no depth to any of the results I quote) but because I think that it
shows some common strands running through a rather long list of facts.
Also it points up the proofs which are {\it not} reducible to simple
applications of ideas in \S512;  for instance, those relating to weak
distributivity.   And, finally, it will provide useful practice for the
ideas of Chapter 52.

I have deliberately arranged the lists of cardinal functions of
topological spaces and Boolean algebras in such a way that the
cardinals of Boolean algebras and their Stone spaces will naturally
correspond.   There are of course important exceptions.   The Maharam
type of a Boolean algebra, and the tightness of a topological space, do
not seem to have significant natural analogues in the other category.
Note that the correspondences depend to a significant degree on the
compactness of Stone spaces.   This is perhaps more important than their
zero-dimensionality.   The point about the open-and-closed algebra of a
zero-dimensional space is that it is order-dense in the regular open
algebra, and that our cardinal functions of Boolean algebras are nearly
all unchanged by Dedekind completion (514Ee).   For arbitrary
topological spaces, we can still investigate their regular open
algebras, and we find that the cardinal functions of a regular open
algebra are much more closely related to those of the topological space
if the space is locally compact (514A, 514H(b)-(c)).

You will not be surprised to recognise some of the results and arguments
of this section as direct generalizations of special cases already
treated;  thus 316B becomes 514Bb, 316E (or 215B(iv)) becomes 514Db,
316I becomes 514Be and 4A1O becomes 514De.

I have to admit that there are rather more pages than ideas in this
section.   What it is really here for is to provide a compendium of
useful facts in the language which I wish to use in the rest of the
volume.   Perhaps I should say `languages', because much of the space is
taken up by repeating results in three forms, as they apply to partially
ordered sets, to Boolean algebras and to topological spaces.   The point
is of course that we frequently find that a fact which is obvious in one
of its three manifestations is a surprise in another.   And some care is
needed in the translations.   The theory of finite-support products of
partially ordered sets (514T-514U), for instance, is supposed to mimic
the theory of products of topological spaces.   But actually it reflects
the theory of $\pi$-bases of topologies rather than the theory of
spaces-with-points.   And while we have straightforward functors between
the {\it categories} of Boolean algebras and topological spaces, with
Boolean homomorphisms corresponding to continuous functions (312Q-312S),
such results as we have concerning functions between partially ordered
sets and their actions on the corresponding
regular open algebras are partial and
delicate (514O-514R).  %514O 514Q 514R

The Tukey classification (513D) and the
regular open algebras of 514N are both attempts to reduce the
multitudinous variety of partially ordered sets to relatively coherent
schemes.   They carry rather different information;  the Tukey
classification tells us about additivity and cofinality (513E) and
precalibers (516C below), while the regular open algebra determines
linking numbers (514N).   It is easy to find partially ordered sets with
the same regular open algebras but different additivity and cofinality
(514Xm), or with the same Tukey classification but different regular
open algebras (514Xn).   The regular open algebras
studied here are primarily of interest in relation to the use of
partially ordered sets in the theory of
forcing;  I hope to return to such questions later in this volume.

%514F relevant only when we have uncountable cellularity

Of the cardinal functions of Boolean algebras defined in \S511, I have
not mentioned Martin numbers or Freese-Nation numbers.   These will
be dealt with at length in \S\S517-518.
}%end of notes

\discrpage

\frfilename{mt515.tex}
\versiondate{23.12.15}
\copyrightdate{2001}

\def\chaptername{Cardinal functions}
\def\sectionname{The Balcar-Fran\v{e}k theorem}

\def\ind{\mathop{\text{ind}}}

\newsection{515}

I interpolate a section to give two basic results on Dedekind complete
Boolean algebras:  the Balcar-Fran\v{e}k theorem (515H) on independent
sets and the Pierce-Koppelberg theorem (515L) on cardinalities.
The concept of `Boolean-independence' (515A) provides a tool for some
useful results on regular open algebras (515N-515Q).

\leader{515A}{Definition} Let $\frak A$ be a Boolean algebra, not
$\{0\}$.

(a) I say that a family $\familyiI{\frak B_i}$ of subalgebras of
$\frak A$ is {\bf Boolean-independent} if $\inf_{i\in J}b_i\ne 0$
whenever $J\subseteq I$ is finite and
$b_i\in\frak B_i^+=\frak B_i\setminus\{0\}$ for every $i\in J$.

(b) I say that a family $\familyiI{a_i}$ in $\frak A$ is
{\bf Boolean-independent} if
$\inf_{j\in J}a_j\Bsetminus\sup_{k\in K}a_k$ is non-zero whenever $J$,
$K\subseteq I$ are disjoint finite sets.   Similarly, a set
$B\subseteq\frak A$ is {\bf Boolean-independent} if
$\inf J\Bsetminus\sup K\ne 0$ for any disjoint finite sets $J$,
$K\subseteq B$.

(c) I say that a family $\familyiI{D_i}$ of partitions of unity in
$\frak A$ is {\bf Boolean-independent} if $\inf_{i\in J}d_i\ne 0$
whenever $J\subseteq I$ is finite and $d_i\in D_i$ for every $i\in J$.

\cmmnt{(Many authors write `independent' rather than
`Boolean-independent', and in the proofs of this section I may do the same.
But in this book as a whole it is more often natural to
read `independent' as `stochastically independent', as in 458L and 525H.)}

\leader{515B}{Lemma}\cmmnt{ (Compare 272D.)} Let $\frak A$ be a Boolean
algebra, not $\{0\}$.

(a) A family $\familyiI{a_i}$ in $\frak A$ is Boolean-independent iff no
$a_i$ is $0$ or $1$ and $\familyiI{\{0,a_i,1\Bsetminus a_i,1\}}$ is a
Boolean-independent family of subalgebras of $\frak A$.

(b) Let $\familyiI{\frak B_i}$ be a family of subalgebras of $\frak A$.
Let $\frak B$ be the free product of $\familyiI{\frak B_i}$, and
$\varepsilon_i:\frak B_i\to\frak B$ the canonical homomorphism for each
$i\in I$\cmmnt{ (315I)}.   Then we have a unique Boolean homomorphism
$\phi:\frak B\to\frak A$ such that $\phi\varepsilon_i(b)=b$ whenever
$i\in I$ and $b\in\frak B_i$, and $\familyiI{\frak B_i}$ is
Boolean-independent iff $\phi$ is injective;  in which case $\frak B$ is
isomorphic to the subalgebra of $\frak A$ generated by
$\bigcup_{i\in I}\frak B_i$.

(c) If $\familyiI{\frak B_i}$ is a Boolean-independent family of
subalgebras of $\frak A$, $\family{j}{J}{I_j}$ is a disjoint family
of subsets of $I$, and $\frak C_j$ is the subalgebra of $\frak A$
generated by $\bigcup_{i\in I_j}\frak B_i$ for each $j$, then
$\family{j}{J}{\frak C_j}$ is Boolean-independent.

(d) Suppose that $B\subseteq\frak A$ is a Boolean-independent set and
that $\family{j}{J}{C_j}$ is a disjoint family of subsets of $B$.   For
$j\in J$ write $\frak C_j$ for the subalgebra of $\frak A$ generated by
$C_j$.   Then $\family{j}{J}{\frak C_j}$ is Boolean-independent.

(e) Suppose that $\familyiI{\frak B_i}$ is a Boolean-independent family
of subalgebras of $\frak A$, and that for each $i\in I$ we have a
Boolean-independent subset $B_i$ of $\frak B_i$.   Then $\familyiI{B_i}$
is disjoint and $\bigcup_{i\in I}B_i$ is Boolean-independent.

(f) Let $\familyiI{D_i}$ be a family of partitions of unity in
$\frak A$, none containing $0$.
For each $i\in I$ let $\frak B_i$ be the order-closed subalgebra of
$\frak A$ generated by $D_i$.   Then
$\familyiI{\frak B_i}$ is Boolean-independent iff $\familyiI{D_i}$ is
Boolean-independent.

\proof{{\bf (a)} The point is just that in 515Aa we need consider only
$b_i\in\frak B_i\setminus\{0,1\}$, while in 515Ab we have

\Centerline{$\inf_{j\in J}a_j\Bsetminus\sup_{k\in K}a_k
=\inf_{j\in J}a_j\Bcap\inf_{i\in K}1\Bsetminus a_k$.}

\medskip

{\bf (b)} 315Jb, applied to the identity maps from the $\frak B_i$ to
$\frak A$, assures us that there is a unique Boolean homomorphism
$\phi:\frak B\to\frak A$ such that $\phi\varepsilon_i(b)=b$ for every
$i\in I$ and $b\in\frak B_i$.

\medskip

\quad{\bf (i)} By 315K(e-ii), $\familyiI{\varepsilon_i[\frak B_i]}$ is a
Boolean-independent family of subalgebras of $\frak B$.   So if $\phi$
is injective,
$\familyiI{\frak B_i}=\familyiI{\pi_i[\varepsilon_i[\frak B_i]]}$ is
Boolean-independent in $\phi[\frak B]$ and therefore in $\frak A$.
In this case, because $\frak B$ is the subalgebra of itself generated by
$\bigcup_{i\in I}\varepsilon_i[\frak B_i]$ (315Ka), the subalgebra of
$\frak A$ generated by $\bigcup_{i\in I}\frak B_i$ is $\phi[\frak B]$
and is isomorphic to $\frak B$.

\medskip

\quad{\bf (ii)} If $\familyiI{\frak B_i}$ is Boolean-independent and
$b\in\frak B^+$, there are a finite $J\subseteq I$ and a
family $\family{j}{J}{b_j}\in\prod_{j\in J}\frak B_j^+$ such
that $b\Bsupseteq\inf_{j\in J}\varepsilon_j(b_j)$ (315Kb).   Now
$\phi(b)\Bsupseteq\inf_{j\in J}b_j$ is non-zero;  as $b$ is arbitrary,
$\phi$ is injective.

\medskip

{\bf (c)} Let $L\subseteq J$ be a finite set and suppose that
$c_j\in\frak C_j^+$ for each $j\in L$.   As observed in (b),
the embeddings $\frak B_i\embedsinto\frak C_j$ identify $\frak C_j$ with
the free product of $\family{i}{I_j}{\frak B_i}$, so 315Kb tells us that
there must be a finite set $K_j\subseteq I_j$ and elements
$b_i\in\frak B_i^+$, for $i\in K_j$, such that
$\inf_{i\in K_j}b_i\Bsubseteq c_j$.   Now
$\inf_{j\in L}c_j\Bsupseteq\inf\{b_i:i\in\bigcup_{j\in L}K_j\}$ is
non-zero.   As $\family{j}{L}{c_j}$ is arbitrary,
$\family{j}{J}{\frak C_j}$ is Boolean-independent.  (Compare 315L.)

\medskip

{\bf (d)} Set $\frak B_b=\{0,b,1\Bsetminus b,1\}$ for $b\in B$, so that
$\family{b}{B}{\frak B_b}$ is Boolean-independent, by (a);  now apply
(c).

\medskip

{\bf (e)(i)} If $i$, $j\in I$ are distinct, $b\in B_i$ and $b'\in B_j$,
then $b\in\frak B_i^+$ and
$1\Bsetminus b'\in\frak B_j^+$, so $b\Bsetminus b'\ne 0$ and
$b\ne b'$.

\medskip

\quad{\bf (ii)} If $J$, $K$ are disjoint finite subsets of
$\bigcup_{i\in I}B_i$, then $J\cap B_i$ and $K\cap B_i$ are disjoint
finite subsets of $B_i$, so that

\Centerline{$b_i=\inf(J\cap B_i)\Bsetminus\sup(K\cap B_i)
\in\frak B_i^+$}

\noindent for each $i\in I$.   Let $L\subseteq I$ be a finite set such
that $J\cup K\subseteq\bigcup_{i\in L}B_i$;  then

\Centerline{$\inf J\Bsetminus\sup K=\inf_{i\in L}b_i\ne 0$.}

\noindent As $J$ and $K$ are arbitrary, $\bigcup_{i\in I}B_i$ is
Boolean-independent.

\medskip

{\bf (f)} Since $D_i\subseteq\frak B_i$, $\familyiI{D_i}$ must be
Boolean-independent if $\familyiI{\frak B_i}$ is.

On the other hand, each $D_i$ is order-dense in $\frak B_i$.   \Prf\ For
$d\in D_i$, the set $\{b:d\Bsubseteq b$ or $d\Bcap b=0\}$ is an
order-closed subalgebra of $\frak A$ including $D_i$, so includes
$\frak B_i$.
If  $b\in\frak B_i^+$, then (because $\sup D_i=1$ in $\frak A$) there
must be a $d\in D_i$ such that $b\Bcap d\ne 0$, in which case
$0\ne d\Bsubseteq b$.   As $b$ is arbitrary, $D_i$ is order-dense in
$\frak B_i$.\ \Qed

Now suppose that $\familyiI{D_i}$ is Boolean-independent, $J\subseteq I$
is finite and $b_i\in\frak B_i^+$ for each $i\in J$.   Then we have
non-zero $d_i\in D_i$ such that $d_i\Bsubseteq b_i$ for each $i$.   So
$\inf_{i\in J}b_i\Bsupseteq\inf_{i\in J}d_i$ is non-zero.   As
$\family{i}{J}{b_i}$ is arbitrary, $\familyiI{\frak B_i}$ is
Boolean-independent.
}%end of proof of 515B

\leader{515C}{Proposition} Let $\frak A$ be a Boolean algebra, not
$\{0\}$, and $\kappa$ a cardinal.

(a) There is a Boolean-independent subset of $\frak A$ with cardinal
$\kappa$ iff there is a subalgebra of $\frak A$ which is isomorphic to
the algebra of open-and-closed subsets of $\{0,1\}^{\kappa}$.

(b) If $\frak A$ is Dedekind complete,
there is a Boolean-independent subset of $\frak A$ with cardinal
$\kappa$ iff there is a subalgebra of $\frak A$ which is isomorphic to
the regular open algebra of $\{0,1\}^{\kappa}$.

\proof{ Set $Z=\{0,1\}^{\kappa}$;  write $\Cal E$ for the algebra of
open-and-closed subsets of $Z$ and $\frak G$ for the regular open
algebra of $Z$.

\medskip

{\bf (a)(i)} Suppose that $\frak A$ has a Boolean-independent subset of
cardinal $\kappa$, enumerated as $\ofamily{\xi}{\kappa}{a_{\xi}}$.
Setting
$\frak A_{\xi}=\{0,a_{\xi},1\Bsetminus a_{\xi},1\}$ for each $\xi$,
$\ofamily{\xi}{\kappa}{\frak A_{\xi}}$ is a Boolean-independent family
of subalgebras of $\frak A$, and the subalgebra $\frak C$ of $\frak A$
generated by $\bigcup_{\xi<\kappa}\frak A_{\xi}$ can be identified with
the free product of $\ofamily{\xi}{\kappa}{\frak A_{\xi}}$ (515Bb).
But since the Stone space of each $\frak A_{\xi}$ has just two points,
the construction of 315I makes it plain that the Stone space of
$\frak C$ is homeomorphic to $Z$, so that $\frak C$ is
isomorphic to $\Cal E$.

\medskip

\quad{\bf (ii)} In the other direction, the sets
$E_{\xi}=\{z:z\in Z,\,z(\xi)=1\}$ are Boolean-independent in $\Cal E$,
so if $\Cal E$ can be embedded in $\frak A$ there must be a
Boolean-independent subset of $\frak A$ with cardinal $\kappa$.

\medskip

{\bf (b)} Now suppose that $\frak A$ is Dedekind complete.   $\Cal E$ is
an order-dense subalgebra of $\frak G$ (314T).   So if $\frak A$ has a
subalgebra isomorphic to $\frak G$ it certainly has one isomorphic to
$\Cal E$.   On the other hand, if $\frak A$ has a subalgebra isomorphic
to $\Cal E$, so that there is an injective Boolean homomorphism
$\pi:\Cal E\to\frak A$, then (because $\frak A$ is Dedekind complete)
$\pi$ has an extension to a Boolean homomorphism
$\pi_1:\frak G\to\frak A$ (314K);
because $\Cal E$ is order-dense in $\frak G$ and $\pi$ is injective,
$\pi_1$ is injective, so that $\pi_1[\frak G]$ is a
subalgebra of $\frak A$ isomorphic to $\frak G$.

Putting this together with (a), we see that $\frak A$ has a
Boolean-independent subset with cardinal $\kappa$ iff it has a subalgebra
isomorphic to $\frak G$.
}%end of proof of 515C

\leader{515D}{Lemma} Let $\frak A$ be a Dedekind complete Boolean
algebra, not $\{0\}$, and $\frak B$ an order-closed subalgebra of
$\frak A$ such that
$\frak A$ is relatively atomless over $\frak B$.   Then there is an
$a^*\in\frak A\setminus\{0,1\}$ such that
$\frak B$ and $\{0,a^*,1\Bsetminus a^*,1\}$ are Boolean-independent
subalgebras of $\frak A$.

\cmmnt{\medskip

\noindent{\bf Remark} Recall from 331A that a Boolean algebra $\frak A$
is `relatively atomless' over an order-closed subalgebra $\frak B$ if
for every $a\in\frak A^+$ there is a $c\Bsubseteq a$ which is not of the
form $a\Bcap b$ for any $b\in\frak B$.
}%end of comment

\proof{ Set

\Centerline{$C
=\{c:c\in\frak A,\,c\ne 0,\,\frak B\cap\frak A_c=\{0\}\}$,}

\noindent where $\frak A_c$ is the principal ideal of $\frak A$
generated by $c$.   Then $C$ is order-dense in $\frak A$.   \Prf\ If
$a\in\frak A^+$, there is a
$c\in\frak A_a\setminus\{a\Bcap b:b\in\frak B\}$.
Set $b_0=\sup\{b:b\in\frak B,\,b\Bsubseteq c\}$;
then $c\Bsetminus b_0\Bsubseteq a$ and $c\Bsetminus b_0\in C$.\ \Qed

For $a\in\frak A$ set
$\upr(a,\frak B)=\inf\{b:a\Bsubseteq b\in\frak B\}$, as in 313S.   Set
$E=\{\upr(c,\frak B):c\in C\}$.   Then $E$ is order-dense in $\frak B$,
because if $b\in\frak B^+$ there is a $c\in C$ such that
$c\Bsubseteq b$, and now $\upr(c,\frak B)$ belongs to
$E$ and is included in $b$.
So there is a partition $D$ of unity in $\frak B$ included in $E$
(313K).   For each $d\in D$ choose $c_d\in C$ such that
$d=\upr(c_d,\frak B)$, and set $a^*=\sup\{c_d:d\in D\}$.   If
$b\in\frak B^+$, there is a $d\in D$ such that
$b\Bcap d\ne 0$, that is, $d\notBsubseteq 1\Bsetminus b$, so
$c_d\notBsubseteq 1\Bsetminus b$ and $b\Bcap c_d\ne 0$;  so
$b\Bcap a^*\ne 0$.   Also, because $c_d\in C$,
$b\Bcap d\notBsubseteq c_d=a^*\Bcap d$, so $b\notBsubseteq a^*$ and
$b\Bcap(1\Bsetminus a^*)\ne 0$.   As $b$ is arbitrary, $\frak B$ and
$\{0,a^*,1\Bsetminus a^*,1\}$ are Boolean-independent.
}%end of proof of 515D

\leader{515E}{Lemma}\cmmnt{ ({\smc Balcar \& Vojt\'a\v{s} 77})} Let
$\frak A$ be a Boolean algebra.   Suppose that
$C\subseteq\frak A^+$ and that $\#(C)<c(\frak A_c)$ for
every $c\in C$\cmmnt{, where $\frak A_c$ is the
principal ideal of $\frak A$ generated by $c$}.   Then there is a
partition $D$ of unity in $\frak A$ such that every member of $C$
includes a non-zero member of $D$.

\proof{ Enumerate $C$ as $\ofamily{\xi}{\kappa}{c_{\xi}}$.   For each
$\xi<\kappa$, let $B_{\xi}$ be a disjoint set in
$\frak A_{c_{\xi}}^+$ with cardinal $\kappa^+$, and set

\Centerline{$A_{\xi}
=\{\eta:\eta<\kappa,\,\#(\{b:b\in B_{\xi},\,b\Bcap c_{\eta}\ne 0\})
  \le\kappa\}$,}

\Centerline{$B'_{\xi}
=B_{\xi}\setminus\bigcup_{\eta\in A_{\xi}}\{b:b\in B_{\xi},
  \,b\Bcap c_{\eta}\ne 0\}$.}

\noindent Then $B'_{\xi}$ is a disjoint set in $\frak A_{c_{\xi}}$,
$\#(B'_{\xi})=\kappa^+$, and
$\{b:b\in B'_{\xi},\,b\Bcap c_{\eta}\ne 0\}$ is empty if
$\eta\in A_{\xi}$ and has cardinal $\kappa^+$ otherwise.   Now define
$A\subseteq\kappa$ inductively by saying that $\xi\in A$ iff
$\xi\in A_{\eta}$ whenever $\eta\in A\cap\xi$, and set
$B=\bigcup_{\xi\in A}B'_{\xi}$.

$B$ is disjoint.   \Prf\ If $\eta$, $\xi\in A$, $\eta\le\xi$,
$b\in B'_{\eta}$, $b'\in B'_{\xi}$ and $b\ne b'$, then either
$\eta=\xi$ and
$b\Bcap b'=0$ because $B'_{\xi}$ is disjoint, or $\eta<\xi$ and
$\xi\in A_{\eta}$ and $b\Bcap b'\Bsubseteq b\Bcap c_{\xi}=0$.\ \QeD\
Also $D_{\xi}=\{b:b\in B,\,b\Bcap c_{\xi}\ne 0\}$ has cardinal
$\kappa^+$ for every $\xi<\kappa$.   \Prf\ If $\xi\in A$, then
$D_{\xi}\supseteq B'_{\xi}$ has cardinal $\kappa^+$.   If $\xi\notin A$
there is some $\eta\in A\cap\xi$ such that $\xi\notin A_{\eta}$ and
$D_{\xi}\supseteq\{b:b\in B'_{\eta},\,b\Bcap c_{\xi}\ne 0\}$ has
cardinal $\kappa^+$.\ \Qed

We can therefore find an injection $\xi\mapsto b_{\xi}:\kappa\to B$ such
that $c_{\xi}\Bcap b_{\xi}\ne 0$ for every $\xi$.   Let $D$ be any
partition of unity including $\{c_{\xi}\Bcap b_{\xi}:\xi<\kappa\}$;
this works.
}%end of proof of 515E

\leader{515F}{Lemma} Let $\frak A$ be a Dedekind complete Boolean
algebra such that $c(\frak A)=\sat(\frak A)$ and $\frak A$ is
cellularity-homogeneous.   Then there is a Boolean-independent family
$\familyiI{D_i}$ of partitions of unity in $\frak A$ such that
$\#(I)=\sup_{i\in I}\#(D_i)=c(\frak A)$.

\proof{ Write $\kappa$ for $c(\frak A)=\sat(\frak A)$.   Choose
$\ofamily{\xi}{\kappa}{D_{\xi}}$ inductively, as follows.
Given $D_{\eta}$ for $\eta<\xi$, let $\frak C_{\xi}$ be the subalgebra
of $\frak A$ generated by $\bigcup_{\eta<\xi}D_{\eta}$;  then
$\#(\frak C_{\xi})<\kappa$.   (Recall from 513Bb that
$\kappa=\sat^{\downarrow}(\frak A^+)$ must be a regular
uncountable cardinal, while of course $\#(D_{\eta})<\kappa$ for every
$\eta$.)   By 515E we have a partition $D$ of unity in $\frak A$, not
containing $\{0\}$, such that every non-zero element of $\frak C_{\xi}$
includes an element of $D$.   For each $d\in D$ the principal ideal of
$\frak A$ generated by $d$ has cellularity $\kappa>\#(\xi)$ so
there is a disjoint family
$\langle b_{d\eta}\rangle_{\eta\le\xi}$ of non-zero elements with supremum
$d$.   Set $b_{\eta}=\sup_{d\in D}b_{d\eta}$ for $\eta\le\xi$, and
$D_{\xi}=\{b_{\eta}:\eta\le\xi\}$;  then $D_{\xi}$ is a partition of
unity in $\frak A$.

The construction ensures that whenever $d\in D_{\xi}$ and
$c\in\frak C_{\xi}^+$ then $d\Bcap c\ne 0$.   It follows
that $\ofamily{\xi}{\kappa}{D_{\xi}}$ is Boolean-independent.   \Prf\ I
show by induction on $\#(J)$ that if $J\subseteq\kappa$ is finite and
$d_{\xi}\in D_{\xi}$ for each $\xi\in J$, then
$\inf_{\xi\in J}d_{\xi}\ne 0$.   If $J$ is empty this is trivial.   For
the inductive step to $\#(J)=n+1$, set $\xi=\max J$ and $J'=\xi\cap J$.
By the inductive hypothesis, $c=\inf_{\eta\in J'}d_{\eta}$ is non-zero;
but $c\in\frak C_{\xi}$, so, by the construction of
$D_{\xi}$, $c\Bcap d_{\xi}=\inf_{\eta\in J}d_{\eta}$ is non-empty.
So the induction proceeds.\ \Qed
}%end of proof of 515F

\leader{515G}{Lemma} Let $\familyiI{\frak A_i}$ be a non-empty family of
Boolean algebras with simple product $\frak A$.   Suppose that
for each $i\in I$ the algebra $\frak A_i$ has a Boolean-independent set
with cardinal $\kappa_i\ge\omega$.   Then $\frak A$ has a
Boolean-independent set with cardinal $\kappa=\#(\prod_{i\in I}\kappa_i)$.

\proof{ For each $i\in I$ let $B_i$ be a Boolean-independent set in
$\frak A_i$ with cardinal $\kappa_i$.
Let $\ofamily{\xi}{\kappa}{a_{\xi}}$ be a family in
$\prod_{i\in I}B_i\subseteq\frak A$ such that for every finite
$J\subseteq\kappa$ there is an $i\in I$ such that
$a_{\xi}(i)\ne a_{\eta}(i)$ whenever $\xi$,
$\eta\in J$ are distinct (5A1L).    Now $\ofamily{\xi}{\kappa}{a_{\xi}}$
is Boolean-independent in $\frak A$.   \Prf\ Suppose that $J$,
$K\subseteq\kappa$ are finite and disjoint.   Then there is an $i\in I$
such that $a_{\xi}(i)\ne a_{\eta}(i)$ whenever $\xi$, $\eta\in J\cup K$
are distinct.   But this means that $\family{\xi}{J\cup K}{a_{\xi}(i)}$
is Boolean-independent in $\frak A_i$, so that, setting
$a=\inf_{\xi\in J}a_{\xi}\Bsetminus\sup_{\xi\in K}a_{\xi}$,

\Centerline{$a(i)
=\inf_{\xi\in J}a_{\xi}(i)\Bsetminus\sup_{\xi\in K}a_{\xi}(i)\ne 0$,}

\noindent and $a\ne 0$.   As $J$ and $K$ are arbitrary,
$\ofamily{\xi}{\kappa}{a_{\xi}}$ is a Boolean-independent family.\ \Qed

Accordingly $\{a_{\xi}:\xi<\kappa\}$ is a Boolean-independent set of
size $\kappa$.
}%end of proof of 515G

\leader{515H}{The Balcar-Fran\v{e}k theorem}\cmmnt{ ({\smc Balcar \&
Fran\v{e}k 82})} Let $\frak A$ be an infinite
Dedekind complete Boolean algebra.   Then there is a Boolean-independent
set $A\subseteq\frak A$ such that $\#(A)=\#(\frak A)$.

\proof{ Set $\kappa=\#(\frak A)$.   For $a\in\frak A$ write $\frak A_a$
for the principal ideal of $\frak A$ generated by $a$.

\medskip

{\bf (a)} Suppose that $\frak A$ is purely atomic.   Then $\frak A$ has
an independent set with cardinal $\kappa$.   \Prf\ Let $B$ be the set of its
atoms;  because
$\frak A$ is infinite, so is $B$;  set $\lambda=\#(B)$, so that

\Centerline{$\frak A\cong\prod_{b\in B}\frak A_b\cong\Cal P\lambda$}

\noindent (315F(iii)), and $\kappa=2^{\lambda}$.   There is a dense
subset $D$ of $\{0,1\}^{\kappa}$ with $\#(D)=\lambda$ (5A4Be);  let
$f:B\to D$ be a surjection.   For $\xi<\kappa$ set

\Centerline{$a_{\xi}=\sup\{b:b\in B,\,f(b)(\xi)=1\}$.}

\noindent If $J$, $K\subseteq\kappa$ are disjoint finite sets, the set

\Centerline{$G=\{x:x\in\{0,1\}^{\kappa},\,x(\xi)=1\Forall\xi\in J,
  \,x(\eta)=0\Forall\eta\in K\}$}

\noindent is a non-empty open set, so there is a $b\in B$ such that
$f(b)\in G$;  but this means that
$\inf_{\xi\in J}a_{\xi}\Bsetminus\sup_{\eta\in K}a_{\eta}\Bsupseteq b$
is non-zero.   As $J$ and $K$ are arbitrary, $\{a_{\xi}:\xi<\kappa\}$ is
an independent subset of $\frak A$ with cardinal $\kappa$.\ \Qed

\medskip


{\bf (b)} Suppose that $\frak A$ is \Mth, and that $\frak B$ is an
order-closed subalgebra of $\frak A$ with Maharam type less than
$\tau(\frak A)$.   Then there is a subalgebra $\frak C$ of $\frak A$,
Boolean-independent of $\frak B$, such that $\frak C$ has a
Boolean-independent subset with cardinal $\tau(\frak A)$.   \Prf\ Let
$B\subseteq\frak B$ be a set with cardinal less than $\tau(\frak A)$ which
$\tau$-generates $\frak B$.   Choose
$\ofamily{\xi}{\tau(\frak A)}{c_{\xi}}$ inductively, as follows.   Given
$\ofamily{\eta}{\xi}{c_{\eta}}$, where $\xi<\tau(\frak A)$, let
$\frak B_{\xi}$ be the order-closed subalgebra of $\frak A$ generated by
$B\cup\{c_{\eta}:\eta<\xi\}$.   If $a\in\frak A^+$, the
order-closed subalgebra
$\frak D=\{a\Bcap b:b\in\frak B_{\xi}\}$ of $\frak A_a$ is
$\tau$-generated by
$\{a\Bcap c_{\eta}:\eta<\xi\}\cup\{a\Bcap b:b\in B\}$ (314Hb), so
$\tau(\frak D)<\tau(\frak A)=\tau(\frak A_a)$ and
$\frak D\ne\frak A_a$.   Thus $\frak A$ is relatively atomless over
$\frak B_{\xi}$;  by 515D, there is a
$c_{\xi}\in\frak A\setminus\{0,1\}$ such that $\frak B_{\xi}$ and
$\{0,c_{\xi},1\Bsetminus c_{\xi},1\}$ are Boolean-independent.
Continue.   Now an easy induction on $\#(J\cup K)$ (as in the last part
of the proof of 515F) shows that if $J$,
$K$ are disjoint finite subsets of $\tau(\frak A)$, and $b\in\frak B$ is
non-zero,
$b\Bcap\inf_{\xi\in J}c_{\xi}\Bsetminus\sup_{\eta\in K}c_{\eta}\ne 0$.
So if we take $\frak C$ to be the subalgebra of $\frak A$ generated by
$C=\{c_{\xi}:\xi<\tau(\frak A)\}$, $\frak C$ and $\frak B$ are
Boolean-independent and $C\subseteq\frak C$ is a Boolean-independent set
with cardinal $\tau(\frak A)$.\ \Qed

\medskip

{\bf (c)} Suppose that $\frak A$ is \Mth\ and that
$c(\frak A)<\sat(\frak A)$.   Then $\frak A$ has a Boolean-independent
subset with cardinal $\kappa$.   \Prf\ Because $\frak A$ is infinite,
$c(\frak A)$ is infinite.   Let $D\subseteq\frak A^+$ be a
disjoint set with cardinal $c(\frak A)$;  adding $1\Bsetminus\sup D$ if
necessary, we may suppose that $D$ is a partition of unity.   For each
$d\in D$, $\frak A_d$ has a Boolean-independent set with cardinal
$\tau(\frak A_d)=\tau(\frak A)$ (apply (b) above to $\frak A_d$, with
$\frak D=\{0,d\}$).   By
315F(iii) again, $\frak A\cong\prod_{d\in D}\frak A_d$;  by 515G,
$\prod_{d\in D}\frak A_d$ has a Boolean-independent subset with cardinal the
cardinal power
$\tau(\frak A)^{\#(D)}=\tau(\frak A)^{c(\frak A)}$, so $\frak A$ also
has.   But

\Centerline{$\kappa
\le\sup_{\lambda<\sat(\frak A)}\tau(\frak A)^{\lambda}
=\tau(\frak A)^{c(\frak A)}$}

\noindent by 514De, so $\frak A$ has a Boolean-independent set of
cardinal $\kappa$.\ \Qed

\medskip

{\bf (d)} Suppose that $\frak A$ is cellularity-homogeneous and
\Mth\ and
$c(\frak A)=\sat(\frak A)$.   Then $\frak A$ has a Boolean-independent
subset with cardinal $\kappa$.

\medskip

\Prf\ {\bf (i)} By 515F, we can find a
Boolean-independent family $\familyiI{D_i}$ of partitions of unity in
$\frak A$ such that $\#(I)=\sup_{i\in I}\#(D_i)=\sat(\frak A)$.   We
know that $\sat(\frak A)\ge\omega_1$, so we can suppose that all the
$D_i$ are infinite.   For each $i\in I$, let $\frak D_i$ be the
order-closed subalgebra of $\frak A$ generated by $D_i$.   By (a) above,
$\frak D_i$ has a Boolean-independent subset $B_i$ with cardinal
$2^{\#(D_i)}$, so that $B=\bigcup_{i\in I}B_i$ has cardinal
$\sup_{\lambda<\sat(\frak A)}2^{\lambda}$.
By 515Df, $\familyiI{\frak D_i}$ is Boolean-independent.   By 515De, $B$
is Boolean-independent.

\medskip

\quad{\bf (ii)} If $\#(B)=\kappa$, we can stop.   Otherwise, let $\frak D$
be the order-closed subalgebra of $\frak A$ generated by
$D=\bigcup_{i\in I}D_i$.   Because

$$\eqalignno{\sup_{\lambda<\sat(\frak A)}\#(B)^{\lambda}
&=\#(B)
\Displaycause{5A1Ff}
<\kappa
\le\sup_{\lambda<\sat(\frak A)}\tau(\frak A)^{\lambda}
\cr}$$

\noindent (514Be), we must have

\Centerline{$\tau(\frak A)>\#(B)=\sup_{\lambda<\sat(\frak A)}2^{\lambda}
\ge\sat(\frak A)=\#(D)\ge\tau(\frak D)$.}

\noindent By (b), we have a subalgebra $\frak C$ of $\frak A$,
Boolean-independent of $\frak D$, such that $\frak C$ has a
Boolean-independent subset $C$ with cardinal $\tau(\frak A)$.
Let $\familyiI{C_i}$ be a disjoint family of subsets of $C$ all with
cardinal $\tau(\frak A)$.

For $i\in I$, let $\frak C_{i0}$ be the subalgebra of $\frak A$ generated
by $D_i$ and $\frak C_{i1}$ the subalgebra generated by $C_i$.
Let $\frak E_i$ be the subalgebra generated by
$\frak C_{i0}\cup\frak C_{i1}$ and $\widehat{\frak E}_i$ its Dedekind
completion (314T-314U).   In $\widehat{\frak E}_i$ we have the partition
of unity $D_i$ and the Boolean-independent set $C_i$ with cardinal
$\tau(\frak A)$.
For each $b\in D_i$, the principal ideal $(\widehat{\frak E}_i)_b$ of
$\widehat{\frak E}_i$ generated by $b$ has a Boolean-independent set
$\{b\Bcap c:c\in C_i\}$ with cardinal $\tau(\frak A)$.   Because
$\widehat{\frak E}_i$ is Dedekind complete, it is isomorphic to
$\prod_{b\in D_i}(\widehat{\frak E}_i)_b$, and has a Boolean-independent
subset with cardinal $\tau(\frak A)^{\#(D_i)}$ (515G again).

Because $\frak A$ is Dedekind complete, the
embedding $\frak E_i\embedsinto\frak A$ extends to a Boolean homomorphism
$\pi_i:\widehat{\frak E}_i\to\frak A$ (314K).
Because $\frak E_i$ is order-dense in $\widehat{\frak E}_i$,
$\pi_i$ is injective.
So $\frak E_i^*=\pi_i[\widehat{\frak E}_i]$ is a
subalgebra of $\frak A$ isomorphic to $\widehat{\frak E}_i$, and has a
Boolean-independent subset $E_i$ with cardinal $\tau(\frak A)^{\#(D_i)}$.

\medskip

\quad{\bf (iii)} By 515Df and 515Dd, $\familyiI{\frak C_{i0}}$ and
$\familyiI{\frak C_{i1}}$ are both Boolean-independent families;  because
$\frak C_{i0}\subseteq\frak D$ and $\frak C_{j1}\subseteq\frak C$ whenever
$i$, $j\in I$, and $\frak D$ and $\frak C$ are Boolean-independent,
$\langle\frak C_{ij}\rangle_{i\in I,j\in\{0,1\}}$ is Boolean-independent,
so $\familyiI{\frak E_i}$ is Boolean-independent (515Dc).   If
$J\subseteq I$ is finite, and $e_i\in(\frak E_i^*)^+$ for each
$i\in J$, then there are $e'_i\in\frak E_i$ such that
$0\ne e'_i\Bsubseteq e_i$ for each $i$.   Now
$\inf_{i\in J}e_i\Bsupseteq\inf_{i\in J}e'_i\ne 0$.   As
$\family{i}{J}{e_i}$ is arbitrary, $\familyiI{\frak E^*_i}$ is
Boolean-independent.   But this means that $E=\bigcup_{i\in I}E_i$ is
Boolean-independent (515De), while

\Centerline{$\#(E)\ge\sup_{i\in I}\tau(\frak A)^{\#(D_i)}
=\sup_{\lambda<\sat(\frak A)}\tau(\frak A)^{\lambda}\ge\kappa$.}

\noindent Of course $\#(E)\le\#(\frak A)=\kappa$, so
we have a Boolean-independent set with cardinal $\kappa$ in this case
also.\ \Qed

\medskip

{\bf (e)} If $\frak A$ is atomless it has a Boolean-independent subset
with cardinal $\kappa$.   \Prf\ Because Maharam type and cellularity are
both order-preserving cardinal functions (514Ed), $\frak A$ is
isomorphic to the product of a family $\familyiI{\frak A_i}$ of \Mth\
cellularity-homogeneous algebras, none of them $\{0\}$ (514Gc).
Now, for each $i$, $\frak A_i$ is an atomless (therefore infinite) \Mth\
cellularity-homogeneous Dedekind complete Boolean algebra, so by (c)-(d)
above has a Boolean-independent set with cardinal $\#(\frak A_i)$.   By
515G once more, $\frak A$ has a Boolean-independent set with cardinal
$\#(\prod_{i\in I}\frak A_i)=\kappa$.\ \Qed

\medskip

{\bf (f)} Finally, for the general case, let $A$ be the set of atoms of
$A$ and set $c=\sup A$, so that the principal ideal $\frak A_c$ is
purely atomic and the principal ideal $\frak A_{1\Bsetminus c}$ is
atomless.   Because $\frak A\cong\frak A_c\times\frak A_{1\Bsetminus c}$
is infinite, one of $\frak A_c$, $\frak A_{1\Bsetminus c}$ has cardinal
$\kappa$, and therefore (by (a) or (f)) has a Boolean-independent subset
with cardinal $\kappa$;  which is now a Boolean independent subset of
$\frak A$ with cardinal $\kappa$.

This completes the proof.
}%end of proof of 515H

\leader{515I}{Corollary} If $\frak A$ is an infinite Dedekind complete
Boolean algebra and $\kappa\le\#(\frak A)$,
$\frak A$ has a subalgebra isomorphic to the regular open algebra of
$\{0,1\}^{\kappa}$.

\proof{ By 515H, $\frak A$ has a
Boolean-independent family $\ofamily{\xi}{\kappa}{a_{\xi}}$.
By 515Cb, $\frak A$ has a subalgebra isomorphic to the regular open
algebra of $\{0,1\}^{\kappa}$.
}%end of proof of 515I

\leader{515J}{Corollary} If $\frak A$ is an infinite Dedekind complete
Boolean algebra with Stone space $Z$, then $\#(Z)=2^{\#(\frak A)}$.

\proof{ Since $Z$ may be identified with the set of uniferent ring
homomorphisms from $\frak A$ to $\Bbb Z_2$ (311E),
$\#(Z)\le 2^{\#(\frak A)}$.   On the other hand, writing
$W=\{0,1\}^{\#(\frak A)}$, we have a subalgebra of $\frak A$ isomorphic to
the algebra $\Cal E$ of open-and-closed subsets of $W$ (515I).
If $\pi:\Cal E\to\frak A$ is an injective Boolean homomorphism,
it corresponds to a surjective continuous function $\psi:Z\to W$
(312Sa), so that $\#(Z)\ge\#(W)=2^{\#(\frak A)}$.
}%end of proof of 515J

\leader{515K}{}\cmmnt{ I extract part of the proof of the next theorem
as a lemma.

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be an infinite Boolean algebra with
the $\sigma$-interpolation property.

(a) Let $\sequencen{a_n}$ be a disjoint sequence in $\frak A$.   Then
$\#(\frak A)\ge\#(\prod_{n\in\Bbb N}\frak A_{a_n})$\cmmnt{, writing
$\frak A_d$ for the principal ideal of $\frak A$ generated by $d$, as
usual}.

(b) Set $\kappa=\#(\frak A)$, and let $I$ be the set of those
$a\in\frak A$ such that $\#(\frak A_a)<\kappa$.   Then $I$ is an ideal
of $\frak A$, and

\quad{\it either} $\frak A/I$ is infinite,

\quad{\it or} there is a set $J\subseteq I$ with cardinal $\kappa$
such that every sequence in $J$ has an upper bound in $J$,

\quad{\it or} $\#(\prod_{n\in\Bbb N}\frak A_{a_n})=\kappa$ for
some sequence $\sequencen{a_n}$ in $I$.

\cmmnt{\medskip

\noindent{\bf Remark} Recall from 466G that $\frak A$ has the
`$\sigma$-interpolation property' if whenever $A$, $B\subseteq\frak A$
are countable and $a\Bsubseteq b$ for every $a\in A$ and $b\in B$, then
there is a $c\in\frak A$ such that $a\Bsubseteq c\Bsubseteq b$ for every
$a\in A$ and $b\in B$.   See also 514Yf above.
}%end of comment

\proof{{\bf (a)} The point is that the map
$a\mapsto\sequencen{a\Bcap a_n}:
\frak A\to\prod_{n\in\Bbb N}\frak A_{a_n}$ is surjective.
\Prf\ If $\sequencen{b_n}\in\prod_{n\in\Bbb N}\frak A_{a_n}$, there must
be an $a\in\frak A$ such that
$b_n\Bsubseteq a\Bsubseteq 1\Bsetminus(a_n\Bsetminus b_n)$ for every
$n$, so that $a\Bcap a_n=b_n$ for every $n$.\ \QeD\  The result follows
at once.

\medskip

{\bf (b)} If $a$, $b\in I$ then $(c,d)\mapsto c\Bcup d$ is a surjection
from $\frak A_a\times\frak A_b$ onto $\frak A_{a\Bcup b}$, so
$a\Bcup b\in I$;  of course $b\in I$ whenever $b\Bsubseteq a\in I$, so
$I$ is an ideal of $\frak A$.

\Quer\ Suppose, if possible, that all three alternatives are false.
Then $\frak A/I$ is finite;  let $v_0,\ldots,v_m$ be its atoms.
Let  $c_0,\ldots,c_m\in\frak A$ be such that $c_i^{\ssbullet}=v_i$ for
every $i$.   Observe that
$\frak A$ is the union of finitely many sets with cardinal $\#(I)$,
so $I$ itself must
have cardinal $\kappa$, and there is a sequence $\sequencen{b'_n}$ in $I$
with no upper bound in $I$;  setting $b_n=b'_n\Bsetminus\sup_{m<n}b'_m$
for each $n$, we get a disjoint sequence $\sequencen{b_n}$ in $I$ with
no upper bound in $I$.
Now there is some $k\le m$ such that
$\sequencen{b_n\Bcap c_k}$ has no upper bound in $I$.   Set
$K=\{d:d\Bsubseteq c_k,\,d\Bcap b_n=0$ for every $n\in\Bbb N\}$.
Then $K\normalsubgroup\frak A_{c_k}$.   If $d\in K$, $c_k\Bsetminus d$
is an upper bound for
$\{b_n\cap c_k:n\in\Bbb N\}$, so does not belong to $I$;  as
$c_k^{\ssbullet}$ is an atom in $\frak A/I$, $d$ must belong to $I$.
Thus $K\subseteq I$.
The function $d\mapsto\sequencen{d\Bcap b_n}:
\frak A_{c_k}\to\prod_{n\in\Bbb N}\frak A_{c_k\Bcap b_n}$ is a Boolean
homomomorphism with kernel $K$, so

\Centerline{$\#(\frak A_{c_k}/K)
\le\#(\prod_{n\in\Bbb N}\frak A_{c_k\Bcap b_n})<\kappa$}

\noindent (since the third alternative is false, and
$\#(\prod_{n\in\Bbb N}\frak A_{c_k\Bcap b_n})\le\kappa$ by (a));  as
$\#(\frak A_{c_k})=\kappa$, $\#(K)=\kappa$.
There is therefore a sequence $\sequencen{d_n}$ in $K$ with no upper
bound in $K$.   But there is a $d\in\frak A$ such that
$d_n\Bsubseteq d\Bsubseteq 1\Bsetminus b_n$ for every $n\in\Bbb N$,
because $\frak A$ has the $\sigma$-interpolation property;   so that
$d\Bcap c_k\in K$ is an upper bound for $\{d_n:n\in\Bbb N\}$.\ \Bang
}%end of proof of 515K

\leader{515L}{Theorem}\cmmnt{ ({\smc Koppelberg 75})} If
$\frak A$ is an infinite Boolean algebra with the $\sigma$-interpolation
property, then $\#(\frak A)$ is equal to the cardinal power
$\#(\frak A)^{\omega}$.

