% 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{mt5a.tex} 
\versiondate{21.11.13} 
\copyrightdate{2007} 
      
\def\chaptername{Appendix} 
\def\sectionname{Introduction} 
      
\gdef\topparagraph{} 
\gdef\bottomparagraph{Appendix to Vol.\ 5 {\it intro.}} 
      
\centerline{\bf Appendix to Volume 5} 
      
\medskip 
      
\centerline{\bf Useful facts} 
      
\medskip 
      
For this volume, the most substantial ideas demanded are, 
naturally enough, in
set theory.   Fragments of general set theory are in \S5A1, with cardinal
arithmetic and infinitary combinatorics.
\S5A2 contains results from Shelah's pcf theory,
restricted to those which are actually used in this book.   \S5A3
describes the language I will use when I discuss forcing constructions;
in essence, I follow {\smc Kunen 80}, but with some variations which need
to be signalled.   

As usual, some bits of general topology are needed;   I give these in
\S5A4, starting with a list of cardinal functions to complement the
definitions in \S511.   There is a tiny piece of real analysis in \S5A5.
In \S5A6 are notes on a few undecidable propositions, mostly standard.   

\discrpage 
      
      
\frfilename{mt5a1.tex}
\versiondate{3.9.20}
\copyrightdate{2006}

\def\chaptername{Appendix}
\def\sectionname{Set theory}

\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\Glsquare{{Gl}$\Square$}
\def\Sing{\mathop{\text{Sing}}}
\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\JechM{{\smc Jech 03}} %........
\def\Jech{{\smc Jech 78}} %.....
%{\smc Devlin 84}} %..... serious problems according to ARDM
%{\smc Kanamori 03} %....
\def\Kunen{{\smc Kunen 80}} %...
%\def\DS{{\smc Drake \& Singh 96}} %...

\newsection{5A1}

As usual, I begin with set theory, continuing from \S\S2A1 and 4A1.
I start with definitions and
elementary remarks filling some minor gaps in the deliberately sketchy
accounts in the earlier volumes (5A1A-5A1E).   %5A1A 5A1B 5A1C 5A1E
I give a relatively solid
paragraph on cardinal arithmetic (5A1F), including an account of
cofinalities of ideals $[\kappa]^{\le\lambda}$.
5A1H-5A1K %5{}A1G 5A1I 5A1J 5A1K
are devoted to infinitary combinatorics, with the Erd\H{o}s-Rado theorem
and Hajnal's Free Set Theorem.   5A1L-5A1P %5{}A1K 5A1M 5A1N 5A1O 5A1P
deal with the existence of `transversals' of various kinds in spaces of
functions, that is, large sets of functions which are well separated on
some combinatorial criterion.   5A1Q is a fragment of finite combinatorics,
5A1R-5A1S introduce `stationary families' of sets and 5A1T is a remarkable
property of the ordering of $\omega_1$.

\leader{5A1A}{Order types (a)} If $X$ is a well-ordered set, its
{\bf order type}
$\otp X$ is the ordinal order-isomorphic to $X$\cmmnt{ (2A1Dg)}.

If $S$ is a set of ordinals, an ordinal-valued function $f$ with
domain $S$ is {\bf regressive} if $f(\xi)<\xi$ for every
$\xi\in S$\cmmnt{ (cf.\ 4A1Cc)}.

\spheader 5A1Ab The non-stationary ideal on a
cardinal $\kappa$ of uncountable cofinality
is $(\cf\kappa)$-additive\prooflet{, because the intersection of fewer than
$\cf\kappa$ closed cofinal sets is a closed cofinal set (4A1Bd)}.

\spheader 5A1Ac If $\kappa$ is a cardinal,
$\lambda<\cf\kappa$ is an infinite regular cardinal and $C\subseteq\kappa$
is a closed cofinal set, then
$S=\{\xi:\xi<\kappa$, $\cf(\xi\cap C)=\lambda\}$ is stationary in $\kappa$.
\prooflet{\Prf\
If $C'\subseteq\kappa$ is a closed cofinal set, let
$\ofamily{\xi}{\otp(C\cap C')}{\gamma_{\xi}}$ be the increasing
enumeration of $C\cap C'$.
Then $\otp(C\cap C')\ge\cf\kappa>\lambda$, so $\gamma_{\lambda}$ is
defined and belongs to $S\cap C'$.\ \Qed}

\spheader 5A1Ad If $\alpha$ is an ordinal and $C\subseteq\alpha$ has
closure $\overline{C}$ for the order topology of $\alpha$, then
$\#(\overline{C})=\#(C)$.   \prooflet{\Prf\ If $C$ is finite this is
trivial.   Otherwise, for any $\beta\in\overline{C}\setminus\{\sup C\}$
set $f(\beta)=\min(C\setminus\beta)$.  If $\beta$,
$\beta'\in\overline{C}$
and $\beta<\beta'<\sup C$ there must be a $\gamma\in C$ such that
$\beta\le\gamma<\beta'$ and $f(\beta)\le\gamma<f(\beta')$ (4A2S(a-ii));
thus $f$ is injective.   So

\Centerline{$\#(\overline{C})
=\#(\overline{C}\setminus\{\sup C\})\le\#(C)\le\#(\overline{C})$.  \Qed}}

\spheader 5A1Ae If $\alpha$ is an ordinal, there is a closed cofinal set
$C\subseteq\alpha$ such that $\otp C=\cf\alpha$.   \prooflet{\Prf\ If
$\cf\alpha$ is finite then $\alpha$ is either $0$ or a successor ordinal and the
result is trivial.   Otherwise, let $C_0$ be a cofinal subset of $\alpha$ with
cardinality $\cf\alpha$.
Then its closure $C$ in the order topology of $\alpha$ is
cofinal and $\#(C)=\cf\alpha$ by (d).   Since any cofinal subset of $C$ is also
cofinal with $\alpha$,

\Centerline{$\cf\alpha\le\cf C\le\#(C)=\cf\alpha$}

\noindent and $C$ has cofinality $\cf\alpha$.\ \Qed}
%5{}A2B

\vleader{72pt}{5A1B}{Ordinal arithmetic (a)}
For ordinals $\xi$, $\eta$ their
{\bf ordinal sum} $\xi+\eta$ is defined inductively by saying that

\inset{$\xi+0=\xi$,

$\xi+(\eta+1)=(\xi+\eta)+1$,

$\xi+\eta=\sup_{\zeta<\eta}\xi+\zeta$ for non-zero limit ordinals
$\eta$.}

\noindent Ordinal addition is associative, and if $\xi\le\zeta<\xi+\eta$
there is a unique $\zeta'<\eta$ such that $\zeta=\xi+\zeta'$.
\cmmnt{(\Kunen, I.7.18;  \Jech, p.\ 18;  \JechM, 2.18;
{\smc Just \& Weese 97}, \S10.2.)}   If we identify $\Bbb N$ with $\omega$, then
the ordinal sum of two finite ordinals corresponds to ordinary addition on
$\Bbb N$.

\spheader 5A1Bb For ordinals
$\xi$, $\eta$ their {\bf ordinal product} $\xi\cdot\eta$ is defined
inductively by saying

\inset{$\xi\cdot 0=0$,

$\xi\cdot(\eta+1)$ is the ordinal sum $\xi\cdot\eta+\xi$,

$\xi\cdot\eta=\sup_{\zeta<\eta}\xi\cdot\zeta$ for non-zero limit ordinals
$\eta$\dvro{.}{}}

\noindent\cmmnt{(\Kunen, I.7.20;  \Jech, p.\ 19;
\JechM, 2.19). }\cmmnt{Note that }$0\cdot\eta=0$ and
$1\cdot\eta=\eta$ for every
$\eta$, and\cmmnt{ that} $\sup_{\zeta\in A}\xi\cdot\zeta=\xi\cdot(\sup A)$ for
every $\xi$ and every
non-empty set $A$ of ordinals.   Ordinal multiplication is
associative\cmmnt{ (\Kunen, I.7.20;  \JechM, 2.21)}.

\spheader 5A1Bc For ordinals $\xi$, $\eta$ the {\bf ordinal power}
$\xi^{\eta}$ is defined inductively by saying that

\inset{$\xi^0=1$,

$\xi^{\eta+1}$ is the ordinal product $\xi^{\eta}\cdot\xi$,

$\xi^{\eta}=\sup_{\zeta<\eta}\xi^{\zeta}$ for non-zero limit ordinals
$\eta$\dvro{.}{}}

\noindent\cmmnt{(\Kunen, I.9.5;  \JechM, 2.20).
{\bf Warning!}  If $\xi$ and $\eta$
happen to be cardinals, this is quite different from the `cardinal power'
of 5A1F below.

}If $\xi$, $\eta$ are ordinals, $\eta\ne 0$ and $\eta$ is greater than or
equal to the ordinal product $\xi\cdot\eta$, then $\eta$ is at least the
ordinal power $\xi^{\omega}$.   \cmmnt{\Prf\ Note first that as
multiplication is
associative, we can induce on $n$ to show that $\xi\cdot\xi^n=\xi^{n+1}$
for every $n$.   Now we are supposing that
$\eta\ge 1=\xi^0$.   If $n\in\Bbb N$ and $\eta\ge\xi^n$, then

\Centerline{$\eta\ge\xi\cdot\eta\ge\xi\cdot\xi^n=\xi^{n+1}$.}

\noindent So $\eta\ge\xi^n$ for every $n$ and $\eta\ge\xi^{\omega}$.\ \Qed}

\leader{5A1C}{Concatenation}\dvAnew{2015}
\cmmnt{It will perhaps be
helpful if I describe in detail a semi-standard notation which I have
already used, in special cases, at many points in Volumes 3 and 4.}
Suppose that
$\sigma$, $\tau$ are two functions with domains $\alpha$, $\beta$ respectively
which are ordinals\cmmnt{ (e.g.,
initial segments of $\Bbb N$, if we think of non-negative integers as finite
ordinals)}.   Then we can form their
{\bf concatenation} $\sigma^{\smallfrown}\tau$, setting

\Centerline{$\dom(\sigma^{\smallfrown}\tau)=\alpha+\beta$}

\noindent(the ordinal sum),

$$\eqalign{(\sigma^{\smallfrown}\tau)(\xi)
&=\sigma(\xi)\text{ if }\xi<\alpha,\cr
(\sigma^{\smallfrown}\tau)(\alpha+\eta)
&=\tau(\eta)\text{ if }\eta<\beta.\cr}$$

The operator $^{\smallfrown}$ is associative\cmmnt{,
that is, if
$\sigma$, $\tau$, $\upsilon$ have ordinal domains, then
$(\sigma^{\smallfrown}\tau)^{\smallfrown}\upsilon
=\sigma^{\smallfrown}(\tau^{\smallfrown}\upsilon)$},
so we can omit brackets and speak of
$\sigma^{\smallfrown}\tau^{\smallfrown}\upsilon$.   The empty function
$\emptyset$ is an identity in the sense that

\Centerline{$\emptyset^{\smallfrown}\sigma=\sigma^{\smallfrown}\emptyset
=\sigma$}

\noindent whenever $\dom\sigma$ is an ordinal.