%Pierce 58 does the Dedekind complete case

\proof{ Induce on $\kappa=\#(\frak A)$.

\medskip

{\bf (a)} If $\kappa\le\frak c$, then (because $\frak A$ is infinite)
there is a disjoint sequence $\sequencen{a_n}$ in $\frak A^+$, so that

\Centerline{$\frak c\le\#(\prod_{n\in\Bbb N}\frak A_{a_n})
\le\#(\frak A)$}

\noindent by 515Ka, and $\kappa=\frak c$.  So
$\kappa^{\omega}=(2^{\omega})^{\omega}=\kappa$.

\medskip

{\bf (b)} For the inductive step to $\kappa>\frak c$, set
$I=\{a:a\in\frak A,\,\#(\frak A_a)<\kappa\}$, as in 515Kb.
It is easy to see that every principal ideal of $\frak A$ has the
$\sigma$-interpolation property, so that
$\#(\frak A_a)^{\omega}\le\max(\frak c,\#(\frak A_a))$ for every
$a\in I$.   Now consider the three possibilities of 515Kb.

\medskip

\quad{\bf case 1} If the quotient algebra $\frak A/I$ is infinite, then
$\kappa^{\omega}=\kappa$.   \Prf\ There is a disjoint sequence
$\sequencen{u_n}$ in $\frak A/I$.   For each $n\in\Bbb N$ take
$a_n\in\frak A$ such that $a_n^{\ssbullet}=u_n$;  now setting
$a'_n=a_n\Bsetminus\sup_{i<n}a_i$ for each $n$, $\sequencen{a'_n}$ is a
disjoint sequence in $\frak A\setminus I$.   So

\Centerline{$\kappa\le\kappa^{\omega}
=\#(\prod_{n\in\Bbb N}\frak A_{a'_n})\le\kappa$}

\noindent by 515Ka again.\ \Qed

\medskip

\quad{\bf case 2} Suppose that there is a set $J\subseteq I$ such that
$\#(J)=\kappa$ and every sequence in $J$ has an upper bound in $J$.
Then $\kappa^{\omega}=\kappa$.   \Prf\

$$\eqalignno{\kappa^{\omega}
&=\#(J^{\Bbb N})
\le\#(\bigcup_{a\in J}\frak A_a^{\Bbb N})\cr
\displaycause{because every sequence in $J$ is included in $\frak A_a$
for some $a\in J$}
&\le\max(\omega,\#(J),\sup_{a\in I}\#(\frak A_a^{\Bbb N}))
\le\max(\kappa,\sup_{a\in J}\#(\frak A_a))
=\kappa\le\kappa^{\omega}. \text{ \Qed}\cr}$$

\medskip

\quad{\bf case 3} Suppose there is a sequence $\sequencen{a_n}$ in $I$
such that $\#(\prod_{n\in\Bbb N}\frak A_{a_n})=\kappa$.   Then
$\kappa^{\omega}=\kappa$.   \Prf\ Set
$L=\{n:n\in\Bbb N,\,\frak A_{a_n}$ is infinite$\}$.   Then

$$\eqalign{\kappa^{\omega}
&=\#((\prod_{n\in\Bbb N}\frak A_{a_n})^{\Bbb N})
=\#(\prod_{n\in\Bbb N\setminus L}\frak A_{a_n}^{\Bbb N}
  \times\prod_{n\in L}\frak A_{a_n}^{\Bbb N})\cr
&\le\#(\frak c\times\prod_{n\in L}\frak A_{a_n})
\le\max(\frak c,\kappa)
=\kappa.  \text{ \Qed}\cr}$$

\noindent Thus in all three cases we have $\kappa^{\omega}=\kappa$, and
the induction proceeds.
}%end of proof of 515L

\leader{515M}{Corollary} (a) If $\frak A$ is an infinite ccc Dedekind
$\sigma$-complete Boolean algebra then $\#(\frak A)=\tau(\frak A)^{\omega}$.

(b) If $\frak A$ is any infinite Dedekind $\sigma$-complete Boolean
algebra, then $\#(L^0(\frak A))=\#(L^{\infty}(\frak A))=\#(\frak A)$.

\proof{{\bf (a)} Of course $\frak A$, being Dedekind $\sigma$-complete, has the
$\sigma$-interpolation property, as noted in 466G.   So by 515L and
514De,

\Centerline{$\tau(\frak A)^{\omega}\le\#(\frak A)^{\omega}
=\#(\frak A)\le\tau(\frak A)^{\omega}$.}

\medskip

{\bf (b)} $a\mapsto\chi a:\frak A\to L^{\infty}(\frak A)$ and
$u\mapsto\family{q}{\Bbb Q}{\Bvalue{u>q}}:L^0(\frak A)\to\frak A^{\Bbb Q}$
are injective, so

\Centerline{$\#(\frak A)\le\#(L^{\infty}(\frak A))
\le\#(L^0(\frak A))\le\#(\frak A)^{\omega}=\#(\frak A)$.}
}%end of proof of 515M

\leader{515N}{}\cmmnt{ It will be convenient later to
know a little more about the regular open algebras of powers of $\{0,1\}$.

\medskip

\noindent}{\bf Proposition\dvArevised{2015}} Let $I$ be a set.
Write $\frak G$ for the regular open algebra $\RO(\{0,1\}^I)$.

(a) $\frak G$ is
ccc and Dedekind complete and isomorphic to the category algebra of
$\{0,1\}^I$.   The algebra of open-and-closed subsets of $\{0,1\}^I$ is
an order-dense subalgebra of $\frak G$.

(b) Let $\frak A$ be a Boolean algebra.
Then $\frak A$ is isomorphic to $\frak G$ iff it is Dedekind complete and
there is a Boolean-independent family
$\familyiI{a_i}$ in $\frak A$ such that the subalgebra
generated by $\{a_i:i\in I\}$ is order-dense in $\frak A$.

(c) If I is infinite, $\frak G$ is homogeneous.

\proof{ Write $\Cal E$ for the algebra of open-and-closed subsets of
$\{0,1\}^I$ and
$e_i=\{x:x\in\{0,1\}^I$, $x(i)=1\}$ for $i\in I$.
Note that the set

\Centerline{$\{e:e\subseteq\{0,1\}^I$
is determined by coordinates in a finite set$\}$}

\noindent is an algebra of subsets of $\{0,1\}^I$ (254Ma)
which contains $e_i$ for every $i$ so includes $\Cal E$.
We also know that every member of $\frak G$ is determined by coordinates in
a countable subset of $I$ (4A2E(b-i)).

\medskip

{\bf (a)} $\frak G$, being a regular open algebra, is Dedekind
complete (314P).   By 4A2E(a-iii), $\{0,1\}^I$ is ccc;  by 514H(b-i),
$\frak G$ is ccc.   By 514Kg, $\Cal E$ is an order-dense subalgebra of
$\frak G$ and $\frak G$ is
isomorphic to the category algebra of $\{0,1\}^I$.

\medskip

{\bf (b)(i)} Because $\{0,1\}^I$ is zero-{\vthsp}dimensional,
$\Cal E$ is order-dense in $\frak G$.   Now
$\familyiI{e_i}$ is independent and generates
$\Cal E$ (cf.\ 315Ka).   So $\frak G$ has the declared properties.

\medskip

\quad{\bf (ii)} If $\frak A$ satisfies the conditions,
then, as in (a-i) of the
proof of 515C, the subalgebra of $\frak A$ generated by
$\{e_i:i\in I\}$ is isomorphic to $\Cal E$.   Now both $\frak A$
and $\frak G$ are Dedekind completions of $\Cal E$, so they
are isomorphic (314U).

\medskip

{\bf (c)(i)} Suppose first that $I$ is countable, and that $a\in\frak G^+$.
Then the principal ideal $\frak G_a$ is atomless and Dedekind complete and
has countable
$\pi$-weight (514Eb), so has a countable order-dense subalgebra $\frak B$.
$\frak B$ is countable, atomless and not $\{0\}$, so is isomorphic to
the algebra of open-and-closed subsets of $\{0,1\}^{\Bbb N}$ and
to $\Cal E$ (316M).
Now any isomorphism between $\frak B$ and $\Cal E$ extends
to an isomorphism between their completions, which by 314Ub can be
identified with $\frak G_a$ and $\frak G$.

\medskip

\quad{\bf (ii)} Now suppose that $I$ is uncountable.   Again take
$a\in\frak G^+$.   Then $a$ is
determined by coordinates in some countable subset $J$ of $I$.
We can of course suppose that $J$ is infinite.
Now we can express $a$ as $b\times\{0,1\}^{I\setminus J}$ where
$b$ is a non-empty open subset of $\{0,1\}^J$,
and in fact is a regular open subset
(use the formulae in 4A2B(g-i) to see that
$\interior\overline{b}\times\{0,1\}^{I\setminus J}
=\interior\overline{a}$).   Let $\frak B$ be the principal ideal of
$\RO(\{0,1\}^J)$ generated by $b$.   By (i) here, $\frak B$ is isomorphic
to $\RO(\{0,1\}^J)$ and there is an independent family
$\family{j}{J}{b_j}$ in $\frak B$ generating an order-dense subalgebra
$\Cal E_1$ of $\frak B$.

Next, we have a independent family $\family{k}{I\setminus J}{c_k}$
in $\RO(\{0,1\}^{I\setminus J})$ generating an order-dense subalgebra
$\Cal E_2$ of $\RO(\{0,1\}^{I\setminus J})$.   Note that, by
4A2B(g-i) again,

\Centerline{$b'\mapsto b'\times\{0,1\}^{I\setminus J}:
\RO(\{0,1\}^I)\to\frak G$,
\quad$c'\mapsto\{0,1\}^J\times c':\RO(\{0,1\}^{I\setminus J}\to\frak G$}

\noindent are injective Boolean homomorphisms.
So if we set

\Centerline{$a_j=b_j\times\{0,1\}^{I\setminus J}$ for $j\in J$,
\quad$a_j=\{0,1\}^J\times c_k$ for $k\in I\setminus J$,}

\noindent $\familyiI{a_i}$ will be an independent family in $\frak G$
and the subalgebra $\Cal E$ it generates will contain $b'\times c'$
for every
$b\in\Cal E_1$ and $c\in\Cal E_2$;  as every member of $\frak G^+$,
being a non-empty
open set in $\{0,1\}^I\cong\{0,1\}^J\times\{0,1\}^{I\setminus J}$,
includes a product $b'\times c'$ where $b'\in\RO(\{0,1\}^I)^+$ and
$c'\in\RO(\{0,1\}^{I\setminus J})^+$, $\Cal E$ is order-dense in
$\frak G_a$.   Accordingly (b) tells us that $\frak G_a\cong\frak G$, as
required.
}%end of proof of 515N
%maybe (c-ii) easier with category algebras?

\cmmnt{\medskip

\noindent{\bf Remark} Algebras of this kind will appear regularly as the
volume proceeds;  they are among the basic building blocks from which
Dedekind complete Boolean algebras are constructed.   I will
occasionally use the phrase
`{\bf Cohen algebra}' indifferently to mean either the category algebra, or
the regular open algebra, of a set $\{0,1\}^I$ where $I$ is infinite.
}

\leader{515O}{}\cmmnt{ We need to know some elementary facts about the
algebra $\RO(\{0,1\}^{\Bbb N})$ which I have not yet spelt out.

\medskip

\noindent}{\bf Proposition}\dvAformerly{5{}46H}
(a) A Boolean algebra\cmmnt{ $\frak A$} is
isomorphic to
$\frak G=\RO(\{0,1\}^{\Bbb N})$ iff it is Dedekind complete, atomless, has
countable $\pi$-weight and is not $\{0\}$.   In particular, the regular
open algebra $\RO(\Bbb R)$ is isomorphic to $\frak G$.

(b) Every atomless order-closed subalgebra of $\frak G$ is
isomorphic to $\frak G$.

\proof{{\bf (a)(i)} By 515N, $\frak G$ is Dedekind complete (514Ic) and has
$\pi$-weight $\omega$.   As in the proof of 515N, the algebra
$\Cal E$ of open-and-closed subsets of $\{0,1\}^{\Bbb N}$ is order-dense in
$\frak G$;  as $\Cal E$ is atomless, so is $\frak G$.

\medskip

\quad{\bf (ii)} If $\frak A$ satisfies the conditions,
let $B$ be a countable order-dense
subset of $\frak A$ and $\frak B$ the subalgebra of $\frak A$ generated
by $B$.   Then $\frak B$ is countable, atomless and not $\{0\}$,
so is isomorphic to $\Cal E$ (316M).   Now any isomorphism between
$\frak B$ and $\Cal E$ extends to an isomorphism
between their completions, which by 314Ub can be identified with
$\frak A$ and $\frak G$ respectively.

\medskip

\quad{\bf (iii)} All regular open algebras are Dedekind complete.
Because $\Bbb R$ is not empty, $\RO(\Bbb R)\ne\{0\}$.
Because $\Bbb R$ is Hausdorff and has no isolated points, $\RO(\Bbb R)$ is
atomless.   Because $\Bbb R$ has countable $\pi$-weight, so has
$\RO(\Bbb R)$ (514H(b-ii)).   So $\RO(\Bbb R)\cong\frak G$
by (ii) here.

\medskip

{\bf (b)} All we have to observe is that any
atomless order-closed subalgebra of $\frak G$
satisfies the conditions of (a) (see 514E).
}%end of proof of 515O

\leader{515P}{}\cmmnt{ There is a more complicated, but still manageable,
characterization of $\RO(\{0,1\}^{\omega_1})$.

\medskip

\noindent}{\bf Proposition}\dvAnew{2015}
A Boolean algebra $\frak A$ is isomorphic to
$\RO(\{0,1\}^{\omega_1})$ iff

%\inset{
($\alpha$) it is non-zero, ccc and Dedekind complete,

($\beta$) every non-zero
principal ideal of $\frak A$ has $\pi$-weight $\omega_1$,

($\gamma$) there is a non-decreasing family
$\ofamily{\xi}{\omega_1}{A_{\xi}}$ of countable subsets of $\frak A$
such that

\inset{each $A_{\xi}$ is order-dense in the order-closed subalgebra of
$\frak A$ which it generates,

$A_{\zeta}=\bigcup_{\xi<\zeta}A_{\xi}$ for every non-zero countable limit
ordinal $\zeta$,

$\bigcup_{\xi<\omega_1}A_{\xi}$ is order-dense in $\frak A$.}
%}

\proof{ Write $\frak G$ for $\RO(\{0,1\}^{\omega_1})$.

\medskip

{\bf (a)(i)} We know from 515N that $\frak G$
satisfies the conditions ($\alpha$) and ($\beta$) above.

\medskip

\quad{\bf (ii)} For $\xi<\omega_1$, let $\Cal E_{\xi}$ be the set of
open-and-closed subsets of $\{0,1\}^{\omega_1}$ which are determined by
coordinates in $\xi$, and $\frak G_{\xi}$ the set of regular open
subsets of $\{0,1\}^{\omega_1}$ which are determined by coordinates in
$\xi$.   Then $\frak G_{\xi}$ is an order-closed subalgebra of
$\frak G$.   \Prf\ Reviewing the formulae of 314O, we see
that the Boolean operations of $\frak G$ can all be expressed in terms of
(arbitrary) unions and intersections, set difference, closure and interior.
Since all of these can be done within the family of subsets of
$\{0,1\}^{\omega_1}$ determined by coordinates in $\xi$ (254Ma,
4A2B(g-i)), $\frak G_{\xi}$ is an order-closed subalgebra of $\frak G$.\
\QeD\   Next, $\Cal E_{\xi}$ is a subalgebra of the algebra $\Cal E$ of
open-and-closed subsets of $\{0,1\}^{\omega_1}$, and is order-dense
in $\frak G_{\xi}$, so $\frak G_{\xi}$ is the order-closed subalgebra of
$\frak G$ generated by $\Cal E_{\xi}$.

Of course $\ofamily{\xi}{\omega_1}{\Cal E_{\xi}}$ is a non-decreasing
family of countable subsets of $\frak G$.   If $\zeta$ is a non-zero
countable limit ordinal, then every member of $\Cal E_{\zeta}$ is
determined by coordinates in a finite subset of $\zeta$, so
$\Cal E_{\zeta}=\bigcup_{\xi<\zeta}\Cal E_{\xi}$.   And
$\bigcup_{\xi<\omega_1}\Cal E_{\xi}=\Cal E$ is order-dense in $\frak G$.
So $\ofamily{\xi}{\omega_1}{\Cal E_{\xi}}$ witnesses that $\frak G$
satisfies condition ($\gamma$).

\medskip

{\bf (b)} Now suppose that $\frak A$ and $\ofamily{\xi}{\omega_1}{A_{\xi}}$
satisfy the conditions ($\alpha$)-($\gamma$).   For each $\xi<\omega_1$ let
$\frak A_{\xi}$ be the order-closed subalgebra generated by $A_{\xi}$.

\medskip

\quad{\bf (i)} We need to know that
$\frak A=\bigcup_{\xi<\omega_1}\frak A_{\xi}$.   \Prf\ Because
$\frak A$ is ccc and $\ofamily{\xi}{\omega_1}{\frak A_{\xi}}$ is a
non-decreasing family of order-closed subalgebras of $\frak A$,
$\bigcup_{\xi<\omega_1}\frak A_{\xi}$ is an order-closed subalgebra
(use 316Fb).   But
$\bigcup_{\xi<\omega_1}\frak A_{\xi}$ includes the order-dense set
$\bigcup_{\xi<\omega_1}A_{\xi}$ so is the whole of $\frak A$.\ \Qed

\medskip

\quad{\bf (ii)} Suppose that $\frak C$ is an order-closed subalgebra of
$\frak C$ with countable $\pi$-weight.   Then there is an
$a\in\frak A$ which is independent of $\frak C$ in the sense
that $c\Bcap a$ and $c\Bsetminus a$ are non-zero for every non-zero
$c\in\frak C$.   \Prf\ Let $\familyiI{(a_i,c_i)}$ be a maximal family such
that $\familyiI{c_i}$ is a disjoint family in $\frak C$ and, for each
$i\in I$, $a_i\Bsubseteq c_i$ is such that $c\Bcap a_i$ and
$c\Bsetminus a_i$ are non-zero for every $c\in\frak C$ such that
$c\Bcap c_i$ is non-zero.   Set $c^*=1\Bsetminus\sup_{i\in I}c_i$.
\Quer\ If $c^*\ne 0$, consider the principal ideal $\frak A_{c^*}$
of $\frak A$ generated
by $c^*$.   By hypothesis, this must have uncountable $\pi$-weight, so
the countable set $\frak C_{c^*}$ cannot be order-dense in
$\frak A_{c^*}$, and there is a non-zero $a'\in\frak A_{c^*}$ not including
any non-zero member of $\frak C_{c^*}$, therefore not including any
non-zero member of $\frak C$.   Set $c'=\upr(a',\frak C)$;  then
$c'$ is non-zero and $c'\Bsubseteq c^*$, so $c'\Bcap c_i=0$ for every
$i\in I$.   If $c\in\frak C$ is non-zero and meets $c'$, then
$c\Bcap c'\notBsubseteq a$ so $c\Bsetminus a\ne 0$, while
$a\notBsubseteq c'\Bsetminus c$ so $c\Bcap a\ne 0$.   But this means that
we ought to have added $(a',c')$ to the family $\familyiI{(a_i,c_i)}$.\
\Bang

Thus $\sup_{i\in I}c_i=1$.   Set $a=\sup_{i\in I}a_i$.   If
$c\in\frak C^+$, there is an $i\in I$ such that
$c\Bcap c_i\ne 0$, in which case $c\Bcap a\Bsupseteq c\Bcap a_i$ and
$c\Bsetminus a\Bsupseteq(c\Bcap c_i)\Bsetminus a_i$ are both non-zero.
Accordingly $a$ is independent of $\frak C$, as required.\ \Qed

\medskip

\quad{\bf (iii)} Suppose that $\frak C$ is an order-closed subalgebra of
$\frak A$ with countable $\pi$-weight, and $b\in\frak A$.   Then there
are an
an $a\in\frak A$ and a $d\in\frak C^+$ such that
$a$ is independent of $\frak C$ and $a\Bcap d\Bsubseteq b$.
\Prf\ By (ii), we have an $a'\in\frak A$ which is independent of
$\frak C$.   If there is a non-zero $d\in\frak C$ such that $d\Bsubseteq b$
we can set $a=a'$.   Otherwise, set $d=\upr(b,\frak C)$ and
$a=b\Bcup(a'\Bsetminus d)$.   If $c\in\frak C$ is non-zero, either
$c\Bcap d\ne 0$ and

\Centerline{$c\Bcap a\Bsupseteq c\Bcap d\Bcap b\ne 0$,
\quad$c\Bsetminus a\Bsupseteq c\Bcap d\Bsetminus b\ne 0$,}

\noindent or $c\Bsetminus d\ne 0$ and

\Centerline{$c\Bcap a\Bsupseteq(c\Bsetminus d)\Bcap a'\ne 0$,
\quad$c\Bsetminus a\Bsupseteq(c\Bsetminus d)\Bsetminus a'\ne 0$.}

\noindent So $a$ is independent of $\frak C$, while of course
$0\ne a\Bcap d\Bsubseteq b$.\ \Qed

\medskip

\quad{\bf (iv)} Suppose that $\zeta<\omega_1$.   Then there are a
$\zeta'\ge\zeta$ and an independent sequence
$\sequence{i}{a_i}$ in $\frak A_{\zeta'}$ such that
$\frak A_{\zeta}$ and the
subalgebra $\frak B$ generated by $\{a_i:i\in\Bbb N\}$ are
independent and
the subalgebra generated by $\frak A_{\zeta}\cup\frak B$ is order-dense in
$\frak G_{\zeta'}$.
\Prf\ We need a couple of bookkeeping devices.   For each $\xi<\omega_1$ le
$\sequencen{b(\xi,n)}$ be a sequence running over $A_{\xi}\setminus\{0\}$,
and let
$\sequence{i}{(j_i,k_i)}$ be an enumeration of $\Bbb N\times\Bbb N$ with
$j_i\le i$ for every $i\in\Bbb N$.    Now choose sequences
$\sequence{i}{\zeta_i}$ in $\omega_1$, $\sequence{i}{a_i}$ in $\frak A$
and $\sequence{i}{\frak D_i}$ inductively so that

\Centerline{$\zeta_i=\min\{\xi:\zeta\le\xi$, $a_j\in\frak A_{\xi}$
for every $j<i\}$}

\noindent (by (i) above, $\zeta_i$ will always be countable, and
$\zeta=\zeta_0\le\zeta_1\le\ldots$)

\Centerline{$\frak D_i$ is the subalgebra of $\frak A$ generated by
$\frak A_{\zeta}\cup\{a_j:j<i\}$}

\noindent (so that $\frak D_i$ is always an order-closed subalgebra of
$\frak A$, by 314Ja),

\Centerline{$a_i$ is independent of $\frak D_i$ and there is a
$d\in\frak D_i$ such that $0\ne a_i\Bcap d\Bsubseteq b(\zeta_{j_i},k_i)$}

\noindent (using (iii)).   At the end of the induction, set
$\zeta'=\sup_{i\in\Bbb N}\zeta_i$.

Because $a_i$ is always independent of $\frak D_i$,
$\sequence{i}{a_i}$ is independent and $\frak B$, the subalgebra generated
by $\{a_i:i\in\Bbb N\}$, is independent of $\frak A_{\zeta}$.   Next, if
$a\in\frak A_{\zeta'}$ is non-zero, there is an
$a'\in A_{\zeta'}$ such that
$0\ne a'\Bsubseteq a$;  there are a $\xi<\zeta'$ such that
$a'\in\ A_{\xi}$,
a $j\in\Bbb N$ such that $\xi\le\zeta_j$ and $a'\in A_{\zeta_j}$, and an
$i\in\Bbb N$ such that $j_i=j$ and $a'=b(\zeta_{j_i},k_i)$.   There is now
a $d\in\frak D_i^+$ such that $a_i\Bcap d\Bsubseteq a'$.
Because $\frak D_i$ is generated by $\frak A_{\zeta}\cup\{a_l:l<i\}$,
there are $c\in\frak A_{\zeta}^+$ and $J\subseteq i$ such that
$a''=c\Bcap\inf_{l\in J}a_l\Bsetminus\sup_{l\in i\setminus J}a_l$ is
included in $a'$.   But now $a''$ belongs to the subalgebra generated by
$\frak A_{\zeta}\cup\frak B$ and is included in $a$.   As $a$ is arbitrary,
the subalgebra generated by $\frak A_{\zeta}\cup\frak B$ is order-dense in
$\frak A_{\zeta'}$, as required.\ \Qed

\medskip

\quad{\bf (v)} We can therefore build inductively families
$\ofamily{\xi}{\omega_1}{\zeta_{\xi}}$ and
$\langle a_{\xi i}\rangle_{\xi<\omega_1,i\in\Bbb N}$ such that, for each
$\xi<\omega_1$,

\Centerline{$\langle a_{\eta i}\rangle_{\eta<\xi,i\in\Bbb N}$ is
independent,}

\Centerline{the subalgebra generated by
$\{a_{\xi'i}:\xi'<\xi$, $i\in\Bbb N\}$ is an
order-dense subset of $\frak A_{\zeta_i}$.}

\noindent\Prf\ Start with $\zeta_0=0$.   For the inductive step to $\xi+1$,
where $\xi<\omega_1$, apply (iv) with $\zeta=\zeta_{\xi}$ to choose
$\sequence{i}{a_{\xi i}}$ and $\zeta'$, and set $\zeta_{\xi+1}=\zeta'$.
If $a\in\frak A_{\zeta_{\xi+1}}^+$ then $a$ includes $c\Bcap b$ for
some non-zero $c\in\frak A_{\zeta}$ and $b$ in the algebra generated by
$\{a_{\xi i}:i\in\Bbb N\}$;  now $c$ includes $c'$ for some non-zero
$c'$ in the subalgebra generated by
$\{a_{\xi'i}:\xi'<\xi$, $i\in\Bbb N\}$, and we see that $c'\Bcap b$ is a
non-zero element of the subalgebra generated by
$\{a_{\xi'i}:\xi'\le\xi+1$, $i\in\Bbb N\}$ and is included in $a$.

For the inductive step to a non-zero limit ordinal $\xi<\omega_1$ set
$\zeta_{\xi}=\sup_{\eta<\xi}\zeta_{\eta}$.   In this case, if
$a\in\frak A_{\zeta_{\xi}}^+$, there is a non-zero
$a'\in A_{\zeta_{\xi}}$ included in $a$, there is an $\eta<\xi$ such that
$a'\in A_{\zeta_{\eta}}$, and there is a non-zero $a''$ in the subalgebra
generated by $\{a_{\eta' i}:\eta'<\eta$, $i\in\Bbb N\}$ such that
$a''\Bsubseteq a'$.   So again the subalgebra generated by
$\{a_{\xi'i}:\xi'<\xi$, $i\in\Bbb N\}$ is order-dense in
$\frak A_{\zeta_{\xi}}$ and the induction proceeds.\ \Qed

\medskip

\quad{\bf (vi)} At the end of the induction,
$\langle a_{\xi i}\rangle_{\xi<\omega_1,i\in\Bbb N}$ is independent,
because all its finite subfamilies are independent.
Because
$a_{\xi 0}\in\frak A_{\zeta_{\xi+1}}\setminus\frak A_{\zeta_{\xi}}$ for
every $\xi$, $\ofamily{\xi}{\omega_1}{\zeta_{\xi}}$ is strictly increasing
and $\sup_{\xi<\omega_1}\zeta_{\xi}=\omega_1$, so that
$\frak A=\bigcup_{\xi<\omega_1}\frak A_{\zeta_{\xi}}$.
Accordingly the subalgebra
generated by $\{a_{\xi i}:\xi<\omega_1$, $i\in\Bbb N\}$ will be order-dense
in $\frak A$.   But 515Nc now tells us that $\frak A\cong\frak G$.
So the conditions listed are sufficient as well as necessary.
}%end of proof of 515P

\leader{515Q}{}\cmmnt{ Concerning closed subalgebras, the position with
$\RO(\{0,1\}^{\omega_1})$ is nearly as straightforward as with
$\RO(\{0,1\}^{\omega})$, though the proof is a good deal deeper.

\medskip

\noindent}{\bf Proposition}\dvAnew{2015} Let $\frak A$ be an
atomless order-closed subalgebra of
$\frak G=\RO(\{0,1\}^{\omega_1})$.   Then $\frak A$ is isomorphic either
to $\RO(\{0,1\}^{\omega})$ or to $\frak G$ or to the simple product
$\RO(\{0,1\}^{\omega})\times\frak G$.

\proof{{\bf (a)} We know that there is a non-decreasing family
$\ofamily{\xi}{\omega_1}{B_{\xi}}$ of countable subsets of $\frak G$ such
that $\bigcup_{\xi<\omega_1}B_{\xi}$ is order-dense in $\frak G$,
$B_{\xi}=\bigcup_{\eta<\xi}B_{\eta}$ for every non-zero countable
limit ordinal $\xi$, every $B_{\xi}$ is order-dense in the order-closed
subalgebra $\frak G_{\xi}$ of $\frak G$ which it generates, and
$\frak G=\bigcup_{\xi<\omega_1}\frak G_{\xi}$.

Consider the set

$$\eqalign{C&=\{\zeta:\zeta<\omega_1\text{ is a
non-zero limit ordinal},\cr&
\mskip100mu
   \upr(b,\frak A)\in\frak G_{\zeta}
   \text{ for every }b\in\bigcup_{\xi<\zeta}B_{\xi}\}.
\cr}$$

\noindent Then $C$ is a closed cofinal subset of $\omega_1$, and
$\upr(b,\frak A)\in\frak G_{\zeta}$ whenever $\zeta\in C$ and
$b\in\bigcup_{\xi<\zeta}B_{\xi}=B_{\zeta}$.
Set $A_{\zeta}=\{\upr(b,\frak A):b\in B_{\zeta}\}$ for each $\zeta\in C$.
Then $\family{\zeta}{C}{A_{\zeta}}$ is a non-decreasing family
of countable subsets of $\frak A$ and
$A_{\zeta}=\bigcup_{\xi\in C\cap\zeta}A_{\xi}$
whenever $\zeta\in C$ is $\sup(C\cap\zeta)$.
Moreover, if $\zeta\in\frak C$, $A_{\zeta}$ is an order-dense subset of
$\frak A\cap\frak G_{\zeta}$.   \Prf\ Surely $A_{\zeta}\subseteq\frak A$,
and $A_{\zeta}\subseteq\frak G_{\zeta}$ by the definition of $C$.
If $a\in(\frak A\cap\frak G_{\zeta})^+$, there is a non-zero
$b\in B_{\zeta}$ such that $b\Bsubseteq a$, and now
$\upr(b,\frak A)\in A_{\zeta}$ and
$0\ne\upr(b,\frak A)\Bsubseteq a$.\ \QeD\  Since
$\frak A\cap\frak G_{\zeta}$ is an order-closed subalgebra of $\frak G$ it
is also an order-closed subalgebra of $\frak A$ and must be the
order-closed subalgebra of $\frak A$ generated by $A_{\zeta}$.

What this means is that $\family{\zeta}{C}{A_{\zeta}}$, suitably
re-indexed, witnesses that $\frak A$ satisfies condition ($\gamma$) of
515P.

\medskip

{\bf (b)} So if $\frak A$ is
$\pi$-weight-homogeneous with $\pi$-weight $\omega_1$, that is, all its
non-zero principal ideals have $\pi$-weight $\omega_1$, then
$\frak A\cong\frak G$.   \Prf\ As surely
it is non-zero, ccc and Dedekind complete (because $\frak G$ is),
it satisfies all three conditions of 515P.\ \Qed

On the other hand, if $\pi(\frak A)$ is countable, then 515O tells us that
$\frak A\cong\RO(\{0,1\}^{\omega})$.   So let us suppose from now on
that $\frak A$ has uncountable $\pi$-weight and has a principal ideal which
has $\pi$-weight different from $\omega_1$.

\medskip

{\bf (c)} Because $\pi$-weight is an order-preserving ordinal function of
Boolean algebras (514Ed), and $\frak A$ is Dedekind complete, it is
isomorphic to a simple product of $\pi$-weight-homogeneous
principal ideals (514Gc), all of which are Dedekind complete and have
$\pi$-weight at most $\pi(\frak A)\le\pi(\frak G)=\omega_1$ (514Eb).
Because $c(\frak A)\le c(\frak G)$ is countable (514Ea), we are dealing
with a countable product.   As $\pi(\frak A)>\omega$, not all the terms
have countable $\pi$-weight, and we are supposing also that not all the
terms have $\pi$-weight $\omega_1$.

The terms with $\pi$-weight $\omega$ can be joined together as a
single principal ideal $\frak B$ with $\pi$-weight $\omega$ (514Ef),
which must now be isomorphic to $\RO(\{0,1\}^{\omega})$.
Similarly the terms with $\pi$-weight $\omega_1$
can be joined together as a single $\pi$-weight-homogeneous
principal ideal $\frak C$ of $\pi$-weight $\omega_1$.   So
$\frak A\cong\frak B\times\frak C$.

Express $\frak C$
as $\frak A_c$ where $c\in\frak A$.   Consider the corresponding
principal ideal $\frak G_c$ of $\frak G$.   This is isomorphic to
$\frak G$ (515Nc) and $\frak C$ is an order-closed subalgebra of
$\frak G_c$.   So (b) tells us that $\frak C\cong\frak G$, and our
decomposition of $\frak A$ is of the required form.
}%end of proof of 515Q

\exercises{\leader{515X}{Basic exercises (a)}
%\spheader 515Xa
Let $\frak A$ be a Boolean algebra, not $\{0\}$, and $\familyiI{D_i}$
a family of partitions of unity in $\frak A$, none containing $0$.
Show that the following
are equiveridical:  (i) $\familyiI{D_i}$ is Boolean-independent;  (ii)
$\familyiI{\frak B_i}$ is Boolean-independent, where $\frak B_i$ is the
subalgebra of $\frak A$ generated by $D_i$ for each $i\in I$.
%515B

\spheader 515Xb\dvAnew{2014}
Give an example of a Boolean algebra $\frak A$ with
Boolean-independent subalgebras $\frak B$, $\frak C$ such that the
order-closed subalgebras generated by $\frak B$ and $\frak C$ are not
Boolean-independent.
%515B

\spheader 515Xc For a Boolean algebra $\frak A$, not $\{0\}$,
write $\ind(\frak A)$
for $\sup\{\#(A):A\subseteq\frak A$ is Boolean-independent$\}$.   (If
$\frak A=\{0\}$, say $\ind(\frak A)=0$.)   (i)
Show that if $\frak B$ is either a subalgebra or a principal ideal or a
homomorphic image of $\frak A$ then $\ind(\frak B)\le\ind(\frak A)$.
(ii) Show that
$\frak A$ is infinite iff $\ind(\frak A)$ is infinite.   (iii) Show that
if $\frak A$ is finite and not $\{0\}$ then $\ind(\frak A)$ is the
largest $n$ such that
$2^{2^n}\le\#(\frak A)$.   (iv) Show that if $\frak A$ is the
finite-cofinite algebra of subsets of an infinite set $X$, then
$\ind(\frak A)=\omega$ but $\frak A$ has no infinite Boolean-independent
set.   (v) Show that if $\frak A$ and $\frak B$ are Boolean algebras
then
$\ind(\frak A\times\frak B)$ is at most the cardinal sum
$\ind(\frak A)+\ind(\frak B)$.
(vi) Show that if $\frak A$ is infinite and has the
$\sigma$-interpolation property then $\ind(\frak A)\ge\frak c$.
%515B

\spheader 515Xd Let $Z$ be an infinite extremally disconnected
compact Hausdorff space.   Show that there is a
continuous surjection from $Z$ onto $\{0,1\}^{w(Z)}$.
%515H

\spheader 515Xe Let $\frak A$ be a Boolean algebra with the
$\sigma$-interpolation property.   Show that any homomorphic image of
$\frak A$ has the $\sigma$-interpolation property.
%515K

\spheader 515Xf Let $\kappa$ be an infinite cardinal.   Show that the
following are equiveridical:  (i) there is a measure algebra with
cardinal $\kappa$;  (ii) there is a measurable algebra with cardinal
$\kappa$;  (iii) $\kappa^{\omega}=\kappa$.
%515M

\leader{515Y}{Further exercises (a)}(i)
%spheader 515Ya
Show that if $\frak A$ is any Boolean algebra, other than $\{0\}$,
with cardinal at most $\omega_1$,
it is isomorphic to a subalgebra of $\Cal P\Bbb N/[\Bbb N]^{<\omega}$.
%514Yf(i)  $\Cal P\Bbb N/[\Bbb N]^{<\omega}$ has $\sigma$-interp propy
%do purely atomic and atomless separately
(ii) Show that an atomless Boolean algebra
with cardinal $\omega_1$ and the $\sigma$-interpolation property is
isomorphic to $\Cal P\Bbb N/[\Bbb N]^{<\omega}$.   (This is a version of
{\bf Parovi\v{c}enko's theorem}.)
%Koppelberg 5.30

\spheader 515Yb Let $\frak A$ be a Dedekind complete Boolean algebra, and
$\kappa\le\#(\frak A)$ a regular uncountable cardinal.   Show
that there is a strictly increasing family
$\ofamily{\xi}{\kappa}{\frak A_{\xi}}$ of subalgebras of $\frak A$ with
union $\frak A$.   (Compare 494Yk.)
%find suitable homomorphism from \frak A to \frak B_{\kappa}
%query:  what about other algebras?
}%end of exercises

\endnotes{
\Notesheader{515}
The material of this section is taken from {\smc Koppelberg 89}, where
you can find a good deal more.   I have picked out the results which are
essential to a proper understanding of measure algebras.   Of course
there are short cuts, using Maharam's theorem (332B), if we know that we
are dealing with a localizable measure algebra;  but I should not like
to leave you with the impression that the theorems here are restricted
to measure algebras.

Any theorem about Boolean algebras is also a theorem about
zero-dimensional compact Hausdorff spaces;  thus 515H and 515Xd have an
equal right to be called the Balcar-Fran\v{e}k theorem.
515D and part (b) of the proof of 515H may be regarded as a simple form
of some of the ideas of \S331.

Clearly some of the ideas of this section can be expressed in terms of
the independence number $\ind(\frak A)$ (515Xc).   But the expression is
complicated by the fact that (like cellularity) the independence number
may not be attained (see 515Xc(iv)), while the theorems here mostly need
actual Boolean-independent families.   Since $\ind(\frak A)=\#(\frak A)$ for
infinite Dedekind complete Boolean algebras (515H), we shall not have to
grapple with these difficulties.

The results of 515O-515Q give us a good grip on the regular open algebras
of $\{0,1\}^{\omega}$ and $\{0,1\}^{\omega_1}$ and their order-closed
subalgebras.   There are serious obstacles in the way of extending these
ideas;  order-closed
subalgebras of the category algebra of $\{0,1\}^{\omega_2}$, for instance,
can be very different in character.   For examples see
{\smc Koppelberg \& Shelah 96} and {\smc Balcar Jech \& Zapletal 97}.
}%end of notes
\discrpage

\frfilename{mt516.tex}
\versiondate{9.10.14}
\copyrightdate{2001}

\def\chaptername{Cardinal functions}
\def\sectionname{Precalibers}

\newsection{516}

In this section I will try to display the elementary connexions between
`precalibers', as defined in 511E, and the cardinal functions we have
looked at so far.
The first step is to generalize the idea of precaliber from partially
ordered sets to supported relations (516A);  the point is that
Galois-Tukey connections give us information on precalibers (516C), and
in particular give quick proofs that partially ordered sets, topological
spaces and Boolean algebras related in the canonical ways explored in
\S514 have many of the same precalibers (516G, 516H, 516M).
Much of the section is taken up with lists of expected facts, but for
some results the hypotheses need to be chosen with care.   I end with a
fundamental theorem on the saturation of product spaces (516T).

\leader{516A}{\bf Definition} If $(A,R,B)$ is a supported relation, a
{\bf precaliber triple} of $(A,R,B)$ is a triple
$\triplepc{\kappa}{\lambda}{\theta}$
where $\kappa$, $\lambda$ and $\theta$ are cardinals and whenever
$\ofamily{\xi}{\kappa}{a_{\xi}}$ is a family in $A$ then there is a set
$\Gamma\in[\kappa]^{\lambda}$ such that $\family{\xi}{\Gamma}{a_{\xi}}$
is $\hbox{$<$}\theta$-linked in the sense of 512Bc\cmmnt{, that is,
for every $I\in[\Gamma]^{<\theta}$ there is a $b\in B$ such that
$(a_{\xi},b)\in R$ for every $\xi\in I$}.   Similarly,
$(\kappa,\lambda,\theta)$ is a precaliber triple of $(A,R,B)$ if
whenever $\ofamily{\xi}{\kappa}{a_{\xi}}$ is a family in $A$ then there
is a set $\Gamma\in[\kappa]^{\lambda}$ such that
$\family{\xi}{\Gamma}{a_{\xi}}$ is $\theta$-linked;  that is,
if $\triplepc{\kappa}{\lambda}{\theta^+}$ is a precaliber triple.

Now $(\kappa,\lambda)$ is a {\bf precaliber pair} of $(A,R,B)$ if
$\triplepc{\kappa}{\lambda}{\omega}$ is a precaliber triple of
$(A,R,B)$, and $\kappa$ is a {\bf precaliber} of $(A,R,B)$ if
$(\kappa,\kappa)$ is a precaliber pair.

\leader{516B}{Elementary remarks}\cmmnt{ I ought perhaps to spell out
the following immediate consequences of the definitions.}
Let $(A,R,B)$ be a supported relation.

\spheader 516Ba If $\kappa'\ge\kappa$, $\lambda'\le\lambda$,
$\theta'\le\theta$ and $\triplepc{\kappa}{\lambda}{\theta}$ is a
precaliber triple of $(A,R,B)$, then
$\triplepc{\kappa'}{\lambda'}{\theta'}$ is a precaliber triple of
$(A,R,B)$.   So if $\kappa'\ge\kappa$, $\lambda'\le\lambda$ and
$(\kappa,\lambda)$ is a precaliber pair of $(A,R,B)$, then
$(\kappa',\lambda')$ is a precaliber pair of $(A,R,B)$.

\spheader 516Bb If $\theta>0$, then $\triplepc{0}{0}{\theta}$ is a
precaliber triple of
$(A,R,B)$ iff $B\ne\emptyset$.   If $A=\emptyset$ then
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of $(A,R,B)$
whenever $\kappa\ge 1$.   If $A\ne\emptyset$ and
$A\ne R^{-1}[B]$\cmmnt{, that
is, $\cov(A,R,B)=\infty$}, then the only precaliber triples of $(A,R,B)$
are of the form $\triplepc{\kappa}{0}{\theta}$.   If $A\ne\emptyset$ and
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of
$(A,R,B)$, then $\lambda\le\kappa$.   $\cov(A,R,B)=\infty$ iff $1$ is
not a precaliber of $(A,R,B)$.

\spheader 516Bc If $(\kappa,\lambda,\lambda)$ is a precaliber triple of
$(A,R,B)$ then $\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber
triple of $(A,R,B)$ for every $\theta$;  in particular,
$(\kappa,\lambda)$ is a precaliber pair of $(A,R,B)$.

\spheader 516Bd If $\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber
triple of $(A,R,B)$, so is $\triplepc{\cf\kappa}{\cf\kappa}{\theta}$.
\prooflet{\Prf\ If $\cf\kappa=\kappa$ there is nothing to prove.   If
$2\le\kappa<\omega$ and $A$ is empty the result is trivial.   If
$2\le\kappa<\omega$ and $A$ is not empty, then $B$ is not empty, so if
$\theta\le 1$ the result is trivial.   If $2\le\kappa<\omega$ and $A$ is
not empty and $\theta>1$, then $R^{-1}[B]=A$ so
$\triplepc{\cf\kappa}{\cf\kappa}{\theta}=\triplepc{1}{1}{\theta}$ is a
precaliber triple of $(A,R,B)$.

If $\kappa>\cf\kappa$ is infinite, let
$\ofamily{\xi}{\cf\kappa}{\gamma_{\xi}}$ be a strictly increasing family
with supremum $\kappa$.   For $\eta<\kappa$, set
$f(\eta)=\min\{\xi:\eta\le\gamma_{\xi}\}$.   If
$\ofamily{\xi}{\cf\kappa}{a_{\xi}}$ is a family in $A$, set
$a'_{\eta}=a_{f(\eta)}$ for each $\eta<\kappa$.   Then there is a
$\Gamma\in[\kappa]^{\kappa}$ such that
$\family{\eta}{\Gamma}{a'_{\eta}}$ is $\hbox{$<$}\theta$-linked.   Set
$\Gamma'=\{f(\eta):\eta\in\Gamma\}$;  then
$\family{\xi}{\Gamma'}{a_{\xi}}$ is $\hbox{$<$}\theta$-linked.   Also
$\Gamma$ must be cofinal with $\kappa$, so $\Gamma'$ is cofinal with
$\cf\kappa$ and $\#(\Gamma')=\cf\kappa$.   As
$\ofamily{\xi}{\cf\kappa}{a_{\xi}}$ is arbitrary,
$\triplepc{\cf\kappa}{\cf\kappa}{\theta}$ is a precaliber triple of
$(A,R,B)$.\ \Qed

}%end of prooflet
In particular, if $\kappa$ is a precaliber of $(A,R,B)$, so is
$\cf\kappa$.

\leader{516C}{Theorem} Suppose that $(A,R,B)$ and $(C,S,D)$ are
supported relations, and that $(A,R,B)\prGT(C,S,D)$.   Then
$\triplepc{\kappa}{\lambda}{\theta}$ or $(\kappa,\lambda,\theta)$ is a
precaliber triple of $(A,R,B)$
whenever it is a precaliber triple of $(C,S,D)$, so $(\kappa,\lambda)$
is a precaliber pair of $(A,R,B)$ whenever it is a precaliber pair of
$(C,S,D)$, and $\kappa$ is a precaliber of $(A,R,B)$ whenever it is a
precaliber of $(C,S,D)$.

\proof{ Let $(\phi,\psi)$ be a Galois-Tukey connection from $(A,R,B)$
to $(C,R,S)$.   If $\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber
triple of
$(C,S,D)$, and $\ofamily{\xi}{\kappa}{a_{\xi}}$ is a family in $A$, then
there is a set $\Gamma\in[\kappa]^{\lambda}$ such that whenever
$I\in[\Gamma]^{<\theta}$ there is a $d\in D$ such that
$(f(a_{\xi}),d)\in S$ for every $\xi\in I$, and now
$(a_{\xi},g(d))\in R$ for every $\xi\in I$.   Thus
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of
$(A,R,B)$.   The results for
precaliber pairs and precalibers follow at once.
}%end of proof of 516C

\leader{516D}{Corollary} If $(A,R,B)\equivGT(C,S,D)$ then $(A,R,B)$ and
$(C,S,D)$ have the same precaliber triples, the same precaliber pairs
and the same precalibers.

\cmmnt{
\leader{516E}{Remark} Because all the definitions in 516A start
from precaliber triples $\triplepc{\kappa}{\lambda}{\theta}$, any
theorem about such precaliber triples is likely to lead at once to
corresponding results concerning precaliber triples
$(\kappa,\lambda,\theta)$, precaliber pairs and precalibers.   In the
rest of this section I shall not always take the space to spell these
out systematically, and when later I wish to use a fact about
precalibers I
may direct you, without comment, to a fact about precaliber triples or
pairs from which it may be deduced.
}%end of comment

\leader{516F}{}\cmmnt{ The next step is to check the connexion between
the definition in 516A and those of \S511.   But this is elementary.

\medskip

\noindent}{\bf Proposition} (a) If $P$ is a partially ordered set,
$\triplepc{\kappa}{\lambda}{\theta}$ or $(\kappa,\lambda,\theta)$ is a
precaliber triple of
$(P,\le,P)$ iff it is an upwards precaliber triple of $P$.

(b) If $\frak A$ is a Boolean algebra, then $\frak A$ and
$(\frak A^+,\Bsupseteqshort,\frak A^+)$ have the same precaliber
triples\cmmnt{, where $\frak A^+=\frak A\setminus\{0\}$}.

(c) If $(X,\frak T)$ is a topological space, then $X$ and
$(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$ have the same precaliber triples.

\proof{ Read the definitions in 511E and 516A.
}%end of proof of 516F

\leader{516G}{Corollary} Let $(P,\le)$ be a partially ordered set.

(a) If $Q$ is a cofinal subset of $P$, then $P$ and $Q$ have the same
upwards precaliber triples.

(b) Let $\frak T^{\uparrow}$ be the up-topology of
$P$\cmmnt{ (definition: 514L)}.    Then
$\triplepc{\kappa}{\lambda}{\theta}$ is an upwards precaliber triple for
$(P,\le)$ iff it is a precaliber triple for $(P,\frak T^{\uparrow})$.

\proof{{\bf (a)} By 513E(d-ii), $(P,\le,P)\equivGT(Q,\le,Q)$.

\medskip

{\bf (b)} By 514Na, $(P,\le,P)
\equivGT(\frak T^{\uparrow}\setminus\{\emptyset\},\supseteq,
\frak T^{\uparrow}\setminus\{\emptyset\})$.
}%end of proof of 516G

\leader{516H}{Corollary} Let $\frak A$ be a Boolean algebra.

(a) If $Z$ is the Stone space of $\frak A$, then $\frak A$ and $Z$ have
the same precaliber triples.

(b) If $\frak B$ is an order-dense subalgebra of $\frak A$, then
$\frak A$ and $\frak B$ have the same precaliber triples.

\proof{{\bf (a)} Write $\frak T$ for the topology of $Z$ and $\Cal E$
for the algebra of open-and-closed sets.   Because $Z$ is
zero-dimensional, $\Cal E^+$ is coinitial with
$\frak T\setminus\{\emptyset\}$, so
$(\frak A^+,\Bsupseteqshort,\frak A^+)
\cong(\Cal E^+,\supseteq,\Cal E^+)$
and $(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$ have the same precaliber triples, by
516Ga, inverted.

\medskip

{\bf (b)} $\frak B^+$ is coinitial with $\frak A^+$, so we can use the
same idea.
}%end of proof of 516H

\leader{516I}{Corollary} Let $(X,\frak T)$ be a topological space.

(a) If $Y$ is an open subspace of $X$, then every precaliber triple of
$X$ is a precaliber triple of $Y$.

(b) If $Y$ is a dense subspace of $X$, then every precaliber triple of
$X$ is a precaliber triple of $Y$.

(c) If $X$ is regular and $Y$ is a dense subspace of $X$, then $X$ and
$Y$ have the same precaliber triples.

(d) Suppose that $Y$ is a topological space, and that there is a
continuous surjection $f:X\to Y$ such that $\interior f[G]\ne\emptyset$
whenever $G\subseteq X$ is a non-empty open set.   Then every precaliber
triple of $X$ is a precaliber triple of $Y$.

\proof{{\bf (a)} Write $\frak S$ for the topology of $Y$.   For
$H\in\frak S\setminus\{\emptyset\}$, set $\phi(H)=H$;  for
$G\in\frak T\setminus\{\emptyset\}$, set $\psi(G)=G\cap Y$ if this is
non-empty, $G$ otherwise.   Then $(\phi,\psi)$ is a Galois-Tukey
connection from
$(\frak S\setminus\{\emptyset\},\supseteq,
\frak S\setminus\{\emptyset\})$ to
$(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$, so 516C and 516Fc give the result.

\medskip

{\bf (b)} Again write $\frak S$ for the topology of $Y$.   For
$H\in\frak S\setminus\{\emptyset\}$, set
$\phi(H)=X\setminus\overline{Y\setminus H}$, where the closure is taken
in $X$;  for $G\in\frak T\setminus\{\emptyset\}$, set $\psi(G)=G\cap Y$.
Then $(\phi,\psi)$ is a Galois-Tukey connection from
$(\frak S\setminus\{\emptyset\},\supseteq,
\frak S\setminus\{\emptyset\})$ to
$(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$, so again we have the result.

\medskip

{\bf (c)} If now $X$ is regular, then for each
$G\in\frak T\setminus\{\emptyset\}$ choose
$V_G\in\frak T\setminus\{\emptyset\}$ such that
$\overline{V}_G\subseteq G$ and set $\psi'(G)=V_G\cap Y$.   Then
$(\psi',\phi)$ is a Galois-Tukey connection from
$(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$ to
$(\frak S\setminus\{\emptyset\},\supseteq,
\frak S\setminus\{\emptyset\})$, so every precaliber triple of $Y$ is a
precaliber triple of $X$.

\medskip

{\bf (d)} Once more writing $\frak S$ for the topology of $Y$, set
$\phi(H)=f^{-1}[H]$ for every $H\in\frak S\setminus\{\emptyset\}$ and
$\psi(G)=\interior f[G]$ for every $G\in\frak T\setminus\{\emptyset\}$;
then again $(\phi,\psi)$ is a Galois-Tukey connection from
$(\frak S\setminus\{\emptyset\},\supseteq,
\frak S\setminus\{\emptyset\})$ to
$(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$.
}%end of proof of 516I

\cmmnt{\medskip

\noindent{\bf Remark} For variations on (b) and (d) here, see 516Xh and
516Oa.}

\leader{516J}{}\cmmnt{ Straightforward counting arguments give us
some connexions between precalibers and other cardinal functions, as
follows.

\medskip

\noindent}{\bf Proposition} Let $(A,R,B)$ be a supported relation.

(a) $\sat(A,R,B)$ is the least cardinal $\kappa$, if there is one, such
that $(\kappa,2)$ is a precaliber pair of $(A,R,B)$;  if there is no
such $\kappa$, $\sat(A,R,B)=\infty$.   In particular, if $\kappa\ge 2$
is a precaliber of $(A,R,B)$, then $\kappa\ge\sat(A,R,B)$.

(b) If $\kappa>\max(\omega,\lambda,\link_{<\theta}(A,R,B))$ then
$\triplepc{\kappa}{\lambda^+}{\theta}$ is a precaliber triple of
$(A,R,B)$.   In particular, if
$\kappa>\max(\omega,\lambda,\cov(A,R,B))$ then
$\triplepc{\kappa}{\lambda^+}{\theta}$ is a precaliber triple of
$(A,R,B)$ for every $\theta$.
%, and if
%$\kappa>\max(\omega,\lambda,\link_{<\omega}(A,R,B))$ then
%$(\kappa,\lambda^+)$ is a precaliber pair of $(A,R,B)$.

(c) If $\cf\kappa>\link_{<\theta}(A,R,B)$ then
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of $(A,R,B)$.
%In particular, if $\cf\kappa>\link_{<\omega}(A,R,B)$ then $\kappa$ is a
%precaliber of $(A,R,B)$.

\proof{{\bf (a)} If $(\kappa,2)$ is a precaliber pair of $(A,R,B)$, and
$\ofamily{\xi}{\kappa}{a_{\xi}}$ is any family in $A$, then there must
be a $\Gamma\in[\kappa]^2$ such that for every finite $I\subseteq\Gamma$
there is a $b\in B$ such that $(a_{\xi},b)\in R$ for every $\xi\in I$.
But this means that if $\Gamma=\{\xi,\eta\}$ then $\xi$, $\eta$ are
distinct members of $\kappa$ such that, for some $b\in B$, both
$(a_{\xi},b)$ and $(a_{\eta},b)$ belong to $R$.
As $\ofamily{\xi}{\kappa}{a_{\xi}}$ is arbitrary,
$\sat(A,R,B)\le\kappa$.

Conversely, any witness that $(\kappa,2)$ is not a precaliber pair of
$(A,R,B)$ will provide a witness that $\sat(A,R,B)>\kappa$.

Now if $\kappa\ge 2$ is a precaliber of $(A,R,B)$, that is,
$(\kappa,\kappa)$ is a precaliber pair, then $(\kappa,2)$ is a
precaliber pair of $(A,R,B)$, by 516Ba, so $\kappa\ge\sat(A,R,B)$.

\medskip

{\bf (b)} Write $\delta$ for $\link_{<\theta}(A,R,B)$, and let
$\ofamily{\eta}{\delta}{A_{\eta}}$ be a cover of $A$ by
$\hbox{$<$}\theta$-linked sets.   Let $\ofamily{\xi}{\kappa}{a_{\xi}}$
be any family in $A$.   For $\eta<\delta$ set
$C_{\eta}=\{\xi:a_{\xi}\in A_{\eta}\}$;  then
$\kappa=\bigcup_{\eta<\delta}C_{\eta}$ so there must be some
$\eta<\delta$ such that $\#(C_{\eta})>\lambda$.   Now if
$\Gamma\subseteq C_{\eta}$ is a set with cardinal $\lambda^+$,
$\{a_{\xi}:\xi\in\Gamma\}$ is $\hbox{$<$}\theta$-linked in $(A,R,B)$.
As $\ofamily{\xi}{\kappa}{a_{\xi}}$ is arbitrary,
$\triplepc{\kappa}{\lambda^+}{\theta}$ is a precaliber triple of
$(A,R,B)$.

The special case is now elementary, if we remember that
$\link_{<\theta}(A,R,B)\le\cov(A,R,B)$ for every $\theta$ (512Bc).

\medskip

{\bf (c)} If $\link_{<\theta}(A,R,B)=0$ then $A=\emptyset$ and the
result is trivial.   Otherwise, $\cf\kappa\ge\omega$.   Choose $\delta$
and $\ofamily{\eta}{\delta}{C_{\eta}}$ as in (b) above.  Let
$\ofamily{\xi}{\kappa}{a_{\xi}}$ be any family in $A$.   Then there must
be some $\eta<\delta$ such that $\#(C_{\eta})=\kappa$, and
$\{a_{\xi}:\xi\in C_{\eta}\}$ is $\hbox{$<$}\theta$-linked in $(A,R,B)$.
As $\ofamily{\xi}{\kappa}{a_{\xi}}$ is arbitrary,
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of $(A,R,B)$.
%Taking $\theta=\omega$, we see that $\kappa$ is a precaliber of
%$(A,R,B)$ if $\cf\kappa>\link_{<\omega}(A,R,B)$.
}%end of proof of 516J

\leader{516K}{}\cmmnt{ For partially ordered sets, we have translations
of the results above, and a further useful fact.

\medskip

\noindent}{\bf Proposition} Let $P$ be a partially ordered set.

(a) $\sat^{\uparrow}(P)$ is the least cardinal $\kappa$ such that
$(\kappa,2)$ is an upwards precaliber pair of $P$.

(b) If $\kappa>\max(\omega,\lambda,\link^{\uparrow}_{<\theta}(P))$ then
$\triplepc{\kappa}{\lambda^+}{\theta}$ is an upwards precaliber triple
of $P$.   In particular, if $\kappa>\max(\omega,\lambda,\cf P)$ then
$\triplepc{\kappa}{\lambda^+}{\theta}$ is an upwards precaliber triple
of $P$ for every $\theta$, and if
$\kappa>\max(\omega,\lambda,\duparrow(P))$ then $(\kappa,\lambda^+)$
is an upwards precaliber pair of $P$.

(c) If $\cf\kappa>\cf P$ then $\triplepc{\kappa}{\kappa}{\theta}$ is an
upwards precaliber triple of $P$ for every $\theta$.   If
$\cf\kappa>\duparrow(P)$ then $\kappa$ is an
up-precaliber of $P$.

(d) If $\sat^{\uparrow}(P)\ge\omega$,
$(\sat^{\uparrow}(P),\omega)$ is an upwards precaliber pair of $P$.

\proof{{\bf (a)-(c)} We need only identify $\cf P$ with
$\cov(P,\le,P)\ge\sup_{\theta}\link_{<\theta}(P,\le,P)$ (512Bc) and
$\duparrow(P)$ with the centering number
$\link_{<\omega}(P,\le\nobreak,P)$, as in 512Ea.

\medskip

{\bf (d)(i)} Set $\kappa=\sat^{\uparrow}(P)$.   By 513Bb, $\kappa$ is a
regular uncountable cardinal.   The first thing to note is that if
$\ofamily{\xi}{\kappa}{p_{\xi}}$ is any family in $P$, then there is a
$\zeta<\kappa$ such that
$\{\xi:\xi<\kappa$,
$p_{\xi}$ and $p_{\zeta}$ are compatible upwards in $P\}$ has cardinal
$\kappa$.   \Prf\Quer\ Otherwise, for each $\zeta<\kappa$ there is an
$\alpha_{\zeta}<\kappa$ such that $p_{\zeta}$ and $p_{\xi}$ are
upwards-incompatible for every $\xi\ge\alpha_{\zeta}$.   Set
$C=\{\xi:\xi<\kappa$, $\alpha_{\eta}\le\xi$ for every $\eta<\xi\}$.   Then
$\#(C)=\kappa$ and $\family{\xi}{C}{p_{\xi}}$ is an up-antichain in $P$,
which is impossible.\ \Bang\Qed

\medskip

\quad{\bf (ii)} Now let $\ofamily{\xi}{\kappa}{p_{\xi}}$ be a
family in $P$.   Choose inductively sets $A_n\in[\kappa]^{\kappa}$,
ordinals $\zeta_n\in A_n$ and
families $\family{\xi}{A_n}{p_{n\xi}}$ in $P$, as follows.   $A_0=\kappa$,
$p_{0\xi}=p_{\xi}$ for each $\xi<\kappa$.   Given
$\family{\xi}{A_n}{p_{n\xi}}$, then by (i) there is a $\zeta_n\in A_n$ such
that

\Centerline{$A_{n+1}=\{\xi:\xi\in A_n$, $\xi\ne\zeta_n$,
$p_{n\xi}$ is compatible upwards with $p_{n,\zeta_n}\}$}

\noindent has cardinal
$\kappa$.   Now, for $\xi\in A_{n+1}$, let $p_{n+1,\xi}$ be an upper bound
of $\{p_{n,\zeta_n},p_{n\xi}\}$;  continue.

At the end of the induction, observe that $\sequencen{p_{n,\zeta_n}}$ is
non-decreasing.   At the same time, we see that $p_{\xi}\le p_{n\xi}$
whenever $n\in\Bbb N$ and $\xi\in A_n$.   So $\{p_{\zeta_n}:n\in\Bbb N\}$
is upwards-centered.   Also the $\zeta_n$ are all different, so
$\Gamma=\{\zeta_n:n\in\Bbb N\}$ is infinite.   As
$\ofamily{\xi}{\kappa}{p_{\xi}}$ is arbitrary,
$(\sat^{\uparrow}(P),\omega)$ is an upwards precaliber pair of $P$.
}%end of proof of 516K

\cmmnt{\medskip

\noindent{\bf Remark} There will be a stronger form of (d) in
517Fa below.
}

\leader{516L}{Corollary} Let $\frak A$ be a Boolean algebra.

(a) $\sat(\frak A)$ is the least cardinal $\kappa$ such
that $(\kappa,2)$ is a precaliber pair of $\frak A$.

(b) If $\kappa>\max(\omega,\lambda,\link_{<\theta}(\frak A))$ then
$\triplepc{\kappa}{\lambda^+}{\theta}$ is a precaliber triple of
$\frak A$.   In particular, if
$\kappa>\max(\omega,\lambda,\pi(\frak A))$ then
$\triplepc{\kappa}{\lambda^+}{\theta}$ is a precaliber triple of
$\frak A$ for every $\theta$, and if
$\kappa>\max(\omega,\lambda,d(\frak A))$ then $(\kappa,\lambda^+)$ is a
precaliber pair of $\frak A$.

(c) If $\cf\kappa>d(\frak A)$ then $\kappa$ is a precaliber
of $\frak A$.

(d) If $\frak A$ is infinite,
$(\sat(\frak A),\omega)$ is a precaliber pair of $\frak A$.

\proof{ Apply 516K, inverted, to $\frak A^+$, recalling that
$\pi(\frak A)=\ci(\frak A^+)$.
}%end of proof  of 516L

\leader{516M}{}\cmmnt{ When we turn to topological spaces, we can
refine the results slightly, using the following elementary facts.

\medskip

\noindent}{\bf Lemma} Let $(X,\frak T)$ be a topological space and
$\RO(X)$ its regular open algebra.   If $\kappa$, $\lambda$ and $\theta$
are cardinals, and $\theta\le\omega$, then the following are
equiveridical:

(i) $\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of
$(X,\frak T)$;

(ii) $\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of
$(\frak T\setminus\{\emptyset\},\ni,X)$;

(iii) $\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of
$\RO(X)$.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} If we set $\phi(G)=G$ and choose a
point $\psi(G)\in G$ for every non-empty open set $G\subseteq X$, then
$(\phi,\psi)$ is a Galois-Tukey connection from
$(\frak T\setminus\{\emptyset\},\ni,X)$ to
$(\frak T\setminus\{\emptyset\},\supseteq,\frak T\setminus\{\emptyset\})$,
so any precaliber triple of the latter is a precaliber triple of the
former.

\medskip

{\bf (b)(ii)$\Rightarrow$(iii)} Assume (ii), and let
$\ofamily{\xi}{\kappa}{G_{\xi}}$ be a family in $\RO(X)^+$.   Then there
is a $\Gamma\in[\kappa]^{\lambda}$ such that $\bigcap_{\xi\in
I}G_{\xi}\ne\emptyset$ for every $I\in[\Gamma]^{<\theta}$.   But in this
case, because $I$ is finite, $\bigcap_{\xi\in I}G_{\xi}$ is a lower
bound for $\{G_{\xi}:\xi\in I\}$ in $\RO(X)^+$.   As
$\ofamily{\xi}{\kappa}{G_{\xi}}$ is arbitrary,
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of $\RO(X)$.

\medskip

{\bf (c)(iii)$\Rightarrow$(i)} Assume (iii), and let
$\ofamily{\xi}{\kappa}{G_{\xi}}$ be a family in
$\frak T\setminus\{\emptyset\}$.   Then there is a
$\Gamma\in[\kappa]^{\lambda}$ such that
$\bigcap_{\xi\in I}\interior\overline{G}_{\xi}\ne\emptyset$ for every
$I\in[\Gamma]^{<\theta}$.   But in this case, because $I$ is finite,
$\bigcap_{\xi\in I}G_{\xi}$ is not empty, and is a lower bound for
$\{G_{\xi}:\xi\in I\}$ in $\frak T\setminus\{\emptyset\}$.   As
$\ofamily{\xi}{\kappa}{G_{\xi}}$ is arbitrary,
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of
$(X,\frak T)$.
}%end of proof of 516M

\leader{516N}{Corollary} Let $X$ be a topological space.

(a) $\sat(X)$ is the least cardinal $\kappa$ such that
$(\kappa,2)$ is a precaliber pair of $X$.

(b) If $\kappa>\max(\omega,\lambda,d(X))$
then $(\kappa,\lambda^+)$ is a precaliber pair of $X$.

(c) If $\cf\kappa>d(X)$
then $\kappa$ is a precaliber of $X$.

(d) If $\sat(X)$ is infinite, then $(\sat(X),\omega)$ is
a precaliber pair of $X$.

\proof{ Here we need to know that $\sat(X)=\sat(\RO(X))$ and
$d(X)\ge d(\RO(X))$ (514H(b-i)).
}%end of proof of 516N

\leader{516O}{}\cmmnt{ The idea of 516M leads to further results about
precalibers of topological spaces.

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T)$ be a topological space.

(a) If $Y$ is a continuous image of $X$ and $\theta\le\omega$, then
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of $Y$
whenever it is a precaliber triple of $X$.
%In particular, every
%precaliber pair of $X$ is a precaliber pair of $Y$.

(b) Suppose that $X$ is the product of a family $\familyiI{X_i}$ of
topological spaces.   If $\triplepc{\kappa}{\kappa}{\theta}$ is a
precaliber triple
of every $X_i$ and {\it either} $I$ is finite {\it or} $\theta\le\omega$
and $\kappa$ is a regular uncountable cardinal, then
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of $X$.
%In particular, a regular uncountable cardinal $\kappa$ is a precaliber
%of $X$ if it is a precaliber of every $X_i$.

\proof{{\bf (a)} Let $f:X\to Y$ be a continuous surjection.   Writing
$\frak S$ for the topology of $Y$, we have a Galois-Tukey connection
$(\phi,f)$ from $(\frak S\setminus\{\emptyset\},\ni,Y)$ to
$(\frak T\setminus\{\emptyset\},\ni,X)$, if we set $\phi(H)=f^{-1}[H]$
for $H\in\frak S\setminus\{\emptyset\}$.   Now if $\theta\le\omega$ and
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of
$(X,\frak T)$, it is a precaliber triple of
$(\frak T\setminus\{\emptyset\},\ni,X)$,
$(\frak S\setminus\{\emptyset\},\ni,Y)$ and $(Y,\frak S)$, using 516M
and 516C.

\medskip

{\bf (b)} If $X=\emptyset$ then
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of $X$ just
because $X=X_i$ for some $i$;  so let us suppose that $X\ne\emptyset$.

\medskip

\quad{\bf (i)} If $I=\{0,1\}$ then $\triplepc{\kappa}{\kappa}{\theta}$
is a precaliber triple of $X$.   \Prf\ Let
$\ofamily{\xi}{\kappa}{W_{\xi}}$ be a family of non-empty open sets in
$X$.   For each $\xi<\kappa$, let $G_{\xi 0}\subseteq X_0$ and
$G_{\xi 1}\subseteq X_1$ be non-empty open sets such that
$G_{\xi 0}\times G_{\xi 1}\subseteq W_{\xi}$.   Because
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of $X_0$,
there is a $\Gamma\in[\kappa]^{\kappa}$ such that
$H_K^{(0)}=\interior(\bigcap_{\xi\in K}G_{\xi 0})$ is non-empty for
every $K\in[\Gamma]^{<\theta}$.   Because
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of $X_1$,
there is a $\Delta\in[\Gamma]^{\kappa}$ such that
$H_K^{(1)}=\interior(\bigcap_{\xi\in K}G_{\xi 1})$ is non-empty for
every $K\in[\Delta]^{<\theta}$.   Now
$\bigcap_{\xi\in K}W_{\xi}\supseteq H_K^{(0)}\times H_K^{(1)}$ has
non-empty interior for every
$K\in[\Delta]^{<\theta}$.   As $\ofamily{\xi}{\kappa}{W_{\xi}}$ is
arbitrary, $\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of
$X$.\ \Qed

\medskip

\quad{\bf (ii)} If $I$ is finite, then
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of $X$.
\Prf\ Induce on $\#(I)$, using (i) for the inductive step.\ \Qed

\medskip

\quad{\bf (iii)} Now suppose that $I$ is infinite, $\kappa$ is regular
and uncountable and $\theta\le\omega$.   Then
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of $X$.
\Prf\ Let
$\ofamily{\xi}{\kappa}{W_{\xi}}$ be a family of non-empty open sets in
$X$.   Let $\Cal V$ be the standard base for the topology of $X$
consisting of sets of the form $\prod_{\xi<\kappa}U_{\xi}$ where
$U_{\xi}\subseteq X_{\xi}$ is open for every $\xi$ and
$\{\xi:U_{\xi}\ne X_{\xi}\}$ is finite.   For each $\xi<\kappa$ let
$W'_{\xi}\subseteq W_{\xi}$ be a non-empty member of $\Cal V$, so that
$W'_{\xi}$ is determined by a coordinates in a finite subset $I_{\xi}$
of $I$.   By the $\Delta$-system Lemma (4A1Db) there is a set
$A\subseteq\kappa$, with cardinal $\kappa$, such that
$\family{\xi}{A}{I_{\xi}}$ is a $\Delta$-system with root $J$ say.
For $\xi\in A$ express $W'_{\xi}$ as $U_{\xi}\cap V_{\xi}$ where
$U_{\xi}$ is determined by coordinates in $J$ and $V_{\xi}$ is
determined by coordinates in $I_{\xi}\setminus J$.   Now $U_{\xi}$ is of
the form $\pi_J^{-1}[H_{\xi}]$ where $H_{\xi}\subseteq\prod_{i\in J}X_i$
is a non-empty open set and $\pi_J:X\to\prod_{i\in J}X_i$ is the
canonical map.   By (ii), $\triplepc{\kappa}{\kappa}{\theta}$ is a
precaliber triple of $\prod_{i\in J}X_i$, so there is a
$\Gamma\in[A]^{\kappa}$ such that $\bigcap_{\xi\in K}H_{\xi}$ is
non-empty whenever $K\in[\Gamma]^{<\theta}$.    Now take any
$K\in[\Gamma]^{<\theta}$.   Then
$U=\pi_J^{-1}[\bigcap_{\xi\in K}H_{\xi}]$ is a non-empty
set determined by coordinates in $J$, while $V_{\xi}$ is a non-empty
open set determined by coordinates in $I_{\xi}\setminus J$ for each
$\xi\in K$;  because the $I_{\xi}\setminus J$ are disjoint and $K$ is
finite, $U\cap\bigcap_{\xi\in J}V_{\xi}$ is non-empty, and
$\bigcap_{\xi\in K}W_{\xi}$ is a non-empty set, necessarily open because
$K$ is finite.   As
$\ofamily{\xi}{\kappa}{W_{\xi}}$ is arbitrary,
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of $X$.\
\Qed
}%end of proof of 516O

\leader{516P}{Corollary} Let $\familyiI{P_i}$ be a family of non-empty
partially ordered sets, with upwards finite-support product
$P=\bigotimes^{\uparrow}_{i\in I}P_i$\cmmnt{ (definition:  514T)}.
If $\triplepc{\kappa}{\kappa}{\theta}$ is an upwards precaliber
triple of every $P_i$ and either $I$ is finite or $\theta\le\omega$ and
$\kappa$ is a regular uncountable cardinal, then
$\triplepc{\kappa}{\kappa}{\theta}$ is an upwards precaliber triple of
$P$.
%In particular, a regular infinite cardinal $\kappa$ is an upwards
%precaliber of $P$ if it is an upwards precaliber of every $P_i$.

\proof{ Suppose first that $\theta$ is countable.
By 516Gb and 516M we can identify the
relevant upwards precaliber triples of each $P_i$ and $P$ with the
precaliber triples of their regular open algebras.   But
$\RO^{\uparrow}(P)\cong\RO(\prod_{i\in I}P_i)$ (514Ua), so 516Ob gives
the result at once.

For finite $I$, $P^*=\prod_{i\in I}P_i$ is a cofinal subset of $P$
(514Ub), so that it has the same upwards precaliber triples (516Ga);  at
the same time, it is easy to see that the up-topology of $P^*$ is just
the product of the up-topologies on the $P_i$.   So this time we do not
need to look at regular open algebras and can use 516Gb and 516Ob
directly.
}%end of proof

\leader{516Q}{}\cmmnt{ For locally compact spaces, as usual, we have
further results.

\medskip

\noindent}{\bf Proposition} Let $X$ be a locally compact Hausdorff
topological space.

(a) $(\kappa,\lambda)$ is a precaliber pair of $X$ iff whenever
$\ofamily{\xi}{\kappa}{G_{\xi}}$ is a family of non-empty open
subsets of $X$, then there is an $x\in X$ such that
$\#(\{\xi:x\in G_{\xi}\})\ge\lambda$.

(b)\dvArevised{2014} Suppose that $\kappa$ is a regular infinite cardinal.
Then $\kappa$ is a precaliber of $X$ iff $\sat(X)\le\kappa$ and whenever
$\ofamily{\xi}{\kappa}{E_{\xi}}$ is a
non-decreasing family of nowhere dense subsets of $X$ then
$\bigcup_{\xi<\kappa}E_{\xi}$ has empty interior.

\proof{{\bf (a)(i)} The condition asserts that
$(\kappa,\lambda,\lambda)$ is a precaliber triple of
$(\frak T\setminus\{\emptyset\},\ni,X)$.   It follows at once that
$(\kappa,\lambda)$ is a precaliber pair of
$(\frak T\setminus\{\emptyset\},\ni,X)$
and therefore of $(X,\frak T)$, by 516M.

\medskip

\quad{\bf (ii)} Now suppose that $(\kappa,\lambda)$ is a precaliber pair
of $X$, and that $\ofamily{\xi}{\kappa}{G_{\xi}}$ is a family of
non-empty open subsets of $X$.   For each $\xi<\kappa$ choose a
non-empty relatively compact open set $H_{\xi}$ such that
$\overline{H}_{\xi}\subseteq G_{\xi}$.   Then there is a
$\Gamma\in[\kappa]^{\lambda}$ such that
$\{H_{\xi}:\xi\in\Gamma\}$ is centered.   In this case,
$\{\overline{H}_{\xi}:\xi\in\Gamma\}$ has the finite intersection
property, so has non-empty intersection.   If $x$ is any point of this
intersection, then $\{\xi:x\in G_{\xi}\}\supseteq\Gamma$ has cardinal at
least $\lambda$.

\medskip

{\bf (b)(i)} Suppose that $\kappa$ is a precaliber of $X$.   Then surely
$\sat(X)\le\kappa$ (516Ja).   If $\ofamily{\xi}{\kappa}{E_{\xi}}$ is a
non-decreasing family of nowhere dense subsets of $X$, take any
non-empty open set $G\subseteq X$.   For each $\xi<\kappa$,
$G_{\xi}=G\setminus\overline{E}_{\xi}$ is a non-empty open set, so by
(a) there is an $x\in X$ such that $\Gamma=\{\xi:x\in G_{\xi}\}$ has
cardinal $\kappa$.  But as $\ofamily{\xi}{\kappa}{G_{\xi}}$ is
non-increasing, this means that $\Gamma=\kappa$ and
$x\in G\setminus\bigcup_{\xi<\kappa}E_{\xi}$.   As $G$ is arbitrary,
$\bigcup_{\xi<\kappa}E_{\xi}$ has empty interior.

\medskip

\quad{\bf (ii)} Now suppose that the condition is satisfied.
Let $\ofamily{\xi}{\kappa}{G_{\xi}}$ be a family of non-empty open
subsets of $X$.   For $\xi<\kappa$ set
$H_{\xi}=\bigcup_{\eta\ge\xi}G_{\eta}$,
$W_{\xi}=X\setminus\overline{H}_{\xi}$.   By 5A4Bd, there is a set
$I\subseteq\kappa$ such that $\#(I)<\sat(X)$ and

\Centerline{$\overline{\bigcupop_{\xi\in I}W_{\xi}}
=\overline{\bigcupop_{\xi<\kappa}W_{\xi}}$.}

\noindent Because $\#(I)<\cf\kappa$, $\zeta=\sup I$ is less than
$\kappa$, and $H_{\zeta}\cap W_{\xi}=\emptyset$ for every $\xi\in I$, so
$H_{\zeta}\cap W_{\xi}=\emptyset$ for every $\xi<\kappa$, that is,
$H_{\zeta}\subseteq\overline{H}_{\xi}$ for every $\xi<\kappa$.

Setting

\Centerline{$E_{\xi}=H_{\zeta}\setminus H_{\xi}
\subseteq\overline{H}_{\xi}\setminus H_{\xi}$}

\noindent for each $\xi$, $\ofamily{\xi}{\kappa}{E_{\xi}}$ is a
non-decreasing family of nowhere dense sets, and cannot cover
$H_{\zeta}$.   If $x\in H_{\zeta}\setminus\bigcup_{\xi<\kappa}E_{\xi}$,
then $x\in H_{\xi}$ for every $\xi<\kappa$, so $\Gamma=\{\eta:x\in
G_{\eta}\}$ is cofinal with $\kappa$.   Because $\kappa$ is regular,
$\Gamma\in[\kappa]^{\kappa}$, and $\bigcap_{\xi\in I}G_{\xi}$ is
non-empty for every $I\in[\Gamma]^{<\omega}$.   As
$\ofamily{\xi}{\kappa}{G_{\xi}}$ is arbitrary, $\kappa$ is a precaliber
of $X$, by (a).
}%end of proof of 516Q

\leader{516R}{}\cmmnt{ We can use the last proposition to give
corresponding characterizations of precaliber pairs of Boolean algebras
in terms of their Stone spaces.

\medskip

\noindent}{\bf Corollary} Let $\frak A$ be a Boolean algebra and $Z$ its
Stone space.

(a) A pair $(\kappa,\lambda)$ of cardinals is a precaliber pair of
$\frak A$ iff whenever $\ofamily{\xi}{\kappa}{G_{\xi}}$ is a family of
non-empty open sets in $Z$ there is a $z\in Z$ such that
$\#(\{\xi:z\in G_{\xi}\})\ge\lambda$.

(b) Suppose that $\kappa\ge\sat(\frak A)$ is a regular infinite
cardinal.   Then $\kappa$ is a precaliber of $\frak A$ iff whenever
$\ofamily{\xi}{\kappa}{E_{\xi}}$ is a
non-decreasing family of nowhere dense subsets of $Z$ then
$\bigcup_{\xi<\kappa}E_{\xi}$ has empty interior.

\proof{ Put 516Ha and 516Q together.
}%end of proof of 516R

\leader{516S}{}\cmmnt{ I collect some further results relating
precalibers to the standard constructions involving Boolean algebras as
considered in 514E.

\medskip

\noindent}{\bf Proposition}\dvArevised{2015}
Let $\frak A$ be a Boolean algebra.

(a) If $\frak B$ is a subalgebra of $\frak A$ and
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of $\frak A$
such that $\theta\le\omega$, then
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of $\frak B$.
In particular,
every precaliber pair of $\frak A$ is a precaliber pair of $\frak B$ and
$\frak B$ will satisfy Knaster's condition if $\frak A$ does.
%548B

(b) If $\frak B$ is a regularly embedded subalgebra of $\frak A$, then
every precaliber triple of $\frak A$ is a precaliber triple of $\frak B$.

(c) If $\frak B$ is a Boolean algebra and $\phi:\frak A\to\frak B$ is a
surjective order-continuous Boolean homomorphism, then every
precaliber triple of $\frak A$ is a precaliber triple of $\frak B$.

(d)\dvAformerly{5{}16Sc}
If $\frak B$ is a principal ideal of $\frak A$ then every
precaliber triple of $\frak A$ is a precaliber triple of $\frak B$.

(e)\dvAformerly{5{}16Sd} If $\frak A$ is the simple product of a family
$\familyiI{\frak A_i}$ of Boolean algebras,
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of $\frak A_i$
for every $i\in I$ and $\cf\kappa>\#(I)$, then
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of $\frak A$.

\proof{{\bf (a)} The Stone space of $\frak B$ is a continuous image of
the Stone space of $\frak A$ (312Sa).   So all we have to do is to put
516Ha and 516Oa together.

Taking $\theta=\omega$ we see that a precaliber pair of $\frak A$ will be a
precaliber pair of $\frak B$.   Taking $\lambda=\kappa=\omega_1$ and
$\theta=3$, we see that if $\frak A$ satisfies Knaster's condition so does
$\frak B$.

\medskip

{\bf (b)} Any subset of $\frak B^+$ with a lower bound in $\frak A^+$ has a
lower bound in $\frak B^+$, so the identity map from $\frak B^+$ to
$\frak A^+$ is a Tukey
function from $(\frak B^+,\Bsupseteqshort)$ to
$(\frak A^+,\Bsupseteqshort)$ and we can put 516C together with
516Fb.

\medskip

{\bf (c)} If $A\subseteq\frak A^+$ has no lower bound in $\frak A^+$, that
is, $\inf A=0$, then $\inf\phi[A]=0$ and $\phi[A]$ has no lower bound in
$\frak B^+$.   Accordingly $\phi\restr\frak A^+$ is a dual Tukey function
from $(\frak A^+,\Bsupseteqshort)$ to $(\frak B^+,\Bsupseteqshort)$,
$(\frak B^+,\Bsupseteqshort)\prT(\frak A^+,\Bsupseteqshort)$ and
we can proceed as in (c).

\medskip

{\bf (d)} As in (b), the identity map from
$\frak B^+$ to $\frak A^+$ is a Tukey
function from $(\frak B^+,\Bsupseteqshort)$ to
$(\frak A^+,\Bsupseteqshort)$.   (Or put 312T and 516Ia together.)

\medskip

{\bf (e)} Let $\ofamily{\xi}{\kappa}{a_{\xi}}$ be a family in $\frak A^+$.
Then for each $\xi<\kappa$ there is an $i\in I$ such that $a_{\xi}(i)$ is
non-zero in $\frak A_i$.   As $\cf\kappa>\#(I)$ there is an $i\in I$ such
that $\Delta=\{\xi:\xi<\kappa$, $a_{\xi}(i)\ne 0\}$ has cardinal $\kappa$.
As $\triplepc{\#(\Delta)}{\lambda}{\theta}$ is a precaliber triple of
$\frak A_i$, there is a $\Gamma\in[\Delta]^{\lambda}$ such that
$\{a_{\xi}(i):\xi\in I\}$ has a non-zero lower bound in $\frak A_i$ for every
$I\in[\Gamma]^{<\theta}$.   But now
$\{a_{\xi}:\xi\in I\}$ has a non-zero lower bound in $\frak A$ for every
$I\in[\Gamma]^{<\theta}$, while $\Gamma\in[\kappa]^{\lambda}$.   As
$\ofamily{\xi}{\kappa}{a_{\xi}}$ is arbitrary,
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of $\frak A$.
}%end of proof of 516S

\leader{516T}{}\cmmnt{ A central problem from the very beginning of
set-theoretic topology concerns the saturation of product spaces.   Here
I describe one of the principal methods of showing that product spaces
have small saturation, in a form adapted to partially ordered sets.

\medskip

\noindent}{\bf Theorem} (a) Let $P$ and $Q$ be partially ordered
sets, and $\kappa$ a cardinal such that
$(\kappa,\sat^{\uparrow}(Q),2)$ is an upwards
precaliber triple of $P$.   Then $\sat^{\uparrow}(P\times Q)\le\kappa$.

(b) Let $\familyiI{P_i}$ be a family of non-empty partially ordered sets
with upwards finite-support product $P$.   Suppose that
$\kappa$ is a regular uncountable cardinal such that
$(\kappa,\kappa,2)$ is an
upwards precaliber triple of every $P_i$.   Then
$\sat^{\uparrow}(P)\le\kappa$.

\proof{{\bf (a)} \Quer\ Otherwise, there is an up-antichain
$\ofamily{\xi}{\kappa}{(p_{\xi},q_{\xi})}$ in $P\times Q$.   Let
$\Gamma\subseteq\kappa$ be a set with cardinal $\sat^{\uparrow}(Q)$ such that
$\{p_{\xi}:\xi\in\Gamma\}$ is upwards-linked.   Then
$\family{\xi}{\Gamma}{q_{\xi}}$ must be an up-antichain in $Q$;  but
this is impossible.\ \Bang

\medskip

{\bf (b)} By 516P, $(\kappa,\kappa,2)$ is an upwards
precaliber triple of $P$.   So $(\kappa,2,2)$ and
$\triplepc{\kappa}{2}{\omega}$ also are (516Ba, 516Bc), and
$\sat^{\uparrow}(P)\le\kappa$ (516Ka).
}%end of proof of 516T

\leader{516U}{}\cmmnt{ It will be useful to be able to quote what
amounts to a simple special case of the above result.

\medskip

\noindent}{\bf Corollary} Let $\frak A$ be a Boolean algebra satisfying
Knaster's condition\cmmnt{ (511Ef)}
and $\frak B$ a ccc Boolean algebra.   Then their free
product $\frak A\otimes\frak B$ is ccc.

\proof{ By 516Ta, inverted,
$(\frak A\setminus\{0\})\times(\frak B\setminus\{0\})$ is downwards-ccc.
But $(a,b)\mapsto a\otimes b$ is an order-preserving bijection between
$(\frak A\setminus\{0\})\times(\frak B\setminus\{0\})$ and an order-dense
(that is, coinitial) subset of $(\frak A\otimes\frak B)\setminus\{0\}$
(315Kb);  so
$(\frak A\otimes\frak B)\setminus\{0\}$ is downwards-ccc
(513Gc, inverted), that is, $\frak A\otimes\frak B$ is ccc.
}%end of proof of 516U

\leader{516V}{}\dvAnew{2015}\cmmnt{ An elementary, but not quite obvious,
fact will turn out to be useful.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be an atomless Boolean algebra
which satisfies Knaster's condition.   Then $\frak A$ has an atomless
order-closed subalgebra with countable Maharam type.
%for 548B

\proof{ For each $a\in\frak A\setminus\{0\}$ choose $a'$ such that
$a'\Bsubseteq a$ and $a'\notin\{0,a\}$.   Define
$\ofamily{\xi}{\omega_1}{\frak B_{\xi}}$ and
$\ofamily{\xi}{\omega_1}{B_{\xi}}$ inductively, as follows.
$\frak B_{\xi}$ will be the order-closed subalgebra of $\frak A$ generated
by $\bigcup_{\eta<\xi}\frak B_{\eta}$;  $B_{\xi}$ will be
$\{b:b$ is an atom of $\frak B_{\xi}\}
\cup\{b':b$ is an atom of $\frak B_{\xi}\}$.
Because $\frak A$ is ccc (511Ef), every $B_{\xi}$ will be countable and
every $\frak B_{\xi}$ will be countably $\tau$-generated.
Of course $\ofamily{\xi}{\omega_1}{\frak B_{\xi}}$ is non-decreasing.

\Quer\ If no $\frak B_{\xi}$ is atomless, then choose an atom
$b_{\xi}$ of $\frak B_{\xi}$
for each $\xi<\omega_1$.   Because $\frak A$ satisfies Knaster's condition,
there is an uncountable set $\Gamma\subseteq\omega_1$ such that
$b_{\xi}\Bcap b_{\eta}\ne 0$ for all $\xi$, $\eta\in\Gamma$.   If
$\xi<\eta$ in $\omega_1$, then $b_{\xi}\in\frak B_{\eta}$ and $b_{\eta}$ is
an atom of $\frak B_{\eta}$ so either $b_{\eta}\Bsubseteq b_{\xi}$ or
$b_{\eta}\Bcap b_{\xi}=0$.   So we see that if $\xi<\eta$ in $\Gamma$, then
$b_{\eta}\Bsubseteq b_{\xi}$.   But we know also that
$b'_{\xi}\in\frak B_{\eta}$, so $b_{\xi}$ is not an atom of
$\frak B_{\eta}$ and $b_{\xi}\Bsetminus b_{\eta}\ne 0$.