In this context, it will often be helpful to have a special notation for
functions with domain the singleton set
$\{0\}=1$;  I will write $\fraction{t}$ for the
function with domain $\{0\}$ and value $t$.

\cmmnt{We can also have infinite concatenations.}   
If $\sequencen{\sigma_n}$ is a
sequence of functions with ordinal domains, we can form the
concatenations

\Centerline{$\sigma_0^{\smallfrown}\sigma_1$,
\quad$\sigma_0^{\smallfrown}\sigma_1^{\smallfrown}\sigma_2$,
\quad$\sigma_0^{\smallfrown}\sigma_1^{\smallfrown}
\sigma_2^{\smallfrown}\sigma_3$,
\quad$\ldots$}

\noindent to get a sequence of functions each extending its predecessors.
The union will be a function with domain the ordinal
$\sup_{n\in\Bbb N}\dom\sigma_0+\ldots+\dom\sigma_n$.   I will generally
denote it
$\sigma_0^{\smallfrown}\sigma_1^{\smallfrown}\sigma_2^{\smallfrown}\ldots$
or in some similar form.

\leader{5A1D}{Well-founded sets}\dvAformerly{5{}A1C}{\bf (a)} A partially ordered set $P$ is {\bf
well-founded} if every non-empty $A\subseteq P$ has a minimal
element\cmmnt{, that
is, a $p\in A$ such that $q\not<p$ for every $q\in A$}.

\spheader 5A1Db If $P$ is a well-founded partially ordered set,
we have a rank function $r:P\to\On$ defined by saying that

\Centerline{$r(p)=\sup\{r(q)+1:q<p\}$}

\noindent for every $p\in P$\cmmnt{ (\Kunen, III.5.7;
\Jech, p.\ 21;  \JechM, 2.27)}.
The {\bf height} of $P$ is the least
ordinal $\zeta$ such that $r(p)<\zeta$ for
every $p\in P$.

\spheader 5A1Dc A partially ordered set $P$ is well-founded iff there is no
sequence $\sequencen{p_n}$ in $P$ such that $p_{n+1}<p_n$ for every
$n\in\Bbb N$.   \prooflet{(If $A\subseteq P$ is non-empty and has no
minimal element, we can choose inductively a strictly decreasing sequence
in $A$.)}

\spheader 5A1Dd If $P$ is a well-founded partially ordered set with height
$\zeta$, $\#(\zeta)\le\#(P)$.
\prooflet{\Prf\ Let $r:P\to\On$ be the rank function of
$P$.   Set $A_{\xi}=\{p:r(p)\ge\xi\}$ for ordinals $\xi$.   If
$A_{\xi}\ne\emptyset$, then $A_{\xi}$ has a minimal element $p$;  now
$r(q)<\xi$ whenever $q<p$, so $r(p)=\xi$.
The height of $P$ is the least ordinal $\zeta$ such that $r[P]\subseteq\zeta$;
now $A_{\xi}\ne\emptyset$ for $\xi<\zeta$, so $\zeta=r[P]$ and
$\#(\zeta)\le\#(P)$.\ \Qed}

\spheader 5A1De\dvAnew{2017}\cmmnt{ The
next result really belongs with the descriptive
set theory of Chapter 42, but I had no reason to call on it in that volume, so
I set it out here.}
If $X$ is a Polish space and $\le$ is a well-founded
relation on $X$ such that $\{(x,y):x<y\}$ is analytic, then the height of
$\le$ is countable.   \prooflet{({\smc Kechris 95}, Theorem 31.1.
This is a form of the `Kunen-Martin theorem'.)}

\leader{5A1E}{Trees }\dvAformerly{5{}A1D}\cmmnt{In \S421 %421N
I introduced trees of sequences.
For this volume a more abstract approach is useful.

\medskip

}{\bf (a)} A {\bf tree} is a partially
ordered set $T$ such that $\{s:s\in T$, $s\le t\}$ is well-ordered for
every $t\in T$\cmmnt{;  alternatively, a well-founded partially ordered
set such that $\{s:s\in T$, $s\le t\}$ is totally ordered for
every $t\in T$}.   \cmmnt{In particular,} $T$ has
a rank function $r:T\to\On$ defined by saying that

\Centerline{$r(t)=\otp\{s:s<t\}=\min\{\xi:r(s)<\xi$ whenever $s<t\}$}

\noindent for every $t\in T$\cmmnt{ (5A1D)}.
\cmmnt{ (Try to avoid using this terminology
in the same sentence as that of 421Ne and 562A.)}

The {\bf levels} of $T$ are now the sets $\{t:r(t)=\xi\}$ for $\xi\in\On$.
A {\bf branch} of $T$ is a maximal totally ordered
subset.   A tree is {\bf well-pruned} if it has at most one minimal element
and whenever $s$, $t\in T$ and $r(s)<r(t)$, there is an $s'\ge s$ such that
$r(s')=r(t)$.
If $T$ is a tree, a {\bf subtree} of $T$ is a set
$T'\subseteq T$ such that $s\in T'$ whenever $s\le t\in T'$;  in this
case, the rank function of $T'$ is the restriction to $T'$ of the rank
function of $T$.

\spheader 5A1Eb{\bf (i)} Let $T$ be a tree in which every level is
finite.   Then
$T$ has a branch meeting every level.   \prooflet{\Prf\ If $T$ is empty,
this is trivial.   Otherwise, let $r$ be the rank function of $T$,
and $\zeta>0$ the height of
$T$;  let $\Cal F$ an ultrafilter on $T$ containing $\{t:r(t)\ge\xi\}$ for
every $\xi<\zeta$.   Set $C=\{t:\coint{t,\infty}\in\Cal F\}$.
Any two elements of $C$ are upwards-compatible, so $C$ is totally ordered,
and $C$ meets every level of $T$;  so $C$ is a branch of the kind we seek.\
\Qed}

\medskip

\quad{\bf (ii)} Let $(T,\preccurlyeq')$ be a tree of height $\omega_1$ in
which every level is countable.    Then there is an ordering
$\preccurlyeq$ of $\omega_1$, included in the usual ordering $\le$ of
$\omega_1$, such that $(T,\preccurlyeq')$ is
isomorphic to $(\omega_1,\preccurlyeq)$.   \prooflet{\Prf\ Let
$\ofamily{\xi}{\omega_1}{T_{\xi}}$ be the levels of $T$.   Let
$\le'_{\xi}$ be a well-ordering of $T_{\xi}$ for each
$\xi<\omega_1$, and define $\le'$ on $T$ by saying that
$s\le't$ if either $r(s)<r(t)$ or $r(s)=r(t)=\xi$ and
$s\le'_{\xi}t$;  then $\le'$ is a well-ordering of $T$
of order type $\omega_1$.   Now the order-isomorphism between
$(T,\le')$ and $(\omega_1,\le)$ copies $\preccurlyeq'$ onto a tree ordering
of $\omega_1$, isomorphic to $\preccurlyeq'$, and included in the usual
ordering.\ \Qed}

\spheader 5A1Ec An {\bf Aronszajn tree} is a tree $T$ of height $\omega_1$
in which every branch and every level is countable.
An Aronszajn tree $T$ is {\bf special} if it is expressible as
$\bigcup_{n\in\Bbb N}A_n$ where no two elements of any $A_n$ are
comparable\cmmnt{, that is, every $A_n$ is an up-antichain}.

\spheader 5A1Ed(i) A {\bf Souslin tree} is a tree $T$ of height $\omega_1$
in which every branch and every up-antichain is countable.

\quad(ii) Every Souslin tree is a non-special Aronszajn tree.

\quad(iii) If $T$ is a Souslin tree,
it has a subtree which is a well-pruned
Souslin tree.   \prooflet{(\Kunen, II.5.11;  \Jech,
p.\ 218;  \JechM, 9.13.)}

\quad(iv) {\bf Souslin's hypothesis} is the assertion

\Centerline{(SH)\quad There are no Souslin trees.}

\leader{5A1F}{Cardinal arithmetic}\dvAformerly{5{}A1E}{\bf (a)(i)} 
An infinite cardinal which is
not regular\cmmnt{ (4A1Aa)} is {\bf singular}.
A cardinal $\kappa$ is a {\bf successor cardinal} if it is of the form
$\lambda^+$\cmmnt{ (2A1Fc, 2A1Kb)};  otherwise it is a {\bf limit cardinal}.
$\kappa$ is a
{\bf strong limit cardinal} if it is uncountable and $2^{\lambda}<\kappa$
for every $\lambda<\kappa$.   It
is {\bf weakly inaccessible} if it is a regular uncountable limit
cardinal;  it is {\bf strongly inaccessible} if moreover it is a strong
limit cardinal.

\medskip

\quad{\bf (ii)} If $\kappa$ is a cardinal,
define $\kappa^{(+\xi)}$, for ordinals $\xi$, by setting

\Centerline{$\kappa^{(+0)}=\kappa$,
\quad$\kappa^{(+\xi)}=\sup_{\eta<\xi}(\kappa^{(+\eta)})^+$ if $\xi>0$,}

\noindent that is, $\kappa^{(+\xi)}=\omega_{\zeta+\xi}$ if
$\kappa=\omega_{\zeta}$.

\medskip

\spheader 5A1Fb{\bf (i)} If $\familyiI{\kappa_i}$ is a
family of cardinals, its {\bf cardinal sum} is
$\#(\{(i,\xi):i\in I$, $\xi<\kappa_i\})$, which is at most
$\max(\omega,\#(I),\sup_{i\in I}\kappa_i)$.

\medskip

\quad{\bf (ii)} For cardinals $\kappa$ and $\lambda$,
the {\bf cardinal product} $\kappa\cdot\lambda$ is
$\#(\kappa\times\lambda)\le\max(\omega,\kappa,\lambda)$.
%5{}12I

\medskip

\quad{\bf (iii)} If $\kappa$ and
$\lambda$ are cardinals there are two natural interpretations
of the formula $\kappa^{\lambda}$:  (i) the set of functions from $\lambda$
to $\kappa$ (ii) the cardinal of this set.   In this volume the latter will
be the usual one, but I will try to signal this by using the phrase {\bf
cardinal power}.   \cmmnt{Of course}
$2^{\lambda}$ is always the cardinal power;
the corresponding set of functions will be denoted
by $\{0,1\}^{\lambda}$.   \cmmnt{We could also think of $\kappa$ and
$\lambda$ as ordinals, and look at the ordinal power $\kappa^{\lambda}$ as
described in 5A1Bc;  but I think I do this exactly three times,
all at the end of \S539.}

\spheader 5A1Fc{\bf (i)} The cardinal power
$\kappa^{\lambda}$ is at most $2^{\max(\omega,\kappa,\lambda)}$ for any
cardinals $\kappa$ and $\lambda$.
\prooflet{(The
set of functions from $\lambda$ to $\kappa$ is a subset
of $\Cal P(\lambda\times\kappa)$.)}

\medskip

\quad{\bf (ii)} $\frak c^{\omega}\cmmnt{=\#(((\{0,1\})^{\Bbb N})^{\Bbb N})
=\#(\{0,1\}^{\Bbb N\times\Bbb N})=\#(\{0,1\}^{\Bbb N})}=\frak c$.

%5{}21N

\spheader 5A1Fd $\cf 2^{\kappa}>\kappa$ for every infinite cardinal
$\kappa$.  \prooflet{(\JechM, 5.11;  \Jech, p.\ 46;
{\smc Erd\H{o}s Hajnal M\'at\'e \& Rado 84}, 6.9;  \Kunen, 10.41;  %\DS, 7.4.6(5);
{\smc Just \& Weese 97}, 11.2.24.   Compare (e-v) below.)}
%5{}22U

\spheader 5A1Fe{\bf (i)} If $\kappa$ and $\lambda$ are infinite cardinals,
then\cmmnt{, defining $[\kappa]^{\le\lambda}$ as in 3A1J,}

$$\eqalign{\cff[\kappa]^{\le\lambda}
&=1\text{ if }\lambda\ge\kappa,\cr
&\ge\kappa\text{ if }\lambda<\kappa.\cr}$$

\medskip

\quad{\bf (ii)} Let $\kappa$, $\lambda$ and $\theta$ be infinite
cardinals such that $\theta\le\lambda\le\kappa$.   Then
$\cff[\kappa]^{\le\theta}
\le\max(\cff[\kappa]^{\le\lambda},\cff[\lambda]^{\le\theta})$.
\prooflet{\Prf\ Let $\Cal A\subseteq[\kappa]^{\lambda}$ be a cofinal set
with cardinal $\cff[\kappa]^{\lambda}=\cff[\kappa]^{\le\lambda}$.   Then
$[\kappa]^{\le\theta}=\bigcup_{A\in\Cal A}[A]^{\le\theta}$, so

\Centerline{$\cff[\kappa]^{\le\theta}
\le\max(\#(\Cal A),\sup_{A\in\Cal A}\cff[A]^{\le\theta})
\le\max(\cff[\kappa]^{\le\lambda},\cff[\lambda]^{\le\theta})$.  \Qed}}

\medskip

\quad{\bf (iii)} Let $\kappa$ and $\lambda$ be infinite cardinals.   Then
the cardinal power $\kappa^{\lambda}$ is
$\max(2^{\lambda},\cff[\kappa]^{\le\lambda})$.   \prooflet{\Prf\
$\kappa^{\lambda}\ge 2^{\lambda}$ because $\kappa\ge 2$;
$\kappa^{\lambda}\ge\#([\kappa]^{\le\lambda})\ge\cff[\kappa]^{\le\lambda}$
because $f\mapsto f[\lambda]$ is a surjection from the family
$F$ of functions from $\lambda$ to $\kappa$ onto
$[\kappa]^{\le\lambda}\setminus\{\emptyset\}$.   In the other direction, if
$\kappa\le\lambda$ then $F\subseteq\Cal P(\lambda\times\kappa)$ so

\Centerline{$\kappa^{\lambda}
=\#(F)\le 2^{\lambda}=\max(2^{\lambda},\cff[\kappa]^{\le\lambda})$.}

\noindent
If $\lambda<\kappa$ let $\Cal A\subseteq[\kappa]^{\le\lambda}$ be a cofinal
family with cardinal $\cff[\kappa]^{\le\lambda}$;  then
$F=\bigcup_{A\in\Cal A}A^{\lambda}$ so

\Centerline{$\#(F)\le\max(\#(\Cal A),\sup_{A\in\Cal A}\#(A^{\lambda}))
=\max(\cff[\kappa]^{\le\lambda},2^{\lambda})$.  \Qed}}

\cmmnt{Putting this together with (i) and (c-ii) above,}

\Centerline{$\cff[\frak c]^{\le\omega}\cmmnt{\mskip5mu=\max(\frak c,\cff[\frak c]^{\le\omega})
=\max(2^{\omega},\cff[\frak c]^{\le\omega})=\frak c^{\omega}}=\frak c$.}
\medskip

\quad{\bf (iv)} If $\lambda$ is an infinite cardinal and
$\lambda\le\kappa<\lambda^{(+\omega)}$, then
$\cff[\kappa]^{\le\omega}\le\max(\kappa,\cff[\lambda]^{\le\omega})$, with
equality if $\kappa>\omega$.
\prooflet{\Prf\
Induce on $n$ to see that
$\cff[\lambda^{(+n)}]^{\le\omega}
\le\max(\lambda^{(+n)},\cff[\lambda]^{\le\omega})$ for every $n\in\Bbb N$.   At
the inductive step to $n>0$,

$$\eqalign{\cff([\lambda^{(+n)}]^{\le\omega})
&=\cff(\bigcup_{\xi<\lambda^{(+n)}}[\xi]^{\le\omega})
\le\max(\lambda^{(+n)},\sup_{\xi<\lambda^{(+n)}}\cff[\xi]^{\le\omega})
  \cr
&\le\max(\lambda^{(+n)},\lambda^{(+(n-1))},\cff[\lambda]^{\le\omega})
=\max(\lambda^{(+n)},\cff[\lambda]^{\le\omega}).
\text{ \Qed}\cr}$$

\noindent If $\kappa$ is uncountable, then $\cff[\kappa]^{\le\omega}\ge\kappa$,
so $\cff[\kappa]^{\le\omega}=\max(\kappa,\cff[\lambda]^{\le\omega})$.
}Consequently the cardinal power $\kappa^{\omega}$ is
$\cmmnt{\max(\frak c,\kappa,\cff[\lambda]^{\le\omega})
=}\max(\kappa,\lambda^{\omega})$.

In particular, if $\omega_1\le\kappa<\omega_{\omega}$
then $\cff[\kappa]^{\le\omega}=\kappa$ and
$\kappa^{\omega}=\max(\frak c,\kappa)$.   \cmmnt{Similarly,}
$(\frak c^+)^{\omega}\cmmnt{\mskip5mu
  =\max(\frak c^+,\frak c)}=\frak c^+$,
$(\frak c^{++})^{\omega}=\frak c^{++}$.
%5{}22T 5{}23J

\medskip

\quad{\bf (v)} If $\kappa$ is a singular infinite cardinal,
then $\cf([\kappa]^{\le\cf\kappa})>\kappa$.
\prooflet{\Prf\ Set $\lambda=\cf\kappa$, and
let $\ofamily{\xi}{\lambda}{\kappa_{\xi}}$
be a strictly increasing
family of cardinals with supremum $\kappa$.   If
$\ofamily{\eta}{\kappa}{A_{\eta}}$ is a family in $[\kappa]^{\le\lambda}$,
then for each $\xi<\lambda$ take
$\alpha_{\xi}\in\kappa\setminus\bigcup_{\eta<\kappa_{\xi}}A_{\eta}$;
set $A=\{\alpha_{\xi}:\xi<\lambda\}\in[\kappa]^{\le\lambda}$;  then
$A\not\subseteq A_{\eta}$ for every $\eta<\kappa$.\ \Qed}
%5{}24P

\spheader 5A1Ff If $\lambda$ is a regular uncountable
cardinal, $\theta\ge 2$ is a cardinal and
$\kappa=\sup_{\delta<\lambda}\theta^{\delta}$, where
$\theta^{\delta}$ is the cardinal power, then

\Centerline{$\#([\kappa]^{<\lambda})
\cmmnt{\mskip5mu=\sup_{\delta<\lambda}\kappa^\delta}=\kappa$.}

\prooflet{\noindent\Prf\ Of course
$\kappa\le\#([\kappa]^{<\lambda})$ because $\lambda\ge 2$, while

\Centerline{$\#([\kappa]^{<\lambda})
=\#(\bigcup_{\delta<\lambda}[\kappa]^{\delta})
\le\max(\lambda,\omega,\sup_{\delta<\lambda}\#([\kappa]^{\delta}))
\le\sup_{\delta<\lambda}\kappa^{\delta}$}

\noindent because $\lambda\le\sup_{\delta<\lambda}2^{\delta}\le\kappa$.
If $\delta<\lambda$ then, because $\lambda$ is regular,

\Centerline{$\kappa^{\delta}
=(\sup_{\zeta<\lambda}\theta^{\zeta})^{\delta}
=\sup_{\zeta<\lambda}(\theta^{\zeta})^{\delta}
\le\sup_{\zeta<\lambda}\theta^{\max(\omega,\zeta,\delta)}
\le\kappa$.  \Qed}}
%5{}14D 5{}15H

In particular, if $\kappa$ is strongly inaccessible then
$\kappa^{\delta}\le\kappa$ for every $\delta<\kappa$.   \prooflet{(Take
$\lambda=\kappa$ and $\theta=2$.)}

\spheader 5A1Fg Let $X$, $Y$ and $Z$ be sets, with $\#(X)\le 2^{\#(Z)}$
and $0<\#(Y)\le\#(Z)$.
Then there is a function $f:X\times Z^{\Bbb N}\to Y$ such that
whenever $\sequencen{x_n}$ is a sequence of distinct elements
of $X$ and $\sequencen{y_n}$ is a sequence in $Y$ there is
a $z\in Z^{\Bbb N}$ such that $f(x_n,z)=y_n$ for every
$n\in\Bbb N$.  \prooflet{\Prf\ We can suppose that $X\subseteq\Cal PZ$ and
$Y\subseteq Z$;  moreover, the case of finite $X$ is trivial, so we can
suppose that $Z$ is infinite.   For each countably infinite set
$I\subseteq Z$, (c-ii) above tells us that there is a surjection
$g_I:I^{\Bbb N}\to(\Cal PI)^{\Bbb N}\times I^{\Bbb N}$.
Now let $f:X\times Z^{\Bbb N}\to Y$ be such that