This means that if we define $f:\Gamma\to\Gamma$ by taking
$f(\xi)=\min(\Gamma\setminus(\xi+1))$ for $\xi\in\Gamma$,
$\family{\xi}{\Gamma}{b_{\xi}\Bsetminus b_{f(\xi)}}$ will be a disjoint
family in $\frak A\setminus\{0\}$, which is impossible.\ \Bang

So one of the $\frak B_{\xi}$ will serve for $\frak B$.
}%end of proof of 516V

\exercises{\leader{516X}{Basic exercises (a)}
%\spheader 516Xa
Let $(A,R,B)$ be a supported relation, and $n\ge 1$ an
integer.   Show that $n$ is a precaliber of $(A,R,B)$ iff
$\add(A,R,B)>n$.
%516B

\spheader 516Xb\dvAnew{2013}
Let $P$ and $Q$ be partially ordered sets, and $f:P\to Q$ a
surjection such that, for any finite set $I\subseteq P$, $I$ is bounded
above in $P$ iff $f[I]$ is bounded above in $Q$.   Show that $P$ and $Q$
have the same upwards precaliber pairs.
%516D

\spheader 516Xc(i) Show that if $P$ is a partially ordered set and
$\kappa>\cf P$ is an infinite cardinal such that $\cf\kappa$ is an
up-precaliber of $P$, then $\kappa$ is an up-precaliber of $P$.  (ii)
Show that if $\frak A$ is a Boolean algebra and
$\kappa>\pi(\frak A)$ is an infinite cardinal such that $\cf\kappa$ is a
precaliber of $\frak A$, then $\kappa$ is a precaliber of $\frak A$.
%516J

\spheader 516Xd Let $P$ be a partially ordered set and $\kappa$
an infinite cardinal.   Show that $\sat^{\uparrow}(P)\le\kappa$ iff
$(\kappa,\omega)$ is an upwards precaliber pair of
$P$.   \Hint{if $\kappa=\sat^{\uparrow}(P)$ and
$\ofamily{\xi}{\kappa}{p_{\xi}}$ is a family in $P$, choose $\xi_n$,
$q_n$ such that $p_{\xi_i}\le q_n$ for $i\le n$ and $\{\xi:q_n$ is
compatible upwards with $p_{\xi}\}$ is always cofinal with $\kappa$.}
%516K

\spheader 516Xe Let $(X,\frak T)$ be a topological space, $\RO(X)$ its
regular open algebra and $\frak G$ its category algebra (definition:
514I).   (i) Show that any precaliber triple of
$(X,\frak T)$ is also a precaliber triple of $\RO(X)$,
$(\frak T\setminus\{\emptyset\},\ni,X)$ and $\frak G$.   (ii) Show that
if $(X,\frak T)$ is regular, then $(X,\frak T)$ and $\RO(X)$ have the
same precaliber triples.   (iii) Show that if $(X,\frak T)$ is locally
compact and Hausdorff, then $(X,\frak T)$ and $\frak G$ have the same
precaliber triples.
%516M

\spheader 516Xf Let $(P,\le)$ be the totally ordered set $\omega_1$,
$\frak T^{\uparrow}$ its up-topology and $\RO^{\uparrow}(P)$ the regular
open algebra of $(P,\frak T^{\uparrow})$.   Show that
$(\omega_1,\omega_1,\omega_1)$ is a precaliber triple of
$\RO^{\uparrow}(P)$ but not of $(P,\le)$ or $(P,\frak T^{\uparrow})$.
%516G, 516M

\spheader 516Xg Let $\familyiI{\frak A_i}$ be a family of Boolean
algebras and $\frak A$ their free product.   Show that if
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber
triple of every $\frak A_i$ and either $I$ is finite or
$\theta\le\omega$ and $\kappa$ is a regular infinite cardinal, then
$\triplepc{\kappa}{\kappa}{\theta}$ is a precaliber triple of $\frak A$.
%516O

\spheader 516Xh Suppose that $X$ is a topological space and $Y$ is a
dense subset of $X$ and $\theta\le\omega$.   Show that
$\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of $Y$  iff
it is a precaliber triple of $X$.
%516O

\spheader 516Xi Let $X$ be a locally compact Hausdorff space, and
$\kappa$ a precaliber of $X$.   Show that whenever
$\ofamily{\xi}{\kappa}{E_{\xi}}$ is a
non-decreasing family of nowhere dense subsets of $X$ then
$\bigcup_{\xi<\kappa}E_{\xi}$ has empty interior.
%516Q

\spheader 516Xj Prove 516Sa-516Sc without mentioning Stone spaces.
%516S

\spheader 516Xk(i) Let $X$ and $Y$ be topological spaces, and $\kappa$
a cardinal such that $(\kappa,\sat(Y),2)$ is a
precaliber triple of $X$.   Show that
$\sat(X\times Y)\le\kappa$.
(ii) Let $\familyiI{X_i}$ be a family of topological spaces
with product $X$.   Suppose that
$\kappa$ is a regular uncountable
cardinal such that $(\kappa,\kappa,2)$ is a
precaliber triple of every $X_i$.   Show that
$\sat(X)\le\kappa$.
%516T

\spheader 516Xl(i) Let $\frak A$ and $\frak B$ be Boolean algebras, and
$\frak A\otimes\frak B$ their free product.   Suppose that $\kappa$ is a
cardinal such that $(\kappa,\sat(\frak B),2)$ is a
precaliber triple of $\frak A$.   Show that
$\sat(\frak A\otimes\frak B)\le\kappa$.
(ii) Let $\familyiI{\frak A_i}$ be a family of Boolean algebras
with free product $\frak A$.   Suppose that
$\kappa$ is a regular uncountable
cardinal such that $(\kappa,\kappa,2)$ is a
precaliber triple of every $\frak A_i$.   Show that
$\sat(\frak A)\le\kappa$.
%516T

\spheader 516Xm Let $X$ and $Y$ be topological spaces.   Show that if
$\triplepc{\kappa}{\kappa'}{\theta}$ is a precaliber triple of $X$ and
$\triplepc{\kappa'}{\lambda}{\theta}$ is a precaliber triple of $Y$,
then $\triplepc{\kappa}{\lambda}{\theta}$ is a precaliber triple of
$X\times Y$.
%516T

\spheader 516Xn\dvAnew{2015}
Suppose that $\frak A$ and $\frak B$ are Boolean algebras
and that there is a surjective Boolean homomorphism from $\frak A$ to
$\frak B$.   Show that if $\triplepc{\kappa}{\lambda}{\theta}$ is a
precaliber triple of $\frak A$ and $\theta$ is countable, then
$\triplepc{\kappa}{\lambda}{\theta}$ is a
precaliber triple of $\frak B$.
%516S out of order query

\leader{516Y}{Further exercises (a)}
%\spheader 516Ya
\dvAnew{2015} Let $\frak A$ be an atomless
Boolean algebra such that $(\kappa,\kappa,2)$ is a precaliber triple of
$\frak A$ for every regular uncountable cardinal $\kappa$.   Show that
there is a countable $B\subseteq\frak A$ such that for every non-zero
$a\in\frak A$ there is a $b\in\frak B$ such that $a$ meets both $b$ and       $1\Bsetminus b$.
%516S mt51bits
}%end of exercises

\endnotes{
\Notesheader{516}
`Precaliber triples' are visibly complex.   With three cardinals in
action, there is a promise of a powerful method of describing special
features of a partially ordered set or Boolean algebra, but at the same
time a threat of alarming demands on our memory.   In fact none of the
arguments in this section are deep, and they are here mainly for
reference.
Some of the results depend in not-quite-obvious ways on the exact
hypotheses, and it will be useful later to have clear statements to
hand.   In the proofs I have emphasized Galois-Tukey connections
whenever possible;  at the cost of possibly tedious repetititions of
such formulae as $(\frak T\setminus\{\emptyset\},\supseteq,
\frak T\setminus\{\emptyset\})$ naming the supported relations involved,
they can save us the trouble of negotiating the quantifiers in the
definition

\Centerline{$\Forall\ofamily{\xi}{\kappa}{a_{\xi}}\in A^{\kappa}
\Exists\Gamma\in[\kappa]^{\lambda}
\Forall I\in[\Gamma]^{<\theta}\Exists b\in B\ldots$.}

\noindent But of course it is a useful exercise to find proofs from
first principles, not mentioning supported relations and not (for
instance) using Stone spaces to deal with Boolean algebras.

`Supported relations' form a materially more various class of structures
than partially ordered sets, topological spaces or Boolean algebras.
But the constructions already developed in this book (Stone spaces,
regular open algebras, up-topologies) give us functorial relations
between the last three categories which mean that from the point of view
of this section they are nearly the same.   So such results as 516T can
be expected to apply to topological spaces and Boolean algebras as well
(516Xk, 516Xl).   (But note 516Xf.)

Precaliber triples belong with saturation and linking numbers as
parameters describing the `breadth' of a topological space or Boolean
algebra;  see {\smc Comfort \& Negrepontis 82}.   In the first place,
they address a classic problem:  when is the product of ccc topological
spaces ccc?   (This is the case $\kappa=\omega_1$ of 516Xk.)   But with
the exception of saturation, there do not
appear to be simple connexions between precalibers and the cardinal
functions we have looked at so far.   Precalibers seem to correspond to
new features of the structures considered here.   When we come to look
at the most important objects of measure theory (in particular, measure
algebras), we shall find that their precalibers are relatively fluid;  I
mean that while cellularity, Maharam types and many linking numbers, for
instance, are determined by simple formulae in ZFC, precalibers are not.
}%end of notes

\discrpage

\frfilename{mt517.tex}
\versiondate{14.11.14}
\copyrightdate{2003}


\def\chaptername{Cardinal functions}
\def\sectionname{Martin numbers}

\newsection{517}

I devote a section to the study of `Martin numbers' of partially ordered
sets and Boolean algebras.   Like additivity and cofinality they
enable us to frame as theorems of ZFC some important arguments which
were first used in special models of set theory, and to pose
challenging questions on the relationships between classical structures
in analysis.   I begin with some general remarks on the Martin numbers
of partially ordered sets (517A-517E);  %517A 517B 517C 517D 517E
most of these are perfectly elementary but the equivalent conditions of
517B, in particular, are useful and not all obvious.   Much of the
importance of Martin numbers comes from their effect on precalibers
(517F, 517H) and hence on saturation of products (517G).   The same
ideas can be expressed in terms of Boolean algebras, with no surprises
(517I).   I have not set out a definition of `Martin number' for a
topological space, but the Nov\'ak number of a locally compact Hausdorff
space is closely related to the Martin numbers of its regular open
algebra and its algebra of open-and-closed sets (517J-517K).
Consequently we have connexions between the Martin number and the weak
distributivity of a Boolean algebra (517L).   A striking fact, which
will have a prominent role in the next chapter, is that
non-trivial countable partially ordered sets all have the same Martin
number $\frakmctbl$ (517P).

\leader{517A}{Proposition} For any partially ordered set $P$,
$\frak m^{\uparrow}(P)\ge\omega_1$.

\proof{ If $\Cal Q$ is a countable family of cofinal subsets of $P$ and
$p_0\in P$, let $\sequencen{Q_n}$ be a sequence running over
$\Cal Q\cup\{P\}$, and choose $\langle p_n\rangle_{n\ge 1}$ inductively
such that $p_{n+1}\ge p_n$ and $p_{n+1}\in Q_n$ for every $n\in\Bbb N$.
Then $\{p_n:n\in\Bbb N\}$ is an upwards-linked subset of $P$ meeting
every member of $\Cal Q$.}

\vleader{72pt}{517B}{Lemma}
Let $P$ be a partially ordered set, and $\kappa$ a
cardinal.   Then the following are equiveridical:

(i) $\kappa<\frak m^{\uparrow}(P)$;

(ii) whenever $p_0\in P$ and $\Cal Q$ is a family of up-open cofinal
subsets of $P$ with $\#(\Cal Q)\le\kappa$, there is an upwards-linked
subset of $P$ which contains $p_0$ and meets every member of $\Cal Q$;

(iii) whenever $p_0\in P$ and $\Cal A$ is a family of maximal
up-antichains in $P$ with $\#(\Cal A)\le\kappa$, there is an
upwards-linked subset of $P$ which contains $p_0$ and meets every member
of $\Cal A$;

(iv) whenever $p_0\in P$ and $\Cal Q$ is a family of cofinal subsets of
$P$ with $\#(\Cal Q)\le\kappa$, there is an upwards-directed subset of
$P$ which contains $p_0$ and meets every member of $\Cal Q$;

(v) whenever $p_0\in P$ and $\Cal Q$ is a family of up-open cofinal
subsets of $P$ with $\#(\Cal Q)\le\kappa$, there is an upwards-directed
subset of $P$ which contains $p_0$ and meets every member of $\Cal Q$;

(vi) whenever $p_0\in P$ and $\Cal A$ is a family of maximal
up-antichains in $P$ with $\#(\Cal A)\le\kappa$, there is an
upwards-directed subset of $P$ which contains $p_0$ and meets every
member of $\Cal A$.

\proof{{\bf (a)} Most of the circuit is elementary.

\medskip

\quad{\bf(vi)$\Rightarrow$(iv)}
because every cofinal subset of $P$ includes a
maximal up-antichain (513Aa).

\medskip

\quad{\bf(iv)$\Rightarrow$(v)$\Rightarrow$(ii)} are trivial.

\medskip

\quad{\bf(ii)$\Rightarrow$(i)} Assuming (ii), let $\Cal Q$ be a family of
cofinal subsets of $P$ with $\#(\Cal Q)\le\kappa$.   For each $Q\in\Cal Q$,
$U_Q=\bigcup_{q\in Q}\coint{q,\infty}$ is up-open and cofinal with $P$
(513Ab).
If $p_0\in P$, (ii) tells us that there is an upwards-linked subset $R_0$ of
$P$ containing $p_0$ and meeting $U_Q$ for every $Q\in\Cal Q$.
Set $R=\bigcup_{p\in R_0}\ocint{-\infty,p}$;  then $R$ is an upwards-linked
subset of $P$ containing $p_0$ and meeting every member of $\Cal Q$.   As
$\Cal Q$ and $p_0$ are arbitrary, (i) is true.

\medskip

\quad{\bf(i)$\Rightarrow$(iii)} Assuming (i), let $\Cal A$ be a family of
maximal up-antichains in $P$ with $\#(\Cal A)\le\kappa$.   For each
$A\in\Cal A$, $U_A=\bigcup_{p\in A}\coint{p,\infty}$ is cofinal with $P$.
So (i)
tells us that if $p_0\in P$ there is an upwards-linked subset $R_0$ of $P$
containing $p_0$ and meeting $U_A$ for every $A\in\Cal A$.   As just
above, set
$R=\bigcup_{p\in R_0}\ocint{-\infty,p}$;  then $R$ is upwards-linked, contains
$p_0$ and meets every member of $\Cal A$.   As $\Cal A$ and $p_0$ are arbitrary,
(iii) is true.

\medskip

{\bf (b)} So we are left with (iii)$\Rightarrow$(vi).
Assume (iii), and take $p_0\in P$
and a family $\Cal A$ of maximal up-antichains in $P$ with
$\#(\Cal A)\le\kappa$.   Let $\Cal C$ be the set of all maximal up-antichains in
$P$.   For $A\in\Cal C$, set
$U_A=\bigcup_{q\in A}\coint{q,\infty}$.   Then $U_A$ is cofinal with
$P$.   Consequently $U_A\cap U_B$ is cofinal with
$P$ for any $A$, $B\in\Cal C$, because if $p\in P$ there are $q\in U_A$
and $r\in U_B$ such that $p\le q\le r$, and now $r\in U_A\cap U_B$.   It
follows that $U_A\cap U_B$ includes a maximal up-antichain $D(A,B)$.

Take any $A_0\in\Cal C$ such that $p_0\in A_0$.
Let $\Cal A^*\subseteq\Cal C$ be such that
$\{A_0\}\cup\Cal A\subseteq\Cal A^*$,
$D(A,B)\in\Cal A^*$ for all $A$, $B\in\Cal A^*$, and
$\#(\Cal A^*)\le\max(\omega,\kappa)$ (5A1Gb).
Then there is an upwards-linked subset
$R_0$ of $P$ containing $p_0$ and meeting every member of $\Cal A^*$.   \Prf\ If
$\#(\Cal A^*)\le\kappa$, this is immediate from (iii);  if
$\#(\Cal A^*)\le\omega$,
it is because $\omega<\frak m^{\uparrow}(P)$, by 517A, and
(i)$\Rightarrow$(iii).\ \Qed

Set $R=R_0\cap\bigcup\Cal A^*$.   Then $R$ contains $p_0$ (because
$p_0\in R_0\cap A_0$) and $R$ meets every member of $\Cal A$.   Also $R$ is
upwards-directed.   \Prf\ If $p$, $q\in R$, take $A$, $B\in\Cal A^*$ such that
$p\in A$ and $q\in B$.   Then $D(A,B)\in\Cal A^*$, so
there is an $r\in R_0\cap D(A,B)$, and $r\in R$.
As $r\in U_A\cap U_B$, there must be $p'\in A$ and $q'\in B$ such that
$p'\le r$ and $q'\le r$.   But $R_0$ is upwards-linked, so

\Centerline{$\emptyset\ne\coint{p,\infty}\cap\coint{r,\infty}\subseteq
\coint{p,\infty}\cap\coint{p',\infty}$;}

\noindent as $A$ is an up-antichain, $p=p'$.   Similarly, $q=q'$ and
$r\in R$ is an upper bound of $\{p,q\}$.   As $p$ and $q$ are arbitrary,
$R$ is upwards-directed.\ \Qed

So we have a set $R$ of the kind required by (vi).
}%end of proof of 517B

\leaveitout{
elementary-submodel version of (iii)=>(vi):

Let $M$ be a suitable elementary submodel with cardinal $\max(\omega,\kappa)$.
Let $R_1$ be an upwards-linked subset of $P$ containing $p_0$ and meeting every
maximal antichain belonging to $M$;  let $R$ be the set of elements of $R_1$
which belong to maximal antichains in $M$.   There is a maximal antichain in $M$
containing $p_0$, so $p_0\in R$.   If $p$, $q\in R$, let $A$, $B$ be maximal
antichains in $M$ containing $p$, $q$.   Then there is a maximal antichain
$C$ in
$M$ refining both $A$ and $B$;  let $r$ be the member of $R_1\cap C$;  then $r$
dominates both $p$ and $q$ and belongs to $R$, so $R$ is upwards-directed.
}%end of leaveitout

\leader{517C}{Lemma} Let $P_0$ and $P_1$ be partially ordered sets, and
suppose that there is a relation $S\subseteq P_0\times P_1$ such that
$S[P_0]$ is cofinal with $P_1$, $S^{-1}[Q]$ is cofinal with $P_0$ for
every cofinal $Q\subseteq P_1$, and $S[R]$ is upwards-linked in $P_1$
for every upwards-linked $R\subseteq P_0$.   Then
$\frak m^{\uparrow}(P_1)\ge\frak m^{\uparrow}(P_0)$.

\proof{ Suppose that $p_1\in P_1$ and that $\Cal Q$ is a family of
cofinal subsets of $P_1$ with $\#(\Cal Q)<\frak m^{\uparrow}(P_0)$.
Then there is a pair $(p_0,p'_1)\in S$ such that $p'_1\ge p_1$.   Now
$S^{-1}[Q]$ is cofinal with $P_0$ for every $Q\in\Cal Q$, so there is an
upwards-linked $R\subseteq P_0$ containing $p_0$ and meeting $S^{-1}[Q]$
for every $Q\in\Cal Q$.   In this case $p'_1\in S[R]$ and
$S[R]$ is upwards-linked, so $\{p_1\}\cup S[R]$ is an
upwards-linked subset of $P_1$ containing $p_1$ and meeting every member
of $\Cal Q$.   As $p_1$ and $\Cal Q$ are arbitrary,
$\frak m^{\uparrow}(P_1)\ge\frak m^{\uparrow}(P_0)$.
}%end of proof of 517C

\vleader{48pt}{517D}{Proposition} (a)
If $P$ is a partially ordered set and $Q$ is a cofinal subset of $P$, then
$\frak m^{\uparrow}(P)=\frak m^{\uparrow}(Q)$.

(b) If $P$ is any partially ordered set and $\RO^{\uparrow}(P)$ is its
regular open algebra when it is given its up-topology, then
$\frak m^{\uparrow}(P)=\frak m(\RO^{\uparrow}(P))$.

(c) If $P$ is a partially ordered set and $p_0\in P$, then
$\frak m^{\uparrow}(\coint{p_0,\infty})\ge\frak m^{\uparrow}(P)$.

\proof{{\bf (a)} Let $P_0$, $P_1$ be cofinal subsets of $P$, and set
$S=\{(p_0,p_1):p_0\in P_0$, $p_1\in P_1$, $p_0\ge p_1\}$.   Then $S$
satisfies the conditions of 517C, so
$\frak m^{\uparrow}(P_1)\ge\frak m^{\uparrow}(P_0)$.   It follows at
once that all cofinal subsets of $P$, including $P$ itself,
have the same Martin number.

\medskip

{\bf (b)(i)} Setting $S=\{(p,G):p\in G\in\RO^{\uparrow}(P)\}$, $S$
satisfies the conditions of 517C with $P_0=(P,\le)$ and
$P_1=(\RO^{\uparrow}(P)^+,\supseteq)$, so
$\frak m(\RO^{\uparrow}(P))\ge\frak m^{\uparrow}(P)$.

\medskip

\quad{\bf (ii)} Setting
$S'=\{(G,p):G\in\RO^{\uparrow}(P)^+,\,
p\in P,\,G\subseteq\overline{\coint{p,\infty}}\}$,
$S'$ satisfies the conditions of 517C with
$P_0=(\RO^{\uparrow}(P)^+,\supseteq)$ and $P_1=(P,\le)$, so
$\frak m^{\uparrow}(P)\ge\frak m(\RO^{\uparrow}(P))$.

\medskip

{\bf (c)} Let $\Cal Q$ be a family of upwards-cofinal subsets of
$\coint{p_0,\infty}$ with $\#(\Cal Q)<\frak m^{\uparrow}(P)$, and
$p_1\in\coint{p_0,\infty}$.   For each $Q\in\Cal Q$, set
$Q'=Q\cup\{p:p\in P,\,
\coint{p,\infty}\cap\coint{p_0,\infty}=\emptyset\}$.   Then every $Q'$
is cofinal with $P$.   So there is an upwards-linked set $R\subseteq P$
containing $p_1$ and meeting $Q'$ for every $Q\in\Cal Q$.   If
$Q\in\Cal Q$ and $r\in R\cap Q'$, then
$\coint{r,\infty}\cap\coint{p_0,\infty}
\supseteq\coint{r,\infty}\cap\coint{p_1,\infty}$ is non-empty, so
$r\in Q$.   Thus $R\cap\coint{p_0,\infty}$ is an upwards-linked subset
of $\coint{p_0,\infty}$ containing $p_1$ and meeting every member of
$\Cal Q$.  As $\Cal Q$ and $p_1$ are arbitrary,
$\frak m^{\uparrow}(\coint{p_0,\infty})\ge\frak m^{\uparrow}(P)$.
}%end of proof of 517D

\leader{517E}{Corollary} Let $P$ be a partially ordered set such that
$\frak m^{\uparrow}(P)$ is not $\infty$.   Then
$\frak m^{\uparrow}(P)\le 2^{\cf P}$.

\proof{ Let $Q_0$ be a
cofinal subset of $P$ with $\#(Q_0)=\cf P$.   Then
$\frak m^{\uparrow}(Q_0)=\frak m^{\uparrow}(P)<\infty$.   So there are
$q_0\in Q_0$ and a family
$\Cal Q$ of cofinal subsets of $Q_0$ such that no upwards-linked subset
of $Q_0$ containing $p_0$ can meet every member of $\Cal Q$.   Now

\Centerline{$\frak m^{\uparrow}(P)
=\frak m^{\uparrow}(Q_0)\le\#(\Cal Q)\le 2^{\#(Q_0)}=2^{\cf P}$.}
}%end of proof of 517E

\leader{517F}{Proposition} Let $P$ be a non-empty partially ordered set.

(a) Suppose that $\kappa$ and $\lambda$ are cardinals such that
$\sat^{\uparrow}(P)\le\cf\kappa$, $\lambda\le\kappa$ and
$\lambda<\frak m^{\uparrow}(P)$.   Then $(\kappa,\lambda)$ is an upwards
precaliber pair of $P$.

(b) In particular, if
$\sat^{\uparrow}(P)\le\cf\kappa\le\kappa<\frak m^{\uparrow}(P)$ then
$\kappa$ is an up-precaliber of $P$.

\proof{{\bf (a)} Since $P$ is not empty, $\sat^{\uparrow}(P)\ge 2$ and
$\kappa$ is infinite.
Write $\theta$ for $\sat^{\uparrow}(P)$.   Let
$\ofamily{\xi}{\kappa}{p_{\xi}}$ be a family in $P$.   For
$I\subseteq\kappa$, set

\Centerline{$U_I=\bigcup_{\xi\in I}\coint{p_{\xi},\infty}$,
\quad$V_I=\{q:q\in P,\,
  \coint{q,\infty}\cap\coint{p_{\xi},\infty}=\emptyset$ for
every $\xi\in I\}$.}

\noindent Then for every $J\subseteq\kappa$ there is an
$I\in[J]^{<\theta}$ such that $V_J\cup U_I$ is cofinal with $P$.   \Prf\
$V_J\cup U_J$ is cofinal with $P$, so there is a maximal up-antichain
$A\subseteq V_J\cup U_J$.   Now
$\#(A\cap U_J)<\sat^{\uparrow}(P)=\theta$, so there is a set
$I\in[J]^{<\theta}$ such that $A\cap U_J\subseteq U_I$, and
$A\subseteq V_J\cup U_I$.   Now
$V_J\cup U_I\supseteq\bigcup_{q\in A}\coint{q,\infty}$, so is cofinal
with $P$.\ \Qed

Next, $Q=\bigcup\{V_{\kappa\setminus I}:I\in[\kappa]^{<\kappa}\}$ is not
cofinal with $P$.   \Prf\Quer\ If it were, there would be a maximal
up-antichain $A\subseteq Q$.   For each $q\in A$, let
$I_q\in[\kappa]^{<\kappa}$ be such that $q\in V_{\kappa\setminus I_q}$.
Because $\#(A)<\theta\le\cf\kappa$,
$\bigcup_{q\in A}I_q\ne\kappa$, and there is a
$\xi\in\kappa\setminus\bigcup_{q\in A}I_q$.   But now
$\coint{q,\infty}\cap\coint{p_{\xi},\infty}=\emptyset$ for every
$q\in A$, and $A$ is not a maximal antichain.\ \Bang\Qed

Let $q_0\in P$ be such that $Q\cap\coint{q_0,\infty}=\emptyset$.
Choose
$\ofamily{\xi}{\lambda}{I_{\xi}}$ inductively in such a way that,
writing $J_{\xi}=\kappa\setminus\bigcup_{\eta<\xi}I_{\eta}$,
$I_{\xi}\in[J_{\xi}]^{<\theta}$ and
$Q_{\xi}=V_{J_{\xi}}\cup U_{I_{\xi}}$ is cofinal with $P$ for every
$\xi<\lambda$.   Because $\lambda<\frak m^{\uparrow}(P)$, there is an
upwards-directed set $R\subseteq P$ containing $q_0$ and meeting every
$Q_{\xi}$.   Set $\Gamma
=\{\eta:\eta<\kappa,\,R\cap\coint{p_{\eta},\infty}\ne\emptyset\}$;  then
$\{p_{\eta}:\eta\in\Gamma\}$ is upwards-centered.   Next,
$\kappa\setminus J_{\xi}=\bigcup_{\eta<\xi}I_{\eta}$ has cardinal less
than $\kappa$ for every $\xi<\lambda$.   \Prf\ If $\theta=\kappa$ or
$\kappa=\omega$, this
is because $\#(\xi)<\kappa$ and $\#(I_{\eta})<\kappa$ for every
$\eta<\xi$ and $\kappa$ is regular (use 513Bb).   Otherwise it's
because $\max(\omega,\theta,\#(\xi))<\kappa$.\ \Qed

This means that $V_{J_{\xi}}\cap\coint{q_0,\infty}=\emptyset$ and
$R\cap V_{J_{\xi}}$
must be empty, for every $\xi<\lambda$.   We must therefore have
$R\cap U_{I_{\xi}}\ne\emptyset$ for each $\xi<\lambda$, so that
$\Gamma\cap I_{\xi}\ne\emptyset$;  as $\ofamily{\xi}{\lambda}{I_{\xi}}$
is disjoint, $\#(\Gamma)\ge\lambda$.

As $\ofamily{\xi}{\kappa}{p_{\xi}}$ is arbitrary, $(\kappa,\lambda)$ is
an upwards precaliber pair of $P$.

\medskip

{\bf (b)} This follows at once, setting $\lambda=\kappa$.
}%end of proof of 517F

\vleader{48pt}{517G}{Corollary} (a)
If $P$ and $Q$ are partially ordered sets
and $\sat^{\uparrow}(Q)<\frak m^{\uparrow}(P)$, then
$\sat^{\uparrow}(P\times Q)$ is at most
$\max(\omega,\sat^{\uparrow}(P),\sat^{\uparrow}(Q))$.

(b) Let $\familyiI{P_i}$ be a family of non-empty partially ordered sets
with upwards finite-support product $P$.   Let $\kappa$ be a regular
uncountable cardinal such that
$\sat^{\uparrow}(P_i)\le\kappa<\frak m^{\uparrow}(P_i)$
for every $i\in I$.   Then $\sat^{\uparrow}(P)\le\kappa$.

\proof{{\bf (a)} Set $\lambda=\sat^{\uparrow}(Q)$,
$\kappa=\max(\omega,\sat^{\uparrow}(P),\sat^{\uparrow}(Q))$.   Then
$\kappa$ is regular (513Bb again), $\lambda\le\kappa$ and
$\lambda<\frak m^{\uparrow}(P)$, so $(\kappa,\lambda)$ is an upwards
precaliber pair of $P$ and $(\kappa,\lambda,2)$ is an upwards
precaliber triple of $P$.   By 516Ta,
$\sat^{\uparrow}(P\times Q)\le\kappa$.

\medskip

{\bf (b)} By 517Fb, $\kappa$ is an up-precaliber of $P_i$ for every $i$,
so $(\kappa,\kappa,2)$ is an upwards precaliber triple of every $P_i$, and
we can use 516Tb.
}%end of proof of 517G

\leader{517H}{Proposition} Let $P$ be a non-empty partially ordered set,
and let $P^*$ be the upwards finite-support product of the family
$\sequencen{P_n}$ where $P_n=P$ for every $n$.   Suppose that
$\kappa<\frak m^{\uparrow}(P^*)$.

(a) Every subset of $P$ with $\kappa$ or fewer members can be covered by
a sequence of upwards-directed sets.

(b) In particular, if $\kappa$ is uncountable then $(\kappa,\lambda)$ is
an upwards precaliber pair of $P$ for every $\lambda<\kappa$, and if
$\kappa$ has uncountable cofinality then $\kappa$ is an up-precaliber of
$P$.

\proof{{\bf (a)} Let $A\in[P]^{<\kappa}$.   For each $p\in A$, set
$Q_p=\{q:q\in P^*,\Exists\ n\in\dom q,\,q(n)=p\}$;  then $Q_p$ is
cofinal with $P^*$.   So there is an upwards-directed set
$R\subseteq P^*$ such that
$R\cap Q_p\ne\emptyset$ for every $p\in A$.   For each $n\in\Bbb N$, set
$R_n=\{q(n):q\in R,\,n\in\dom q\}$.   Then
$A\subseteq\bigcup_{n\in\Bbb N}R_n$.   If $n\in\Bbb N$ and $r$,
$r'\in R_n$, there are $q$, $q'\in R$ such that $q(n)=r$ and $q'(n)=r'$.
Now there is a $q''\in R$ such that $q''\ge q$ and $q''\ge q'$, in which
case $q''(n)$ belongs to $R_n\cap\coint{r,\infty}\cap\coint{r',\infty}$.
As $r$ and $r'$ are arbitrary, $R_n$ is upwards-directed.   Thus
$\sequencen{R_n}$ is an appropriate sequence.

\medskip

{\bf (b)} Suppose that $\kappa$ is uncountable and that either
$\lambda<\kappa$ or $\cf\kappa>\omega$ and $\lambda=\kappa$.   Let
$\ofamily{\xi}{\kappa}{p_{\xi}}$ be any family in $P$.   Let
$\sequencen{R_n}$ be a sequence of upwards-directed sets covering
$\{p_{\xi}:\xi<\kappa\}$, and for each $n\in\Bbb N$ set
$\Gamma_n=\{\xi:p_{\xi}\in R_n\}$.   There must be some $n$ such that
$\#(\Gamma_n)\ge\lambda$, and $\{p_{\xi}:\xi\in\Gamma_n\}$ is
upwards-centered.
}%end of proof of 517H

\leader{517I}{Proposition} Let $\frak A$ be a Boolean algebra.

(a) If $\frak B$ is a regularly embedded subalgebra of $\frak A$, then
$\frak m(\frak B)\ge\frak m(\frak A)$.

(b) If $\frak B$ is a principal ideal of $\frak A$, then
$\frak m(\frak B)\ge\frak m(\frak A)$.

(c) If $\frak B$ is an order-dense subalgebra of $\frak A$, then
$\frak m(\frak B)=\frak m(\frak A)$.

(d) If $\widehat{\frak A}$ is the Dedekind completion of $\frak A$, then
$\frak m(\widehat{\frak A})=\frak m(\frak A)$.

(e) If $D\subseteq\frak A$ is non-empty and $\sup D=1$, then
$\frak m(\frak A)=\min_{d\in D}\frak m(\frak A_d)$, where $\frak A_d$ is
the principal ideal generated by $d$.

(f) If $\frak A$ is the simple product of a non-empty family
$\familyiI{\frak A_i}$ of Boolean algebras, then
$\frak m(\frak A)=\min_{i\in I}\frak m(\frak A_i)$.

(g) Suppose that $\kappa$ and $\lambda$ are infinite cardinals such that
$\sat(\frak A)\le\cf\kappa$, $\lambda\le\kappa$ and
$\lambda<\frak m(\frak A)$.   Then $(\kappa,\lambda)$ is a
precaliber pair of $\frak A$.

\proof{{\bf (a)} Setting
$S=\{(a,b):a\in\frak A^+,\,a\Bsubseteq b\in\frak B\}$,
$S$ satisfies the conditions of 517C for
$P_0=(\frak A^+,\Bsupseteqshort)$
and $P_1=(\frak B^+,\Bsupseteqshort)$.  \Prf\ The
only non-trivial part is the check that if $Q$ is coinitial with $\frak B^+$
then $S^{-1}[Q]$ is coinitial with $\frak A^+$.   But $\sup Q=1$ in $\frak B$;
as $\frak B$ is regularly embedded in $\frak A$, $\sup Q=1$ in $\frak A$.  So if
$a\in\frak A^+$, there is a $b\in Q$ such that $a\Bcap b\ne 0$, and now
$a\Bcap b\in S^{-1}[Q]$ and $a\Bcap b\Bsubseteq a$.   As $a$ is arbitrary,
$S^{-1}[Q]$ is coinitial with $\frak A^+$.\ \QeD\  So
$\frak m(\frak B)\ge\frak m(\frak A)$.

\medskip

{\bf (b)} If $\frak A_a$ is the principal ideal generated by
$a\in\frak A^+=\frak A\setminus\{0\}$, we have

$$\eqalignno{\frak m(\frak A)
&=\frak m^{\downarrow}(\frak A^+)\le\frak m^{\downarrow}(\ocint{0,a})\cr
\displaycause{by 517Dc, inverted}
&=\frak m(\frak A_a).\cr}$$

\noindent On my definitions the trivial ideal $\{0\}$ also is a
principal ideal, but of course
$\frak m(\{0\})=\infty\ge\frak m(\frak A)$.

\medskip

{\bf (c)} Apply 517Da (inverted) to $\frak A^+$ and $\frak B^+$.

\medskip

{\bf (d)} This follows from (c), because $\frak A$ is order-dense in
$\widehat{\frak A}$.

\medskip

{\bf (e)} By (b), $\frak m(\frak A)\le\frak m(\frak A_d)$ for every $d$.
In the other direction, let $\Cal Q$ be a family of coinitial subsets of
$\frak A^+$ such that $\#(\Cal Q)<\min_{d\in D}\frak m(\frak A_d)$, and
take any $c\in\frak A^+$.   Then there is a $d\in D$ such that
$c\Bcap d\ne 0$.   For $Q\in\Cal Q$ set
$Q'=\{a\Bcap d:a\in Q\}\setminus\{0\}$;  then $Q'$ is coinitial with
$\frak A_d^+$.   Since $\#(\{Q':Q\in\Cal Q\})<\frak m(\frak A_d)$, there
is a downwards-linked set $R'\subseteq\frak A_d^+$ meeting every $Q'$
and containing $c\Bcap d$.   Set $R=\{a:a\in\frak A,\,a\Bcap d\in R'\}$;
then $R$ is a downwards-linked subset of $\frak A^+$ meeting every
member of $\Cal Q$ and containing $c$.   As $c$ and $\Cal Q$ are
arbitrary, $\frak m(\frak A)\ge\min_{d\in D}\frak m(\frak A_d)$.

\medskip

{\bf (f)} This is, in effect, a special case of (e), since we can
identify the $\frak A_i$ with principal ideals of $\frak A$ (315E).

\medskip

{\bf (g)} Apply 517Fa (inverted) to $\frak A^+$.
}%end of proof of 517I

\leader{517J}{Proposition} Let $X$ be a locally compact Hausdorff space,
and $\kappa$ a cardinal.   Then the following are equiveridical:

(i) $\kappa<\frak m(\RO(X))$, where $\RO(X)$ is the regular open algebra
of $X$;

(ii) $X\cap\bigcap\Cal G$ is dense in $X$ whenever $\Cal G$ is a family
of dense open subsets of $X$ and $\#(\Cal G)\le\kappa$;

(iii) $\kappa<n(H)$ for every non-empty open set $H\subseteq X$.

\proof{{\bf (i)$\Rightarrow$(iii)} Suppose that
$\kappa<\frak m(\RO(X))$.   Let $H\subseteq X$ be a non-empty open set
and $\Cal E$ a family of nowhere dense subsets of $H$ with
$\#(\Cal E)\le\kappa$.   Note that every member of $\Cal E$ is nowhere
dense in $X$.   Because $X$ is locally compact and regular, we have a
non-empty regular open set $H_0$ such that $K=\overline{H}_0$ is compact
and included in $H$.   For each $E\in\Cal E$, set
$\Cal G_E=\{G:G\in\RO(X)^+,\,\overline{G}\cap E=\emptyset\}$;  then
$\Cal G_E$ is coinitial with $\RO(X)^+$.   Because
$\kappa<\frak m^{\downarrow}(\RO(X)^+)$, there is a
centered $\Cal G\subseteq\RO(X)^+$ containing $H_0$ and meeting every
$\Cal G_E$.   But in this case
$\{K\}\cup\{\overline{G}:G\in\Cal G\}$ is a
family of closed sets in $X$ containing the compact set $K$ and with the
finite intersection property, so has non-empty intersection $F$, which
is included in $H\setminus\bigcup\Cal E$.   As $H$ and $\Cal E$ are
arbitrary, (iii) is true.

\medskip

{\bf (iii)$\Rightarrow$(ii)} This is easy.   If (iii) is true, $\Cal G$
is a family of dense open subsets of $X$ with $\#(\Cal G)\le\kappa$, and
$H\subseteq X$ is a non-empty open set, then
$\Cal E=\{H\setminus G:G\in\Cal G\}$ is a family of nowhere dense
subsets of $H$, so cannot cover $H$, and
$H\cap\bigcap\Cal G\ne\emptyset$.   As $\Cal G$ and $H$ are arbitrary,
(ii) is true.

\medskip

{\bf (ii)$\Rightarrow$(i)} Suppose that (ii) is true.   Take
$H\in\RO(X)^+$ and a family $\frak G$ of coinitial subsets of $\RO(X)^+$
with $\#(\frak G)\le\kappa$.   For each $\Cal G\in\frak G$,
$\bigcup\Cal G$ is a dense open subset of $X$.   Accordingly there is a
point $x\in H\cap\bigcap_{\Cal G\in\frak G}\bigcup\Cal G$.   Set
$R=\{G:G\in\RO(X),\,x\in G\}$.   Then $R$ is a
downwards-linked subset of $\RO(X)^+$ containing $H$ and meeting every
member of $\frak G$.   As $H$ and $\frak G$ are arbitrary,
$\kappa<\frak m(\RO(X))$.
}%end of proof of 517J

\leader{517K}{Corollary} Let $\frak A$ be a Boolean algebra with Stone
space $Z$.

(a) $\frak m(\frak A)=\frak m(\RO(Z))$.

(b) For any cardinal $\kappa$, the following are equiveridical:

\inset{(i) $\kappa<\frak m(\frak A)$;

(ii) $Z\cap\bigcap\Cal G$ is dense in $Z$ whenever $\Cal G$ is a family
of dense open subsets of $Z$ and $\#(\Cal G)\le\kappa$;

(iii) $\kappa<n(H)$ for every non-empty open set $H\subseteq Z$.}

\proof{{\bf (a)} $\frak A$ is isomorphic to the algebra of
open-and-closed subsets of $Z$, which is an order-dense subalgebra of
$\RO(Z)$ (314Ta).   So $\frak m(\frak A)=\frak m(\RO(Z))$ by 517Ic.

\medskip

{\bf (b)} now follows from 517J.
}%end of proof of 517K

\leader{517L}{}\cmmnt{ These identifications make the following
results easy.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Boolean algebra.

(a) $\wdistr(\frak A)\le\frak m(\frak A)$.

(b) If $\wdistr(\frak A)$ is a precaliber of $\frak A$ then
$\wdistr(\frak A)<\frak m(\frak A)$.

\proof{{\bf (a)} Let $Z$ be the Stone space of $\frak A$ and $\CalNwd$
the ideal of nowhere dense subsets of $Z$.   Then
$\wdistr(\frak A)=\add\CalNwd$ (514Be), while $\frak m(\frak A)$ is the
least cardinal of any subset of $\CalNwd$ covering a non-empty open
subset of $Z$, if there is one (517Kb).   Since no non-empty open
subset of $Z$ can belong to $\CalNwd$,
$\wdistr(\frak A)\le\frak m(\frak A)$.

\medskip

{\bf (b)} Because $\wdistr(\frak A)=\add\CalNwd\ge\omega$, it is a
regular infinite cardinal (513C(a-i)).   If
$\ofamily{\xi}{\wdistr(\frak A)}{G_{\xi}}$ is a family of dense open
subsets of $Z$, and $H\subseteq Z$ is open and not empty, then
$H_{\xi}=H\cap\interior(\bigcap_{\eta<\xi}G_{\eta})$ is
non-empty for every $\xi<\wdistr(\frak A)$.
So if also $\wdistr(\frak A)$ is a
precaliber of $\frak A$ and therefore of $Z$ (516Ha),
there is a point $z$ of $Z$ such that
$\{\xi:z\in H_{\xi}\}$ has cardinal $\wdistr(\frak A)$ (516Qb) and is
therefore cofinal with $\wdistr(\frak A)$;  which means that
$z\in H\cap\bigcap_{\xi<\wdistr(\frak A)}G_{\xi}$.   Thus
$n(H)>\wdistr(\frak A)$;  as $H$ is arbitrary,
$\frak m(\frak A)>\wdistr(\frak A)$, by 517Kb again.
}%end of proof of 517L

\leader{517M}{}\cmmnt{ It is worth extracting an idea from the proofs
just above as a general result.

\medskip

\noindent}{\bf Proposition} Let $X$ be any topological space.   Then the
Nov\'ak number $n(X)$ of $X$\cmmnt{ (5A4Af)} is at most
$\sup\{\frak m(\RO(G)):G\subseteq X$ is open and not empty$\}$, where
$\RO(G)$ is the regular open algebra of $G$.

\proof{{\bf (a)} If there is a non-empty open subset $G$ of $X$ such
that $\frak m(\RO(G))=\infty$, the result is trivial;  suppose
otherwise.   Set
$\kappa=\sup\{\frak m(\RO(G)):G\subseteq X$ is open and not empty$\}$.
Then for any non-empty open set $G\subseteq X$ there is a family
$\ofamily{\xi}{\kappa}{E_{\xi}}$ of nowhere dense sets such that
$\#(\Cal E)\le\kappa$ and
$G\cap\interior(\bigcup_{\xi<\kappa}E_{\xi})\ne\emptyset$.   \Prf\ We
have a family $\ofamily{\xi}{\kappa}{\Cal Q_{\xi}}$ of order-dense
subsets of $\RO(G)^+$ and an $H\in\RO(G)^+$ such that there is no
downwards-directed family in $\RO(G)^+$ containing $H$ and meeting every
$\Cal Q_{\xi}$.   Set
$E_{\xi}=G\setminus\bigcup\Cal Q_{\xi}$ for each $\xi$;
then $E_{\xi}$ must be nowhere dense in the topological sense because
any open set meeting $G$ at all must meet some member of $\Cal Q_{\xi}$.
If $x\in H$, then $\Cal R=\{U:U\in\RO(G)$, $x\in U\}$ is a
downwards-directed family in $\RO(G)^+$ containing $H$, so does not meet
every $\Cal Q_{\xi}$, and there must be a $\xi<\kappa$ such that
$x\notin\bigcup\Cal Q_{\xi}$, that is, $x\in E_{\xi}$.   As $x$ is
arbitrary, $G\cap\interior(\bigcup_{\xi<\kappa}E_{\xi})\supseteq H$ is
not empty.\ \Qed

\medskip

{\bf (b)} Let $\familyiI{H_i}$ be a maximal disjoint family of non-empty
open sets in $X$ such that every $H_i$ can be covered by a family of at
most $\kappa$ nowhere dense sets.   By (a), $\bigcup_{i\in I}H_i$ is
dense.   For each $i\in I$, let $\ofamily{\xi}{\kappa}{E_{i\xi}}$ be a
family of nowhere dense sets covering $H_i$.   Set
$E_{\xi}=\bigcup_{i\in I}H_i\cap E_{i\xi}$ for each $\xi<\kappa$;  then
$E_{\xi}$ is nowhere dense (5A4Ea).   Also
$\bigcup_{\xi<\kappa}E_{\xi}=\bigcup_{i\in I}H_i$ is a dense open set,
so that $\{E_{\xi}:\xi<\kappa\}\cup(X\setminus\bigcup_{i\in I}H_i)$ is a
cover of $X$ by nowhere dense sets, and $n(X)\le\kappa$.   (Of course
$\kappa$ is infinite, by 517A, except in the trivial case $X=\emptyset$.)
}%end of proof of 517M

\leader{517N}{Corollary} If $\frak A$ is a Martin-number-homogeneous
Boolean algebra with Stone space $Z$, then $\frak m(\frak A)=n(Z)$.