\inset{\noindent
whenever $z\in Z^{\Bbb N}$ is such that $I=z[\Bbb N]$ is
infinite, $g_I(z)=(\sequencen{a_n},\sequencen{y_n})$ and $x\in X$ is such
that there is just one $n$ for which $a_n=x\cap I$, then $f(x,z)=y_n$.}

\noindent In this case, if $\sequencen{x_n}$ is a sequence of distinct
members of $X$ and $\sequencen{y_n}$ is a sequence in $Y$, let $I$ be a
countably infinite subset of $Z$ containing every $y_n$ and such that
$x_m\cap I\ne x_n\cap I$ for $m<n$;  let $z\in I^{\Bbb N}$ be such that
$g_I(z)=(\sequencen{x_n\cap I},\sequencen{y_n})$;  we shall have
$f(x_n,z)=y_n$ for every $n$.\ \Qed}
%5{}25M

\spheader 5A1Fh If $\kappa$ is an infinite cardinal, then
$2^{\kappa}$ is at most the cardinal power
$(\sup_{\lambda<\kappa}2^{\lambda})^{\cf\kappa}$.   \prooflet{\Prf\
Let $\ofamily{\xi}{\cf\kappa}{\alpha_{\xi}}$ be a family in $\kappa$ with
supremum $\kappa$.   Set $D=\bigcup_{\alpha<\kappa}\Cal P\alpha$;  then

\Centerline{$\#(D)\le\max(\kappa,\sup_{\alpha<\kappa}2^{\#(\alpha)})
=\sup_{\lambda<\kappa}2^{\lambda}$.}

\noindent Let $F$ be the set of functions from $\cf\kappa$ to $D$;  we
have an injection $A\mapsto\ofamily{\xi}{\cf\kappa}{A\cap\alpha_{\xi}}$
from $\Cal P\kappa$ to $F$, so $2^{\kappa}\le\#(F)$.\ \Qed}

If $\omega\le\lambda<\kappa$ and
$2^{\theta}=2^{\lambda}$ for $\lambda\le\theta<\kappa$ but
$2^{\kappa}>2^{\lambda}$ then $\kappa$ is regular.   \prooflet{\Prf\
$2^{\kappa}\le(2^{\lambda})^{\cf\kappa}=2^{\max(\lambda,\cf\kappa)}$.\
\Qed}

\leader{5A1G}{Three fairly simple facts}\dvAformerly{5{}A1F} (a) There 
is a family
$\langle a_I\rangle_{I\subseteq\Bbb N}$ of infinite subsets of $\Bbb N$
such that $a_I\cap a_J$ is finite whenever $I$, $J\subseteq\Bbb N$
are distinct.

(b) Let $X$ be a set,
$f:[X]^{<\omega}\to[X]^{\le\omega}$ a function, and $Y\subseteq X$.
Then there is a $Z\subseteq X$ such that $Y\subseteq Z$,
$f(I)\subseteq Z$ for every $I\in[Z]^{<\omega}$, and
$\#(Z)\le\max(\omega,\#(Y))$.

(c) Let $\kappa\ge\frak c$ be a cardinal and
$\Cal A$ a family of countable subsets of $\kappa$ such that $\#(\Cal A)$
is less than the cardinal power $\kappa^{\omega}$.   Then there is a
countably infinite $K\subseteq\kappa$ such that $I\cap K$ is finite for
every $I\in\Cal A$.

\proof{{\bf (a)} For each $n\in\Bbb N$, set
$K_n=\{i:2^n\le i<2^{n+1}\}$, and let $f_n:\Cal Pn\to K_n$ be a bijection;
set $a_I=\{f_n(I\cap n):n\in\Bbb N\}$.   (Or apply 5A1Nc below with
$\kappa=\omega$.)

\medskip

{\bf (b)} Define $\sequencen{Z_n}$ inductively
by setting $Z_0=Y$ and
$Z_{n+1}=Z_n\cup\bigcup\{f(I):I\in[Z_n]^{<\omega}\}$
for each $n$.   Then $\#(Z_n)\le\max(\omega,\#(Y))$ for each $n$, so
setting $Z=\bigcup_{n\in\Bbb N}Z_n$ we still have
$\#(Z)\le\max(\omega,\#(Y))$, while $f(I)\subseteq Z$ for every
$I\in[Z]^{<\omega}$.

\medskip

{\bf (c)} If $\#(\Cal A)<\kappa$ this is trivial, as we can take
$K\subseteq\kappa\setminus\bigcup\Cal A$.   Otherwise, let
$\lambda\le\kappa$ be the least cardinal such that
$\#(\Cal A)<\lambda^{\omega}$.   Then $\cf\lambda=\omega$.   \Prf\Quer\
Otherwise,

\Centerline{$\lambda^{\omega}
=\max(\lambda,\sup_{\theta<\lambda}\theta^{\omega})
\le\#(\Cal A)$.  \Bang\Qed}

\noindent Let $\sequencen{\lambda_n}$ be a strictly increasing sequence of
cardinals with supremum $\lambda$, starting from $\lambda_0=0$ and
$\lambda_1=\omega$ (of course $\lambda>\omega$ because
$\#(\Cal A)\ge\frak c$).   For $n\in\Bbb N$ let
$\phi_n:[n\times\lambda_n]^{<\omega}\to\lambda_{n+1}\setminus\lambda_n$ be
an injective function.   For $f:\Bbb N\to\lambda$ define
$C_f\subseteq\lambda$ by setting

\Centerline{$C_f=\{\phi_n(f\cap(n\times\lambda_n)):n\in\Bbb N\}$.}

\noindent If $f$, $g\in\lambda^{\Bbb N}$ are distinct, then there are an
$i\in\Bbb N$ such that $f(i)\ne g(i)$ and an $m>i$ such that both $f(i)$
and $g(i)$ are less than $\lambda_m$, so that
$f\cap(n\times\lambda_n)\ne g\cap(n\times\lambda_n)$ for every $n\ge m$ and
$C_f\cap C_g$ is finite.   It follows that for any $I\in\Cal A$ the set
$B_I=\{f:f\in\lambda^{\Bbb N}$, $C_f\cap I$ is infinite$\}$ has cardinal at
most $\frak c$.   Since
$\frak c\le\#(\Cal A)<\lambda^{\omega}$, there must be an
$f\in\lambda^{\Bbb N}$ such that $C_f\cap I$ is finite for every
$I\in\Cal A$, and we can set $K=C_f$.
}%end of proof of 5A1G

\leader{5A1H}{Partition calculus}\dvAformerly{5{}A1G}
{\bf (a) The Erd\H{o}s-Rado theorem}
Let $\kappa$ be an infinite cardinal.   Set $\kappa_1=\kappa$,
$\kappa_{n+1}=2^{\kappa_n}$ for $n\ge 1$.   If $n\ge 1$,
$\#(B)\le\kappa$,
$\#(A)>\kappa_n$ and
$f:[A]^n\to B$ is a function, there is a
$C\in[A]^{\kappa^+}$ such that $f$ is constant on $[C]^n$.
\prooflet{({\smc Erd\H{o}s Hajnal M\'at\'e \& Rado 84}, 16.5;
{\smc Kanamori 03}, 7.3;  {\smc Just \& Weese 97}, 15.13.)}

\spheader 5A1Hb Let $\kappa$ be a cardinal of uncountable cofinality, and
$Q\subseteq[\kappa]^2$.   Then {\it either} there is a stationary
$A\subseteq\kappa$ such that $[A]^2\subseteq Q$ {\it or} there
is an infinite closed $B\subseteq\kappa$ such that
$[B]^2\cap Q=\emptyset$.
\prooflet{\Prf\ (Cf.\ {\smc Erd\H{o}s Hajnal M\'at\'e \& Rado 84}, 11.3.)   Let $C\subseteq\kappa$ be a closed
cofinal set with $\otp(C)=\cf\kappa$ (5A1Ae).
Let $S_0$ be $\{\alpha:\alpha\in C$, $\cf\alpha=\omega\}$, so that
$S_0$ is stationary (5A1Ac).   For each $\alpha\in S_0$ let
$\sequencen{f_{\alpha}(n)}$
be a strictly increasing sequence in $\alpha$ with supremum $\alpha$.
Set

\Centerline{$\Cal I_{\alpha}
=\{I:I\subseteq\alpha\cap C$, $[I\cup\{\alpha\}]^2\cap Q=\emptyset$,
$\#(I\cap f_{\alpha}(n))\le n$ for every $n\in\Bbb N\}$.}

\noindent Let $I_{\alpha}$ be a maximal member of $\Cal I_{\alpha}$.
If there is any $\alpha$ such that $I_{\alpha}$ is infinite, we have the
second alternative, witnessed by $B=I_{\alpha}\cup\{\alpha\}$,
and we can stop.   Otherwise, there is an $n\in\Bbb N$
such that $S_1=\{\alpha:\alpha\in S_0$,
$I_{\alpha}\subseteq f_{\alpha}(n)\}$ is stationary.   As
$f_{\alpha}(n)<\alpha$ for every $\alpha\in S_1$, the Pressing-Down
Lemma (4A1Cc) tells us that there is a $\gamma<\kappa$ such that
$S_2=\{\alpha:\alpha\in S_1$, $f_{\alpha}(n)=\gamma\}$ is stationary.
Because

\Centerline{$\#([\gamma\cap C]^{<\omega})\le\max(\omega,\#(\gamma\cap C))
<\cf\kappa$,}

\noindent there is an $I\subseteq\gamma\cap C$
such that $A=\{\alpha:\alpha\in S_2$, $I_{\alpha}=I\}$ is stationary.

\Quer\ Suppose, if possible, that $[A]^2\not\subseteq Q$.  Take
$\alpha$, $\beta\in A$ such that $\alpha<\beta$ and
$\{\alpha,\beta\}\notin Q$.   We know that
$[I\cup\{\alpha\}]^2$ and $[I\cup\{\beta\}]^2$ are both disjoint from $Q$.
So $[J\cup\{\beta\}]^2$ is disjoint from $Q$, where $J=I\cup\{\alpha\}$.
If $m\le n$,

\Centerline{$f_{\beta}(m)\le f_{\beta}(n)=\gamma
=f_{\alpha}(n)<\alpha$,}

\noindent so
$\#(J\cap f_{\beta}(m))=\#(I_{\beta}\cap f_{\beta}(m))\le m$;
while if $m>n$ then

\Centerline{$\#(J\cap f_{\beta}(m))
\le\#(I)+1
=\#(I_{\alpha})+1
=\#(I_{\alpha}\cap f_{\alpha}(n))+1
\le n+1
\le m$.}

\noindent So $J\in\Cal I_{\beta}$;
but $J$ properly includes $I=I_{\beta}$, so this is
impossible.\ \Bang

Thus $[A]^2\subseteq Q$ and we have the first alternative.\ \Qed}
%5{}31S

\leader{5A1I}{$\Delta$-systems and free sets:  Proposition}\dvAformerly{5{}A1H}
Let $\kappa$ and $\lambda$ be infinite cardinals
and $\ofamily{\xi}{\kappa}{I_{\xi}}$ a family of
sets with cardinal less than $\lambda$.

(a) If $\cf\kappa>\lambda$, there are a
$\Gamma\in[\kappa]^{\kappa}$ and a set $J$ of
cardinal less than $\kappa$ such that $I_{\xi}\cap I_{\eta}\subseteq J$
for all distinct $\xi$, $\eta\in\Gamma$.
%5{}25J \family{\xi}{\Gamma}{I_{\xi}} is a Delta-nebula with
%  root-cover  J

(b) If $\kappa>\lambda$ is regular and
the cardinal power $\theta^{\delta}$ is less than $\kappa$ whenever
$\theta<\kappa$ and $\delta<\lambda$, then
there is a $\Gamma'\in[\kappa]^{\kappa}$ such
that $\family{\xi}{\Gamma'}{I_{\xi}}$ is a $\Delta$-system\cmmnt{ (definition:
4A2A)}.
%5{}25N

(c) If $\kappa>\lambda$ there is a $\Gamma''\in[\kappa]^{\kappa}$ such
that $\eta\notin I_{\xi}$ for any distinct $\xi$, $\eta\in\Gamma''$.
%5{}31K

\proof{{\bf (a)} \Quer\ Otherwise, choose
$\ofamily{\alpha}{\lambda}{\Gamma_{\alpha}}$ and
$\ofamily{\alpha}{\lambda}{J_{\alpha}}$ as follows.
$J_{\alpha}=\bigcup_{\beta<\alpha}\bigcup_{\xi\in\Gamma_{\beta}}I_{\xi}$.
Given $J_{\alpha}$, let $\Gamma_{\alpha}\subseteq\kappa$ be maximal subject
to the requirement that $I_{\xi}\cap I_{\eta}\subseteq J_{\alpha}$ for all
distinct $\xi$, $\eta\in\Gamma_{\alpha}$.   Then we see by induction that
$\#(J_{\alpha})<\kappa$ so $\#(\Gamma_{\alpha})<\kappa$ for every
$\alpha<\lambda$;  because $\cf\kappa>\lambda$,
$\bigcup_{\alpha<\lambda}\Gamma_{\alpha}$ cannot be the whole of $\kappa$.

Take any $\xi\in\kappa\setminus\bigcup_{\alpha<\lambda}\Gamma_{\alpha}$.
As $\#(I_{\xi})<\lambda$, there must be an $\alpha<\lambda$ such that
$I_{\xi}\cap J_{\alpha}=I_{\xi}\cap J_{\alpha+1}$.
As $\xi\notin\Gamma_{\alpha}$,
there is an $\eta\in\Gamma_{\alpha}$ such that
$I_{\xi}\cap I_{\eta}\not\subseteq J_{\alpha}$;  but now
$I_{\xi}\cap I_{\eta}\setminus J_{\alpha}
\subseteq I_{\xi}\cap J_{\alpha+1}\setminus J_{\alpha}$.\ \Bang

\medskip

{\bf (b)} Let $J$ and $\Gamma$ be as in (a).   Because $\cf\kappa>\lambda$,
there must be some cardinal $\delta<\lambda$ such that
$\Gamma_1=\{\xi:\xi\in\Gamma$, $\#(I_{\xi}\cap J)\le\delta\}$ has cardinal
$\kappa$.   Now
$\#([J]^{\le\delta})\le\max(2,\#(J)^{\delta})<\cf\kappa$, so there must be a
$K\subseteq J$ such that
$\Gamma'=\{\xi:\xi\in\Gamma_1$, $I_{\xi}\cap J=K\}$
has cardinal $\kappa$;  and $\family{\xi}{\Gamma'}{I_{\xi}}$ is a
$\Delta$-system with root $K$.

\medskip

{\bf (c)} It is enough to consider the case in which $\xi\in I_{\xi}$ for
every $\xi<\kappa$.

\medskip

\quad{\bf (i)} If $\cf\kappa>\lambda$,
take $\Gamma$ and $J$ from (a).   Then we can choose
$\ofamily{\delta}{\kappa}{\xi_{\delta}}$ inductively so that

\Centerline{$\xi_{\delta}
\in\Gamma\setminus(J\cup\bigcup_{\beta<\delta}I_{\xi_{\beta}})$}

\noindent for every $\delta<\kappa$;  and $\{\xi_{\delta}:\delta<\kappa\}$
will serve for $\Gamma''$.

\medskip

\quad{\bf (ii)} If $\cf\kappa=\theta\le\lambda$, let
$\ofamily{\alpha}{\theta}{\kappa_{\alpha}}$ be a strictly increasing family
of regular cardinals with supremum $\kappa$, starting from
$\kappa_0\ge\lambda^{++}$.   For each $\alpha<\theta$, (i) tells us that there is
an $A_{\alpha}\in[\kappa_{\alpha}]^{\kappa_{\alpha}}$ such that
$\eta\notin I_{\xi}$ for any distinct $\xi$, $\eta\in A_{\alpha}$.   Set

\Centerline{$B_{\alpha}
=A_{\alpha}\setminus\bigcup_{\beta<\alpha}
  (B_{\beta}\cup\bigcup_{\xi\in A_{\beta}}I_{\xi})$;}

\noindent then $\#(B_{\alpha})=\kappa_{\alpha}$
for each $\alpha<\theta$.   Choose
$\langle C_{\alpha\gamma}\rangle_{\alpha<\theta,\gamma<\lambda^+}$ and
$\ofamily{\alpha}{\theta}{\zeta_{\alpha}}$
inductively, as follows.   Given that
$\langle C_{\beta\gamma}\rangle_{\beta<\alpha,\gamma<\lambda^+}$ is
disjoint, then for each $\xi\in B_{\alpha}$ there is a $\zeta<\lambda^+$ such
that $I_{\xi}\cap\bigcup_{\beta<\alpha}C_{\beta\gamma}$ is empty for every
$\gamma\ge\zeta$;  because $\lambda^+<\cf\kappa_{\alpha}$, there is a
$\zeta_{\alpha}<\lambda^+$ such that

\Centerline{$B'_{\alpha}=\{\xi:\xi\in B_{\alpha}$,
$I_{\xi}\cap C_{\beta\gamma}=\emptyset$ whenever $\beta<\alpha$ and
$\zeta_{\alpha}\le\gamma<\lambda^+\}$}

\noindent has cardinal $\kappa_{\alpha}$.   Let
$\ofamily{\gamma}{\lambda^+}{C_{\alpha\gamma}}$ be a partition of
$B'_{\alpha}$ into sets with cardinal $\kappa_{\alpha}$, and continue.

At the end of the induction, $\gamma=\sup_{\alpha<\theta}\zeta_{\alpha}$ is
less than $\lambda^+$.   Set
$\Gamma''=\bigcup_{\alpha<\theta}C_{\alpha\gamma}$.   Then
$\#(\Gamma'')=\kappa$.   If $\xi$, $\eta$ are
distinct members of $\Gamma''$, let $\alpha$, $\beta<\theta$ be such that
$\xi\in C_{\alpha\gamma}$ and $\eta\in C_{\beta\gamma}$.   If
$\alpha<\beta$ then $\xi\in A_{\alpha}$ and $\eta\in B_{\beta}$ so
$\eta\notin I_{\xi}$.   If $\alpha=\beta$ then both $\xi$ and $\eta$ belong
to $A_{\alpha}$ so $\eta\notin I_{\xi}$.   If $\beta<\alpha$ then
$\eta\in C_{\beta\gamma}$ while $\gamma\ge\zeta_{\alpha}$ and
$\xi\in B'_{\alpha}$, so $\eta\notin I_{\xi}$.   So $\Gamma''$ will serve.
}%end of proof of 5A1I

\cmmnt{\medskip

\noindent{\bf Remark} (c) above is Hajnal's Free Set Theorem.}

\leader{5A1J}{Remarks}\dvAformerly{5{}A1I} {\bf (a)}\cmmnt{ I spell out the 
applications of these results which are used in this volume.}
Let $\kappa$ be an infinite cardinal and
$\ofamily{\xi}{\kappa}{I_{\xi}}$ a family of countable sets.

\quad(i) If $\cf\kappa\ge\omega_2$, there are a
$\Gamma\in[\kappa]^{\kappa}$ and a set $J$ with cardinal less than $\kappa$
such that $I_{\xi}\cap I_{\eta}\subseteq J$ for all distinct $\xi$,
$\eta\in\Gamma$.

\quad(ii) If $\kappa$ is regular and
the cardinal power $\lambda^{\omega}$ is less than $\kappa$ for every
$\lambda<\kappa$, there is a $\Gamma'\in[\kappa]^{\kappa}$ such
that $\family{\xi}{\Gamma'}{I_{\xi}}$ is a $\Delta$-system.
\prooflet{(Of course $\kappa$ cannot be $\omega_1$, so we can apply 5A1Ib
with $\lambda=\omega_1$.)}
%5{}25N

\quad(iii) If $\kappa\ge\omega_2$ there is a $\Gamma''\in[\kappa]^{\kappa}$ such
that $\eta\notin I_{\xi}$ for any distinct $\xi$, $\eta\in\Gamma''$.
%5{}31K