\proof{ By 517Kb(i)$\Rightarrow$(iii), $\frak m(\frak A)\le n(Z)$.
In the other direction, given $a\in\frak A$, write $\widehat{a}$ for the
open-and-closed subset of $Z$ corresponding to $a$, and $\frak A_a$ for
the principal ideal generated by $a$.   If $G\subseteq Z$ is a non-empty
regular open set, let $a\in\frak A\setminus\{0\}$ be such that
$\widehat{a}\subseteq G$.   Then

$$\eqalignno{\frak m(\RO(G))
&\le\frak m(\RO(\widehat{a}))\cr
\displaycause{by 517Ib, because $\RO(\widehat{a})$ can be regarded as a
principal ideal of $\RO(G)$}
&=\frak m(\frak A_a)\cr
\displaycause{because we can identify $\widehat{a}$ with the Stone space
of $\frak A_a$, by 312T, and use 517Ka}
&=\frak m(\frak A).\cr}$$

\noindent By 517M, $n(Z)\le\frak m(\frak A)$.
}%end of proof of 517N

\leader{517O}{Martin cardinals (a)} For any class $\Cal P$ of partially
ordered sets, we have an associated cardinal

\Centerline{$\frak m^{\uparrow}_{\Cal P}
=\min\{\frak m^{\uparrow}(P):P\in\Cal P\}$.}

\noindent Much the most important of these is the cardinal

\Centerline{$\frak m=\min\{\frak m^{\uparrow}(P):P$ is upwards-ccc$\}$.}

\noindent Others of great interest are

\Centerline{$\frak p
=\min\{\frak m^{\uparrow}(P):P$ is $\sigma$-centered upwards$\}$,}

\Centerline{$\frak m_{\text{K}}
=\min\{\frak m^{\uparrow}(P):P$ satisfies Knaster's condition
upwards$\}$,}

\Centerline{$\frakmctbl=\min\{\frak m^{\uparrow}(P):P$ is a countable
partially ordered set$\}$.}

\noindent Two more which are worth examining are

\Centerline{$\frak m_{\sigma\text{-linked}}
=\min\{\frak m^{\uparrow}(P):P$ is $\sigma$-linked upwards$\}$,}

\Centerline{$\frak m_{\text{pc}\omega_1}
=\min\{\frak m^{\uparrow}(P):\omega_1$ is an up-precaliber of $P\}$.}

\wheader{517O}{0}{0}{0}{72pt}
\spheader 517Ob These cardinals are related as follows:

\def\tmphrule{\hskip0.4em\raise
2.5pt\hbox{\leaders\hrule\hskip1.7em\hfil}\hskip0.4em}
\def\tmpstrut{\vrule height10.5pt depth5.5pt width0pt}

$$\vbox{\offinterlineskip
\halign{\hfil#\hfil&\hfil#\hfil&\hfil#\hfil&\hfil#\hfil
  &\hfil#\hfil&\hfil#\hfil&\hfil#\hfil
  &\hfil#\hfil&\hfil#\hfil&\hfil#\hfil&\hfil#\hfil\cr
&\tmpstrut&&&$\frak m_{\sigma\text{-linked}}$&\tmphrule
  &$\frak p$&\tmphrule&$\frakmctbl$&\tmphrule&$\frak c$\cr
&\tmpstrut&&&\vrule&&\vrule\cr
$\tmpstrut\omega_1$&\tmphrule&$\frak m$&\tmphrule&$\frak m_{\text{K}}$
&\tmphrule&$\frak m_{\text{pc}\omega_1}$\cr
}}$$

\noindent The numbers here increase from bottom left to
top right;  that is,

\Centerline{$\omega_1\le\frak m\le\frak m_{\text{K}}
\le\frak m_{\text{pc}\omega_1}\le\frak p\le\frakmctbl\le\frak c$,}

\Centerline{$\frak m_{\text{K}}
\le\frak m_{\sigma\text{-linked}}\le\frak p$.}

\prooflet{
\noindent From 517A we see that $\omega_1\le\frak m$.   For the proof
that $\frakmctbl\le\frak c$, see 517P below.   As for the intermediate
inequalities involving Martin cardinals, they follow directly from
inclusions between the corresponding classes of partially ordered set.
These are all immediate from the definitions;  I give references to the
general results of this chapter which cover the relevant facts, as
follows.

\medskip

\quad{\bf (i)} Every partially ordered set satisfying Knaster's
condition upwards is ccc.   \prooflet{(If $(\omega_1,2)$ is an upwards
precaliber pair of $P$, then $\sat^{\uparrow}(P)\le\omega_1$ (516Ka).)}

\medskip

\quad{\bf (ii)} If $\omega_1$ is an up-precaliber of $P$, then $P$
satisfies Knaster's condition upwards.   \prooflet{(If
$\triplepc{\omega_1}{\omega_1}{\omega}$ is a triple precaliber of $P$,
so is $\triplepc{\omega_1}{2}{\omega}$, by 516Ba.)}

\medskip

\quad{\bf (iii)} If $P$ is $\sigma$-linked upwards, it satisfies
Knaster's condition upwards.  \prooflet{(As
$\omega_1>\max(\omega,\omega,\link^{\uparrow}(P))$,
$\triplepc{\omega_1}{\omega_1}{3}$ is an upwards precaliber triple of
$P$ (516Kb), so $\triplepc{\omega_1}{2}{3}$ and
$\triplepc{\omega_1}{2}{\omega}$ also are, by 516Ba again.)}

\medskip

\quad{\bf (iv)} If $P$ is $\sigma$-centered upwards, it is
$\sigma$-linked upwards.
\prooflet{($\link(P)\le\link_{<\omega}(P)$, by 511Hb.)}

\medskip

\quad{\bf (v)} If $P$ is $\sigma$-centered upwards, $\omega_1$ is an
up-precaliber of $P$.   \prooflet{(As
$\omega_1>\max(\omega,\omega,\link_{<\omega}^{\uparrow}(P))$,
$(\omega_1,\omega_1)$ is an upwards precaliber pair
of $P$, by 516Kb again.)}

\medskip

\quad{\bf (vi)} If $P$ is countable, it is $\sigma$-centered upwards.
\prooflet{(Singleton subsets are centered.)}
}%end of prooflet

\spheader 517Oc\cmmnt{ I should note a special feature of the bottom
row of this diagram.}   In the chain
$\omega_1\le\frak m\le\frak m_{\text{K}}\le\frak m_{\text{pc}\omega_1}$,
at most one of the inequalities can be strict.   \prooflet{\Prf\ Suppose
that $P$ is upwards-ccc and
$\frak m^{\uparrow}(P)>\omega_1$.   Then $\omega_1$ is an up-precaliber
of $P$ (517Fb), so
$\frak m^{\uparrow}(P)\ge\frak m_{\text{pc}\omega_1}$.   So if, for
instance, $\frak m_{\text{K}}>\omega_1$ and $P$ satisfies Knaster's
condition upwards, $\frak m^{\uparrow}(P)>\omega_1$ and
$\frak m^{\uparrow}(P)\ge\frak m_{\text{pc}\omega_1}$;  as $P$ is
arbitrary, $\frak m_{\text{K}}\ge\frak m_{\text{pc}\omega_1}$.
Similarly, if $\frak m>\omega_1$ then
$\frak m=\frak m_{\text{pc}\omega_1}$.\ \Qed}

\spheader 517Od Now {\bf Martin's Axiom} is the assertion

\Centerline{`$\frak m=\frak c$'.}

\noindent From the diagram above, we see that this is a consequence of
the continuum hypothesis (`$\omega_1=\frak c$'), and fixes all the
intermediate cardinals.

\cmmnt{\spheader 517Oe All the partially ordered sets considered in
(b) are ccc, which is why $\frak m$ appears at bottom left.   The same
idea can be applied to larger classes, e.g.\ `proper' or
`stationary-set-preserving' partial orders.   For the moment I will not
even define these classes;  I mention them only for the sake of readers
who are already familiar with them and may be expecting a reference
here.   There is an important difference, however, in that if the
cardinal which we might call

\Centerline{$\frak m_{\text{proper}}
=\min\{\frak m^{\uparrow}(P):P$ is upwards-proper$\}$}

\noindent is greater than $\omega_1$, then
$\frak c=\frak m_{\text{proper}}=\omega_2$ ({\smc Veli\v{c}kovi\'c 92}, or
{\smc Moore 05});  so that we have
only to say whether the Proper Forcing Axiom
(`$\frak m_{\text{proper}}>\omega_1$') is true or false to
determine the value of $\frak m_{\text{proper}}$.
}%end of comment

\leader{517P}{}\cmmnt{ All the cardinals here have special features,
but the ones I will concentrate on just now are the two largest,
$\frakmctbl$ and $\frak p$.

\medskip

\noindent}{\bf Proposition} (a) $\omega_1\le\frakmctbl\le\frak c$.

(b) Let $\frak A$ be a Boolean algebra with countable $\pi$-weight.
If $\frak A$ is purely atomic, then $\frak m(\frak A)=\infty$;
otherwise, $\frak m(\frak A)=\frakmctbl$.

(c) If $P$ is a partially ordered set of countable cofinality and
$\frak m^{\uparrow}(P)$ is not $\infty$, then
$\frak m^{\uparrow}(P)=\frakmctbl$.

(d)(i) Let $X$ be a topological space such that its category algebra is
atomless and has countable $\pi$-weight.   Then $n(X)\le\frakmctbl$.

\quad (ii) If $X$ is a non-empty locally compact Hausdorff space with
countable $\pi$-weight and no isolated points, then $n(X)=\frakmctbl$.

\quad (iii) If $X$ is a non-empty Polish space with no isolated points,
then $n(X)=\frakmctbl$.

\proof{ Let $\frak B$ be the algebra of open-and-closed subsets of
$\{0,1\}^{\Bbb N}$.   The argument will go more smoothly if I prove
(a)-(c) with $\frak m(\frak B)$ in place of $\frakmctbl$, and at an
appropriate moment point out that I have shown that the two are equal.

\medskip

{\bf (a)} $\frak m(\frak B)=\frak m^{\downarrow}(\frak B^+)$ is
uncountable, by 517A.    To see that
$\frak m(\frak B)\le\frak c$, let $\Cal Q$ be the set of all
coinitial subsets of $\frak B^+$;  then $\#(\Cal Q)\le\frak c$ because
$\frak B$ is countable.   \Quer\ If $\frak m(\frak B)>\frak c$, there
must be a linked set $R\subseteq\frak B^+$ meeting every member of
$\Cal Q$.   But now consider $Q=\frak B^+\setminus R$.   If
$a\in\frak B^+$, there are disjoint non-zero $a'$, $a''\Bsubseteq a$
which cannot both belong to $R$, so at least one belongs to $Q$.   But
this means that $Q$ is order-dense in $\frak B$ and ought to meet $R$.\
\BanG\  (Compare 517E.)

\medskip

{\bf (b)(i)} If $\frak A$ is purely atomic, $\frak m(\frak A)=\infty$,
by 511If.

\medskip

\quad{\bf (ii)} Suppose that $\frak A$ is not purely atomic.   Because
$\pi(\frak A)$ is countable, there is a countable order-dense set
$C\subseteq\frak A$.   Let $\frak C$ be the subalgebra of $\frak A$
generated by $C$, so that $\frak C$ is a countable order-dense
subalgebra of $\frak A$, and is not purely atomic.
Consider the free product $\frak C\otimes\frak B$ (315N).
This is a countable atomless Boolean algebra (use 315O),
so is isomorphic to
$\frak B$ (316M).   Also we have an injective
order-continuous Boolean homomorphism from $\frak C$ to
$\frak C\otimes\frak B$ (315K), so that $\frak C$ is isomorphic to a
regularly embedded subalgebra of $\frak B$ and
$\frak m(\frak C)\ge\frak m(\frak B)$ (517Ia).

Next, $\frak C$ has a non-trivial atomless principal ideal $\frak C_a$
say.   Because $\frak C_a$ is still countable, it is itself isomorphic
to $\frak B$.   So 517Ib tells us that
$\frak m(\frak C)\le\frak m(\frak C_a)=\frak m(\frak B)$, and
$\frak m(\frak C)=\frak m(\frak B)$.

Finally, $\frak m(\frak A)=\frak m(\frak C)$ by 517Ic.

\medskip

{\bf (c)} We know that
$\frak m^{\uparrow}(P)=\frak m(\RO^{\uparrow}(P))$  (517Db) and that
$\pi(\RO^{\uparrow}(P))\le\cf P$ (514Nb) is countable.   Let
$D\subseteq\RO^{\uparrow}(P)^+$ be a countable order-dense set, and
$\frak A$ the subalgebra of $\RO^{\uparrow}(P)$ generated by $D$.   Then
$\frak m(\RO^{\uparrow}(P))=\frak m(\frak A)$ by 517Ic, and $\frak A$ is
countable.   By (b), $\frak m^{\uparrow}(P)=\frak m(\frak A)$ is either
$\infty$ or $\frak m(\frak B)$;  since the former is ruled out by
hypothesis, we are left with the latter.

What this shows, however, is that

$$\eqalignno{\frak m(\frak B)
&\le\min\{\frak m^{\uparrow}(P):\cf P\le\omega\}
\le\min\{\frak m^{\uparrow}(P):\#(P)\le\omega\}\cr
&=\frakmctbl
\le\frak m^{\downarrow}(\frak B^+)
=\frak m(\frak B)\cr}$$

\noindent so that $\frakmctbl=\frak m(\frak B)$ and we can rewrite the
results so far in the forms given in the statement of the proposition.

\medskip

{\bf (d)(i)} Consider first the case in which $X$ is a non-empty Baire
space, so that its category algebra is isomorphic to $\RO(X)$ (514If).
Since $\RO(X)$ is atomless and not $\{\emptyset\}$, and in particular is
not purely atomic, but has countable $\pi$-weight,
$\frak m(\RO(X))=\frakmctbl$, by (b).    The same applies to any
non-empty open subset $G$ of $X$, recalling that the category algebra of
$G$ can be identified with a principal ideal of the category algebra of
$X$ (514Id).   So $n(X)\le\frakmctbl$ by 517M.

If $X$ is not a Baire space, then it has a smallest comeager regular
open set $H$, which is itself a Baire space (4A3Sa\formerly{4{}A3Ra}), and $X$ and $H$
have isomorphic category algebras (514Ic), so we see from the argument
just above that $n(H)\le\frakmctbl$.   But $X\setminus H$ is a countable
union of nowhere dense subsets of $X$, and every subset of $H$ which is
nowhere dense in $H$ is also nowhere dense in $X$, so
$n(X)\le\max(\omega,n(H))\le\frakmctbl$.

\medskip

\quad{\bf (ii)} Because $X$ is Hausdorff and has no isolated points,
$\RO(X)$ is atomless.   Next, $\pi(\RO(X))\le\pi(X)$ is countable
(514H(b-i)), and $\RO(X)$ is isomorphic to the category algebra of $X$, by
Baire's theorem.   So the first part of the proof of (i) tells us that
$n(X)\le\frakmctbl=\frak m(\RO(X))$.   From 517J we now see that

\Centerline{$\frak m(\RO(X))
=\min\{n(H):H\subseteq X$ is a non-empty open set$\}
\le n(X)$,}

\noindent so $n(X)=\frakmctbl$ exactly.

\medskip

\quad{\bf (iii)} Now suppose that $X$ is a non-empty Polish space
without isolated points.   As in (ii), the category algebra of $X$ is
atomless and has countable $\pi$-weight, so $n(X)\le\frakmctbl$.   In
the other direction, suppose that $\kappa<\frakmctbl$ and that
$\ofamily{\xi}{\kappa}{E_{\xi}}$ is a family of nowhere dense subsets of
$X$.   Let $\rho$ be a metric defining the topology of $X$ under which
$X$ is complete, and $\Cal U$ a countable base for the topology of $X$,
not containing $\emptyset$.   For $\xi<\kappa$, set
$\Cal Q_{\xi}=\{U:U\in\Cal U$, $\overline{U}\cap E_{\xi}=\emptyset\}$;
for $n\in\Bbb N$ set $\Cal Q'_n=\{U:U\in\Cal U$, $\diam U\le 2^{-n}\}$.
Then every $\Cal Q_{\xi}$ and every $\Cal Q'_n$ is coinitial with
$\Cal U$.   By (c) above,

\Centerline{$\frak m^{\downarrow}(\Cal U)\ge\frakmctbl
>\max(\kappa,\omega)$,}

\noindent so there is a downwards-directed $\Cal V\subseteq\Cal U$
meeting every $\Cal Q_{\xi}$ and every $\Cal Q_n$.   Now
$\{\overline{V}:V\in\Cal V\}$ is a downwards-directed set containing
sets of arbitrarily small diameter, so generates a Cauchy filter and
(because $(X,\rho)$ is complete) has non-empty intersection.   Take any
$x\in\bigcap_{V\in\Cal V}\overline{V}$.   Because $\Cal V$ meets every
$\Cal Q_{\xi}$, $x\notin\bigcup_{\xi<\kappa}E_{\xi}$ and
$\ofamily{\xi}{\kappa}{E_{\xi}}$ does not cover $X$.  As
$\ofamily{\xi}{\kappa}{E_{\xi}}$ is arbitrary, $n(X)\ge\frakmctbl$ and
the two are equal.
}%end of proof of 517P

\leader{517Q}{Lemma} If $P$ is any partially ordered set,
$\frak m^{\uparrow}(P)\ge\min(\add_{\omega}P,\frakmctbl)$.

\proof{ (For the definition of $\add_{\omega}P$, see 513H.)   Take
$\kappa<\min(\add_{\omega}P,\frakmctbl)$,
$p_0\in P$ and a family $\ofamily{\xi}{\kappa}{Q_{\xi}}$ of cofinal
subsets of $P$.   Choose $\sequencen{R_n}$ and
$\langle Q_{n\xi}\rangle_{n\in\Bbb N,\xi<\kappa}$ as follows.
$R_0=\{p_0\}$.   Given that $R_n\subseteq P$ is countable, then for each
$\xi<\kappa$ choose a countable set $Q_{n\xi}\subseteq Q_{\xi}$ such
that for every $p\in R_n$ there is a $q\in Q_{n\xi}$ such that $p\le q$.
Now, because $\add_{\omega}P>\kappa$ (and, of course,
$\add_{\omega}P>\omega$, as noted in 513Ib), we can find a countable set
$R_{n+1}\subseteq P$ such that whenever
$q\in\bigcup_{\xi<\kappa}Q_{n\xi}$ there is an $r\in R_{n+1}$ such that
$q\le r$.   This will ensure that whenever $p\in R_n$ and $\xi<\kappa$
there are $q\in Q_{\xi}$ and $p'\in R_{n+1}$ such that
$p\le q\le p'$.

At the end of the induction, consider the countable partially ordered
set $R=\bigcup_{n\in\Bbb N}R_n$.   For $\xi<\kappa$ set

\Centerline{$Q'_{\xi}=\{r:r\in R,\Exists q\in Q_{\xi},\,q\le r\}$;}

\noindent then $Q'_{\xi}$ is cofinal with $R$.   Because
$\kappa<\frakmctbl$, there is an upwards-linked subset
$S$ of $R$ meeting every $Q'_{\xi}$ and containing $p_0$.   But now
$\{p:p\in P,\Exists s\in S,\,p\le s\}$ is an upwards-linked subset of
$p$ containing $p_0$ and meeting every $Q_{\xi}$.   As $p_0$ and
$\ofamily{\xi}{\kappa}{Q_{\xi}}$ are arbitrary,
$\frak m^{\uparrow}(P)\ge\min(\add_{\omega}P,\frakmctbl)$.
}%end of proof of 517Q

\leader{517R}{Proposition} (a)\cmmnt{ (`Booth's Lemma';  see
{\smc Booth 70}}) Suppose that $\Cal A$ is a
family of subsets of $\Bbb N$ such that $\#(\Cal A)<\frak p$ and
$\bigcap\Cal J$ is infinite for every finite $\Cal J\subseteq\Cal A$.
Then there is an infinite $I\subseteq\Bbb N$ such that $I\setminus A$ is
finite for every $A\in\Cal A$.

(b) $2^{\kappa}\le\frak c$ for every $\kappa<\frak p$.

(c)\dvAformerly{5{}55K} Suppose that $X$ is a set and $\#(X)<\frak p$.
Then there is a
countable set $\Cal A\subseteq\Cal PX$ such that $\Cal PX$ is the
$\sigma$-algebra generated by $\Cal A$.

\proof{{\bf (a)} Let $P$ be $[\Bbb N]^{<\omega}\times[\Cal A]^{<\omega}$,
ordered by saying that $(K,\Cal J)\le(K',\Cal J')$ if
$K\subseteq K'\subseteq K\cup\bigcap\Cal J$ and
$\Cal J\subseteq\Cal J'$.   If $(K,\Cal J)\le(K',\Cal J')\le(K'',\Cal J'')$
then of course $K\subseteq K''$ and $\Cal J\subseteq\Cal J''$;  also

\Centerline{$K''\subseteq K'\cup\bigcap\Cal J'
\subseteq K\cup\bigcap\Cal J\cup\bigcap\Cal J'
\subseteq K\cup\bigcap\Cal J$.}

\noindent So $\le$ is a partial ordering of $P$.   For any
$K\in[\Bbb N]^{<\omega}$, $\{(K,\Cal J):J\in[\Cal A]^{<\omega}\}$ is
upwards-centered;  so $P$ is $\sigma$-centered upwards.

For each $A\in\Cal A$, set
$Q_A=\{(K,\Cal J):(K,\Cal J)\in P$, $A\in\Cal J\}$;  since
$(K,\Cal J)\le(K,\Cal J\cup\{A\})$ whenever $(K,\Cal J)\in P$, $Q_A$ is
cofinal with $P$.   For $n\in\Bbb N$, set
$Q'_n=\{(K,\Cal J):(K,\Cal J)\in P$, $K\not\subseteq n\}$.   If
$(K,\Cal J)\in P$, $\bigcap\Cal J$ must be infinite, and
there is an $m\in\Bbb N\cap\bigcap\Cal J\setminus n$;  now
$(K,\Cal J)\le(K\cup\{m\},\Cal J)\in Q'_n$.   So $Q'_n$ is cofinal with
$P$.

Because $\max(\omega,\#(\Cal A))<\frak p$ (517Ob),
there is an upwards-linked
$R\subseteq P$ meeting every $Q_A$ and every $Q'_n$.   Set
$I=\bigcup\{K:(K,\Cal J)\in R\}$.   If $n\in\Bbb N$, there is a
$(K,\Cal J)\in R\cap Q'_n$;  now $K\not\subseteq n$ and $K\subseteq I$, so
$I\not\subseteq n$;  as $n$ is arbitrary, $I$ is infinite.   If
$A\in\Cal A$, there is $(K_0,\Cal J_0)\in R\cap Q_A$.   \Quer\ If
$I\not\subseteq K_0\cup A$, there is a $(K,\Cal J)\in R$ such that
$K\not\subseteq K_0\cup A$.   Now there is a $(K',\Cal J')\in P$ such
that $(K,\Cal J)\le(K',\Cal J')$ and $(K_0,\Cal J_0)\le(K',\Cal J')$.   But
in this case

\Centerline{$K\subseteq K'\subseteq K_0\cup\bigcap\Cal J_0
\subseteq K_0\cup A$.\ \Bang}

\noindent So $I\setminus A\subseteq K_0$ is finite.   As $A$ is arbitrary,
we have a suitable $I$.

\medskip

{\bf (b)} We may suppose that $\kappa$ is infinite.   By 515H, or
otherwise, there is a Boolean-independent family
$\ofamily{\xi}{\kappa}{J_{\xi}}$ in $\Cal P\Bbb N$.
Note that
$I=\bigcap_{\xi\in K}J_{\xi}\setminus\bigcup_{\xi\in L}J_{\xi}$ must be
infinite whenever $K$, $L\subseteq\kappa$ are disjoint finite sets, because
$\family{\xi}{\kappa\setminus(K\cup L)}{I\cap J_{\xi}}$ is
Boolean-independent.   For $C\subseteq\kappa$ set

\Centerline{$\Cal A_C
=\{J_{\xi}:\xi\in C\}\cup\{\Bbb N\setminus J_{\xi}:
  \xi\in\kappa\setminus C\}$.}

\noindent By (a), there is an infinite $I_C\subseteq\Bbb N$ such that
$I_C\setminus A$ is finite for every $A\in\Cal A_C$.   If
$C$, $D\subseteq\kappa$ and $\xi\in C\setminus D$, then
$I_C\setminus J_{\xi}$ and $I_D\cap J_{\xi}$ are
finite, so $I_C\cap I_D$ is finite and $I_C\ne I_D$.   Thus
$C\mapsto I_C$ is injective and $2^{\kappa}\le\frak c$.

\medskip

{\bf (c)} Let
$\family{x}{X}{I_x}$ be a family of infinite subsets of $\Bbb N$ such that
$I_x\cap I_y$ is finite for all distinct $x$, $y\in X$ (5A1Ga).   Set
$A_n=\{x:n\in I_x\}$ for $n\in\Bbb N$.

Take any $A\subseteq X$ and set
$P_A=\Fn_{<\omega}(\Bbb N;\{0,1\})\times[X\setminus A]^{<\omega}$,
partially ordered by saying that

\doubleinset{
$(f,J)\le(f',J')$ if $f'$ extends $f$, $J'\supseteq J$ and
whenever $x\in J$ and
$i\in I_x\cap\dom f'\setminus\dom f$, then $f'(i)=0$.}

\noindent Then $P_A$ is $\sigma$-centered upwards because
$\{(f,J):J\in[X\setminus A]^{<\omega}\}$ is upwards-centered for every
$f\in\Fn_{<\omega}(\Bbb N;\{0,1\})$.   For $x\in A$ and $m\in\Bbb N$ set

\Centerline{$Q_{xm}=\{(f,J):(f,J)\in P_A$, $f(i)=1$ for some
$i\in I_x\setminus m\}$;}

\noindent for $x\in X\setminus A$ set

\Centerline{$Q'_x=\{(f,J):(f,J)\in P_A$, $x\in J\}$.}

\noindent Then every $Q_{xm}$ and every $Q'_x$ is cofinal with $P_A$.
Because $\#(X)<\frak p$, there is an upwards-directed $R\subseteq P_A$
meeting every $Q_{xm}$ and every $Q'_x$.   Set
$L=\bigcup_{(f,J)\in R}\{i:f(i)=1\}$.   Now

\inset{----- if $x\in A$ and $m\in\Bbb N$ then $L\cap I_x\setminus m$ is
non-empty, so $L\cap I_x$ is infinite,

----- if $x\in X\setminus A$, there is a pair $(f_0,J_0)\in R$ such that
$x\in J_0$;  now $f(i)=0$ whenever $(f,J)\in R$ and
$i\in\dom f\setminus\dom f_0$, so $L\cap I_x\subseteq\dom f_0$.}

\noindent Accordingly

\Centerline{$A=\{x:x\in X$, $I_x\cap L$ is infinite$\}
=\bigcap_{n\in\Bbb N}\bigcup_{m\in L\setminus n}A_m$}

\noindent belongs to the $\sigma$-algebra generated by
$\Cal A=\{A_n:n\in\Bbb N\}$, and we have a suitable family.
}%end of proof of 517R

\leader{517S}{Proposition} Let $P$ be a partially ordered set which
satisfies Knaster's condition upwards.   If $A\subseteq P$ and
$\#(A)<\frak m_{\text{K}}$, then $A$ can be covered by a sequence of
upwards-directed subsets of $P$.

\proof{ By 516P, the upwards finite-support product $P^*$ of countably
many copies of $P$ also satisfies Knaster's condition upwards.   So we
can use 517Ha.
}%end of proof of 517S

\exercises{\leader{517X}{Basic exercises (a)}
%\spheader 517Xa
Let $P$ be a partially ordered set and $\kappa$ a
cardinal.   Show that the following are equiveridical:  (i)
$\kappa<\frak m^{\uparrow}(P)$;  (ii) whenever $p_0\in P$ and $\Cal Q$
is a family of cofinal subsets of $P$ with $\#(\Cal Q)\le\kappa$, there
is an upwards-centered subset of $P$ which contains $p_0$ and meets
every member of $\Cal Q$;
(iii) whenever $p_0\in P$ and $\Cal Q$ is a family of up-open cofinal
subsets of $P$ with $\#(\Cal Q)\le\kappa$, there is an upwards-centered
subset of $P$ which contains $p_0$ and meets every member of $\Cal Q$;
(iv) whenever $p_0\in P$ and $\Cal A$ is a family of maximal
up-antichains in $P$ with $\#(\Cal A)\le\kappa$, there is an
upwards-centered subset of $P$ which contains $p_0$ and meets every
member of $\Cal A$.
%517B

\spheader 517Xb Let $P$ be a partially ordered set and $A$ a maximal
up-antichain in $P$.   Show that

\Centerline{$\frak m^{\uparrow}(P)
=\min_{p\in A}\frak m^{\uparrow}(\coint{p,\infty})$.}
%517D

\spheader 517Xc(i) Let $\frak A$ be a Boolean algebra, not $\{0\}$.
For $a\in\frak A$ let $\frak A_a$ be the corresponding principal ideal.
Show that there is an $a\in\frak A^+$ such that
$\frak m(\frak A_b)=\frak m(\frak A_a)$ whenever $0\ne b\Bsubseteq a$.
(ii) Show that any Dedekind complete Boolean algebra is isomorphic to a
simple product of Martin-number-homogeneous Boolean algebras.
%517Xb 517D

\sqheader 517Xd Let $P$ be a partially ordered set.   Show that
$\frak m^{\uparrow}(P)=\infty$ iff
$\{p:\coint{p,\infty}$ is upwards-linked$\}$ is cofinal with $P$.
%517D

\spheader 517Xe Let $\familyiI{\frak A_i}$ be a family of Boolean
algebras and $\frak A$ its free product;  suppose that $\kappa$ is a
regular uncountable cardinal such that
$\sat(\frak A_i)\le\kappa<\frak m(\frak A_i)$ for every $i\in I$.   Show
that $\sat(\frak A)\le\kappa$.
%517I 517G

\spheader 517Xf Let $\frak A$ be a Boolean algebra and $\frak C$ the
free product of a sequence of copies of $\frak A$.   Suppose that
$\kappa<\frak m(\frak C)$.   (i) Show if $A\in[\frak A^+]^{\le\kappa}$
then $A$ can be covered by a sequence of centered subsets of
$\frak A^+$.   (ii) Show that if $\cf\kappa\ge\omega_1$ then $\kappa$ is
a precaliber of $\frak A$.
%517I 517H

\spheader 517Xg Let $\frak A$ be a Boolean algebra and $Z$ its Stone
space.   Show that $\frak m(\frak A)
=\min\{n(Y):Y\subseteq Z$ is a non-meager set with the
Baire property$\}$.
%517K

\sqheader 517Xh Let $P$ be a non-empty partially ordered set and $P^*$
the upwards finite-support product of a sequence of copies of $P$.
Show that if $\frak m^{\uparrow}(P^*)>\omega_1$ then $P$ must be
upwards-ccc.
%517H

\spheader 517Xi Let $X$ be any topological space.   Show that
$\frak m(\RO(X))\ge\min\{n(G):G\subseteq X$ is a non-empty open set$\}$.
%517J

\spheader 517Xj Let $X$ be a locally compact Hausdorff space such that
$\RO(X)$ is Martin-number-homogeneous.   Show that
$\frak m(\RO(X))=n(X)$.
%517N 517J

\spheader 517Xk(i) Let $P$ be a partially ordered set which is
$\sigma$-linked upwards.   Show that if $A\subseteq P$ and
$\#(A)<\frak m_{\sigma\text{-linked}}$,
then $A$ can be covered by a sequence of
upwards-directed subsets of $P$.
(ii) Let $P$ be a partially ordered set such that $\omega_1$ is an
up-precaliber of $P$.   Show that if $A\subseteq P$ and
$\#(A)<\frak m_{\text{pc}\omega_1}$,
then $A$ can be covered by a sequence of
upwards-directed subsets of $P$.
(iii) Let $P$ be a partially ordered set which is
$\sigma$-centered upwards.   Show that if $A\subseteq P$ and
$\#(A)<\frak p$, then $A$ can be covered by a sequence of
upwards-directed subsets of $P$.
%517S

\leader{517Y}{Further exercises (a)}
%\spheader 517Ya
For a partially ordered set $P$,
write $A_P$ for the family of upwards-linked subsets of $P$, $B_P$ for
the family of cofinal subsets of $P$, and $T_P$ for
$\{(R,Q):R\in A_P,\,Q\in B_P,\,R\cap Q=\emptyset\}$.   (i) Show that
$\frak m^{\uparrow}(P)=\min_{p\in P}
\cov(A_{\coint{p,\infty}},T_{\coint{p,\infty}},B_{\coint{p,\infty}})$.
(ii) Show that if $Q$ is another partially ordered set and there is a
relation $S\subseteq P\times Q$ with the properties described in 517C,
then for every $q\in Q$ there is a $p\in P$ such that
$(A_{\coint{p,\infty}},T_{\coint{p,\infty}},B_{\coint{p,\infty}})
\prGT(A_{\coint{q,\infty}},T_{\coint{q,\infty}},B_{\coint{q,\infty}})$.
%517C

\spheader 517Yb Show that for every infinite regular
cardinal $\kappa$ there is a partially ordered set with Martin number
$\kappa^+$.
% \bigcup_{\xi<\kappa}(\kappa^+)^{\xi}

\spheader 517Yc Show that
$\frak m(\Cal P\Bbb N/[\Bbb N]^{<\omega})\ge\frak p$.
% order-dense sets lead to mad families

\spheader 517Yd\dvAnew{2023} (A.Szyma\'nski, 1981) (i) Suppose that $\Cal A$, 
$\Cal B\in[\Cal P\Bbb N]^{<\frak p}$, $\Cal A$ is downwards-directed
and $A\cap B$ is infinite for all $A\in\Cal A$, $B\in\Cal B$.   Show that there
is a set $D\subseteq\Bbb N$ such that $D\setminus A$ is finite for every 
$A\in\Cal A$ and $D\cap B$ is infinite for every $B\in\Cal B$.
(ii) Show that $\frak p$ is regular.
%517R
}%end of exercises

%Note:  \cov\Cal N  is a Martin number

\endnotes{
\Notesheader{517} The study of `Martin numbers' is a natural extension
of investigations into consequences of Martin's axiom.   Most of the
results here are straightforward
expressions of techniques developed for deducing consequences from
$\frak m=\frak c$ or
$\frak m>\omega_1$.   In particular, 517F, 517G and 517H correspond to
the now-classical theorems that if $\frak m>\omega_1$ then $\omega_1$ is
a precaliber of every ccc partially ordered set, the product of any
family of ccc topological spaces is ccc, and a ccc partially ordered set
with cardinal $\omega_1$ is a countable union of directed sets (see
{\smc Fremlin 84a}, \S41).
For those familiar with the use of Martin's axiom there are no surprises
here, though some refinements in the arguments are necessary.   The
cardinal $\frakmctbl$ is probably most commonly known as the Nov\'ak
number of $\Bbb R$ (517Pd), the covering number of the ideal of meager
subsets of $\Bbb R$.   In countable partially ordered sets, most of the
arguments above short-circuit to some degree;  precalibers become
trivial, finite-support products are automatically ccc, and directed
sets have cofinal totally ordered subsets, so that the ideas take on new
colours.

In {\smc Fremlin 84a} I found that focusing on the cardinals
$\frak p$, $\frak m_{\text{K}}$ and $\frak m$
broke the arguments up into reasonably balanced chapters.   Within the
chapter on $\frak m_{\text{K}}$, however, there is a natural division
between
arguments applying to $\frak m_{\text{pc}\omega_1}$ and those applying
to $\frak m_{\sigma\text{-linked}}$, which in the present book I intend
to make explicit.   The notation $\frak p$ is the standard name for the
cardinal $\frak m_{\sigma\text{-centered}}$;
its special position comes in part from the fact that it had been
studied under a different, combinatorial, definition for a decade before
M.G.Bell showed that it could also be described by the definition here
({\smc Bell 81}, or {\smc Fremlin 84a}, 14C).

}%end of notes

\discrpage

\frfilename{mt518.tex}
\versiondate{24.12.14}
\copyrightdate{2003}

\def\doubleheadrightarrow{{\hbox{$\rightarrow$}}\hskip0.2em\llap
  {\hbox{$\rightarrow$}}}
\def\sk{\mathop{\text{sk}}}

\def\chaptername{Cardinal functions}
\def\sectionname{Freese-Nation numbers}

\newsection{518}

\allowmorestretch{468}{
I run through those elements of the theory of Freese-Nation numbers, as
developed by S.Fuchino, S.Geschke, S.Koppelberg, S.Shelah and L.Soukup,
which seem relevant to questions concerning measure spaces and measure
algebras.   The first part of the section
(518A-518K)  %518A 518B 518C 518D 518F 518G 518H 518I 518J 518K
examines the calculation of Freese-Nation numbers of familiar partially
ordered sets and Boolean algebras.   In
518L-518S % 518L 518M 518N 518P 518Q 518R 518S
I look at `tight filtrations', which are of interest to us because of
their use in lifting theorems (518L, \S535).
}%end of allowmorestretch

\cmmnt{For the definitions of `Freese-Nation number' and
`Freese-Nation index' see 511Bi and 511Dh.}

\leader{518A}{Proposition}\cmmnt{
({\smc Fuchino Koppelberg \& Shelah 96})} Let $P$
be a partially ordered set.

(a) $\FN(P)\le\max(3,\#(P))$.

(b) $\FN(P,\ge)=\FN(P,\le)$.

(c) If $P$ has no maximal element, then $\add P\le\FN(P)$.

\proof{{\bf (a)(i)} Suppose first that $P$ is finite and totally
ordered.
If $\#(P)\le 2$, set $f(p)=P$ for every $p\in P$.   Otherwise, take
$p_0\in P$ such that $\ooint{-\infty,p_0}$ and $\ooint{p_0,\infty}$ are
both non-empty, and set $f(p)=\coint{p_0,\infty}$ if $p\ge p_0$,
$\ocint{-\infty,p_0}$ if $p<p_0$;  then $f$ is a Freese-Nation
function witnessing that $\FN(P)\le\#(P)$.

\medskip

\quad{\bf (ii)} Next suppose that $P$ is finite and not totally ordered.
For $p\in P$ set $A_p=\ocint{-\infty,p}\cup\coint{p,\infty}$, and take
$B=\{p:A_p=P\}$;  then $B\ne P$.
Set $f(p)=A_p$ for $p\in P\setminus B$, $B$ for $p\in B$;  then $f$ is a
Freese-Nation function so again $\FN(P)\le\#(P)$.

\medskip

\quad{\bf (iii)} If $P$ is infinite, enumerate it as
$\ofamily{\xi}{\#(P)}{p_{\xi}}$ and set
$f(p_{\xi})=\{p_{\eta}:\eta\le\xi\}$ for each $\xi$;  once more we have
a Freese-Nation function witnessing that $\FN(P)\le\#(P)$.

\medskip

{\bf (b)} A function $f:P\to\Cal PP$ is a Freese-Nation function for $\le$
iff it is a Freese-Nation function for the reverse ordering $\ge$.

\medskip

{\bf (c)} Set $\kappa=\FN(P)$.   Then we have a Freese-Nation
function $f:P\to[P]^{<\kappa}$.

\medskip

\quad{\bf (i)} I had better sort out the trivial cases.   If $P$ is empty,
then $\kappa=\add P=0$.   Otherwise, $p\in f(p)$ for every $p\in P$, so
$\kappa\ge 2$;  if $\add P\le 2$ we can stop.   So we may suppose that
$\add P>2$, that is, that $P$ is upwards-directed.

\medskip

\quad{\bf (ii)} \Quer\ If $\kappa<\add P$, choose
$\langle p_{\xi}\rangle_{\xi\le\kappa}$ inductively, as follows.   Given
$\ofamily{\eta}{\xi}{p_{\eta}}$, where $\xi\le\kappa$, then
$\bigcup_{\eta<\xi}f(p_{\eta})$
has an upper bound $p'_{\xi}$ in $P$.   \Prf\ If $\kappa$ is infinite, this
is because $\#(\bigcup_{\eta<\xi}f(p_{\eta}))\le\kappa<\add P$.
If $\kappa$ is finite, it is because
$\#(\bigcup_{\eta<\xi}f(p_{\eta}))<\omega\le\add P$.\ \Qed

As $P$ has no maximal element, we
can find $p_{\xi}>p'_{\xi}$, and continue.   At the end of the induction,
we have $p_{\xi}<p_{\kappa}$, so there is a
$q_{\xi}\in f(p_{\xi})\cap f(p_{\kappa})\cap[p_{\xi},p_{\kappa}]$, for each
$\xi<\kappa$.   If $\eta<\xi<\kappa$, then

\Centerline{$q_{\eta}\le p'_{\xi}<p_{\xi}\le q_{\xi}$}

\noindent and $q_{\eta}\ne q_{\xi}$.   But this means that
$f(p_{\kappa})\supseteq\{q_{\xi}:\xi<\kappa\}$ has at least $\kappa$
elements.\ \Bang
}%end of proof of 518A

\leader{518B}{Proposition} Let $P$ be a partially ordered set and $Q$ a
subset of $P$.

(a) If $Q$ is order-convex\cmmnt{ (that is, $[q,q']\subseteq Q$
whenever $q$, $q'\in Q$)}, then $\FN(Q)\le\FN(P)$.

(b) If $Q$ is a retract of $P$\cmmnt{ (that is, there is an
order-preserving $h:P\to Q$ such that $h(q)=q$ for every $q\in Q$)},
then $\FN(Q)\le\FN(P)$.

(c) If $Q$ is\cmmnt{, in itself,} Dedekind complete\cmmnt{ (that is,
every non-empty subset of $Q$ with an upper bound in $Q$ has a supremum
in $Q$ for the induced ordering)}, then $\FN(Q)\le\FN(P)$.

\proof{{\bf (a)} If $f:P\to\Cal PP$ is a Freese-Nation function on $P$,
then $q\mapsto Q\cap f(q):Q\to\Cal PQ$ is a Freese-Nation function on
$Q$.

\medskip

{\bf (b)} If $f$ is a Freese-Nation function on $P$, then
$q\mapsto h[f(q)]$ is a Freese-Nation function on $Q$.

\medskip

{\bf (c)} Set $Q_1=\bigcup_{q,q'\in Q}[q,q']$, so that $Q_1$ is an
order-convex subset of $P$ and $\FN(Q_1)\le\FN(P)$.   For $p\in Q_1$,
set $h(p)=\sup(Q\cap\ocint{-\infty,p})$, the supremum being taken in
$Q$;  then $h:Q_1\to Q$ is a retraction, so $\FN(Q)\le\FN(Q_1)$.
}%end of proof of 518B

\leader{518C}{Corollary} (a) If $\frak A$ is an infinite Dedekind
$\sigma$-complete Boolean algebra then
$\FN(\frak A)\ge\FN(\Cal P\Bbb N)$.

(b) Let $\frak A$ be an infinite Dedekind complete Boolean algebra.   Then

\Centerline{$\FN(\RO(\{0,1\}^{\#(\frak A)}))
\le\FN(\frak A)\le\FN(\Cal P(\link(\frak A)))
\le\max(3,2^{\link(\frak A)})$.}

(c) Let $\frak A$ be a Dedekind complete Boolean algebra.   If $\frak B$
is either an order-closed subalgebra or a principal ideal of $\frak A$,
then $\FN(\frak B)\le\FN(\frak A)$.

\proof{{\bf (a)} Take any disjoint sequence $\sequencen{a_n}$ in
$\frak A$;  then $I\mapsto\sup_{n\in I}a_n$ is an embedding of the
partially ordered set $\Cal P\Bbb N$ into $\frak A$.   As $\Cal P\Bbb N$
is Dedekind complete, 518Bc tells us that
$\FN(\Cal P\Bbb N)\le\FN(\frak A)$.

\medskip

{\bf (b)(i)} By 515I, $\frak A$ has a subalgebra $\frak B$
isomorphic to the regular open algebra $\RO(\{0,1\}^{\#(\frak A)})$;
by 518Bc, $\FN(\frak B)\le\FN(\frak A)$.

\medskip

\quad{\bf (ii)} By 514Cb, we have a subset $Q$ of
$\Cal P(\link(\frak A))$ which is order-isomorphic to $\frak A$,
and 518Bc tells us that $\FN(Q)\le\FN(\Cal P(\link(\frak A)))$.

\medskip

\quad{\bf (iii)} By 518Aa,
$\FN(\Cal P(\link(\frak A)))\le 2^{\link(\frak A)}$
except in the trivial case $\frak A=\{0,1\}$.

\medskip

{\bf (c)} Immediate from 518Bc.
}%end of proof of 518C

\vleader{60pt}{518D}{Corollary} The following sets all have the same
Freese-Nation number:

(i) $\Cal P\Bbb N$;

(ii) $\BbbN^{\Bbb N}$, with its usual ordering $\le$;

(iii) any infinite $\sigma$-linked Dedekind complete Boolean algebra;

(iv) the family of open subsets of any infinite Hausdorff
second-countable topological space.