\spheader 5A1Jb\cmmnt{ If, in 5A1Ic,
we are willing to settle for a weaker result,
there is an easier proof which generalizes to more complex systems.}
Let $\lambda$ be an infinite cardinal.   Then there is a $\kappa_0$ such
that for every cardinal $\kappa\ge\kappa_0$, every $n\in\Bbb N$
and every function $f:[\kappa]^n\to[\kappa]^{<\lambda}$ there is an
$A\in[\kappa]^{\lambda^+}$ such that $\xi\notin f(I)$ whenever
$I\in[A]^n$ and $\xi\in A\setminus I$.
\prooflet{\Prf\ By the Erd\H{o}s-Rado theorem (5A1Ha),
there is a $\kappa_0$ such that for every $\kappa\ge\kappa_0$, $n\ge 1$ and
function $g:[\kappa]^n\to\Bbb N$ there is an $A\in[\kappa]^{\lambda^+}$
such that $g$ is constant on $[A]^n$.   Now, given $n\in\Bbb N$,
$\kappa\ge\kappa_0$
and $f:[\kappa]^n\to[\kappa]^{<\lambda}$, define $g:[\kappa]^{n+1}\to\Bbb N$
by saying that if $J=\{\xi_0,\ldots,\xi_n\}$ with
$\xi_0<\xi_1<\ldots<\xi_n$, then
$g(J)=\min(\{n+1\}\cup\{j:j\le n$, $\xi_j\in f(J\setminus\{\xi_j\})\}$.
Let $A\in[\kappa]^{\lambda^+}$ be such that $g$ is constant on
$[A]^{n+1}$.   We can suppose that $A$ has order type $\lambda^+$.
\Quer\ Suppose that the constant value of $g$ in $[A]^{n+1}$ is $j\le n$.
Let $B$ be the set of the first $\lambda$ members of $A$, $I_0$ the set of
the first $j$ members of $A$ and $I_1$ the set of the first $n-j$ members
of $A\setminus B$.   Then we have $g(I_0\cup\{\xi\}\cup I_1)=j$ for every
$\xi\in B\setminus I_0$, so that
$B\setminus I_0\subseteq f(I_0\cup I_1)$;  but
$\#(f(I_0\cup I_1))<\lambda$.\ \BanG\
So the constant value of $g$ on $[A]^{n+1}$ is $n+1$, and $A$ satisfies the
required condition.\ \Qed}

\spheader 5A1Jc\cmmnt{ In the same complex of ideas,
we have an elementary
fact about the case $\lambda<\kappa=\omega$.}   If $n\in\Bbb N$ and
$\sequence{i}{K_i}$ is a sequence in $[\Bbb N]^{\le n}$,
there is an infinite
$\Gamma\subseteq\Bbb N$ such that $\family{i}{\Gamma}{K_i}$ is a
$\Delta$-system.   \prooflet{\Prf\
Let $K\subseteq\Bbb N$ be a maximal set such that
$I=\{i:K\subseteq K_i\}$ is infinite;  then
$\{i:i\in I$, $K_i\cap L\ne\emptyset\}$ is finite for every finite
$L\subseteq\Bbb N\setminus K$, so we can choose $\Gamma$ inductively
by saying that
$\Gamma=\{i:i\in I$, $K_i\cap K_j=K$ whenever $j\in\Gamma\cap i\}$.\ \Qed}

\spheader 5A1Jd\dvAnew{2016}\dvro{ If}{ In 5A1Ic, and in (a-iii) here,
we have a system
$\ofamily{\xi}{\kappa}{I_{\xi}}$ of sets and are looking for large sets
$\Gamma''$ which are `free' in the sense that $\eta\notin I_{\xi}$ for distinct
$\xi$, $\eta\in\Gamma''$.   If we identify
$\ofamily{\xi}{\kappa}{I_{\xi}}$ with the set $R=\{(\xi,\eta):\xi<\kappa$,
$\eta\in I_{\xi}\}$ then we are asking that $(\xi,\eta)$ should not belong to
$R$ for distinct $\xi$, $\eta\in\Gamma''$.   It will be useful to apply the same
idea to other kinds of relation.
In particular, if} $R\subseteq X\times X$ is
an equivalence relation on a set $X$ I will say that a set $A\subseteq X$ is
{\bf $R$-free} if\cmmnt{ $(x,y)\notin R$ for all distinct $x$, $y\in A$,
that is, if} $A$ meets each equivalence class for $R$ in at most one point.

\spheader 5A1Je\dvAnew{2021}\cmmnt{ Concerning free sets for equivalence
relations it will help to have some elementary facts in quotable form.}   Let
$X$ be a set and $R$ an equivalence relation on $X$.

\medskip

\quad{\bf (i)} For any cardinal $\kappa$, there is a partition
$\ofamily{\xi}{\kappa}{X_{\xi}}$ of $X$ into $R$-free sets iff 
every $R$-equivalence class has cardinal at most $\kappa$.   
\prooflet{\Prf\ Write $\Cal K$ for the set of equivalence classes under $R$.
If $\ofamily{\xi}{\kappa}{X_{\xi}}$ is a partition
of $X$ into $R$-free sets and $K\in\Cal K$, then
$\#(K\cap X_{\xi})\le 1$ for every $\xi<\kappa$ and $\#(K)\le\kappa$.   If
$\#(K)\le\kappa$ for every $K\in\Cal K$, then for each $K\in\Cal K$ choose
an injective function $h_K:K\to\kappa$;  set $h(x)=h_K(x)$ whenever 
$x\in K\in\Cal K$;  then $\ofamily{\xi}{\kappa}{h^{-1}[\{\xi\}]}$ is a partition
of $X$ into $R$-free sets.\ \Qed}

\medskip

\quad{\bf (ii)} If $A\subseteq X$ is $R$-free then $R[B]\cap R[C]=\emptyset$
whenever $B$, $C\subseteq A$ are disjoint.   
\prooflet{\Prf\Quer\ If $x\in R[B]\cap R[C]$
there are $b\in B$, $c\in C$ such that $(b,x)\in R$ and $(c,x)\in R$, so
$(b,c)\in R$;  but $b$ and $c$ are distinct members of $A$, which is supposed to
be $R$-free.\ \Bang\Qed}

\leader{5A1K}{Lemma}\dvAformerly{5{}A1J} Suppose that $\theta$, $\lambda$ and $\kappa$
are cardinals, with $\theta<\lambda<\cf\kappa$, and that $S$
is a stationary subset of
$\kappa$.   Let $\langle I_{\xi}\rangle_{\xi\in S}$ be a family in
$[\lambda]^{\le\theta}$.
Then there is a set $M\subseteq\lambda$ such that
$\cf(\#(M))\le\theta$
and $\{\xi:\xi\in S,\,I_{\xi}\subseteq M\}$ is stationary in $\kappa$.
%5{}43 5A1N

\proof{ For $M\subseteq\lambda$, set
$S_M=\{\xi:\xi\in S,\,I_{\xi}\subseteq M\}$.
Let $M\subseteq\lambda$ be a set of minimal cardinality such
that $S_M$ is stationary in $\kappa$.   Set $\delta=\#(M)$.  \Quer\  If
$\cf\delta>\theta$, enumerate $M$ as
$\langle\alpha_{\eta}\rangle_{\eta<\delta}$. For each $\xi\in S_M$,
set $\beta_{\xi}=\sup\{\eta:\alpha_{\eta}\in I_{\xi}\}$;
because $\#(I_{\xi})\le\theta<\cf\delta$,
$\beta_{\xi}<\delta$.   Because $\delta\le\lambda<\cf\kappa$,
there is a $\beta<\delta$ such that
$S'=\{\xi:\xi\in S_M$, $\beta_{\xi}=\beta\}$ is stationary in $\kappa$
(5A1Ab).   Consider $M'=\{\alpha_{\eta}:\eta\le\beta\}$;  then
$\#(M')<\#(M)$  but $S_{M'}\supseteq S'$ so is stationary in
$\kappa$, contrary to the choice of $M$.\ \Bang

Thus $M$ will serve.
}%end of proof of 5A1K

\leader{5A1L}{Lemma}\dvAformerly{5{}A1K} Let $\familyiI{X_i}$ be a non-empty family of
infinite sets, with product $X$.   Then there is a
set $Y\subseteq X$, with $\#(Y)=\#(X)$, such that for every finite
$L\subseteq Y$ there is an $i\in I$ such that
$x(i)\ne y(i)$ for any distinct $x$, $y\in L$.
%5{}15G  sort of transversals

\proof{ Set $\kappa=\#(X)$.

\medskip

{\bf (a)} We can well-order $I$ in such a way that
$\#(X_i)\le\#(X_j)$ whenever $i\le j$ in $I$.   It will therefore be
enough to deal with the case in which $I=\delta$ is an ordinal and
$\#(X_{\alpha})\le\#(X_{\beta})$ whenever $\alpha\le\beta<\delta$.
I proceed by induction on $\delta$.

\medskip

{\bf (b)} If $\delta$ is finite then
$\kappa=\max_{\alpha<\delta}\#(X_{\alpha})$ and the result is trivial,
since we can take the $x_{\xi}$ to be all different at a single
coordinate.

\medskip

{\bf (c)} Suppose there is a $\gamma<\delta$ such that
$\#(\delta\setminus\gamma)<\#(\delta)$.  Then, in particular, the order
type of $\delta\setminus\gamma$ is less than the order type of
$\delta$.   Set $I_0=\gamma$, $I_1=\delta\setminus\gamma$ and
$Y_j=\prod_{\alpha\in I_j}X_{\alpha}$ for both $j$.   Then
$X\cong Y_0\times Y_1$, so $\kappa=\max(\#(Y_0),\#(Y_1))$;  say
$\kappa=\#(Y_j)$.   By the inductive hypothesis, there is a family
$\ofamily{\xi}{\kappa}{y_{\xi}}$ in $Y_j$ such that for any
$L\in[\kappa]^{<\omega}$ there is an $\alpha\in I_j$ such that
$\xi\mapsto y_{\xi}(\alpha):L\to X_{\alpha}$ is injective.   Taking
$x_{\xi}$ to be any member of $X$ extending $y_{\xi}$, for each
$\xi<\kappa$, we have a suitable family
$\ofamily{\xi}{\kappa}{x_{\xi}}$
in $X$, and the induction proceeds.

\medskip

{\bf (d)} Suppose that $\delta$ is infinite and that
$\#(\delta\setminus\gamma)=\#(\delta)=\lambda$ for every
$\gamma<\delta$.   Enumerate $[\delta]^{<\omega}$ as
$\ofamily{\zeta}{\lambda}{J_{\zeta}}$, and choose
$\ofamily{\zeta}{\lambda}{\alpha_{\zeta}}$ such that

\Centerline{$J_{\zeta}
\subseteq\alpha_{\zeta}\in\delta\setminus\{\alpha_{\eta}:\eta<\zeta\}$}

\noindent for each $\alpha<\lambda$.   We have

\Centerline{$\#(X_{\alpha_{\zeta}})
\ge\max(\omega,\sup_{\beta\in J_{\zeta}}\#(X_{\beta}))
\ge\#(\prod_{\beta\in J_{\zeta}}X_{\beta})$,}

\noindent so there is an injective function
$f_{\zeta}:\prod_{\beta\in J_{\zeta}}X_{\beta}\to X_{\alpha_{\zeta}}$
for each $\zeta<\lambda$.
Let $\ofamily{\xi}{\kappa}{z_{\xi}}$ be any enumeration of $X$.
Because all the $\alpha_{\zeta}$ are distinct, we can find
$x_{\xi}\in X$, for each $\xi<\kappa$, such that
$x_{\xi}(\alpha_{\zeta})=f_{\zeta}(z_{\xi}\restr J_{\zeta})$ for every
$\zeta$.   Now if $L\in[\kappa]^{<\omega}$ there must be a
$\zeta<\lambda$ such that
$z_{\xi}\restr J_{\zeta}\ne z_{\eta}\restr J_{\zeta}$
for any distinct $\xi$, $\eta\in L$;  so that
$\xi\mapsto x_{\xi}(\alpha_{\zeta})$ is injective on $L$.   Thus
$\ofamily{\xi}{\kappa}{x_{\xi}}$ is a suitable family in $X$ and the
induction proceeds in this case also.
}%end of proof of 5A1L

\leader{5A1M}{Definitions}\dvAformerly{5{}A1L}{\bf (a)} Let $X$ and $Y$ be sets and
$\Cal I$ an ideal of subsets of $X$.   Write
$\Tr_{\Cal I}(X;Y)$ for the {\bf transversal number}

\Centerline{$\sup\{\#(F):F\subseteq Y^X$,
$\{x:f(x)=g(x)\}\in\Cal I$ for all distinct $f,\,g\in F\}$.}

\spheader 5A1Mb Let $\kappa$ be a cardinal.  Write
$\Tr(\kappa)$ for

\Centerline{$\Tr_{[\kappa]^{<\kappa}}(\kappa;\kappa)
=\sup\{\#(F):F\subseteq\kappa^{\kappa}$,
$\#(f\cap g)<\kappa\text{ for all distinct }f,\,g\in F\}$.}

\leader{5A1N}{Lemma}\dvAformerly{5{}A1M} (a) For any infinite cardinal $\kappa$,

\Centerline{$\kappa^+\le\Tr(\kappa)\le 2^{\kappa}$.}

(b) For any infinite cardinal $\kappa$,

\Centerline{$\max(\Tr(\kappa),\sup_{\delta<\kappa}2^{\delta})
\ge\min(2^{\kappa},\kappa^{(+\omega)})$.}

(c) If $\kappa$ is such that $2^{\delta}\le\kappa$ for
every $\delta<\kappa$, then $\Tr(\kappa)=2^{\kappa}$, and in fact there is
an $F\subseteq\kappa^{\kappa}$ such that $\#(F)=2^{\kappa}$ and
$\#(f\cap g)<\kappa\text{ for all distinct }f,\,g\in F$.

(d) If $X$ and $Y$ are sets and $\Cal I$ is a maximal proper
ideal of $\Cal PX$, then there is
an $F\subseteq Y^X$ such that $\#(F)=\Tr_{\Cal I}(X;Y)$ and
$\{x:f(x)=g(x)\}\in\Cal I$ for all distinct $f,\,g\in F$.

\proof{{\bf (a)} We can build inductively a family
$\langle f_{\alpha}\rangle_{\alpha<\kappa^+}$ in $\kappa^{\kappa}$, as
follows. Given $\langle f_{\alpha}\rangle_{\alpha<\beta}$, where
$\beta<\kappa^+$, let $\theta:\beta\to\kappa$ be any injection.
Now choose $f_{\beta}:\kappa\to\kappa$ so that

\Centerline{$f_{\beta}(\xi)\ne f_{\alpha}(\xi)$
whenever $\alpha<\beta$ and $\theta(\alpha)\le\xi$.}

\noindent This will mean that if $\alpha<\beta$, then

\Centerline{$\{\xi:f_{\alpha}(\xi)=f_{\beta}(\xi)\}
\subseteq\theta(\alpha)$}

\noindent has cardinal less than $\kappa$.   So at the
end of the induction, $F=\{f_{\alpha}:\alpha<\kappa^+\}$
will witness that $\Tr(\kappa)\ge\kappa^+$.   On the other hand,
$\Tr(\kappa)\le\#(\kappa^{\kappa})=2^{\kappa}$.

\medskip

{\bf (b)}\Quer\ If not, then take
$\lambda=\max(\Tr(\kappa),\sup_{\delta<\kappa}2^{\delta})
<\min(2^{\kappa},\kappa^{(+\omega)})$.
For each $\xi<\kappa$ take an
injective function $\phi_{\xi}:\Cal P\xi\to\lambda$.
Because $\lambda<2^{\kappa}$, we have an injective function
$h:\lambda^+\to\Cal P\kappa$.   For $\alpha<\lambda^+$ set
$g_{\alpha}(\xi)=\phi_{\xi}(h(\alpha)\cap\xi)$
for every $\xi<\kappa$;  then
$\langle g_{\alpha}\rangle_{\alpha<\lambda^+}$
is a family in $\lambda^{\kappa}$ such that
$\#(g_{\alpha}\cap g_{\beta})<\kappa$ whenever $\alpha\ne\beta$.

Apply 5A1K with $S=\lambda^+$,
$I_{\alpha}=g_{\alpha}[\kappa]$ to see that there is a set
$M\subseteq\lambda$ with $\cf(\#(M))\le\kappa$ and
$S_1=\{\alpha:\alpha<\lambda^+$, $g_{\alpha}[\kappa]\subseteq M\}$
stationary in $\lambda^+$.   Because
$\lambda<\kappa^{(+\omega)}$, we must have $\#(M)\le\kappa$.   If
$f:M\to\kappa$ is any injection,
$\langle fg_{\alpha}\rangle_{\alpha\in S_1}$ will witness
that $\Tr(\kappa)\ge\#(S_1)=\lambda^+$;  which is impossible.\ \Bang

\wheader{5A1N}{4}{2}{2}{36pt}

{\bf (c)} For each $\xi<\kappa$, let $\phi_{\xi}:\Cal P\xi\to\kappa$
be injective.   For $A\subseteq\kappa$, define
$f_A\in\kappa^{\kappa}$ by writing

\Centerline{$f_A(\xi)=\phi_{\xi}(A\cap\xi)$ for every $\xi<\kappa$.}

\noindent Then $F=\{f_A:A\subseteq\kappa\}$ has the required property, and
$\Tr(\kappa)\ge 2^{\kappa}$;  by (a), we have equality.

\medskip

{\bf (d)} Take any maximal set $F\subseteq Y^X$ such that
$\{x:f(x)=f'(x)\}\in\Cal I$ for all distinct $f$, $f'\in F$.   Then
$\#(F)=\Tr_{\Cal I}(X;Y)$.   \Prf\ Of course $\#(F)\le\Tr_{\Cal I}(X;Y)$.
Suppose that $G\subseteq Y^X$ is such that
$\{x:g(x)=g'(x)\}\in\Cal I$ for all distinct $g$, $g'\in G$.   For each
$g\in G$, there must be an $f_g\in F$ such that
$\{x:g(x)=f_g(x)\}\notin\Cal I$;  because $\Cal I$ is maximal,
$\{x:g(x)\ne f_g(x)\}\in\Cal I$.   If $g$, $h\in G$ are distinct,
then

$$\eqalign{\{x:f_g(x)=f_h(x)\}&
\subseteq\{x:f_g(x)\ne g(x)\}\cup\{x:g(x)=h(x)\}\cup\{x:h(x)\ne f_h(x)\}\cr&
\in\Cal I\cr}$$

\noindent so $f_g\ne f_h$.   Thus we have an injective function from $G$ to
$F$ and $\#(G)\le\#(F)$.   As $G$ is arbitrary,
$\Tr_{\Cal I}(X;Y)\le\#(F)$ and we have equality.\ \Qed
}%end of proof of 5A1N
%5{}42 %5{}38Xk

\leader{5A1O}{Almost-square-sequences:  Lemma}\dvAformerly{5{}A1N}
Let $\lambda$, $\kappa$ be regular infinite cardinals, with
$\kappa>\max(\omega_1,\lambda)$.
Then we can find a stationary set $S\subseteq\kappa^+$ and a family
$\langle C_{\alpha}\rangle_{\alpha\in S}$ of sets such that

(i) for each $\alpha\in S$, $C_{\alpha}$ is a closed cofinal set in
$\alpha$ of order type $\lambda$;

(ii) if $\alpha$, $\beta\in S$ and $\gamma$ is a limit
point of both $C_{\alpha}$ and $C_{\beta}$ then
$C_{\alpha}\cap\gamma=C_{\beta}\cap\gamma$.

\cmmnt{\medskip

\noindent{\bf Remark} Compare the axiom $\square_{\kappa}$ of 5A6D below.}

\proof{{\bf (a)} For each $\gamma<\kappa^+$ fix an injection
$f_{\gamma}:\gamma\to\kappa$.   Let $S_0$ be the set of ordinals
$\alpha<\kappa^+$ of cofinality $\lambda$;  then $S_0$ is stationary
in $\kappa^+$ (5A1Ac).   For each $\alpha\in S_0$ choose a non-decreasing
family $\langle N_{\alpha\delta}\rangle_{\delta<\kappa}$ of subsets of
$\kappa^+$ such that

\inset{($\alpha$) $N_{\alpha0}$ is a cofinal subset of
$\alpha$ with cardinal $\lambda$;

($\beta$) if $\delta<\kappa$ then

\Centerline{$N_{\alpha,\delta+1}
=\bigcup\{f_{\gamma}[N_{\alpha\delta}]\cup f_{\gamma}^{-1}[\delta]:
    \gamma\in N_{\alpha\delta}\}
  \cup\overline{N}_{\alpha\delta}\cup\delta$}

\noindent (taking the closure $\overline{N}_{\alpha\delta}$ in the
order topology of $\kappa^+$);

($\gamma$) if $\delta<\kappa$ is a non-zero limit
ordinal then
$N_{\alpha\delta}=\bigcup_{\delta'<\delta}N_{\alpha\delta'}$.}

\noindent Then
$\#(N_{\alpha\delta})\le\max(\lambda,\#(\delta))<\kappa$
for each $\delta<\kappa$ (using 5A1Ad).   Because $\kappa$ is regular,
$\sup(N_{\alpha\delta}\cap\kappa)<\kappa$ for every $\delta$.
It follows that
$\{\delta:\delta<\kappa$, $N_{\alpha\delta}\cap\kappa=\delta\}$ is a
closed cofinal set in $\kappa$, and in particular
contains an ordinal of cofinality $\omega_1$,
for every $\alpha\in S_0$.   Let $\delta<\kappa$ be
such that $\cf\delta=\omega_1$ and

\Centerline{$S_1=\{\alpha:\alpha\in S_0$,
$N_{\alpha\delta}\cap\kappa=\delta\}$}

\noindent is stationary in $\kappa^+$.   For $\alpha\in S_1$, set
$C^*_{\alpha}=\alpha\cap\overline{N}_{\alpha\delta}$;
then $C^*_{\alpha}$ is a closed cofinal set in $\alpha$ and
$\#(C^*_{\alpha})<\kappa$ so $\otp(C^*_{\alpha})<\kappa$.   Let
$\zeta<\kappa$ be such that

\Centerline{$S=\{\alpha:\alpha\in S_1$, $\otp(C^*_{\alpha})=\zeta\}$}

\noindent is stationary in $\kappa^+$.   Observe that
as $\cf C^*_{\alpha}=\cf\alpha=\lambda$ for each $\alpha\in S$,
$\cf\zeta=\lambda$.