\proof{{\bf (a)} The map $I\mapsto\chi I:\Cal P\Bbb N\to\BbbN^{\Bbb N}$
is an order-preserving embedding;  because $\Cal P\Bbb N$ is Dedekind
complete, $\FN(\Cal P\Bbb N)\le\FN(\BbbN^{\Bbb N})$ (518Bc).

\medskip

{\bf (b)} The map
$f\mapsto\{(i,j):j\le f(i)\}:
\BbbN^{\Bbb N}\to\Cal P(\Bbb N\times\Bbb N)$ is an order-preserving
embedding;  because $\BbbN^{\Bbb N}$ is Dedekind complete,

\Centerline{$\FN(\BbbN^{\Bbb N})\le\FN(\Cal P(\Bbb N\times\Bbb N))
=\FN(\Cal P\Bbb N)$.}

\medskip

{\bf (c)} Now let $\frak A$ be an infinite $\sigma$-linked Dedekind
complete Boolean algebra.   By 518Ca,
$\FN(\Cal P\Bbb N)\le\FN(\frak A)$; by 518Cb,
$\FN(\frak A)\le\FN(\Cal P\Bbb N)$.

\medskip

{\bf (d)} Let $(X,\frak T)$ be an infinite Hausdorff second-countable
space.   ($\alpha$) Because $X$ is Hausdorff and infinite, it has a
disjoint sequence $\sequencen{G_n}$ of non-empty open sets;  now
$I\mapsto\bigcup_{n\in I}G_n$ is an embedding of $\Cal P\Bbb N$ in
$\frak T$, so $\FN(\Cal P\Bbb N)\le\FN(\frak T)$.   ($\beta$) Let
$\Cal U$ be a countable base for $\frak T$.   Then
$G\mapsto\{U:U\in\Cal U$, $U\subseteq G\}$ is an embedding of $\frak T$
in $\Cal P\Cal U$;  as
$\frak T$, regarded as a partially ordered set, is Dedekind complete,
$\FN(\frak T)\le\FN(\Cal P\Cal U)=\FN(\Cal P\Bbb N)$.
}%end of proof of 518D

\leader{518E}{}\cmmnt{ There is a simple
result in general topology which will be used a couple of times in the next
chapter.

\medskip

\noindent}{\bf Lemma} Let $(X,\frak T)$ be a T$_1$ topological
space without isolated points, and $\CalNwd(X)$ the ideal of nowhere dense
sets.   Then there is a set $A\subseteq X$, with cardinal $\cov\CalNwd(X)$,
such that $\#(A\cap F)<\FN^*(\frak T)$ for every $F\in\CalNwd(X)$.

\cmmnt{\medskip

\noindent{\bf Remark} Perhaps I should say here that $\FN^*(\frak T)$ is
the regular Freese-Nation number of the partially ordered set
$(\frak T,\subseteq)$.
}%end of comment

\proof{ As $X$ has no isolated points, $\cov\CalNwd(X)\le\#(X)$.
Set $\kappa=\cov\CalNwd(X)$ and $\lambda=\FN^*(\frak T)$.
If $\kappa<\lambda$ the result
is trivial and we can stop.   Otherwise, let
$f:\frak T\to[\frak T]^{<\lambda}$ be a Freese-Nation function.
Then we can choose $\ofamily{\xi}{\kappa}{x_{\xi}}$
inductively so that
whenever $\eta<\xi$ and $G\in f(X\setminus\{x_{\eta}\})$ is dense, then
$x_{\xi}\in G$.   \Prf\ When we come to choose $x_{\xi}$,
set $\theta
=\#(\bigcup_{\eta<\xi}\{G:G\in f(X\setminus\{x_{\eta}\})$ is dense$\})$.
If $\lambda<\kappa$ then $\theta\le\max(\#(\xi),\omega,\lambda)<\kappa$.
If $\lambda=\kappa$ then $\kappa$ is regular and infinite
and $\#(f(X\setminus\{x_{\eta}\})<\kappa$ for every $\eta<\xi$ so again
$\theta<\kappa$.
So we have fewer than $\cov\CalNwd(X)$ dense open sets and can find a
point $x_{\xi}$ in all of them.\ \Qed

Note that as $X\setminus\{x_{\eta}\}$ is itself dense for every $\eta<\xi$,
and $H\in f(H)$ for every $H\in\frak T$, all the $x_{\xi}$ must be
distinct, and $A=\{x_{\xi}:\xi<\kappa\}$ has cardinal $\kappa$.
Now suppose that $F\in\CalNwd(X)$ and set
$B=\{\xi:\xi<\kappa$, $x_{\xi}\in F\}$.   For each $\xi\in B$,
$X\setminus\overline{F}\subseteq X\setminus\{x_{\xi}\}$, so there is a
$G_{\xi}\in f(X\setminus\overline{F})\cap f(X\setminus\{x_{\xi}\})$ such
that $X\setminus\overline{F}\subseteq G_{\xi}$ and $x_{\xi}\notin G_{\xi}$.
If $\eta$, $\xi\in B$ and $\eta<\xi$, then
$G_{\eta}\in f(X\setminus\{x_{\eta}\})$ is dense, so contains $x_{\xi}$,
and cannot be equal to $G_{\xi}$.   Thus $\xi\mapsto G_{\xi}$ is an
injective function from $B$ to $f(X\setminus\overline{F})$, and
$\#(B)<\lambda$.   Thus we have an appropriate set $A$.
}%end of proof of 518E

\leader{518F}{Lemma} Let $\frak A$ be a Boolean algebra, $\frak B$ a
subalgebra of $\frak A$ and $\kappa$ an infinite cardinal.

(a) If $\cf(\frak B\cap[0,a])<\kappa$ for every $a\in\frak A$, then the
Freese-Nation index of $\frak B$ in $\frak A$ is at most $\kappa$.

(b) Suppose that $I\in[\frak A]^{<\cf\kappa}$ and $\frak B_I$ is the
subalgebra of $\frak A$ generated by $\frak B\cup I$.   If the
Freese-Nation index of $\frak B$ in $\frak A$ is less than or equal to
$\kappa$, so is the Freese-Nation index of $\frak B_I$.

(c) If $\frak B$ is expressible as the
union of fewer than $\kappa$ order-closed subalgebras of $\frak A$,
each of them Dedekind complete in itself, then
the Freese-Nation index of $\frak B$ in $\frak A$ is at most $\kappa$.

\proof{{\bf (a)} For any $a\in\frak A$,

\Centerline{$\ci(\frak B\cap[a,1])
=\cf(\frak B\cap[0,1\Bsetminus a])<\kappa$.}

\medskip

{\bf (b)(i)} Suppose first that $I$ is a singleton $\{d\}$.   In this
case

\Centerline{$\frak B_I
=\{(b\Bcap d)\Bcup(c\Bsetminus d):b$, $c\in\frak B\}$.}

\noindent Take $a\in\frak A$.   Then there are sets $B$,
$C\subseteq\frak B$, with cardinal less than
$\kappa$, which are cofinal in

\Centerline{$\frak B\cap[0,a\Bcup(1\Bsetminus d)]$,
\quad$\frak B\cap[0,a\Bcup d]$}

\noindent respectively.   Set
$D=\{b\cap d:b\in B\}\cup\{c\setminus d:c\in C\}$, so that
$D\subseteq\frak B_I\cap[0,a]$ and $\#(D)<\kappa$.   If $b$,
$c\in\frak B$ and $(b\Bcap d)\Bcup(c\Bsetminus d)\Bsubseteq a$, then
$b\Bsubseteq a\Bcup(1\Bsetminus d)$ and $c\Bsubseteq a\Bcup d$, so there
are $b'\in B$, $c'\in C$ such that $b\Bsubseteq b'$ and
$c\Bsubseteq c'$;  now

\Centerline{$(b\cap d)\Bcup(c\Bsetminus d)
\Bsubseteq(b'\cap d)\Bcup(c'\Bsetminus d)\in D$.}

\noindent Thus $D$ witnesses that
$\cf(\frak C\cap[0,a])<\kappa$.   By (a), this is enough to show that
the Freese-Nation index of $\frak B_I$ in $\frak A$ is at most $\kappa$.

\medskip

\quad{\bf (ii)} An elementary induction now shows that the Freese-Nation
index of $\frak B_I$ in $\frak A$ is at most $\kappa$ for every finite
subset $I$ of $\frak A$.   If $\omega\le\#(I)<\cf\kappa$ and
$a\in\frak A$, then
$\frak B_I=\bigcup\{\frak B_J:J\in[I]^{<\omega}\}$.   For each
$J\in[I]^{<\omega}$, let $B_J$ be a cofinal subset of
$\frak B_J\cap[0,a]$ with cardinal less than $\kappa$.   Then
$B=\bigcup\{B_J:J\in[I]^{<\omega}\}$ is cofinal in $\frak B_I\cap[0,a]$;
and as $\#([I]^{<\omega})<\cf\kappa$, $\#(B)<\kappa$.   So again we have
$\cf(\frak B_I\cap[0,a])<\kappa$ for every $a\in\frak A$, and the
Freese-Nation index of $\frak B_I$ in $\frak A$ is at most $\kappa$.

\medskip

{\bf (c)} Suppose that $\ofamily{\xi}{\lambda}{\frak B_{\xi}}$ is a family
of order-closed subalgebras with union $\frak B$, where $\lambda<\kappa$.
If $a\in\frak A$, then $b_{\xi}=\sup(\frak B_{\xi}\cap[0,a])$ is defined in
$\frak B_{\xi}$, and belongs to $[0,a]$,
for each $\xi<\lambda$, and $\{b_{\xi}:\xi<\lambda\}$ is cofinal with
$\frak B\cap[0,a]$;  so we can apply (a).
}%end of proof of 518F

\leader{518G}{Lemma}\cmmnt{ ({\smc Fuchino Koppelberg \& Shelah 96})} Let
$P$ be a partially ordered set, $\zeta$ an ordinal, and
$\ofamily{\xi}{\zeta}{A_{\xi}}$ a family
with union $P$;  set $P_{\alpha}=\bigcup_{\xi<\alpha}A_{\xi}$ for each
$\alpha\le\zeta$.   Let $\kappa$ be a regular infinite cardinal such
that, for each $\alpha<\zeta$, $\FN(P_{\alpha+1})\le\kappa$ and the
Freese-Nation index of $P_{\alpha}$ in $P_{\alpha+1}$ is at most
$\kappa$.   Then $\FN(P)\le\kappa$.

\proof{ For each $\alpha<\zeta$ set
$A'_{\alpha}=A_{\alpha}\setminus P_{\alpha}$ and choose a Freese-Nation
function $f_{\alpha}:P_{\alpha+1}\to[P_{\alpha+1}]^{<\kappa}$.   For
$p\in P$, let $\gamma(p)$ be that $\alpha<\zeta$ such that
$p\in A'_{\alpha}$, and let
$D_p\subseteq P_{\gamma(p)}$ be a set with cardinal less than $\kappa$ such
that $D_p\cap\ocint{-\infty,p}$ is cofinal with
$P_{\gamma(p)}\cap\ocint{-\infty,p}$ and $D_p\cap\coint{p,\infty}$ is
cofinal with $P_{\gamma(p)}\cap\coint{p,\infty}$.
Define $g$ inductively, on each $A'_{\alpha}$ in turn, by setting
$g(p)=f_{\gamma(p)}(p)\cup\bigcup_{q\in D_p}g(q)$ for every $p\in P$.
Because $\kappa$ is regular, $g$ is a function from $P$ to
$[P]^{<\kappa}$.

Now $g$ is a Freese-Nation function on $P$.   \Prf\ I induce on $\alpha$
to show that if $p$, $q\in P$ and $\max(\gamma(p),\gamma(q))=\alpha$
then $g(p)\cap g(q)\cap[p,q]\ne\emptyset$.   For the inductive step to
$\alpha<\zeta$, if $\gamma(p)=\gamma(q)=\alpha$ then

\Centerline{$g(p)\cap g(q)\cap[p,q]
\supseteq f_{\alpha}(p)\cap f_{\alpha}(q)\cap[p,q]\ne\emptyset$.}

\noindent If $\gamma(p)<\gamma(q)=\alpha$, then there is an $r\in D_q$
such that $p\le r\le q$;  now $\max(\gamma(p),\gamma(r))<\alpha$, so

\Centerline{$g(p)\cap g(q)\cap[p,q]
\supseteq g(p)\cap g(r)\cap[p,r]
\ne\emptyset$}

\noindent by the inductive hypothesis.   The same argument works if
$\gamma(q)<\gamma(p)$.\ \Qed
}%end of proof of 518G

\leader{518H}{Lemma} Suppose that $\kappa$ is an uncountable cardinal
of countable cofinality such that $\square_{\kappa}$ is true and
$\cff[\lambda]^{\le\omega}\le\lambda^+$ for every $\lambda\le\kappa$.
Then there are families
$\langle M_{\alpha n}\rangle_{\alpha<\kappa^+,n\in\Bbb N}$,
$\ofamily{\alpha}{\kappa^+}{M_{\alpha}}$ of sets and a function $\sk$
such that

(i) $\#(M_{\alpha n})<\kappa$ whenever $\alpha<\kappa^+$ and
$n\in\Bbb N$;

(ii) $\sequencen{M_{\alpha n}}$ is non-decreasing for each
$\alpha<\kappa^+$;

(iii) $\ofamily{\alpha}{\kappa^+}{M_{\alpha}}$ is a non-decreasing
family, $M_{\alpha}=\bigcup_{\beta<\alpha}M_{\beta}$ for every
non-zero limit ordinal $\alpha<\kappa^+$, and
$\kappa^+\subseteq\bigcup_{\alpha<\kappa^+}M_{\alpha}$;

(iv) if $\alpha<\kappa^+$ has uncountable cofinality,
$M_{\alpha}=\bigcup_{n\in\Bbb N}M_{\alpha n}$;

(v) $X\subseteq\sk(X)$ for every set $X$;

(vi) $\sk(X)$ is countable whenever $X$ is countable;

(vii) $A\subseteq\sk(X)$ whenever $A\in\sk(X)$ is countable;

(viii) $\sk(X)\subseteq\sk(Y)$ whenever $X\subseteq\sk(Y)$;

(ix) for every $\alpha<\kappa^+$ of uncountable cofinality there is an
$m\in\Bbb N$ such that whenever $n\ge m$ and $A\subseteq M_{\alpha n}$
is countable there is a countable set $D\in M_{\alpha n}$ such that
$A\subseteq\sk(D)$;

(x) $\bigcup_{\alpha<\kappa^+}M_{\alpha}\cap[\kappa]^{\le\omega}$ is
cofinal with $[\kappa]^{\le\omega}$.

\proof{{\bf (a)} There is a strictly increasing sequence
$\sequencen{\kappa_n}$ of infinite cardinals with supremum $\kappa$;  since

\Centerline{$\kappa_n^+\le\cff[\kappa_n^+]^{\le\omega}
\le\max(\kappa_n^+,\cff[\kappa_n]^{\le\omega})=\kappa_n^+$}

\noindent for each $n$ (5A1F(e-iv)), we can suppose that in fact
$\cff[\kappa_n]^{\le\omega}=\kappa_n$ for
every $n$.   Take $\ofamily{\alpha}{\kappa^+}{C_{\alpha}}$
witnessing $\square_{\kappa}$, so that

\inset{for every $\alpha<\kappa^+$, $C_{\alpha}\subseteq\alpha$ is a
closed cofinal set in $\alpha$ of order type at most $\kappa$,

whenever $\delta<\alpha<\kappa^+$ and
$\delta=\sup(\delta\cap C_{\alpha})$, then
$C_{\delta}=\delta\cap C_{\alpha}$}

\noindent (5A6D(a-ii)).   For $\alpha<\kappa^+$ set

\Centerline{$C'_{\alpha}
=\{\delta:\delta<\alpha$, $\delta=\sup(\delta\cap C_{\alpha})\}
\subseteq C_{\alpha}$}

\noindent and

\Centerline{$C'_{\alpha n}
=\{\delta:\delta\in C'_{\alpha},
\,\otp(\delta\cap C_{\alpha})<\kappa_n\}$}

\noindent for each $n$.   Because $\otp(C_{\alpha})\le\kappa$,
$C'_{\alpha}=\bigcup_{n\in\Bbb N}C'_{\alpha n}$, while
$\#(C'_{\alpha n})\le\kappa_n$ for each $n$.    Note that if $\alpha$ has
uncountable cofinality, $C'_{\alpha}$ will be cofinal with $\alpha$.

\medskip

{\bf (b)} Let $g:\kappa^+\to[\kappa]^{\le\omega}$ be such that
$g[\kappa^+]$ is cofinal with $[\kappa]^{\le\omega}$.
For each non-zero $\alpha<\kappa^+$, fix on a surjective
function $f_{\alpha}:\kappa\to\alpha$.   For each $n\in\Bbb N$, let
$g_n:\kappa_n\to[\kappa_n]^{\le\omega}$ be such that $g_n[\kappa_n]$ is
cofinal with $[\kappa_n]^{\le\omega}$.   For each $\alpha<\kappa^+$, let
$h_{\alpha}:\#(C_{\alpha})\to C_{\alpha}$ be a bijection.
Now, for any set $X$, write
$\sk(X)$ for the smallest set including $X$ and such that

\inset{$g(\alpha)\in\sk(X)$ whenever $\alpha\in\sk(X)\cap\kappa^+$,

$f_{\alpha}(\xi)\in\sk(X)$ whenever $0<\alpha<\kappa^+$,
$\xi<\kappa$ and $\alpha$, $\xi\in\sk(X)$,

$g_n(\xi)\in\sk(X)$ whenever $n\in\Bbb N$ and
$\xi\in\kappa_n\cap\sk(X)$,

$h_{\alpha}[A]\in\sk(X)$ whenever $\alpha\in\sk(X)\cap\kappa^+$ and
$A\in\sk(X)$,

$A\cup B\in\sk(X)$ whenever $A$, $B\in\sk(X)$,

$A\subseteq\sk(X)$ whenever $A\in\sk(X)$ is countable.
}%end of inset

\noindent  Of course we always have

\Centerline{$\sk(\sk(X))=\sk(X)=\bigcup\{\sk(I):I\in[X]^{<\omega}\}$}

\noindent and $\#(\sk(X))\le\max(\omega,\#(X))$, so (v)-(viii) are all
true.

\medskip

{\bf (c)} For each $\alpha<\kappa^+$ and $n\in\Bbb N$, set
$M_{\alpha n}=\sk(\kappa_n\cup C'_{\alpha n})$ and
$M_{\alpha}=\sk(\kappa\cup\alpha)$.   Then

\Centerline{$\#(M_{\alpha n})\le\max(\omega,\kappa_n,\#(C'_{\alpha n}))
<\kappa$.}

\noindent Also $\ofamily{\alpha}{\kappa^+}{M_{\alpha}}$ is
non-decreasing and $M_{\alpha}=\bigcup_{\beta<\alpha}M_{\beta}$ whenever
$\alpha<\kappa$ is a non-zero limit ordinal, while
$\kappa^+\subseteq\bigcup_{\alpha<\kappa^+}M_{\alpha}$.  This deals with
(i)-(iii).

\medskip

{\bf (d)} Now for (iv):
$M_{\alpha}=\bigcup_{n\in\Bbb N}M_{\alpha n}$ whenever
$\alpha<\kappa^+$ has uncountable cofinality.   \Prf\ Of course
$M_{\alpha n}\subseteq M_{\alpha}$ for every $n$ just because
$\sk(X)\subseteq\sk(Y)$ whenever $X\subseteq Y$.   On the other hand,
if $\beta<\alpha$, take $\delta\in C'_{\alpha}$ such that
$\beta<\delta$, and $\xi<\kappa$ such that $f_{\delta}(\xi)=\beta$;
then if $n\in\Bbb N$ is such that $\xi<\kappa_n$ and
$\otp(\delta\cap C_{\alpha})<\kappa_n$,
$\beta$ will be in $M_{\alpha n}$.   So
$\bigcup_{n\in\Bbb N}M_{\alpha_n}\supseteq\alpha$.   Moreover,

\Centerline{$\kappa=\bigcup_{n\in\Bbb N}\kappa_n
\subseteq\bigcup_{n\in\Bbb N}M_{\alpha n}$.}

\noindent So
$\alpha\cup\kappa\subseteq\bigcup_{n\in\Bbb N}M_{\alpha n}$ and
$M_{\alpha}$ must be exactly $\bigcup_{n\in\Bbb N}M_{\alpha n}$.\ \Qed

\medskip

{\bf (e)} Again suppose that $\alpha<\kappa^+$ has uncountable cofinality.
Then there must be an $m\in\Bbb N$ such that $C'_{\alpha m}$ is cofinal
with $\alpha$.   Suppose that $n\ge m$ and
$A\subseteq M_{\alpha n}$ is countable.
Then there must be a countable set $C\subseteq\kappa_n\cup C'_{\alpha n}$
such that $A\subseteq\sk(C)$.   Let $\delta\in C'_{\alpha m}$ be such that
$C\cap\alpha\subseteq\delta$.   Then
$C_{\delta}=\delta\cap C_{\alpha}$ has cardinal less than
$\kappa_m\le\kappa_n$, so $(C\cap\kappa_n)\cup h_{\delta}^{-1}[C]$
is a countable subset of $\kappa_n$ and is included in $g_n(\xi)$ for
some $\xi<\kappa_n$.
Now $\xi$ and $\delta$ belong to $M_{\alpha n}$, so $g_n(\xi)$
and $h_{\delta}[g_n(\xi)]$ and $D=g_n(\xi)\cup h_{\delta}[g_n(\xi)]$ all
belong to $M_{\alpha n}$, and are countable.   But $C\subseteq D$, so
$A\subseteq\sk(D)$, as required by (ix).

\medskip

{\bf (f)} Finally, if $A\subseteq\kappa$ is countable, there is a
$\beta<\kappa^+$ such that $g(\beta)\supseteq A$, and now
$g(\beta)\in M_{\beta+1}$.   So (x) is true.
}%end of proof of 518H

\leader{518I}{Theorem}\cmmnt{ ({\smc Fuchino \& Soukup 97})} Let
$\frak A$ be a ccc Dedekind complete Boolean algebra.   Suppose that

\inset{($\alpha$) $\cff[\lambda]^{\le\omega}\le\lambda^+$ for every
cardinal $\lambda\le\tau(\frak A)$,

($\beta$) $\square_{\lambda}$ is true for every uncountable cardinal
$\lambda\le\tau(\frak A)$ of countable cofinality.}

\noindent Let $\frak A$ be a ccc Dedekind complete Boolean algebra, and
$\kappa$ a regular uncountable cardinal such that
$\FN(\frak B)\le\kappa$
for every countably generated order-closed subalgebra $\frak B$ of
$\frak A$.    Then $\FN(\frak A)\le\kappa$.

\proof{ Induce on the Maharam type $\tau(\frak A)$ of
$\frak A$.

\medskip

{\bf (a)} If $\tau(\frak A)\le\omega$ the result is trivial.

\medskip

{\bf (b)} For the inductive step to $\tau(\frak A)=\lambda$, where
$\lambda$ is an infinite cardinal of uncountable cofinality, let
$\ofamily{\xi}{\lambda}{a_{\xi}}$ enumerate a $\tau$-generating subset
of $\frak A$.   For each $\beta<\lambda$, let $\frak B_{\beta}$ be the
order-closed subalgebra of $\frak A$ generated by
$\{a_{\xi}:\xi<\beta\}$, and for $\alpha\le\lambda$ set
$\frak A_{\alpha}=\bigcup_{\beta<\alpha}\frak B_{\beta}$.   By the
inductive hypothesis, $\FN(\frak B_{\beta})\le\kappa$ for every
$\beta<\lambda$.   Also, for $\alpha<\kappa$, either $\alpha$ has
uncountable cofinality, in which case (because $\frak A$ is ccc)
$\frak A_{\alpha}=\frak B_{\alpha}$ is order-closed, or $\alpha$ has
countable cofinality, in which case $\frak A_{\alpha}$ is a countable
union of order-closed subalgebras.   In either case, the Freese-Nation
index of $\frak A_{\alpha}$ in $\frak A_{\alpha+1}$ is countable (518Fc).
Because $\cf\lambda>\omega$, $\frak A=\frak A_{\lambda}$.   By
518G, $\FN(\frak A)\le\kappa$.

\medskip

{\bf (c)(i)} For the inductive step to $\tau(\frak A)=\lambda$, where
$\lambda$ is an uncountable cardinal of countable cofinality,
we may use the method of Lemma 518H to construct
$\langle M_{\alpha n}\rangle_{\alpha<\lambda^+,n\in\Bbb N}$,
$\ofamily{\alpha}{\lambda^+}{M_{\alpha}}$ and $\sk$ as described there.
Enumerate a $\tau$-generating set in $\frak A$ as
$\ofamily{\xi}{\lambda}{a_{\xi}}$, and for any set $X$ write $\frak B_X$
for the order-closed subalgebra of $\frak A$ generated by
$\{a_{\xi}:\xi\in X\cap\lambda\}$.   For each $\alpha<\lambda^+$ set
$\frak E_{\alpha}=\bigcup\{\frak B_{\sk(D)}:D\in M_{\alpha}$ is countable$\}$.

\medskip

\quad{\bf (ii)} If $\alpha<\lambda^+$ has uncountable cofinality, then
$\frak E_{\alpha}$ is the union of a non-decreasing sequence of order-closed
subalgebras of $\frak A$ with Maharam type less than $\lambda$.   \Prf\ By
(ix) of 518H,
there is an $m\in\Bbb N$ such that whenever $n\ge m$ and
$\Cal D\subseteq M_{\alpha n}$ is countable there is a countable set
$F\in M_{\alpha n}$ such that $\Cal D\subseteq\sk(F)$.   For each $n\ge m$,
set
$\frak C_n=\bigcup\{\frak B_{\sk(D)}:D\in M_{\alpha n}$ is countable$\}$.
Then for any countable set $C\subseteq\frak C_n$, there is a countable
set $\Cal D$ of countable sets belonging to $M_{\alpha n}$ such that
$C\subseteq\bigcup_{D\in\Cal D}\frak B_{\sk(D)}$.   So there is a
countable set $F\in M_{\alpha n}$ such that $\Cal D\subseteq\sk(F)$;  by
518H(vii), $D\subseteq\sk(F)$ and
$\frak B_{\sk(D)}\subseteq\frak B_{\sk(F)}$ (518H(viii))
for every $D\in\Cal D$.
But this means that $C\subseteq\frak B_{\sk}(F)\subseteq\frak C_n$,
while $\frak B_{\sk}(F)$ is an order-closed subalgebra of $\frak A$.
Because $\frak A$ is ccc, this is enough to show that $\frak C_n$ is an
order-closed subalgebra of $\frak A$;  by 518H(ii) and 518H(iv),
$\langle\frak C_n\rangle_{n\ge m}$ is non-decreasing and has union
$\frak E_{\alpha}$.   Each $\frak C_n$ is $\tau$-generated by

\Centerline{$\{a_{\eta}:$ there is a countable
$D\in M_{\alpha n}$ such that $\eta\in\sk(D)\cap\lambda^+\}$,}

\noindent so (using 518H(vi))

\Centerline{$\tau(\frak C_n)\le\max(\omega,\#(M_{\alpha n}))<\lambda$.
\Qed}

It follows from 518Fc again that the Freese-Nation index of $\frak E_{\alpha}$ in
$\frak A$ is countable, and from the inductive hypothesis we see also
that $\FN(\frak C_n)\le\kappa$ for every $n\ge m$,
so that (using 518G, as usual) $\FN(\frak E_{\alpha})\le\kappa$.

\medskip

\quad{\bf (iii)} If $\alpha<\lambda^+$ is the union of a sequence
$\sequencen{\alpha_n}$ of ordinals of uncountable cofinality, then
$M_{\alpha}=\bigcup_{n\in\Bbb N}M_{\alpha_n}$ (518H(iii)), so
$\frak E_{\alpha}=\bigcup_{n\in\Bbb N}\frak E_{\alpha_n}$ is again a countable union
of order-closed subalgebras of $\frak A$, and the Freese-Nation index of
$\frak E_{\alpha}$ in $\frak A$ is at most $\omega$.   Moreover, because
$\FN(\frak E_{\alpha_n})\le\kappa$ for each $n$, $\FN(\frak E_{\alpha})\le\kappa$.

\medskip

\quad{\bf (iv)} If $a\in\frak A$, there is a countable set
$D\subseteq\lambda$ such that $a\in\frak B_D$.   But now there is an
$\alpha<\lambda^+$, of uncountable cofinality, such that $D\subseteq D'$
for some countable $D'\in M_{\alpha}$ (518H(x)), and

\Centerline{$a\in\frak B_D\subseteq\frak B_{D'}
\subseteq\frak B_{\sk(D')}\subseteq \frak E_{\alpha}$,}

\noindent by 518H(v).

\medskip

\quad{\bf (v)} Let $F\subseteq\lambda^+$ be the set of ordinals which
are either of uncountable cofinality, or the union of a sequence of such
ordinals;  so that $F$ is a closed cofinal
set in $\lambda^+$, and $\frak E_{\alpha}$
has countable Freese-Nation index in $\frak A$ for every $\alpha\in F$.
By (iv), $\bigcup_{\alpha\in F}\frak E_{\alpha}=\frak A$.   So if we enumerate
$F$ in ascending order as $\ofamily{\xi}{\lambda^+}{\alpha_{\xi}}$ and
set $P_{\xi}=\frak E_{\alpha_{\xi}}$ for each $\xi$, $P_{\lambda^+}=\frak A$
then $\langle P_{\xi}\rangle_{\xi\le\lambda^+}$ satisfies the conditions
of 518G, so $\FN(\frak A)\le\kappa$, and the induction proceeds.
}%end of proof of 518I

\leader{518J}{Lemma} Let $\lambda$ be an infinite cardinal and $\frak G$
the regular open algebra of $\{0,1\}^{\lambda}$.   Suppose
that $\kappa$ is the least cardinal of uncountable cofinality greater
than or equal to $\FN(\frak G)$.   Then $\kappa\le\frak c^+$ and we have
a family $\Cal V\subseteq[\lambda]^{\le\frak c}$, cofinal with
$[\lambda]^{\le\frak c}$, such that $\#(\{A\cap V:V\in\Cal V\})<\kappa$
for every countable set $A\subseteq\lambda$.

\proof{ Actually it is more convenient to work with
$\frak G=\RO(\{0,1\}^{\lambda\times\Bbb N})$;  of course this makes no
difference.

\medskip

{\bf (a)} I will use the phrase `cylinder set' to mean a subset
of $X=\{0,1\}^{\lambda\times\Bbb N}$ of the form
$\{x:x\restr J=z\}$, where $J\subseteq\lambda\times\Bbb N$ is finite.
For $I\subseteq\lambda$, let $\frak G_I$ be the
order-closed subalgebra of $\frak G$ consisting of those regular open
sets determined by coordinates in $I\times\Bbb N$.   For $G\in\frak G$,
there is a smallest subset $J(G)$ of $\lambda$ such that
$G\in\frak G_{J(G)}$
(use 4A2B(g-ii)).  Recall that $J(G)$ is always countable (use
4A2E(b-i)), so that $\#(\frak G_I)\le\frak c$ whenever
$\#(I)\le\frak c$.

\medskip

{\bf (b)} The function $G\mapsto\frak G_{J(G)}$ is a Freese-Nation
function.  \Prf\ Suppose that $G_1\subseteq G_2$ in $\frak G$.   Set
$K=J(G_1)$ and $L=J(G_2)$, and let
$\phi:X\to\{0,1\}^{L\times\Bbb N}$ be the
canonical map, so that $\phi^{-1}[\phi[G_2]]=G_2$.   Set
$H=\phi^{-1}[\interior\overline{\phi[G_1]}]$;  because $\phi$ is
continuous and open (4A2B(f-i)),
$H=\interior\overline{\phi^{-1}[\phi[G_1]]}$ (4A2B(f-ii)).   In
particular, $H$ is a regular open set;  at the same time,
$H\supseteq G_1$ and
$H\subseteq\interior\overline{\phi^{-1}[\phi[G_2]]}=G_2$ and $H$ is
determined by coordinates in $L\times\Bbb N$, so $H\in\frak G_L$.
Next, $\phi[G_1]\subseteq\{0,1\}^{L\times\Bbb N}$ is determined by
coordinates in $(K\cap L)\times\Bbb N$, so
$\interior\overline{\phi[G_1]}$ also is (4A2B(g-i)) and $H$ is
determined by coordinates in $K\times\Bbb N$.   Thus
$H\in\frak G_{J(G_1)}\cap\frak G_{J(G_2)}$, which is what we need.\ \Qed

Since $\#(\frak G_{J(G)})\le\frak c$ for every $G$,
$\FN(\frak G)\le\frak c^+$;  as $\cf\frak c^+$ is surely uncountable,
$\kappa\le\frak c^+$.

\medskip

{\bf (c)} Now let $f:\frak G\to[\frak G]^{<\kappa}$
be a Freese-Nation function.  Let $\Cal V$ be the family of those sets
$V\in[\lambda]^{\le\frak c}$ such that $f(G)\subseteq\frak G_V$ for
every $G\in\frak G_V$;  because $\#(f(G))\le\frak c$ for every $G$, and
$\#(\frak G_V)\le\frak c$ whenever $V\in[\lambda]^{\le\frak c}$,
$\Cal V$ is cofinal with $[\lambda]^{\le\frak c}$.

\medskip

{\bf (d)} Fix a countable set $A\subseteq\lambda$ and
$\zeta\in A$ for the moment.   Let
$\family{\xi}{A}{C_{\xi}}$ be a disjoint family of
non-empty cylinder sets determined by coordinates in
$\{\zeta\}\times\Bbb N$;  for each $\xi\in A$, set
$C'_{\xi}=\{x:x\in X$, $x(\xi,0)=1\}$.   Set

\Centerline{$G^*=\sup_{\xi\in A}C_{\xi}\cap C'_{\xi}\in\frak G_A$.}

\noindent Next, for $V\in\Cal V$, set

\Centerline{$G_V=\sup_{\xi\in A\cap V}C_{\xi}\cap C'_{\xi}$,
\quad$G'_V=\sup\{H:H\in\frak G_V$, $H\subseteq G^*\}$}

\noindent so that $G_V\subseteq G^*$ and $G'_V\in\frak G_V$.
Now if $\zeta\in V\in\Cal V$, $G_V=G'_V$.   \Prf\
Since $C_{\xi}\cap C'_{\xi}\in\frak G_V$ for every $\xi\in V\cap A$,
$G_V\in\frak G_V$ and $G_V\subseteq G'_V$.   \Quer\ Suppose, if
possible, that $G_V\ne G'_V$.   Then $G'_V\setminus\overline{G}_V$ is a
non-empty set belonging to $\frak G_V$, so includes a non-empty cylinder
set $D$ determined by coordinates in $V\times\Bbb N$.   Express $D$ as
$D'\cap D''$, where $D'$ is determined by coordinates in
$(V\cap A)\times\Bbb N$ and $D''$ by coordinates in
$(V\setminus A)\times\Bbb N$.   As
$D'\cap D''\subseteq G^*\in\frak G_A$, $D'\subseteq G^*$, so
$D'\cap C_{\xi}\subseteq C'_{\xi}$ for $\xi\in A$.

If $\xi\in A\setminus V$, $D\cap C'_{\xi}$ is determined by coordinates in
a set not containing $\{(\xi,0)\}$, but is included in $C'_{\xi}$, so must
be empty.   Thus

\Centerline{$D\subseteq D'
=\sup_{\xi\in A\cap V}D'\cap C_{\xi}\cap C'_{\xi}
\subseteq G_V$,}

\noindent which is impossible.\ \BanG\  Accordingly
$G_V=G'_V$, as claimed.\ \Qed

Note next that if $V$, $V'\in\Cal V$ and $V\cap A\ne V'\cap A$, then
$G_V\ne G_{V'}$, because if $\xi\in A\cap(V\symmdiff V')$ then
$C_{\xi}\cap C'_{\xi}\subseteq G_V\symmdiff G_{V'}$.

At this point, consider $f(G^*)$.   For each $V\in\Cal V$ such that
$\zeta\in V$, there must be some $H_V\in f(G^*)\cap f(G_V)$ such that
$G_V\subseteq H_V\subseteq G^*$.   By the definition of $\Cal V$,
$H_V\in\frak G_V$ so $H_V\subseteq G'_V=G_V$ and $H_V=G_V$.   But this
shows that

\Centerline{$\#(\{V\cap A:\zeta\in V\in\Cal V\})
\le\#(\{G_V:\zeta\in V\in\Cal V\})
\le\#(f(G^*))
<\kappa$.}

\medskip

{\bf (e)} Now take any countable $A\subseteq\lambda$.   By (d), we see that
$\#(\{A\cap V:\zeta\in V\in\Cal V\})<\kappa$ for every $\zeta\in A$.
But now

\Centerline{$\{A\cap V:V\in\Cal V\}
\subseteq\{\emptyset\}
  \cup\bigcup_{\zeta\in A}\{A\cap V:\zeta\in V\in\Cal V\}$}

\noindent has size less than $\kappa$, because $\cf\kappa>\omega$.
This completes the proof.
}%end of proof of 518J

\vleader{36pt}{518K}{Theorem}\cmmnt{
({\smc Fuchino Geschke Shelah \& Soukup 01})}
Suppose that $\lambda>\frak c$ is a cardinal of countable cofinality
such that CTP$(\lambda^+,\lambda)$ is true\cmmnt{ (definition: 5A6F)}.
Let $\frak A$ be a Dedekind
complete Boolean algebra with cardinal at least $\lambda$.   Then
$\FN(\frak A)\ge\omega_2$.

\proof{{\bf (a)} By 518Cb and 518Cc, it is enough to show that
$\FN(\frak G)\ge\omega_2$, where $\frak G$ is the regular open algebra
of $\{0,1\}^{\lambda}$.

\medskip

{\bf (b)} \Quer\ Suppose, if possible, that $\FN(\frak G)\le\omega_1$.
Let $\Cal V\subseteq[\lambda]^{\le\frak c}$ be as in 518J, with
$\kappa=\omega_1$.   Note first that if $\Cal V'\subseteq\Cal V$ and
$\#(\Cal V')\le\lambda$ then there is an
$A\in[\lambda]^{\le\omega}$ such that $A\not\subseteq V$ for
every $V\in\Cal V'$.   \Prf\ Let $\sequencen{\lambda_n}$ be a sequence
of cardinals less than $\lambda$ with supremum $\lambda$.   Express
$\Cal V'$ as
$\bigcup_{n\in\Bbb N}\Cal V_n$ where $\#(\Cal V_n)\le\lambda_n$ for each
$n$.   For each $n\in\Bbb N$,
$\#(\bigcup\Cal V_n)<\lambda$, so we can find
an $\alpha_n\in\lambda\setminus\bigcup\Cal V_n$;  now
$A=\{\alpha_n:n\in\Bbb N\}$ is not included in any member of $\Cal V'$.\
\Qed

Choose $\ofamily{\xi}{\lambda^+}{A_{\xi}}$ and
$\ofamily{\xi}{\lambda^+}{V_{\xi}}$ inductively, as follows.   Given
$V_{\eta}\in\Cal V$ for $\eta<\xi$, choose
$A_{\xi}\in[\lambda]^{\le\omega}$ such that
$A_{\xi}\not\subseteq V_{\eta}$ for every $\eta<\xi$;  now take
$V_{\xi}\in\Cal V$ such that $A_{\xi}\subseteq V_{\xi}$, and continue.

Because CTP$(\lambda^+,\lambda)$ is true, there is an uncountable set
$B\subseteq\lambda^+$ such that $C=\bigcup_{\xi\in B}A_{\xi}$ is
countable (5A6F(b-ii)).   If $\eta$, $\xi\in B$ and $\eta<\xi$, then
$A_{\xi}=A_{\xi}\cap C\cap V_{\xi}\not\subseteq V_{\eta}$, so
$C\cap V_{\xi}\ne C\cap V_{\eta}$.   But this means that
$\{C\cap V:V\in\Cal V\}$ is uncountable, contrary to the choice of
$\Cal V$.\ \Bang

Thus $\FN(\frak G)\ge\omega_2$, and the proof is complete.
}%end of proof of 518K

\cmmnt{\medskip

\noindent{\bf Remark} Compare {\smc Fuchino \& Soukup 97}, Theorem 12,
where it is shown that if the generalized continuum hypothesis and
CTP$(\omega_{\omega+1},\omega_{\omega})$ are both true the
Freese-Nation number of $[\omega_{\omega}]^{\le\omega}$ is greater than
$\omega_1$, and also {\smc Fuchino Geschke Shelah \& Soukup 01}, Theorem
4.2, where a different special axiom is used to find a ccc Dedekind
complete Boolean algebra with cardinal $\omega_{\omega+1}$ with Freese-Nation
number greater than $\omega_1$.
}%end of comment

\leader{518L}{}\cmmnt{ I turn now to the associated idea of `tight
filtration' (511Di).   Before discussing conditions ensuring the existence
of such filtrations, I give the application of the idea which is most
important for this book.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra, $\frak B$ a tightly $\omega_1$-filtered Boolean algebra,
and $\pi:\frak A\to\frak B$ a surjective sequentially order-continuous
Boolean homomorphism;  suppose that $\frak B\ne\{0\}$.   Then there is a
Boolean homomorphism $\theta:\frak B\to\frak A$ such that $\pi\theta b=b$
for every $b\in\frak B$.

\proof{ Let $\ofamily{\xi}{\zeta}{b_{\xi}}$ be a tight
$\omega_1$-filtration in $\frak B$;  for $\alpha\le\zeta$,
write $\frak C_{\alpha}$ for the
subalgebra of $\frak B$ generated by $\{b_{\xi}:\xi<\alpha\}$.   Define
Boolean homomorphisms $\theta_{\alpha}:\frak C_{\alpha}\to\frak A$
inductively, as follows.   Start with $\frak C_0=\{0,1\}$,
$\theta_00=\emptyset$, $\theta_01=1$.   Given $\theta_{\alpha}$, let
$B$, $B'\subseteq\frak C_{\alpha}$ be countable sets such that $B$ is
a cofinal subset of
$\{b:b\in\frak C_{\alpha}$, $b\Bsubseteq b_{\alpha}\}$ and
$B'$ is a cofinal subset of
$\{b:b\in\frak C_{\alpha}$, $b\Bsubseteq 1\Bsetminus b_{\alpha}\}$.
Choose any $a\in\frak A$ such that $\pi a=b_{\alpha}$ and set

\Centerline{$a_{\alpha}
=(a\Bcup\sup_{b\in B}\theta_{\alpha}b)
  \Bsetminus\sup_{b\in B'}\theta_{\alpha}b$.}

\noindent Because $B$ and $B'$ are both countable and $\frak A$ is Dedekind
$\sigma$-complete, $a_{\alpha}$ is defined in $\frak A$.
Because $B$ and $B'$ are cofinal with
$\{b:b\in\frak C_{\alpha}$, $b\Bsubseteq b_{\alpha}\}$ and
$\{b:b\in\frak C_{\alpha}$, $b\Bsubseteq 1\Bsetminus b_{\alpha}\}$
respectively,
$\theta b\subseteq a_{\alpha}$ whenever
$b\in\frak C_{\alpha}$ and $b\Bsubseteq b_{\alpha}$,
while $\theta b\Bcap a_{\alpha}=\emptyset$ whenever
$b\in\frak C_{\alpha}$ and $b\Bsubseteq 1\Bsetminus b_{\alpha}$.
This means that we can define a Boolean homomorphism
$\theta_{\alpha+1}:\frak C_{\alpha+1}\to\frak A$ by setting

\Centerline{$\theta_{\alpha+1}((b\Bcap b_{\alpha})
  \Bcup(c\Bsetminus b_{\alpha}))
=(\theta_{\alpha}b\Bcap a_{\alpha})
  \Bcup(\theta_{\alpha}c\Bsetminus a_{\alpha})$}

\noindent for all $b$, $c\in\frak C_{\alpha}$ (312O).

This is the inductive step to a successor ordinal.   For the inductive
step to a non-zero limit ordinal $\alpha\le\zeta$,
$\frak C_{\alpha}=\bigcup_{\xi<\alpha}\frak C_{\xi}$ and we can define
$\theta_{\alpha}$ by setting $\theta_{\alpha}a=\theta_{\xi}a$ whenever
$\xi<\alpha$ and $a\in\frak C_{\xi}$.

An easy induction (using the hypothesis that $\pi$ is sequentially
order-continuous) now shows that $c=\pi\theta_{\alpha}c$
whenever $\alpha\le\zeta$ and $c\in\frak C_{\alpha}$, so that
$\pi\theta_{\zeta}$ is the identity homomorphism on
$\frak C_{\zeta}=\frak B$.
}%end of proof of 518L

\leader{518M}{Theorem} Let $\frak A$ be a Boolean algebra and $\kappa$
a regular infinite cardinal such that $\FN(\frak A)\le\kappa$ and
$\#(\frak A)\le\kappa^+$.   Then $\frak A$ is tightly
$\kappa$-filtered.

\proof{{\bf (a)} Let $\ofamily{\xi}{\kappa^+}{a_{\xi}}$ run over
$\frak A$, and let $f:\frak A\to[\frak A]^{<\kappa}$ be a Freese-Nation
function.   For each $\alpha<\kappa^+$, let $\frak A_{\alpha}$ be the
smallest subalgebra of $\frak A$ containing $a_{\xi}$ for every
$\xi<\alpha$ and such that $f(a)\subseteq\frak A_{\alpha}$ for every
$a\in\frak A_{\alpha}$.   Then
$\ofamily{\alpha}{\kappa^+}{\frak A_{\alpha}}$ is a non-decreasing
family with union $\frak A$, and
$\#(\frak A_{\alpha})\le\kappa$ for every $\alpha<\kappa^+$.

\medskip

{\bf (b)(i)} If $\alpha<\kappa^+$, the Freese-Nation index of
$\frak A_{\alpha}$ in $\frak A$ is at most $\kappa$.   \Prf\ If
$a\in\frak A$, then whenever $b\in\frak A_{\alpha}$ and $b\Bsubseteq a$,
there is a $c\in f(a)\cap f(b)\cap[b,a]$.   Now
$c\in f(a)\cap\frak A_{\alpha}$.   This shows that
$f(a)\cap\frak A_{\alpha}\cap[0,a]$ is cofinal with
$\frak A_{\alpha}\cap[0,a]$, so that
$\cf(\frak A_{\alpha}\cap[0,a])<\kappa$.   By 518Fa, this is what we
need to know.\ \Qed

\medskip

\quad{\bf (ii)} If $\alpha<\kappa^+$, $I\in[\frak A]^{<\kappa}$ and
$\frak B$ is the subalgebra of $\frak A$ generated by
$\frak A_{\alpha}\cup I$, then the Freese-Nation index of $\frak B$ in
$\frak A$ is at most $\kappa$, by 518Fb.

\medskip

{\bf (c)} For each $\alpha<\kappa^+$ enumerate
$\frak A_{\alpha+1}\setminus\frak A_{\alpha}$ as
$\ofamily{\xi}{\kappa_{\alpha}}{a_{\alpha\xi}}$, where
$\kappa_{\alpha}\le\kappa$.   Well-order $\frak A$ by setting
$a\preccurlyeq a'$ if {\it either} there is some $\alpha<\kappa^+$ such
that $a\in\frak A_{\alpha}$ and $a'\notin\frak A_{\alpha}$ {\it or}
there are $\alpha<\kappa^+$ and $\xi\le\eta<\kappa_{\alpha}$ such that
$a=a_{\alpha\xi}$ and $a'=a_{\alpha\eta}$.
Let $\zeta\in\On$ be the order type of this well-ordering and
$\ofamily{\xi}{\zeta}{b_{\xi}}$ the corresponding enumeration of
$\frak A$.   For each $\beta\le\zeta$, let
$\frak B_{\beta}$ be the subalgebra of $\frak A$ generated by
$\{b_{\xi}:\xi<\beta\}$.   Then the Freese-Nation index of
$\frak B_{\beta}$ in $\frak A$ is at most $\kappa$.    \Prf\ If
$\beta<\zeta$, there is a largest $\alpha<\kappa^+$ such that
$\frak A_{\alpha}\subseteq B_{\beta}$, and in this case
$\frak A_{\alpha}=\frak B_{\gamma}$ for some $\gamma\le\beta$, while
$\frak A_{\alpha+1}=\frak B_{\gamma'}$ for some $\gamma'>\beta$;
moreover, $\#(\beta\setminus\gamma)<\kappa_{\alpha}\le\kappa$, because
$\otp(\gamma'\setminus\gamma)=\kappa_{\alpha}$.   But this means that
$\frak B$, which is the subalgebra of $\frak A$ generated by
$\frak A_{\alpha}\cup\{b_{\xi}:\gamma\le\xi<\beta\}$, has Freese-Nation
index at most $\kappa$, by (b-ii) above.\ \Qed

Thus $\ofamily{\xi}{\zeta}{b_{\xi}}$ is a tight $\kappa$-filtration of
$\frak A$.
}%end of proof of 518M

\vleader{72pt}{518N}{Definition} Let $\frak A$ be a Boolean algebra and
$\kappa$ a cardinal.   Then a {\bf $\kappa$-Geschke system} for
$\frak A$ is a family $\Bbb G$ of subalgebras of $\frak A$ such that

\inset{($\alpha$) every element of $\frak A$ belongs to an element of
$\Bbb G$ with cardinal less than $\kappa$;

($\beta$) for any $\Bbb G_0\subseteq\Bbb G$, the subalgebra of $\frak A$
generated by $\bigcup\Bbb G_0$ belongs to $\Bbb G$;

($\gamma$) whenever $\frak B_1$, $\frak B_2\in\Bbb G$, $a\in\frak B_1$,
$b\in\frak B_2$ and $a\Bsubseteq b$, then there is a
$c\in\frak B_1\cap\frak B_2$ such that $a\Bsubseteq c\Bsubseteq b$.}

\cmmnt{\noindent (Of course ($\gamma$) can be rephrased as
`$\frak B_1\cap\frak B_2\cap[0,b]$ is cofinal with $\frak B_1\cap[0,b]$
whenever $\frak B_1$, $\frak B_2\in\Bbb G$ and $b\in\frak B_2$'.)}

\leader{518O}{Lemma}\dvArevised{2014}
Let $\frak A$ be a Boolean algebra,
$\kappa$ a cardinal and $\Bbb G$ a $\kappa$-Geschke system for
$\frak A$.   Suppose that $\lambda\ge\kappa$ is a regular uncountable
cardinal and that
$f:[\frak A]^{<\omega}\to[\frak A]^{<\lambda}$ is a function.
Then there is a $\frak B\in\Bbb G$ such that
$\#(\frak B)<\lambda$ and $f(I)\subseteq\frak B$ whenever
$I\in[\frak B]^{<\omega}$.

\proof{ Enlarging $f$ if necessary, we may suppose that $f(I)$ always
includes the subalgebra of $\frak A$ generated by $I$, and that
$f(\{a\})$ includes a member of $\Bbb G$, with cardinal less than $\kappa$ and
containing $a$, for every $a\in\frak A$.   If now we take
$A_0=\emptyset$ and $A_{n+1}=\bigcup\{f(I):I\in[A_n]^{<\omega}\}$ for each
$n\in\Bbb N$,
$\frak B=\bigcup_{n\in\Bbb N}A_n$ will be a subalgebra of $\frak A$, of
size less than $\lambda$, and a union of members of $\Bbb G$, so belongs
to $\Bbb G$;  while $f(I)\subseteq\frak B$ for every
$I\in[\frak B]^{<\omega}$.
}%end of proof of 518O

\leader{518P}{Lemma}\cmmnt{ ({\smc Geschke 02})} Let $\kappa$ be a
regular uncountable cardinal and $\frak A$ a Boolean algebra.   Then
$\frak A$ is tightly
$\kappa$-filtered iff there is a $\kappa$-Geschke system for $\frak A$.

\proof{{\bf (a)} Suppose that $\frak A$ is tightly $\kappa$-filtered.

\medskip

\quad{\bf (i)} Let $\ofamily{\xi}{\zeta}{a_{\xi}}$ be a tight
$\kappa$-filtration of $\frak A$.   For
$I\subseteq\zeta$ let $\frak A_I$ be the subalgebra of $\frak A$
generated by $\{a_{\xi}:\xi\in I\}$.   For $\alpha<\zeta$, there must be
subsets $U_{\alpha}$, $V_{\alpha}$ of $\frak A_{\alpha}$, with cardinal less
than $\kappa$, such that $U_{\alpha}$ is cofinal with
$\frak A_{\alpha}\cap[0,a_{\alpha}]$ and $V_{\alpha}$ is cofinal with
$\frak A_{\alpha}\cap[0,1\Bsetminus a_{\alpha}]$.   Let
$K_{\alpha}\in[\alpha]^{<\kappa}$ be such that
$U_{\alpha}\cup V_{\alpha}\subseteq\frak A_{K_{\alpha}}$.

Write $\Cal M$ for the family of those subsets $M$ of $\zeta$ such that
$K_{\alpha}\subseteq M$ for every $\alpha\in M$.

\medskip

\quad{\bf (ii)} If $M$, $N\in\Cal M$, $\gamma\le\zeta$,
$a\in\frak A_{M\cap\gamma}$, $b\in\frak A_{N\cap\gamma}$ and
$a\Bsubseteq b$, then there is a $c\in\frak A_{M\cap N\cap\gamma}$ such
that $a\Bsubseteq c\Bsubseteq b$.   \Prf\ Induce on $\gamma$.

\medskip

\qquad\grheada\ If $\gamma=0$ then

\Centerline{$\frak A_{M\cap\gamma}=\frak A_{N\cap\gamma}
=\frak A_{M\cap N\cap\gamma}=\{0,1\}$}

\noindent and the result is trivial.

\medskip

\qquad\grheadb\ For the inductive step to $\gamma=\alpha+1$, consider
the following cases.

\quad\qquad{\bf case 1} If $\alpha\notin M\cup N$ then
$a\in\frak A_{M\cap\alpha}$ and
$b\in\frak A_{N\cap\alpha}$, so the inductive hypothesis gives us a
$c\in\frak A_{M\cap N\cap\alpha}$ such that $a\Bsubseteq c\Bsubseteq b$.

\quad\qquad{\bf case 2} If $\alpha\in N\setminus M$, then
$a\in\frak A_{M\cap\alpha}$ and
$b$ is of the form $(b'\Bcap a_{\alpha})\Bcup(b''\Bsetminus a_{\alpha})$
where $b'$, $b''\in\frak A_{N\cap\alpha}$.   Now
$a\Bsetminus b'\in\frak A_{\alpha}$ and
$a\Bsetminus b'\Bsubseteq 1\Bsetminus a_{\alpha}$, so there is a
$v\in V_{\alpha}$ such that $a\Bsetminus b'\Bsubseteq v$.   Since
$K_{\alpha}\subseteq N\cap\alpha$, $v\in\frak A_{N\cap\alpha}$.
Similarly, there
is a $u\in U_{\alpha}\subseteq\frak A_{N\cap\alpha}$ such that
$a\Bsetminus b''\Bsubseteq u$.   We have

\Centerline{$a\Bsubseteq(u\Bcap b')\Bcup(v\Bcap b'')\Bcup(b'\Bcap b'')
\in\frak A_{N\cap\alpha}$,}

\noindent so the inductive hypothesis tells us that there is a
$c\in\frak A_{M\cap N\cap\alpha}$ such that

\Centerline{$a\Bsubseteq c
\Bsubseteq(u\Bcap b')\Bcup(v\Bcap b'')\Bcup(b'\Bcap b'')
\Bsubseteq b$.}

\quad\qquad{\bf case 3} Similarly, if $\alpha\in M\setminus N$,
then we express $a$ as
$(a'\Bcap a_{\alpha})\Bcup(a''\Bsetminus a_{\alpha})$ where $a'$,
$a''\in\frak A_{M\cap\alpha}$, and find
$v\in V_{\alpha}\subseteq\frak A_{M\cap\alpha}$,
$u\in U_{\alpha}\subseteq\frak A_{M\cap\alpha}$,
$c\in\frak A_{M\cap N\cap\alpha}$ such that

\Centerline{$a'\Bsetminus b\Bsubseteq v$,
\quad$a''\Bsetminus b\Bsubseteq u$,}

\Centerline{$a\Bsubseteq c
\Bsubseteq(b\Bcup u)\Bcap(b\Bcup v)\Bsubseteq b$.}

\quad\qquad{\bf case 4} Finally, if $\alpha\in M\cap N$, express $a$ as
$(a'\Bcap a_{\alpha})\Bcup(a''\Bsetminus a_{\alpha})$ and
$b$ as $(b'\Bcap a_{\alpha})\Bcup\penalty-100(b''\Bsetminus a_{\alpha})$
where $a'$,
$a''$ belong to $\frak A_{M\cap\alpha}$ and $b'$, $b''$ belong to
$\frak A_{N\cap\alpha}$.   As $a'\Bsetminus b'$ belongs to
$\frak A_{\alpha}$ and is included in $1\Bsetminus a_{\alpha}$, there is
a $v\in V_{\alpha}$ such that $a'\Bsetminus b'\Bsubseteq v$;  as
$K_{\alpha}\subseteq M\cap N\cap\alpha$, $v\in\frak A_{M\cap N\cap\alpha}$.
Now $a'\Bsetminus v\in\frak A_{M\cap\alpha}$,
$b'\Bsetminus v\in\frak A_{N\cap\alpha}$ and
$a'\Bsetminus v\Bsubseteq b'\Bsetminus v$, so the inductive hypothesis
tells us that there is a $c'\in\frak A_{M\cap N\cap\alpha}$ such that
$a'\Bsetminus v\Bsubseteq c'\Bsubseteq b'\Bsetminus v$;  in which case
$c'\Bcap a_{\alpha}\in\frak A_{M\cap N\cap\gamma}$ and

\Centerline{$a'\Bcap a_{\alpha}
=a'\Bcap a_{\alpha}\Bsetminus v\Bsubseteq c'\Bcap a_{\alpha}
\Bsubseteq b'\Bcap a_{\alpha}\Bsetminus v
=b'\Bcap a_{\alpha}$.}

\noindent Similarly, there are $u\in\frak A_{M\cap N\cap\alpha}$,
$c''\in\frak A_{M\cap N\cap\gamma}$ such that

\Centerline{$a''\Bsetminus b''\Bsubseteq v$,
\quad$a''\Bsetminus u\Bsubseteq c''\Bsubseteq b''\Bsetminus u$,
\quad$a''\Bsetminus a_{\alpha}\Bsubseteq c''\Bsetminus a_{\alpha}
\Bsubseteq b''\Bsetminus a_{\alpha}$.}

\noindent Putting these together,
$c=(c'\Bcap a_{\alpha})\Bcup(c''\Bsetminus a_{\alpha})$ belongs to
$\frak A_{M\cap N\cap\gamma}$ and $a\Bsubseteq c\Bsubseteq b$.

Thus the induction proceeds to a successor ordinal $\gamma$.

\medskip

\qquad\grheadc\ If $\gamma>0$ is a limit ordinal,
$a\in\frak A_{M\cap\gamma}$ and $b\in\frak A_{N\cap\gamma}$ and
$a\Bsubseteq b$, then there is some $\alpha<\gamma$ such that $a\in\frak
A_{M\cap\alpha}$ and $b\in\frak A_{N\Bcap\alpha}$, so the inductive
hypothesis gives us a
$c\in\frak A_{M\cap N\cap\alpha}\subseteq\frak A_{M\cap N\cap\gamma}$
with $a\Bsubseteq c\Bsubseteq b$, and again the induction proceeds.\
\Qed

\medskip

\quad{\bf (ii)} Now set $\Bbb G=\{\frak A_M:M\in\Cal M\}$, and consider
the conditions ($\alpha$)-($\gamma$) of 518N.

\medskip

\qquad\grheada\ For any $a\in\frak A$, there is a finite set
$I\subseteq\zeta$ such that $a\in\frak A_I$.   Let $M$ be the smallest
element of $\Cal M$ including $I$;  then (because $\kappa$ is regular
and uncountable) $\#(M)<\kappa$, so $\#(\frak A_M)<\kappa$, while
$a\in\frak A_M\in\Bbb G$.

\medskip

\qquad\grheadb\ If $\Bbb G_0\subseteq\Bbb G$, consider
$\Cal M^*=\{M:M\in\Cal M$, $\frak A_M\in\Bbb G_0\}$.   Then
$M^*=\bigcup\Cal M^*$ belongs to $\Cal M$, and $\frak A_{M^*}\in\Bbb G$
must be the subalgebra of $\frak A$ generated by $\bigcup\Bbb G_0$.

\medskip

\qquad\grheadc\ Finally, condition ($\gamma$) is just (ii) above with
$\gamma=\zeta$.

So $\Bbb G$ is a $\kappa$-Geschke system for $\frak A$.

\medskip

{\bf (b)} Suppose that $\frak A$ has a $\kappa$-Geschke system
$\Bbb G$.   I seek to use the ideas of the proof of 518M.

\medskip

\quad{\bf (i)} Enumerate $\frak A$ as $\ofamily{\xi}{\lambda}{a_{\xi}}$,
and for each $\xi<\lambda$ let $\frak C_{\xi}\in\Bbb G$ be such that
$a_{\xi}\in\frak C_{\xi}$ and $\#(\frak C_{\xi})<\kappa$.   For
$\alpha\le\lambda$ let $\frak A_{\alpha}$ be the subalgebra of $\frak A$
generated by $\bigcup_{\xi<\alpha}\frak C_{\xi}$, so that
$\frak A_{\alpha}\in\Bbb G$.   Set
$C_{\alpha}=\frak C_{\alpha}\setminus\frak A_{\alpha}$ for each
$\alpha<\lambda$.   An easy induction shows that, for any
$\alpha\le\lambda$, $\frak A_{\alpha}$ is the subalgebra of $\frak A$
generated by $\bigcup_{\xi<\alpha}C_{\xi}$.

\medskip

\quad{\bf (ii)} If $\alpha\le\lambda$, the Freese-Nation index of
$\frak A_{\alpha}$ in $\frak A$ is at most $\kappa$.   \Prf\ For any
$\xi<\lambda$ and $b\in\frak A_{\alpha}\cap[0,a_{\xi}]$ there must be a
$c\in\frak A_{\alpha}\cap\frak C_{\xi}$ such that
$b\Bsubseteq c\Bsubseteq a_{\xi}$, because both $\frak A_{\alpha}$ and
$\frak C_{\xi}$ belong to $\Bbb G$;  so
$\frak C_{\xi}\cap\frak A_{\alpha}\cap[0,a_{\xi}]$ is cofinal with
$\frak A_{\alpha}\cap[0,a_{\xi}]$ and
$\cf(\frak A_{\alpha}\cap[0,a_{\xi}])\le\#(\frak C_{\xi})<\kappa$.
Similarly, $\frak C_{\xi}\cap\frak A_{\alpha}\cap[a_{\xi},1]$ is
coinitial with $\frak A_{\alpha}\cap[a_{\xi},1]$ and
$\ci(\frak A_{\alpha}\cap[a_{\xi},1])<\kappa$.\ \Qed

\medskip

\quad{\bf (iii)} List $\bigcup_{\alpha<\lambda}C_{\alpha}$ as
$\ofamily{\xi}{\zeta}{b_{\xi}}$, where $\zeta$ is an ordinal, in such a
way that whenever $\xi\le\eta<\zeta$, $b_{\xi}\in C_{\alpha}$ and
$b_{\eta}\in C_{\beta}$, then $\alpha\le\beta$.   Then
$\{b_{\xi}:\xi<\zeta\}$ generates $\frak A$.   If $\beta<\zeta$ and
$\frak B_{\beta}$ is the subalgebra of $\frak A$ generated by
$\{b_{\xi}:\xi<\beta\}$, let $\alpha$ be such that
$b_{\xi}\in C_{\alpha}$;  then $\frak A_{\alpha}=\frak B_{\gamma}$ for
some $\gamma\le\beta$, $\#(\beta\setminus\gamma)<\#(C_{\alpha})<\kappa$
and $\frak B_{\beta}$ is the subalgebra of $\frak A$ generated by
$\frak A_{\alpha}\cup\{b_{\xi}:\gamma\le\xi<\beta\}$, so has
Freese-Nation index at most $\kappa$ in $\frak A$, by 518Fb.   This
shows that $\ofamily{\xi}{\zeta}{b_{\xi}}$ is a tight
$\kappa$-filtration of $\frak A$, and $\frak A$ is tightly
$\kappa$-filtered.
}%end of proof of 518P

\leader{518Q}{Corollary} Let $\kappa$ be a regular uncountable cardinal
and $\frak A$ a tightly $\kappa$-filtered Boolean algebra.

(a) If $\frak C$ is a retract of $\frak A$\cmmnt{ (that is, $\frak C$ is a
subalgebra of $\frak A$ and there is a
Boolean homomorphism $\pi:\frak A\to\frak C$ such that $\pi c=c$ for
every $c\in\frak C$)}, then $\frak C$ is tightly $\kappa$-filtered.

(b) If $\frak C$ is a subalgebra of $\frak A$ which is\cmmnt{ (in
itself)} Dedekind complete, then $\frak C$ is tightly $\kappa$-filtered.

\proof{{\bf (a)} By 518P there is a $\kappa$-Geschke system $\Bbb G$ for
$\frak A$.   Let $\Bbb G_1$ be the set of those $\frak B\in\Bbb G$ such
that $\pi[\frak B]\subseteq\frak B$.   Then $\Bbb G_1$ is a
$\kappa$-Geschke system.   \Prf\ Of course $\Bbb G_1$ satisfies
($\gamma$) of 518N, just because $\Bbb G_1\subseteq\Bbb G$.   As for
($\beta$), if $\Bbb G_0\subseteq\Bbb G_1$ and $\frak B^*$ is the
subalgebra generated by $\bigcup\Bbb G_0$, then $\frak B^*\in\Bbb G$ and
$\pi[\frak B^*]$ must be the subalgebra generated by
$\bigcup_{\frak B\in\Bbb G_0}\pi[\frak B]\subseteq\frak B^*$, so
$\pi[\frak B^*]\subseteq\frak B^*$ and $\frak B^*\in\Bbb G_1$.
Finally, if $a\in\frak A$, 518O (taking $\lambda=\kappa$ and
$f(I)=\{a\}\cup\pi[I]$) tells us that there
is a $\frak B\in\Bbb G_1$ containing $a$ and with cardinal less than
$\kappa$.\ \Qed

Observe next that because $\pi c=c$ for every $c\in\frak C$,
$\pi[\frak B]=\frak B\cap\frak C$ for every $\frak B\in\Bbb G_1$.   Set
$\Bbb H=\{\frak B\cap\frak C:\frak B\in\Bbb G_1\}$.   Then $\Bbb H$ is a
$\kappa$-Geschke system for $\frak C$.   \Prf\ ($\alpha$) If
$c\in\frak C$ there is a $\frak B\in\Bbb G_1$ such that $c\in\frak B$
and $\#(\frak B)<\kappa$;  now $c\in\frak B\cap\frak C\in\Bbb H$ and
$\#(\frak B\cap\frak C)<\kappa$.
($\beta$) If
$\Bbb H'\subseteq\Bbb H$, set
$\Bbb G_1'=\{\frak B:\frak B\in\Bbb G_1$, $\frak B\cap\frak C\in\Bbb H'\}$.
Then the subalgebra $\frak B^*$ generated by $\bigcup\Bbb G_1'$ belongs to
$\Bbb G_1$, and $\pi[\frak B^*]\in\Bbb H$ is the subalgebra generated by
$\bigcup\{\pi[\frak B]:\frak B\in\Bbb G'_1\}=\bigcup\Bbb H$.
($\gamma$) If
$b_1\in\frak D_1\in\Bbb H$, $b_2\in\frak D_2\in\Bbb H$ and
$b_1\Bsubseteq b_2$, express $\frak D_1$, $\frak D_2$ as
$\frak B_1\cap\frak C$, $\frak B_2\cap\frak C$ where $\frak B_1$ and
$\frak B_2$ belong to $\Bbb G_1$.   Then there is a
$b\in\frak B_1\cap\frak B_2$ such that $b_1\Bsubseteq b\Bsubseteq b_2$;
in which case $\pi b\in\frak D_1\cap\frak D_2$ and

\Centerline{$b_1=\pi b_1\Bsubseteq\pi b\Bsubseteq\pi b_2=b_2$.}

\noindent Thus
$\Bbb H$ satisfies ($\gamma$) of 518N and is a $\kappa$-Geschke system
for $\frak C$.   By 518P in the other direction, $\frak C$ is tightly
$\kappa$-filtered.\ \Qed

\medskip

{\bf (b)} In this case, the identity map from $\frak C$ to itself
extends to a Boolean homomorphism from $\frak A$ to $\frak C$ (314K), so
we can use (a).
}%end of proof of 518Q

\leader{518R}{Lemma} (a) Let $I$ be a set and $\frak G$ the regular open
algebra of $\{0,1\}^I$.   For $J\subseteq I$ let $\frak G_J$ be the
order-closed subalgebra of $\frak G$ consisting of regular open sets
determined by coordinates in $J$.   Suppose that $J$ and $K$ are
disjoint subsets of $I$, and $\family{q}{\Bbb Q}{a_q}$,
$\family{q}{\Bbb Q}{b_q}$ disjoint families in
$\frak G_J\setminus\{\emptyset\}$ and
$\frak G_K\setminus\{\emptyset\}$ respectively.   For $t\in\Bbb R$ set
$w_t=\sup_{p,q\in\Bbb Q,p\le t\le q}a_q\cap b_p$, the supremum being
taken in $\frak G$;  set $w=\sup_{p,q\in\Bbb Q,p\le q}a_q\cap b_p$.
If $w'\subseteq w$ belongs to the subalgebra of $\frak G$ generated by
$\frak G_{I\setminus K}\cup\frak G_{I\setminus J}$, then
$\{t:w_t\subseteq w'\}$ is finite.

(b) If $I=\omega_3$ then $\frak G$ is not tightly
$\omega_1$-filtered.

\proof{{\bf (a)(i)} I had better explain why each $\frak G_J$ is an
order-closed subalgebra;  the point is just that if
$A\subseteq\{0,1\}^I$ is determined by coordinates in $J\subseteq I$
then so are its closure and interior (4A2B(g-i) again),
so that the operations
$\Cal H\mapsto\interior(\bigcap\Cal H)$,
$\Cal H\mapsto\interior\overline{\bigcup\Cal H}$ take subsets of
$\frak G_J$ to members of $\frak G_J$.

\medskip

\quad{\bf (ii)} $w'$ must be expressible in the form
$\sup_{i<n}u_i\cap v_i$ where $u_i\in\frak G_{I\setminus K}$ and
$v_i\in\frak G_{I\setminus J}$ for each $i$.   \Quer\ Suppose, if
possible, that there are $t_0<t_1<\ldots<t_n$ in $\Bbb R$ such that
$w_{t_j}\subseteq\sup_{i<n}u_i\cap v_i$ for every $j$.   Take rational
numbers $q_j$ and $q'_j$, for $j\le n$, such that
$q_0\le t_0\le q'_0<q_1\le t_1\le q'_1<\ldots<q_n\le t_n\le q'_n$.   Set
$e_{-1}=\{0,1\}^I$.    Choose $i_j$, $e_j$, $c_j$, $c'_j$ and $c''_j$
inductively, for $j\le n$, as follows.   Given that
$e_{j-1}\in\frak G_{I\setminus(J\cup K)}$ is non-empty, where $j\le n$,
then $a_{q'_j}$, $b_{q_j}$ and $e_{j-1}$ are non-empty sets determined
by coordinates in $J$, $K$ and $I\setminus(J\cup K)$ respectively, so
have non-empty intersection;  also
$a_{q'_j}\cap b_{q_j}\subseteq w_{t_j}\subseteq\sup_{i<n}u_i\cap v_i$.
There is therefore an $i_j<n$ such that
$a_{q'_j}\cap b_{q_j}\cap e_{j-1}\cap u_{i_j}\cap v_{i_j}$ is non-empty,
and includes a basic cylinder set $c_j$ say.   Now we can express $c_j$
as $c'_j\cap c''_j\cap e_j$ where $c'_j$ is determined by coordinates in
$J$, $c''_j$ by coordinates in $K$ and $e_j$ by coordinates in
$I\setminus(J\cup K)$;  note that $e_j\subseteq e_{j-1}$, and continue.

At the end of this process, there must be $j<k\le n$ such that
$i_j=i_k=i$ say.   Now $q'_j<q_k$, so

\Centerline{$a_{q'_j}\cap b_{q_k}\cap u_i\cap v_i
\subseteq a_{q'_j}\cap b_{q_k}\cap w=\emptyset$.}

\noindent (Recall that $\family{q}{\Bbb Q}{a_q}$ and
$\family{q}{\Bbb Q}{b_q}$ are disjoint.)
On the other hand, $c'_j\cap c''_j\cap e_j\subseteq u_i$,
which is determined by coordinates in $I\setminus K$, so
$c'_j\cap e_j\subseteq u_i$;  similarly, $c''_k\cap e_k\subseteq v_i$;
so

\Centerline{$cap c'_j\cap e_j\cap c''_k\cap e_k
\Bsubseteq a_{q'_j}\cap b_{q_k}\cap u_i\cap v_i
=\emptyset$.}

\noindent But $c'_j$, $c''_k$ and
$e_j\cap e_k$ are all non-empty and determined by coordinates in
$J$, $K$ and $I\setminus(J\cup K)$ respectively, so this is impossible.\
\Bang

Thus $\{t:w_t\subseteq w'\}$ has at most $n$ members, and is finite.

\medskip

{\bf (b)} As in 518J, I will work with $I=\omega_3\times\Bbb N$.

\medskip

\quad{\bf (i)} Note that every member of $\frak G$
belongs to $\frak G_J$ for some countable $J$ (4A2E(b-i) again), so we can
choose for each $c\in\frak G$ a countable $J(c)\subseteq\omega_3$ such
that $c\in\frak G_{J(c)\times\Bbb N}$ for some countable $I$.   For each
$\xi<\omega_3$, let $\family{q}{\Bbb Q}{a_{\xi q}}$ be a disjoint family
of non-zero elements of $\frak G_{\{\xi\}\times\Bbb N}$.   For
$t\in\Bbb R$, $\xi<\omega_3$ set
$c'_{\xi t}=\sup_{q\in\Bbb Q,q\le t}a_{\xi q}$,
$c''_{\xi t}=\sup_{q\in\Bbb Q,q\ge t}a_{\xi q}$.
Let $T\subseteq\Bbb R$ be a set with cardinal $\omega_1$.   For
$D\subseteq\frak G$ set $\tilde J(D)=\bigcup_{c\in D}J(c)$.

\medskip

\quad{\bf (ii)} \Quer\ Suppose, if possible, that $\frak G$ is tightly
$\omega_1$-filtered.   Then it has an $\omega_1$-Geschke system $\Bbb B$
say (518P).   By 518O, with $\lambda=\omega_3$ and

\Centerline{$f(\emptyset)=\{c''_{\xi t}:\xi<\omega_2$, $t\in T\}=C$}

\noindent say, there is a $\frak B_1\in\Bbb B$ such that
$C\subseteq\frak B_1$ and $\#(\frak B_1)\le\omega_2$;
take $\xi\in\omega_3\setminus\tilde J(\frak B_1)$, and let
$\frak B_2\in\Bbb B$ be such that $\frak B_2$ is countable and
$a_{\xi p}\in\frak B_2$ for every $p\in\Bbb Q$.
Then $\tilde J(\frak B_2)$ is countable,
so there is an $\eta\in\omega_2\setminus\tilde J(\frak B_2)$.

Set $w=\sup_{p,q\in\Bbb Q,p\le q}a_{\xi p}\Bcap a_{\eta q}$, and for
$t\in T$ set
$w_t=c'_{\xi t}\Bcap c''_{\eta t}$.   Then $w$
belongs to a countable $\frak B_0\in\Bbb B$, while the subalgebra
$\frak B^*$ of $\frak G$ generated by $\frak B_1\cup\frak B_2$ belongs
to $\Bbb B$.   But if we set $J=\{\xi\}\times\Bbb N$,
$K=\{\eta\}\times\Bbb N$ then we see that
$\frak B_1\subseteq\frak G_{(\omega_3\times\Bbb N)\setminus J}$ and
$\frak B_2\subseteq\frak G_{(\omega_3\times\Bbb N)\setminus K}$.   So
(a) tells us that any member of $\frak B^*$ included in $w$ can include
only finitely many $w_t$, while $w_t\in\frak B^*\cap[0,w]$.   Thus
$\cf(\frak B^*\cap[0,w])\ge\omega_1$.   On the other hand,
by ($\gamma$) of 518N, the countable set
$\frak B_0\cap\frak B^*\cap[0,w]$ is cofinal with $\frak B^*\cap[0,w]$.\
\Bang

This contradiction proves the result.
}%end of proof of 518R

\leader{518S}{Theorem}\cmmnt{ ({\smc Geschke 02})} If $\frak A$ is a
tightly $\omega_1$-filtered
Dedekind complete Boolean algebra then $\#(\frak A)\le\omega_2$.

\proof{ \Quer\ Otherwise, by 515I, $\frak A$ has a subalgebra $\frak C$
isomorphic to
the regular open algebra of $\{0,1\}^{\omega_3}$.   By
518Rb, $\frak C$ is not tightly $\omega_1$-filtered;  by 518Qb, nor is
$\frak A$.\ \Bang
}%end of proof of 518S

\exercises{\leader{518X}{Basic exercises (a)}
%\spheader 518Xa
Let $\frak A$ be a Boolean algebra and $\frak B$ a
principal ideal of $\frak A$.   Show that $\FN(\frak B)\le\FN(\frak A)$.
%518B

\spheader 518Xb Show that $\FN(\alpha)=\#(\alpha)$ for every infinite
ordinal $\alpha$.
%518A out of order query

\spheader 518Xc\dvAnew{2014} Show that if $P$ and $Q$
are partially ordered sets, then $\FN(P\times Q)$ is at most the
cardinal product $\FN(P)\cdot\FN(Q)$.
%518A

\sqheader 518Xd Show that $\FN(\Cal P\Bbb N)\ge\omega_1$.
%518C 522U

\spheader 518Xe Show that $\FN(\Bbb Q)=\omega$ and
$\FN(\Bbb R)=\omega_1$.
%518A 518Xd

\spheader 518Xf Show that
$\FN^*(\Cal P\Bbb N/[\Bbb N]^{<\omega})=\FN^*(\Cal P\Bbb N)$.
%518D

\spheader 518Xg Let $P$ be a partially ordered set and $Q$ a subset of
$P$ with Freese-Nation index $\kappa$ in $P$.   Show that if
$\lambda\ge\max(\kappa,\FN(P))$ is a regular infinite cardinal then
$\FN(Q)\le\kappa$.
%518F

\spheader 518Xh\dvAnew{2014}
Let $P$ be a partially ordered set and $\ofamily{\xi}{\zeta}{P_{\xi}}$ a
non-decreasing family of subsets of $P$ such that
$P_{\xi}=\bigcup_{\eta<\xi}P_{\eta}$ for every non-zero limit ordinal
$\xi\le\zeta$.   Suppose that $\kappa$ is a regular infinite cardinal such
that the Freese-Nation index of $P_{\xi}$ in $P_{\xi+1}$ is at most
$\kappa$ for every $\xi<\zeta$.   Show that the Freese-Nation index of
$P_0$ in $P_{\zeta}$ is at most $\kappa$.
%518G out of order query

\spheader 518Xi\dvArevised{2014}
Let $\frak A$ be a Boolean algebra, $\kappa$ a
regular infinite cardinal and $\ofamily{\xi}{\zeta}{a_{\xi}}$ a
family in
$\frak A$.   For each $\alpha\le\zeta$ let $\frak A_{\alpha}$ be the
subalgebra of $\frak A$ generated by $\{a_{\xi}:\xi<\alpha\}$.   Suppose
that $\frak A_{\zeta}=\frak A$ and that the Freese-Nation index of
$\frak A_{\alpha}$ in $\frak A_{\alpha+1}$ is at most $\kappa$ for every
$\alpha<\zeta$.   Show that $\ofamily{\xi}{\zeta}{a_{\xi}}$ is a tight
$\kappa$-filtration of $\frak A$.
%518Xh 518G

\spheader 518Xj Suppose that $\frak c=\omega_1$.   Show
that any Dedekind complete ccc Boolean algebra with cardinal at most
$\frak c^+=\omega_2$ is tightly $\omega_1$-filtered.
%518M

\spheader 518Xk\dvAnew{2014}
Let $\frak A$ be a Boolean algebra, $\kappa\le\lambda$ cardinals and
$\Bbb G$ a $\kappa$-Geschke system for $\frak A$.   Show that
$\Bbb G$ is a $\lambda$-Geschke system for $\frak A$.
%518N out of order query

\spheader 518Xl Let $\kappa\le\frak c$ be a regular uncountable
cardinal.   Show that if $\frak A$ is a tightly $\kappa$-filtered
Dedekind complete Boolean algebra then $\#(\frak A)\le\kappa^+$.
%518S

\leader{518Y}{Further exercises (a)}
%\spheader 518Ya
Show that if $P$ is a finite partially ordered set then
$\FN(P)\le 2+\bover12\#(P)$.
%518A Fremlin & Penman 2D

\spheader 518Yb Show that $\FN(\Cal PI)>\#(I)$ for every infinite
set $I$.
\Hint{{\smc Fuchino Koppelberg \& Shelah 96}.}
% finite sets $I$ not completely checked.   In Fremlin & Penman we have
% $\FN(\Cal P3)=5$ and  $(2/sqrt3)^n<\FN(\Cal Pn)$.

\spheader 518Yc(i) Let $\frak A$ be an infinite Boolean algebra.   Show
that $\FN(S(\frak A))=\FN(\frak A)$.
(ii)\dvArevised{2014} Let $\frak A$ be a
Dedekind complete Boolean algebra.   Show that
$\FN(\frak A)\le\FN(L^0(\frak A))\le\FN(\frak A^{\Bbb N})$.
}%end of exercises

\endnotes{
\Notesheader{518} `Freese-Nation numbers' are a relatively recent topic,
beginning with the investigation of partially ordered sets with
Freese-Nation numbers at most $\omega$ (the `Freese-Nation property') in
{\smc Freese \& Nation 78} and those with Freese-Nation numbers at most
$\omega_1$ (the `weak Freese-Nation property') in {\smc Fuchino
Koppelberg \& Shelah 96}.   There are interesting puzzles concerning
the Freese-Nation numbers of finite and countable partially ordered sets
which I pass over here.   Unlike most of the cardinals discussed in this
chapter, Freese-Nation numbers refer to the internal, rather than
cofinal, structure of a partially ordered set.

The Freese-Nation number $\FN(\Cal P\Bbb N)$ appears in many contexts
besides the identifications of 518D.    I will mention it again in
522U.   I do not know whether it is consistent to suppose that its
cofinality is countable.

Of the special axioms used in 518I,
%($\alpha$) $\cff[\lambda]^{\le\omega}\le\lambda^+$ for every
%cardinal $\lambda\le\tau(\frak A)$,
%($\beta$) $\square_{\lambda}$ is true for every uncountable cardinal
%$\lambda\le\tau(\frak A)$ of countable cofinality.}
($\alpha$) has a more familiar
aspect;  for instance, it is a consequence of GCH, regardless of the
value of $\tau(\frak A)$ (5A6Ab).   ($\beta$) is believed not to be
a consequence of GCH
(see 555Yf), but is true in `ordinary' models of set theory (5A6Db, 5A6Bc).
In 518K I call on a form of Chang's transfer principle;
this is {\it false} in ordinary models of set
theory (5A6Fc), but is believed to be relatively consistent with ZFC + GCH
(5A6Fa).   Freese-Nation numbers are therefore a little exceptional among
those appearing in measure theory, in that they are not fixed by the
generalized continuum hypothesis.
}%end of notes

\discrpage

\references{
\frfilename{mt5r.tex}
\versiondate{28.4.23}
\copyrightdate{2008}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}

\centerline{\bf References for Volume 5}

%use subparagraph details (5A1Aa) only for X. Y. Z.
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\medskip%A

%Antonovskii M.Ja.\ \& Chudnovsky D.V.\ [76] `Some questions of general
%topology and Tikhonov semifields II', Russian Math.\ Surveys 31 (1976)
%69-128.

Argyros S.A.\ \& Kalamidas N.D.\ [82] `The $K_{\alpha n}$ property on
spaces with strictly positive measure', Canadian J.\ Math.\ 34 (1982)
1047-1058.
\cmmnt{[525T.]}

Argyros S.A.\ \& Tsarpalias A.\ [82] `Calibers of compact spaces',
Trans.\ Amer.\ Math.\ Soc.\ 270 (1982) 149-162.
\cmmnt{[525N.]}

\medskip%B

Babiker A.G.\ [76] `On uniformly regular topological measure spaces',
Duke Math.\ J.\ 43 (1976) 775-789.
\cmmnt{[533Xd.]}

Balcar B.\ \& Fran\v{e}k F.\ [82] `Independent families in complete Boolean
algebras', Trans.\ Amer.\ Math.\ Soc.\ 274 (1982) 607-618.
\cmmnt{[515H.]}

Balcar B., Jech T.\ \& Paz\'ak T.\ [05]
`Complete ccc Boolean algebras, the order sequential topology,
and a problem of von Neumann',
Bull.\ London Math.\ Soc.\ 37 (2005) 885-898.
\cmmnt{[\S539 {\it intro.}, 539E, 539L, 539N, \S539 {\it notes\/}, 555K.]}
 % arXiv.math.LO/0312473

Balcar B., Jech T.\ \& Zapletal J.\ [97] `Semi-Cohen Boolean algebras',
Ann.\ Pure \& Applied Logic 87 (1997) 187-208.
\cmmnt{[\S515 {\it notes\/}, \S547 {\it notes\/}.]}

Balcar B.\ \& Vojt\'a\v{s} P.\ [77] `Refining systems on Boolean algebras',
pp.\ 45-58 in {\smc Lachlan Srebny \& Zarach 77}.
\cmmnt{[515E.]}

Bar-Hillel Y.\ [70] {\it Mathematical Logic and Foundations of Set Theory.}
North-Holland, 1970.

Bartoszy\'nski T.\ [84] `Additivity of measure implies additivity of
category', Trans.\ Amer.\ Math.\ Soc.\ 281 (1984) 209-213.
\cmmnt{[Chap.\ 52 {\it intro.}, 522Q, \S524 {\it notes\/}.]}

Bartoszy\'nski T.\ [87] `Combinatorial aspects of measure and category',
Fund.\ Math.\ 127 (1987) 225-239.
\cmmnt{[522S.]}
% http://matwbn.icm.edu.pl/ksiazki/fm/fm127/fm127118.pdf

% Bartoszy\'nski T.\ [88] `On covering of real line by null sets',
% Pacific J.\ Math.\ 131 (1988) 1-12.

% Bartoszy\'nski T.\ [p97] `Invariants of measure and category', in
% preparation for {\smc Foreman Kanamori \& Magidor ??}

% Bartoszy\'nski T.\ [00] `A note on duality between measure and category',
% Proc.\ Amer.\ Math.\ Soc.\ 128 (2000) 2745-2748.
   % (2^N,+,meager) not iso to (2^N,+,negligible)

Bartoszy\'nski T.\ \& Judah H.\ [89] `On the cofinality of the smallest
covering of the real line by meager sets', J.\ Symbolic Logic 54 (1989) 828-832.
\cmmnt{[522V.]}

Bartoszy\'nski T.\ \& Judah H.\ [95] {\it Set Theory:  on the structure
of the real line}.  A.K.Peters, 1995.
\cmmnt{[522R, \S522 {\it notes\/},
526M, \S528 {\it notes\/}, 534Q, \S534 {\it notes\/}, 546I,
553C, 554Ya, 5A1Q.]}

%Bartoszy\'nski T., Judah H.\ \& Shelah S.\ [93] `The Cicho\'n diagram',
%J.\ Symbolic Logic 58 (1993) 401-423.

%Bartoszy\'nski T.\ \& Scheepers M.\ [96] (eds.) {\it Set Theory (BEST, 1992-1994).}
%Amer.\ Math.\ Soc., 1996 (Contemporary Mathematics 192).
% ISBN 0821803069

Bartoszy\'nski T.\ \& Shelah S.\ [92] `Closed measure zero sets', Ann.\
Pure \& Applied Logic 58 (1992) 93-110.
\cmmnt{[526M.]}

Baumgartner J.E., Taylor A.D.\ \& Wagon S.\ [77] `On splitting
stationary subsets of large cardinals',
J.\ Symbolic Logic 42 (1977) 203-214.
\cmmnt{[\S541 {\it notes\/}, \S544 {\it notes\/}.]}

Bell M.G.\ [81] `On the combinatorial principle $P(c)$',
Fundamenta Math.\ 114 (1981) 149-157.
\cmmnt{[\S517 {\it notes\/}.]}

Benedikt M.\ [98] `Ultrafilters which extend measures',
J.\ Symbolic Logic 63 (1998) 638-662.
\cmmnt{[538M, 538Xm, \S538 {\it notes\/}.]}

Benedikt M.\ [99] `Hierarchies of measure-theoretic ultrafilters',
Annals of Pure and Applied Logic 97 (1999) 203-219.
\cmmnt{[538H.]}

Blair C.E.\ [77] `The Baire category theorem implies the principle of
dependent choice', Bull.\ Acad.\ Polon.\ Sci.\ (Math.\ Astron.\ Phys.) 25
(1977) 933-934.
\cmmnt{[\S566 {\it notes\/}.]}

%Blass A.\ [88] `Selective ultrafilters and homogeneity', Ann.\ Pure and
%Applied Logic 38 (1988) 252-285.

%Blass A.\ [p01b] `Combinatorial cardinal characteristics of the continuum',
%pp. ?? in {\smc Foreman Kanamori \& Magidor ??}.

Blass A.\ \& Laflamme C.\ [89] `Consistency results about filters and the
number of inequivalent growth types', J.\ Symbolic Logic 54 (1989) 50-56.
% u<g  filter dichotomy
\cmmnt{[5A6Ib, 5A6J.]}

Booth D.\ [70] `Ultrafilters on a countable set',
Ann.\ Math.\ Logic 2 (1970) 1-24.
\cmmnt{[517R.]}

Borel E.\ [1919] `Sur la classification des ensembles de mesure nulle', Bull.\ Soc.\ Math.\ France 47 (1919) 97-125.
\cmmnt{[\S534 {\it intro.}]}

Brendle J.\ [00] {\it Inner model theory and large cardinals.}
Research report, March 2000.
\cmmnt{[528N, 529F-529H.]} %529F 529G 529H

%Brendle J.\ [01] `Cardinal invariants of the continuum --- a survey',
%J.\ Research Institute for Mathematical Sciences, Kyoto University
%(JRIMS Kokyuroku) 1202 (2001) 7-32.

Brendle J.\ [06] `Cardinal invariants of the continuum and combinatorics
on uncountable cardinals',
Annals of Pure and Applied Logic 144 (2006) 43-72.
\cmmnt{[529F, 529H.]}

Burke M.R.\ [93] `Liftings for Lebesgue measure', pp.\ 119-150 in
{\smc Judah 93}.
\cmmnt{[\S535 {\it notes\/}.]}

%Burke M.R.\ [n95] `Consistent liftings', privately circulated, 1995.

%Burke M.R.\ [n96] `A theorem of Gitik and Shelah on disjoint refinements
%of sequences of subsets of the real line', note of 23.8.96.
%\cmmnt{[5x46J, \S547 {\it intro.}]}

Burke M.R.\ [n05] `Non-null sets of minimal cofinality', note of 2.3.05.
\cmmnt{[523K.]}

Burke M.R.\ \& Magidor M.\ [90] `Shelah's pcf theory and
its applications', Ann.\ Pure and Applied Logic 50 (1990) 207-254.
\cmmnt{[513J, \S5A2 {\it intro.}.]}

Burke M.R.\ \& Shelah S.\ [92]
`Linear liftings for non-complete probability spaces',
Israel J.\ Math.\ 79 (1992) 289-296.
\cmmnt{[\S535 {\it notes\/}.]}

\medskip%C

Carlson T.J.\ [84] `Extending Lebesgue measure to infinitely many sets',
Pacific J.\ Math.\ 115 (1984) 33-45.
\cmmnt{[552N, \S552 {\it notes.}]}

Carlson T.J.\ [93] `Strong measure zero and strongly meager sets',
Proc.\ Amer.\ Math.\ Soc.\ 118 (1993) 577-586.
\cmmnt{[534R.]}

Carlson T., Frankiewicz R.\ \& Zbierski P.\ [94]
`Borel liftings of the measure
algebra and the failure of the continuum hypothesis', Proc.\ Amer.\ Math.\ Soc.\
120 (1994) 1247-1250.
\cmmnt{[\S535 {\it notes\/}, 554I.]}

Choksi J.R.\ \& Fremlin D.H.\ [79] `Completion regular measures on
product spaces', Math.\ Ann.\ 241 (1979) 113-128.
\cmmnt{[532I.]}

%Cicho\'n J., Kamburelis T.\ \& Pawlikowski J.\ [85] `On dense subsets of
%the measure algebra', Proc.\ Amer.\ Math.\ Soc.\ 94 (1985) 142-146.

Cicho\'n J.\ \& Pawlikowski J.\ [86]
`On ideals of subsets of the plane and on Cohen
reals', J.\ Symbolic Logic 51 (1986) 560-569.
\cmmnt{[527F.]}

%Cohen P.\ [63] `The independence of the continuum hypothesis',
% Proc.\ Nat.\ Acad.\ Sci.\ U.S.A.\ 50 (1963) 1143-1148.

Comfort W.W.\ \& Negrepontis S.\ [74] {\it The Theory of Ultrafilters.}
Springer, 1974.
\cmmnt{[538F, 538Yb.]}

Comfort W.W.\ \& Negrepontis S.\ [82] {\it Chain Conditions in
Topology.}   Cambridge U.P., 1982.
\cmmnt{[516 {\it notes\/}.]}

Cummings J.\ [92] `A model in which GCH holds at successors but fails at
limits',
Trans.\ Amer.\ Math.\ Soc.\ 329 (1992) 115-142.
\cmmnt{[525Z.]}

\medskip%D

Dellacherie C., Meyer P.A.\ \& Weil M.\ (eds.) [73]
{\it S\'eminaire de Probabilit\'es VII.}   Springer, 1973
(Springer Lecture Notes in Mathematics 321).

Devlin K.J.\ [84] {\it Constructibility.} Springer, 1984.
\cmmnt{[5A6D.]}

%Dow A.\ \& Hart K.P.\ [p98] `The measure algebra does not always embed',
%preprint, 1998

Dow A.\ \& Stepr\=ans J.\ [94] `The $\sigma$-linkedness of the measure
algebra', Canad.\ Math.\ Bulletin 37 (1994) 42-45.
\cmmnt{[524L.]}

%Drake F.\ [74] {\it Set Theory.}  North-Holland, 1974.

%Drake F.R.\ \& Singh D.\ [96] {\it Intermediate Set Theory.}
%Wiley, 1996.  \cmmnt{[\S5A1.]}
% ISBN 0 471 96496 4

D\v{z}amonja M.\ \& Kunen K.\ [93] `Measures on compact HS spaces',
Fundamenta Math.\ 143 (1993) 41-54.
\cmmnt{[\S531 {\it notes\/}.]}

%D\v{z}amonja M.\ \& Kunen K.\ [95] `Properties of the class of
%measure-separable compact spaces', Fundamenta Math.\ 147 (1995) 261-277.

D\v{z}amonja M.\ \& Plebanek G.\ [04]
`Precalibre pairs of measure algebras',
Topology Appl.\ 144 (2004) 67-94.
\cmmnt{[525L.]}

\medskip%E

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).
\cmmnt{[\S5A4.]}

%Engelking R.\ \& Kar{\l}owicz M.\ [65]
%`Some theorems of set theory and their topological consequences',
% Fund.\ Math.\ 57 (1965) 275-285.

Erd\H{o}s P., Hajnal A., M\'at\'e A.\ \& Rado R.\ [84] {\it Combinatorial
Set Theory:  Partition Relations for Cardinals.}   Akad\'emiai Kiad\'o,
1984 (Disquisitiones Math.\ Hung.\ 13).
\cmmnt{[5A1F, 5A1H, 5A6F.]}

\medskip%F

Farah I.\ [00] {\it Analytic Quotients.}
Mem.\ Amer.\ Math.\ Soc.\ 148 (2000).
\cmmnt{[5A6H.]}

Farah I.\ [03] `How many algebras $\Cal P(\Bbb N)/\Cal I$ are there?',
Illinois J.\ Math.\ 46 (2003) 999-1033.
\cmmnt{[\S556 {\it notes\/}.]}

Farah I.\ [06] `Analytic Hausdorff gaps II: the density zero ideal',
Israel J.\ Math.\ 154 (2006) 235-246.
\cmmnt{[556S.]}

Farah I.\ \& Veli\v{c}kovi\'c B.\ [06]
`Von Neumann's problem and large cardinals',
Bull.\ London Math.\ Soc.\ 38 (2006) 907-912.
\cmmnt{[539Q.]}

Feferman S. \& L\'evy A. [63] `Independence results in set theory by Cohen's
method', Notices Amer.\ Math.\ Soc.\ 10 (1963) 53.
\cmmnt{[561A.]}

Fleissner W.G.\ [91]
`Normal measure axiom and Balogh's theorems', Topology and its Appl.\
39 (1991) 123-143.
\cmmnt{[555N.]}

% Fleissner W.G., Hansell R.W.\ \& Junnila H.J.K.\ [82] `PMEA implies
% Proposition P', Topology and its Appl.\ 13 (1982) 255-262.

% Foreman M., Kanamori A.\ \& Magidor M.\ [??] {\it Handbook of Set
% Theory}.   Kluwer, to appear.

%Foreman M., Magidor M.\ \& Shelah S.\ [88] `Martin's maximum, saturated ideals, and non-regular ultrafilters', Annals of Math.\ (2) 127 (1988) 1-47.

Foreman M.\ [10] `Ideals and generic elementary embeddings',  pp.\ 885-1147
in {\smc Foreman \& Kanamori 10}, vol.\ 2.
\cmmnt{[547Z.]}

Foreman M.\ \& Kanamori A.\  [10] {\it Handbook of Set Theory.}
Springer, 2010.

Foreman M.\ \& Wehrung F.\ [91] `The Hahn-Banach theorem implies the
existence of a non-Lebesgue measurable set', Fundamenta Math.\ 138
(1991) 13-19.
\cmmnt{[563A.]}

Foreman M.\ \& Woodin W.\ [91] `The generalized continuum hypothesis can
fail everywhere', Annals of Math.\ 133 (1991) 1-35.
\cmmnt{[525Z.]}

Fossy J.\ \& Morillon M.\ [98] `The Baire category property and some
notions of compactness', J.\ London Math.\ Soc. (2) 57 (1998) 1-19.
\cmmnt{[\S566 {\it notes\/}.]}

% Frankiewicz R.\ \& Plebanek G.\ [94] `Nonaccessible filters in
% measure algebras and functionals on $L^{\infty}(\lambda)^*$', Studia
% Math.\ 108 (1994) 191-200.

%Frankiewicz R.\ \& Plebanek G.\ [96] `On asymptotic density and
%uniformly distributed sequences', Studia Math.\ 119 (1996) 17-26.

Freese R. \& Nation J.B.\ [78] `Projective lattices',
Pacific J.\ Math.\ 75 (1978) 93-106.
\cmmnt{[\S518 {\it notes\/}.]}

Freiling C.\ [86] `Axioms of symmetry;  throwing darts at the real number
line', J.\ Symbolic Logic 51 (1986) 190-200.
\cmmnt{[537K.]}

Fremlin D.H.\ [74] {\it Topological Riesz Spaces and Measure Theory.}
Cambridge U.P., 1974.
\cmmnt{[564Xc.]}

Fremlin D.H.\ [77]
`Uncountable powers of {\bf R} can be almost Lindel\"of',
Manuscripta Math.\ 22 (1977) 77-85.
\cmmnt{[533J.]}

Fremlin D.H.\ [84a] {\it Consequences of Martin's Axiom.}
Cambridge U.P., 1984.
\cmmnt{[\S511 {\it notes\/}, \S517 {\it notes\/}, \S531 {\it notes\/}.]}

Fremlin D.H.\ [84b] `On the additivity and cofinality of Radon measures',
Mathematika 31 (1984) 323-335.
\cmmnt{[\S524 {\it notes\/}.]}

Fremlin D.H.\ [87] {\it Measure-additive Coverings and
Measurable Selectors.} Dissertationes Math.\ 260 (1987).
\cmmnt{[551A.]}

Fremlin D.H.\ [88] `Large correlated families of positive
random variables',
Math.\ Proc.\ Cambridge Phil.\ Soc.\ 103 (1988) 147-162.
\cmmnt{[525S.]}

Fremlin D.H.\ [91] `The partially ordered sets of measure
theory and Tukey's ordering,' Note di Matematica  11 (1991) 177-214.
[523Yc, \S524 {\it notes\/}, 526B, 526I, 527J, 529C, 529D, 534L.]

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304
in {\smc Judah 93}.
\cmmnt{[\S537 {\it notes\/}, \S541 {\it notes\/}, \S544 {\it notes\/},
\S545 {\it notes\/}, \S5A2 {\it intro.}.]}

Fremlin D.H.\ [97] `On compact spaces carrying Radon measures of
uncountable Maharam type', Fundamenta Math.\  154 (1997) 295-304.
\cmmnt{[531T.]}

%Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces',
%Fundamenta Math.\ 165 (2000) 67-94.

Fremlin D.H.\ [03] `Skew products of ideals',
J.\ Applied Analysis 9 (2003) 1-18.
\cmmnt{[\S527 {\it notes\/}.]}
%n01910.tex

%Fremlin D.H.\ [n97] `A theorem of Gitik and Shelah on disjoint
%refinements of sequences of subsets of the real line', note of 4.4.97.
%n96j01.tex

%Fremlin D.H.\ [n99] `Embedding the Lebesgue measure algebra into
%$\Cal P\Bbb N/\Cal Z$', note of 25.5.99.
%n99525.tex

%Fremlin D.H.\ [n99] `Notes on {\smc Farah p99}', note of 26.7.99.
%n99612.tex

%Fremlin D.H.\ [n99] `Notes on {\smc Solecki p96}', note of 17.6.99.
%n99617.tex

%Fremlin D.H.\ [n02] `Games and the Tukey classification', note of 29.9.02.
%n02929.tex

Fremlin D.H.\ \& Grekas S.\ [95] `Products of completion regular
measures', Fundamenta Math.\ 147 (1995) 27-37.
\cmmnt{[532D.]}

Fremlin D.H.\ \& Miller A.W.\ [88]
`On some properties of Hurewicz, Menger and Rothberger',
Fund.\ Math.\ 129 (1988) 17-33.
\cmmnt{[534Q.]}

Fremlin D.H., Natkaniec T.\ \& Rec{\l}aw I.\ [00]
`Universally Kuratowski-Ulam spaces', Fundamenta Math.\ 165 (2000) 239-247.
\cmmnt{[\S527 {\it notes\/}.]}

%Friedman H.\ [80]
%`A consistent Fubini-Tonelli theorem for nonmeasurable functions',
% Illinois J.\ Math.\ 24 (1980) 390-395.

Friedman S.D. \& Koepke P.\ [97] `An elementary approach to the fine
structure of L', Bull.\ Symbolic Logic 3 (1997) 453-468.
\cmmnt{[5A6D.]}

Fuchino S.,  Geschke S., Shelah S.\ \& Soukup L.\ [01]
`On the weak Freese-Nation property of complete Boolean algebras',
Ann.\ Pure Appl.\ Logic 110 (2001) 89-105.
\cmmnt{[518K.]}

Fuchino S., Geschke S.\ \& Soukup L.\ [01] `On the weak Freese-Nation
property of $\Cal P(\omega)$', Arch.\ Math.\ Logic 40 (2001) 425-435.
\cmmnt{[522U, \S522 {\it notes\/}.]}

Fuchino S., Koppelberg S.\ \& Shelah S.\ [96] `Partial orderings with
the weak Freese-Nation property',
Ann.\ Pure and Applied Logic 80 (1996) 35-54.
\cmmnt{[518A, 518G, 518Yb, \S518 {\it notes\/}, 522U.]}

Fuchino S.\ \& Soukup L.\ [97] `More set theory around the weak
Freese-Nation property', Fundamenta Math.\ 154 (1997) 159-176.
\cmmnt{[518I, 518K.]}

\medskip%G

Galvin F.\ [80] `Chain conditions and products',
Fundamenta Math.\ 108 (1980) 33-48.
\cmmnt{[\S553 {\it notes\/}.]}

Galvin F., Mycielski J.\ \& Solovay R.M.\ [79] `Strong measure zero sets',
Notices Amer.\ Math.\ Soc.\ 26 (1979) A280.
%(abstract 79T-E25);  often assigned to year 1973
\cmmnt{[534K.]}

Gandy R.O.\ \& Hyland J.M.E.\ [77](eds.) {\it Logic Colloquium '76.}
North-Holland, 1977.

Geschke S.\ [02] `On tightly $\kappa$-filtered Boolean algebras',
Algebra Universalis 47 (2002) 69-93.
\cmmnt{[518P, 518S, \S535 {\it notes\/}.]}

Gitik M.\ \& Shelah S.\ [89] `Forcings with ideals and
simple forcing notions', Israel J.\ Math.\ 68 (1989) 129-160.
\cmmnt{[543E, 547F.]}

Gitik M.\ \& Shelah S.\ [93]
`More on simple forcing notions and forcings with ideals',
Annals of Pure and Applied Logic 59 (1993) 219-238.
\cmmnt{[542E, 543E, 547F.]}

Gitik M.\ \& Shelah S.\ [01]
`More on real-valued measurable cardinals and forcing with ideals',
Israel J.\ Math. 124 (2001) 221-242.
\cmmnt{[548E.]}
%GiSh582

G\l\'owczy\'nski W.\ [91] `Measures on Boolean algebras',
Proc.\ Amer.\ Math.\ Soc.\ 111 (1991) 845-849.
\cmmnt{[555K.]}

G\l\'owczy\'nski W.\ [08] `Outer measure on Boolean algebras', Acta Univ.\
Carolinae (Math.\ et Phys.) 49 (2008) 3-8.
\cmmnt{[555K.]}

Goldstern M., Judah H.\ \& Shelah S.\ [93] `Strong measure zero sets
without Cohen reals',
J.\ Symbolic Logic 58 (1993) 1323-1341.
\cmmnt{[\S534 {\it notes\/}.]}

\medskip%H

% Hart J.\ \& Kunen K.\ [98] `Weak Measure Extension Axioms',
% Topology and its Applications 85 (1998) 219-246.
% http://www.math.wisc.edu/~kunen/me.ps

Haydon R.G.\ [77] `On Banach spaces which contain $\ell^1(\tau)$ and
types of measures on compact spaces', Israel J.\ Math.\ 28 (1977) 313-324.
\cmmnt{[531E, 531L.]}

%Haydon R.G.\ [78] `On dual $L^1$-spaces and injective bidual Banach
%spaces,' Israel J.\ Math.\ 31 (1978) 142-152.

Hodges W., Hyland M., Steinhorn C.\ \& Truss J.\ [96] (eds.)
{\it Logic: from Foundations to Applications,
European Logic Colloquium, 1993.}
Clarendon, 1996.
%[ISBN 0-19-853862-6]

Humke P.D.\ \& Laczkovich M.\ [05] `Symmetrically approximately
continuous functions, consistent density theorems, and Fubini type
inequalities', Trans.\ Amer.\ Math.\ Soc.\ 357 (2005) 31-44.
\cmmnt{[537Q, \S537 {\it notes\/}.]}

\medskip%I

Ihoda J.I.\ [88] `Strong measure zero sets and rapid filters',
J.\ Symbolic Logic 53 (1988) 393-402.
%model of Borel conjecture + Ramsey filter
\cmmnt{[534Q.]}

\medskip%J

Jech T.\ [73] {\it The Axiom of Choice.}  North-Holland, 1973.
\cmmnt{[561A, 561Xc, 561Yc, 561Yi, 564Yd.]}

Jech T.\ [78] {\it Set Theory.}  Academic, 1978 (ISBN 0123819504).
\cmmnt{[{\it Intro.}, 553H,
555O, \S555 {\it notes\/}, \S562 {\it notes\/}, \S5A1, 5A6B.]}
%5A1B 5A1D 5A1E 5A1F B
%ASIN B0095LMZ9O

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2003
(ISBN 3540440852).
\cmmnt{[\S521 {\it notes\/}, \S555 {\it notes\/},
\S562 {\it notes\/}, \S567 {\it notes\/}, \S5A1, 5A3N, 5A3P, 5A6B.]}
%5A1B 5A1D 5A1E 5A1F B

%Jin R.\ \& Shelah S.\ [p96] `Compactness of Loeb spaces', preprint, 1996.

%Jovanovi\'c A.\ [p01]
%`Real valued measurability, some set-theoretic aspects',
%pp.\ ?? in {\smc Pap 0?}

Judah H.\ [93] (ed.) {\it Proceedings of the Bar-Ilan Conference on Set
Theory and the Reals, 1991.} Amer.\ Math.\ Soc.\ (Israel Mathematical
Conference Proceedings 6), 1993.

Judah H.\ \& Repick\'y M.\ [95] `Amoeba reals', J.\ Symbolic Logic 60 (1995) 1168-1185.
\cmmnt{[528N.]}

%Juh\'asz I.\ [71]  {\it Cardinal Functions in Topology.}  North-Holland,
%1975 (Math.\ Centre Tracts 34).

%Juh\'asz I.\ [80]  {\it Cardinal Functions in Topology -- Ten Years Later.}  Mathematisch Centrum, Amsterdam, 1980 (Math.\ Centre Tracts 123).   ISBN 9061961963.

Just W.\ [92] `A modification of Shelah's oracle-c.c., with
applications', Trans.\ Amer.\ Math.\ Soc.\ 329 (1992) 325-356.
\cmmnt{[527M.]}

%Just W.\ \& Krawczyk A.\ [84] `On certain Boolean algebras
%$\Cal P(\omega)/I$', Trans.\ Amer.\ Math.\ Soc.\ 285 (1984) 411-429.

Just W.\ \& Weese M.\ [96] {\it Discovering Modern Set Theory I.}
Amer.\ Math.\ Soc., 1996 (Graduate Studies in Mathematics 8).
\cmmnt{[{\it Intro.}.]}

Just W.\ \& Weese M.\ [97] {\it Discovering Modern Set Theory II.}
Amer.\ Math.\ Soc., 1997 (Graduate Studies in Mathematics 18).
\cmmnt{[5A1B, 5A1F, 5A1H.]}

\medskip%K

%Kada M. \& Yuasa Y.\ [96] `Cardinal invariants about shrinkability of unbounded
%sets', Topology and its Applications 74 (1996) 215-223.

%Kamburelis A.\ [89] `Iterations of Boolean algebras with measure',
%Arch.\ Math.\ Logic 29 (1989) 21-28.
% chargeable algebras

%Kamburelis A.\ [90] `A new proof of the Gitik-Shelah theorem', Israel
%J.\ Math.\ 72 (1990) 373-380.

Kanamori A.\ [03] {\it The Higher Infinite.}   Springer, 2003.
\cmmnt{[\S541 {\it notes\/}, \S555 {\it notes\/}, \S567 {\it notes\/},
5A1H, 5A6B, 5A6F.]}

Kechris A.S.\ [95] {\it Classical Descriptive Set Theory.}  Springer,
1995.
\cmmnt{[5A1D, \S567 {\it notes\/}.]}
%QA248.K4

Keisler H.J.\ \& Tarski A.\ [64] `From accessible to
inaccessible cardinals', Fundamenta Math.\ 53 (1964) 225-308;  errata
Fundamenta Math.\ 57 (1965) 119.
\cmmnt{[\S541 {\it intro.\/}, \S541 {\it notes\/}.]}

Kelley J.L.\ [50] `The Tychonoff product theorem implies the axiom of
choice', Fundamenta Math.\ 37 (1950) 75-76.
\cmmnt{[561D.]}

Koppelberg S.\ [75]
`Homomorphic images of $\sigma$-complete Boolean algebras',
Proc.\ Amer.\ Math.\ Soc.\ 51 (1975) 171-175.
\cmmnt{[515L.]}

Koppelberg S.\ [89] {\it General Theory of Boolean
Algebras\/}, vol.\ 1 of {\smc Monk 89}.
\cmmnt{[\S515 {\it notes\/}.]}

Koppelberg S.\ \& Shelah S.\ [96]
`Subalgebras of Cohen algebras need not be Cohen',
pp.\ 261-275 in {\smc Hodges Hyland Steinhorn \& Truss 96}.
\cmmnt{[\S515 {\it notes\/}, \S547 {\it notes\/}.]}

Kraszewski J.\ [01] `Properties of ideals on the generalized Cantor spaces',
J.\ Symbolic Logic 66 (2001) 1303-1320.
\cmmnt{[523G, 523H, 523J, \S523 {\it notes\/}.]}

%Krawczyk A.\ [n85] `Dominating reals add random reals', 1985.

Kumar A.\ [13] `Avoiding rational distances', Real Analysis Exchange 38
(2012/13), 493-498.
%{\tt http://www.math.wisc.edu/\~{}akumar/ARD.pdf}.
%{\tt
%https:/{\bsp}projecteuclid.org/journals/real-analysis-exchange/volume-38%
%/issue-2/Avoiding-Rational-{\vthsp}Distances/rae/1403894908.full}
\cmmnt{[548Xb.]}

%Kumar A.\ [n16] `On a theorem of Gitik and Shelah', note of January 2016
%({\tt http:/{\bsp}www.math.huji.ac.il/\~{}akumar/otgs.pdf}).

Kumar A.\ \& Shelah S.\ [17] `A transversal of full outer measure',
Advances in Math.\ 321 (2017) 475-485.
\cmmnt{[547P, 547Q, \S548 {\it intro.\/}, 548C.]}
%http://home.iitk.ac.in/~krashu/tfom.pdf;  printed version not seen

%Kunen K.\ [75] `Some points in $\beta N$',
% Math.\ Proc.\ Cambridge Phil.\ Soc.\ 80 (1975) 385-398.

Kunen K.\ [80] {\it Set Theory.}  North-Holland, 1980.
\cmmnt{[{\it Intro.}, 511A, \S521 {\it notes\/}, \S522 {\it notes\/},
\S541 {\it notes\/}, 551Q, \S551 {\it notes\/}, 556F,
\S562 {\it notes}, Appendix {\it intro.}, \S5A1, \S5A3, 5A6B.]}

Kunen K.\ [81] `A compact $L$-space under CH', Topology and its Appl.\
12 (1981) 283-287.
\cmmnt{[\S531 {\it notes\/}.]}

Kunen K.\ [84] `Random and Cohen reals', pp.\ 887-911 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[\S554 {\it notes}.]}

Kunen K.\ [n70] `$\Pi^1_1$ reflection at the continuum', note of January
1970.
\cmmnt{[543C, 544C, 544E, 544F.]}

Kunen K.\ \& Mill J.van [95] `Measures on Corson compact spaces',
Fundamenta Math.\ 147 (1995) 61-72.
\cmmnt{[531O.]}
%http://www.math.wisc.edu/~kunen/corson.ps

Kunen K.\ \& Vaughan J.E.\ [84] (eds.) {\it Handbook of Set-Theoretic
Topology.}   North-Holland, 1984.

Kuratowski K.\ [66] {\it Topology,} vol.\ I.   Academic, 1966.
\cmmnt{[562G.]}

\medskip%L

Lachlan A., Srebny M.\ \& Zarach A.\ [77] (eds.) {\it Set Theory and Hierarchy Theory.}  Springer, 1977 (Lecture Notes in Math.\ 619).

%Laczkovich M.\ \& Miller A.W.\ [95] `Measurability of functions with
%approximately continuous vertical sections and measurable horizontal
%sections', Colloquium Math.\ 69 (1995) 299-308.

Larson P.\ [09] `The filter dichotomy and medial limits',
J.\ Math.\ Logic 9 (2009) 159-165.
\cmmnt{[538S.]}

Larson P., Neeman I.\ \& Shelah S.\ [10] `Universally measurable sets in
generic extensions', Fundamenta Math.\ 208 (2010) 173-192.
\cmmnt{[553O.]}
	
%Larson P.\ \& Shelah S.\ [p09]
%`There may be just continuum many universally measurable sets', 2009.
%LrSh:947

Laver R.\ [76] `On the consistency of Borel's conjecture', Acta Math.\
137 (1976) 151-169.
\cmmnt{[534Q.]}

Laver R.\ [87] `Random reals and Souslin trees',
Proc.\ Amer.\ Math.\ Soc.\ 100 (1987) 531-534.
\cmmnt{[553M.]}

Levinski J.-P., Magidor M.\ \& Shelah S.\ [90] `Chang's conjecture for
$\aleph_{\omega}$', Israel J.\ Math.\ 69 (1990) 161-172.
\cmmnt{[5A6F.]}

Levy A.\ [71] `The sizes of the indescribable cardinals',
pp.\ 205-218 in {\smc Scott 71}.
\cmmnt{[\S541 {\it notes\/}, \S544 {\it notes\/}.]}

Lipecki Z.\ [09]
`Semivariations of an additive function on a Boolean ring',
Colloquium Math.\ 117 (2009) 267-279.
\cmmnt{[552Xe.]}

%Loeb P.A.\ \& Talvila E.\ [p00]
%`Covering theorems and Lebesgue integration', preprint, June 2000.

%Losada M. \& Todor\v{c}evi\'c S. [p98] `Chains and antichains in Boolean
%algebras', note of 5.12.98.

%Louveau A.\ \& Veli\v{c}kovi\'c B.\ [99] `Analytic ideals and cofinal
%types', Ann.\ Pure and Applied Logic 99 (1999) 171-195.
%\ell^1\not\prT\CalNwd

\medskip%M

Maharam D.\ [1947] `An algebraic characterization of measure algebras',
Ann.\ Math.\ 48 (1947) 154-167.
\cmmnt{[539P.]}

Makowsky J.A.\ \& Ravve E.V.\ (eds.) [98]
{\it Logic Colloquium '95.}  Springer, 1998 (Lecture Notes in Logic 11).
%ISBN 3-540-63994-2

%Malliaris M.\ \& Shelah S.\ [16] `Cofinality spectrum theorems in
%model theory, set theory and general topology',
%J.\ Amer.\ Math.\ Soc.\ 29 (2016) 237-297.

Martin D.A.\ [70] `Measurable cardinals and analytic games',
Fundamenta Math.\ 66 (1970) 287-291.
\cmmnt{[567N.]}

Martin D.A.\ [75] `Borel determinacy', Ann.\ of Math.\ (2) 102 (1975)
363-371.
\cmmnt{[\S567 {\it notes\/}.]}

Martin D.A.\ \& Solovay R.M.\ [70] `Internal Cohen
extensions', Ann.\ Math.\ Logic 2 (1970) 143-178.
\cmmnt{[Chap.\ 52 {\it intro.}, \S528 {\it intro.}, \S528 {\it notes\/}.]}

%Mathias A.R.D.\ [78] `0\# and the $p$-point problem', pp.\ 375-384 in
%{\smc M\"uller \& Scott 78}.

M\'atrai T.\ [p09] `More cofinal types of definable directed orders', 2009
({\tt https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.414.3938}).
\cmmnt{[526L, \S526 {\it notes\/}.]}
% books/matrai.p09.pdf

%Mercourakis S.\ [96] `Some remarks on countably determined measures and
%uniform distribution of sequences', Monatschefte f\"ur Math.\ 121 (1996)
%79-101.

Meyer P.-A.\ [73] `Limites m\'ediales d'apr\`es Mokobodzki',
pp.\ 198-204 in {\smc Dellacherie Meyer \& Weil 73}.
\cmmnt{[538Yl.]}

Meyer P.A.\ [75] (ed.) {\it S\'eminaire de Probabilit\'es IX.}
Springer, 1975 (Lecture Notes in Mathematics 465).

Miller A.W.\ [80] `There are no Q-points in Laver's model for the Borel
conjecture', Proc.\ Amer.\ Math.\ Soc.\ 78 (1980) 103-106.
\cmmnt{[\S538 {\it notes\/}.]}
%actually, no rapid ultrafilters, therefore no rapid filters

Miller A.W.\ [81] `Some properties of measure and category',
Trans.\ Amer.\ Math.\ Soc.\ 266 (1981) 93-114.
\cmmnt{[522J.]}

Miller A.W.\ [82] `The Baire category theorem and cardinals of countable
cofinality', J.\ Symbolic Logic 47 (1982) 275-288.
\cmmnt{[552G, 552Xb.]}

%Miller A.W.\ [83] `Mapping a set of reals onto the reals',
%J.\ Symbolic Logic 48 (1983) 575-584.

Mokobodzki G.\ [75] `R\'el\`evement bor\'elien compatible avec une
classe d'ensembles n\'egligeables.  Application \grv{a} la
d\'esint\'egration des mesures', pp.\ 437-442 in {\smc Meyer 75}.
\cmmnt{[535I.]}

Mokobodzki G.\ [7?] `D\'esintegration des mesures et rel\`evements
Boreliens de sous-espaces de $L^{\infty}(X,\Cal B,\mu)$',
Lab.\ d'Analyse Fonctionnelle, Univ.\ Paris VI (?).
\cmmnt{[535E, \S535 {\it notes\/}.]\footnote{I have not been able to
locate this paper;  I believe it was a seminar report.   I took notes from
it in 1977.}}
% "On two theorems of Mokobodzki", 23.6.77.

Monk J.D.\ [89] (ed.) {\it Handbook of Boolean Algebra.}
North-Holland, 1989.

%Monk J.D.\ [96] {\it Cardinal Invariants on Boolean Algebras.}
%Birkhauser, 1996.
% QA266

Moore J.T.\ [05] `Set mapping reflection', J.\ Math.\ Logic 5 (2005) 87-98.
\cmmnt{[517O.]}

Moschovakis Y.N.\ [70] `Determinacy and prewellorderings of the continuum',
pp.\ 24-62 in {\smc Bar-Hillel 70}.
\cmmnt{[567M.]}

%M\"uller G.H.\ \& Scott D.S.\ [78] (eds.) {\it Higher Set Theory.}
%Springer, 1978 (Lecture Notes in Mathematics 669).

Mycielski J.\ [64] `On the axiom of determinateness',
Fund.\ Math.\ 53 (1964) 205-224.
\cmmnt{[567D.]}

%Mycielski J.\ \& Steinhaus H.\ [62]
%`A mathematical axiom contradicting the axiom of choice',
%Bull.\ Acad.\ Polon.\ Sci.\ (Math., Astron., Phys.) 10 (1962) 1-3.

Mycielski J.\ \& \'Swierczkowski S.\ [64]
`On the Lebesgue measurability and the axiom of determinateness',
Fund.\ Math.\ 54 (1964) 67-71.
\cmmnt{[567F.]}

\medskip%N

Naimark M.A. [70] {\it Normed Rings.}  Wolters-Noordhoff, 1970.
\cmmnt{[\S561 {\it notes\/}.]}

Neumann J.von [1931] `Algebraische Repr\"asentanten der Funktionen ``bis
auf eine Menge vom Masse Null''\nobreak', Crelle's J.\ Math.\ 165 (1931)
109-115.
\cmmnt{[535G.]}

Neumann J.von \& Stone M.H.\ [1935] `The determination of representative
elements in the residual classes of a Boolean algebra', Fundamenta
Math.\ 25 (1935) 353-378.
\cmmnt{[\S535 {\it notes\/}.]}

%Normann D.\ [76] `Martin's axiom and medial functions', Math.\
%Scandinavica 38 (1976) 167-176.

%Nyikos P.J.\ [80] `A provisional solution to the normal Moore space
%problem', Proc.\ Amer.\ Math.\ Soc.\ 78 (1980) 429-435.

\medskip%O\medskip%Pa

%Pap E.\ [0?] (ed.) {\it Handbook of Measure Theory.}  North-Holland, 200?.

Pawlikowski J.\ [86] `Why Solovay real produces Cohen real',
J.\ Symbolic Logic 51 (1986) 957-968.
\cmmnt{[552G, 552H.]}

%Pb%Pc%Pd%Pe

Pe{\l}czy'nski A.\ [68] `On Banach spaces containing $L^1(\mu)$, Studia
Math.\ 30 (1968) 231-246.
\cmmnt{[531Ye.]}

Perovi\'c \v{Z}.\ \& Veli\v{c}kovi\'c B.\ [18] `Ranks of Maharam algebras',
Advances in Math.\ 330 (2018) 253-279.
\cmmnt{[539V.]}

%Pf%Pg%Ph%Pu%Pi

% Pierce R.S.\ [58] `A note on complete Boolean algebras',
% Proc.\ Amer.\ Math.\ Soc.\ 9 (1958) 892-896.
% the Dedekind complete case of 515L

%Pj%Pk%Pl

%Plebanek G.\ [93] `Remarks on measurable Boolean algebras and
%sequential cardinals', Fundamenta Math.\ 143 (1993) 11-22.

Plebanek G.\ [95] `On Radon measures on first-countable compact spaces',
Fundamenta Math.\ 148 (1995) 159-164.
\cmmnt{[531O.]}

Plebanek G.\ [97] `Non-separable Radon measures and small compact
spaces', Fundamenta Math.\ 153 (1997) 25-40.
\cmmnt{[531L, 531M, 531V.]}

Plebanek G.\ [00]
`Approximating Radon measures on first-countable compact spaces',
Colloquium Math.\ 86 (2000) 15-23.
\cmmnt{[533H, 533Yc.]}

Plebanek G.\ [02]
`On compact spaces carrying Radon measures of large Maharam type',
Acta Univ.\ Carolinae 43 (2002) 87-99.
\cmmnt{[531U.]}

Plebanek G.\ \& Sobota D.\ [15] `Countable tightness in the spaces of
regular probability measures', Fund.\ Math.\ 229 (2015) 159-169.
\cmmnt{[531U.]}

%Pm%Pn%Po

Pol R.\ [82] `Note on the spaces $P(S)$ of regular probability measures
whose topology is determined by countable subsets', Pacific J.\ Math.\
100 (1982) 185-201.
\cmmnt{[533Ya.]}

%Pp%Pq%Pr

Prikry K.\ [75] `Ideals and powers of cardinals', Bull.\
Amer.\ Math.\ Soc.\ 81 (1975) 907-909.
\cmmnt{[555N.]}

%Ps%Pt%Pu%Pv%Pw%Px%Py%Pz
\medskip%Q

Quickert S.\ [02] `CH and the Saks property', Fundamenta Math.\ 171
(2002) 93-100.
\cmmnt{[539L.]}

\medskip%R

Raisonnier J.\ \& Stern J.\ [85] `The strength of measurability hypotheses', Israel J.\ Math.\ 50 (1985) 337-349.
\cmmnt{[Chap.\ 52 {\it intro.}, 522Q.]}

%Roitman J.\ [79] `Adding a random or a Cohen real;  topological
%consequences and the effect on Martin's axiom', Fundamenta Math.\ 103
%(1979) 47-60.

%Ros{\l}anowski A.\ \& Shelah S.\ [p00] `Measured creatures', preprint, 7.10.00

Rothberger F.\ [1938a] `Eine \"Aquivalenz zwischen der kontinuumhypothese
unter der Existenz der Lusinschen und Sierpinschischen Mengen',
Fundamenta Math.\ 30 (1938) 215-217.
\cmmnt{[522G.]}

Rothberger F.\ [1938b] `Eine Versch\"arfung der Eigenschaft C',
Fundamenta Math.\ 30 (1938) 50-55.
\cmmnt{[\S534 {\it notes\/}.]}

Rothberger F.\ [1941] `Sur les familles ind\'enombrables de suites de nombres naturels et les probl\`emes concernant la propri\'et\'e {\bf C}', Proc.\ Cambridge Phil.\ Soc.\ 37 (1941) 109-126.
\cmmnt{[534R.]}

\medskip%S

Scott D.S.\ [71] (ed.) {\it Axiomatic Set Theory.}  Amer.\
Math.\ Soc., 1971
(Proceedings of Symposia in Pure Mathematics XIII, vol.\ 1).

Shelah S.\ [82] {\it Proper Forcing.}  Springer, 1982
(Lecture Notes in Mathematics 940).
\cmmnt{[\S538 {\it notes\/}.]}

Shelah S.\ [83] `Lifting problem of the measure algebra', Israel J.\
Math.\ 45 (1983) 90-96.   % Sh185
\cmmnt{[\S535 {\it notes\/}.]}

Shelah S.\ [84] `Can you take Solovay inaccessible away?', Israel J.\ Math.\ 48 (1984) 1-47.
\cmmnt{[\S522 {\it notes\/}.]}

Shelah S.\ [92] `Cardinal arithmetic for skeptics',
Bull.\ Amer.\ Math.\ Soc.\ 26 (1992) 197-210.
% Sh400a
\cmmnt{[5A2D.]}

Shelah S.\ [94] {\it Cardinal Arithmetic.}  Oxford U.P., 1994.
% Sh400
\cmmnt{[5A2D, 5A2G.]}

Shelah S.\ [96] `Further cardinal arithmetic', Israel J.\ Math.\ 95
(1996) 61-114.
\cmmnt{[542I.]}
%Sh 430

Shelah S.\ [98a] {\it Proper and Improper Forcing.}  Springer, 1998.
\cmmnt{[\S538 {\it notes\/}.]}
%http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pl/1235419814

Shelah S.\ [98b] `There may be no nowhere dense ultrafilter', pp.\ 305-324
in {\smc Makovsky \& Ravve 98}.
\cmmnt{[538H, \S538 {\it notes\/}.]}
% Sh594;  arXiv:math.LO/9611221.

% Shelah S.\ [99] `Special subsets of $^{\cf(\mu)}\mu$, Boolean
% algebras and Maharam measure algebras',
% Topology and its Applications 99 (1999) 135-235.
 % Sh620

Shelah S.\ [00] `Covering of the null ideal may have countable
cofinality', Fundamenta Math.\ 166 (2000) 109-136.
\cmmnt{[\S522 {\it notes\/}.]}
% Sh592

Shelah S.\ [03] `The null ideal restricted to some non-null set may be
$\aleph_1$-saturated',  Fundamenta Math.\ 179 (2003) 97-129.
\cmmnt{[ \S548 {\it notes\/}.]}
% Sh619

%Shelah S.\ \& Fremlin D.H.\ [93] `Pointwise compact and stable sets of
%measurable functions', J.\ Symbolic Logic 58 (1993) 435-455.
  %FrSh 406

Shelah S.\ \& Stepr\=ans J.\ [05]
`Comparing the uniformity invariants of null sets for different measures',
Advances in Math.\ 192 (2005) 403-426.
\cmmnt{[534Za.]}
% Con(\non\Cal N(\mu_H)<\non\Cal N)

%Shelah S.\ \& Woodin H.\ [90] `Large cardinals imply that every
%reasonably definable set of reals is Lebesgue measurable', Israel J.\
%Math.\ 70 (1990) 381-394.

Shipman J.\ [90] `Cardinal conditions for strong Fubini
theorems', Trans.\ Amer.\ Math.\ Soc.\ 321 (1990) 465-481.
\cmmnt{[537K.]}

%Solecki S.\ [06] `Amenability, free subgroups, and Haar null sets in
%non-locally-compact groups',
%Proc.\ London Math.\ Soc.\ 93 (2006) 693-722.

Solecki S.\ \& Todor\v{c}evi\'c S.\ [04]
`Cofinal types of topological directed orders',
Ann.\ Inst.\ Fourier 54 (2004) 1877-1911.
\cmmnt{[513K, 513O, 513Yi.]}

Solecki S.\ \& Todor\v{c}evi\'c S.\ [10]
`Avoiding families and Tukey functions on the nowhere-dense ideal',
J.\ Inst.\ Math.\ Jussieu 9 (2010) 1-31.
\cmmnt{[526L.]}

Solovay R.M.\ [66] `New proof of a theorem of Gaifman and Hales',
Bull.\ Amer.\ Math.\ Soc.\ 72 (1966) 282-284.
\cmmnt{[514Xi.]}

Solovay R.M.\ [70] `A model of set theory in which every set of reals
is Lebesgue measurable', Annals of Math.\ 92 (1970) 1-56.
\cmmnt{[\S522 {\it notes\/}.]}

Solovay R.M.\ [71] `Real-valued measurable cardinals', pp.\ 397-428 in
{\smc Scott 71}.
\cmmnt{[541J, 541P, 541Ya, \S541 {\it notes\/}, 555D, 555O.]}

%Solovay R.M., Reinhardt W.N.\ \& Kanamori A.\ [78]
%`Strong axioms of infinity and elementary embeddings',
%Annals of Mathematical Logic 13 (1978) 73-116.

%Solovay R.M.\ \& Tennenbaum S.\ [71]
%`Iterated Cohen extensions and Souslin's problem',
%Ann.\ Math.\ (2) 94 (1971) 201-245.

%Stepr\=ans J.\ [96] `Cardinal invariants associated with Hausdorff
%capacities', pp.\ 147-184 in {\smc Bartoszy\'nski \& Scheepers 96}.

%Szyma\'nski A.\ [88] `Some remarks on real-valued
%measurable cardinals', Proc.\ Amer.\ Math.\ Soc.\ 104 (1988)
%596-602.

\medskip%T\medskip

Talagrand M.\ [84] {\it Pettis Integral and Measure Theory.}
Mem.\ Amer.\ Math.\ Soc.\  307 (1984).
\cmmnt{[536C.]}

Talagrand M.\ [08] `Maharam's problem', Annals of Math.\ 168 (2008)
981-1009.
\cmmnt{[539A.]}

Tarski A.\ [1945] `Ideale in volst\"andigen Mengenk\"orpen II', Fundamenta Math.\ 33 (1945) 51-65.
\cmmnt{[541P.]}

Todor\v{c}evi\'c S.\ [85] `Remarks on chain conditions in products',
Compositio Math.\ 55 (1985) 295-302.
\cmmnt{[537G.]}

Todor\v{c}evi\'c S.\ [87] `Partitioning pairs of countable ordinals',
Acta Math.\ 159 (1987) 261-294.
\cmmnt{[554Yc.]}

Todor\v{c}evi\'c S.\ [00] `A dichotomy for $P$-ideals of countable sets',
Fundamenta Math.\ 166 (2000) 251-267.
\cmmnt{[5A6G.]}

T\"ornquist A.\ [11] `On the pointwise implementation of near-actions',
Trans.\ Amer.\ Math.\ Soc.\ 363 (2011) 4929-4944.
%{\tt http://arxiv.org/abs/0909.3869}
\cmmnt{[535Yd.]}

Truss J.K.\ [77] `Sets having calibre $\aleph_1$',
pp.\ 595-612 in {\smc Gandy \& Hyland 77}.
\cmmnt{[522J.]}

Truss J.K.\ [88] `Connections between different amoeba algebras',
Fund.\ Math.\ 130 (1988) 137-155.
\cmmnt{[Chap.\ 52 {\it intro.}, 528A, 528Da, 528K, \S528 {\it notes\/}.]}

Tukey J.W.\ [1940] {\it Convergence and Uniformity in Topology.}
Princeton U.P., 1940 (Ann.\ Math.\ Studies 1).
\cmmnt{[513F, \S513 {\it notes}.]}

%U
\medskip

Ulam S.\ [1930] `Zur Masstheorie in der allgemeinen
Mengenlehre', Fundamenta Math.\ 16 (1930) 140-150.
\cmmnt{[\S541 {\it intro.\/}, \S541 {\it notes\/}.]}

%V
\medskip

Veli\v{c}kovi\'c B.\ [92] `Forcing axioms and stationary sets', Advances in
Math.\ 94 (1992) 256-284.
\cmmnt{[517O.]}

Veli\v{c}kovi\'c B.\ [05] `CCC forcing and splitting reals',
Israel J.\ Math.\ 147 (2005) 209-221.
\cmmnt{[\S539 {\it intro.}, 539E, 539N.]}

Vojt\'a\v{s} P.\ [93] `Generalized Galois-Tukey connections between
explicit relations on classical objects of real analysis', pp.\ 619-643
in {\smc Judah 93}.
\cmmnt{[512A.]}

%W
\medskip

Wimmers E.\ [82] `The Shelah $P$-point independence theorem',
Israel J.\ Math.\ 43 (1982) 28-48.
\cmmnt{[\S538 {\it notes\/}.]}

\medskip%X\medskip%Y\medskip%Z

Zakrzewski P.\ [92] `Strong Fubini theorems from measure extension
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\cmmnt{[544J.]}

%Zindulka O.\ [00] `Residual measures in locally compact spaces', Topology Appl.\ 108 (2000) 253-265.

%Zindulka O.\ [p98] `Killing residual measures', preprint, 1998.

\discrpage

}%end of references

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\end