\medskip

{\bf (b)} Take any closed cofinal set $C\subseteq\zeta$ of order
type $\lambda$ and for each $\alpha\in S$ let $C_{\alpha}$ be the image
of $C$ in $C^*_{\alpha}$ under the order-isomorphism between
$\zeta$ and $C^*_{\alpha}$.   Then $C_{\alpha}$ will be a closed
cofinal subset of $\alpha$ of order type $\lambda$.

I claim that if $\alpha$, $\beta\in S$ and $\gamma$
is a common limit point of $C_{\alpha}$, $C_{\beta}$ then
$C_{\alpha}\cap\gamma=C_{\beta}\cap\gamma$.

\Prf\ {\bf case 1} Suppose $\lambda=\omega$.   In this case the only
limit point of $C_{\alpha}$
will be $\alpha$ itself, and similarly for $\beta$, so that in this
case
we have $\alpha=\beta$ and there is nothing more to do.

{\bf case 2} Suppose $\cf\gamma=\omega<\lambda$.
Then $\gamma$ is a limit point of
$C_{\alpha}\subseteq C^*_{\alpha}\subseteq\overline{N}_{\alpha\delta}$, so
there is an increasing sequence in $N_{\alpha\delta}$ with supremum
$\gamma$;  as
$N_{\alpha\delta}=\bigcup_{\delta'<\delta}N_{\alpha\delta'}$ and
$\cf\delta=\omega_1$, this sequence lies entirely within
$N_{\alpha\delta'}$ for some $\delta'<\delta$, and
$\gamma\in\overline{N}_{\alpha\delta'}\subseteq N_{\alpha,\delta'+1}$.
Now, for $\delta'+1\le\xi<\delta$, $N_{\alpha,\xi+1}
\supseteq f_{\gamma}^{-1}[\xi]\cup f_{\gamma}[N_{\alpha\xi}]$;
consequently

\Centerline{$N_{\alpha\delta}\cap\gamma
=f_{\gamma}^{-1}[N_{\alpha\delta}\cap\kappa]
=f_{\gamma}^{-1}[\delta]$.}

\noindent Similarly,
$N_{\beta\delta}\cap\gamma=f_{\gamma}^{-1}[\delta]$.   Now

\Centerline{$C^*_{\alpha}\cap\gamma
=\overline{N}_{\alpha\delta}\cap\gamma
=\overline{f_{\gamma}^{-1}[\delta]}\cap\gamma
=C^*_{\beta}\cap\gamma$.}

\noindent Accordingly the increasing enumerations of $C^*_{\alpha}$ and
$C^*_{\beta}$ must agree on
$C^*_{\alpha}\cap\gamma=C^*_{\beta}\cap\gamma$, and
$C_{\alpha}\cap\gamma=C_{\beta}\cap\gamma$.

{\bf case 3} Suppose that $\cf\gamma>\omega$ and $\lambda>\omega$.
Because $\gamma=\sup(C_{\alpha}\cap\gamma)=\sup(C_{\beta}\cap\gamma)$,

\Centerline{$D
=\{\gamma':\gamma'<\gamma$ is a limit point of both $C_{\alpha}$ and
$C_{\beta}$, $\cf\gamma'=\omega\}$}

\noindent is cofinal with $\gamma$, and

\Centerline{$C_{\alpha}\cap\gamma
=\bigcup_{\gamma'\in D}C_{\alpha}\cap\gamma'=C_{\beta}\cap\gamma$,}

\noindent using case 2.\ \Qed

Thus $S$ and $\langle C_{\alpha}\rangle_{\alpha\in S}$ have
the required properties.
}%end of proof of 5A1O
%5{}A1O

\leader{5A1P}{Corollary}\dvAformerly{5{}A1O} Let $\kappa$, $\lambda$ be regular infinite
cardinals with $\lambda>\max(\omega_1,\kappa)$.
Then we can find a stationary subset $S$ of $\lambda^+$ and a family
$\langle g_{\alpha}\rangle_{\alpha\in S}$ of functions
from $\kappa$ to $\lambda^+$ such that, for all distinct $\alpha$,
$\beta\in S$,

(i) $g_{\alpha}[\kappa]\subseteq\alpha$,

(ii) $\#(g_{\alpha}\cap g_{\beta})<\kappa$,

(iii) if $\theta<\kappa$ is a limit ordinal and
$g_{\alpha}(\theta)=g_{\beta}(\theta)$ then
$g_{\alpha}\restr\theta=g_{\beta}\restr\theta$.

\proof{ Take $\langle C_{\alpha}\rangle_{\alpha\in S}$ from
5A1O and let $g_{\alpha}$ be the increasing enumeration of $C_{\alpha}$.
}%end of proof of 5A1P
%5{}43E

\leader{5A1Q}{}\dvAnew{2015}\cmmnt{ A fragment of finite combinatorics turns out to be
a basis for some interesting measure theory (546I).

\medskip

\noindent}{\bf Lemma} Let $I$ and $J$ be non-empty finite sets, and
$R\subseteq I\times J$ a relation such that $R[I]=J$.   Set

\Centerline{$k=\max_{x\in I}\#(R[\{x\}])$,
\quad$l=\min_{y\in J}\#(R^{-1}[\{y\}])$.}

\noindent Then there is a $K\subseteq I$ such that $R[K]=J$ and
$\#(K)\le\Bover{1+\ln k}{l}\#(I)$.

\proof{\cmmnt{ ({\smc Bartoszy\'nski \& Judah 95}, 3.3.10)}
{\bf (a)} Choose $\langle I_j\rangle_{j<k}$,
$\langle J_j\rangle_{j\le k}$ inductively, as follows.   For each
$j\le k$, set $J_j=J\setminus R[\bigcup_{i<j}I_i]$, and if $j<k$
take a maximal set $I_j\subseteq I$ such that

\Centerline{$\#(R[\{x\}]\cap J_j)=k-j$}

\noindent for every $x\in I_j$,

\Centerline{$R[\{x\}]\cap R[\{x'\}]\cap J_j=\emptyset$}

\noindent whenever $x$, $x'\in I_j$ are distinct.
At the end of the induction set $K=\bigcup_{i<k}I_i$.

Now $\#(R[\{x\}]\cap J_j)\le k-j$ for every $j\le k$ and
$x\in I$.   \Prf\ Induce on $j$.   Start with $J_0=J$ and
$\#(R[\{x\}])\le k$ for every $x$, by the definition of $k$.
For the inductive step to $j+1\le k$, if $x\in I$ then

\Centerline{$\#(R[\{x\}]\cap J_{j+1})\le\#(R[\{x\}]\cap J_j)
\le k-j$}

\noindent by the inductive hypothesis, and if
$\#(R[\{x\}]\cap J_j)=k-j$ then there is an
$x'\in I_j$ such that $R[\{x\}]\cap R[\{x'\}]\cap J_j$ is
non-empty, so that $R[\{x\}]\cap J_{j+1}$ is strictly smaller than
$R[\{x\}]\cap J_j$ and cannot have more than $k-j-1$ members.\ \Qed

In particular, $R[\{x\}]\cap J_k=\emptyset$ for every $x\in I$, so

\Centerline{$R[I]=R[I]\setminus J_k=J\setminus J_k=R[K]$.}

\medskip

{\bf (b)} For each $j<k$, $\family{x}{I_j}{R[\{x\}]\cap J_j}$
is a partition of $J_j\setminus J_{j+1}$ into sets with cardinal $k-j$, so that
$\#(J_j)-\#(J_{j+1})=(k-j)\#(I_j)$.   Next, if we set
$R_j=R\cap(I\times J_j)$ we see that

\Centerline{$\#(R_j[\{x\}])=\#(R_j[\{x\}]\cap J_j)\le k-j$,
\quad$\#(R_j^{-1}[\{y\}])=\#(R^{-1}[\{y\}])\ge l$}

\noindent whenever $x\in I$ and $y\in J_j$, so

\Centerline{$l\#(J_j)\le\#(R_j)\le(k-j)\#(I)$}

\noindent and $\#(J_j)\le\Bover{k-j}{l}\#(I)$.   Accordingly

$$\eqalign{\#(K)
&\le\sum_{j=0}^{k-1}\#(I_j)
=\sum_{j=0}^{k-1}\Bover1{k-j}(\#(J_j)-\#(J_{j+1}))\cr&
=\sum_{j=1}^{k-1}\#(J_j)(\Bover1{k-j}-\Bover1{k-(j-1)})
   +\Bover1k\#(J_0)-\#(J_k)\cr&
\le\Bover1k\#(J_0)+\sum_{j=1}^{k-1}\Bover{\#(J_j)}{(k-j)(k-j+1)}
\le\Bover{\#(I)}{l}(1+\sum_{j=1}^{k-1}\Bover{k-j}{(k-j)(k-j+1)})\cr&
=\Bover{\#(I)}{l}(1+\sum_{j=1}^{k-1}\Bover1{k-j+1})
=\Bover{\#(I)}{l}(1+\sum_{j=2}^k\Bover1j)
\le\Bover{1+\ln k}{l}\#(I)\cr}$$

\noindent as required.
}%end of proof of 5A1Q

\leader{5A1R}{Stationary families of sets\dvrocolon{}}\cmmnt{ For the sake of
a calculation which will be needed in \S547,
%547R prf g
I introduce a kind of
generalization of the idea of `stationary' subset of an ordinal.

\medskip

\noindent}{\bf Definition}\dvAnew{2015}
If $I$ is a set and $\Cal A$ is a family of sets, I will
say that $\Cal A$ is {\bf stationary over $I$} if for every function
$f:[I]^{<\omega}\to[I]^{\le\omega}$ there is an $A\in\Cal A$ such that
$f(J)\subseteq A$ for every $J\in[A\cap I]^{<\omega}$.

\leader{5A1S}{Elementary remarks }\dvAnew{2015}{\bf (a)} If $\Cal A$ is
stationary over $I$,
then $\{A\cap I:A\in\Cal A\}$ is stationary over $I$.

\spheader 5A1Sb If
$\Cal A$ is stationary over $I$, and for every $A\in\Cal A$ we are given
a family $\Cal B_A$ which is
stationary over $A$, then $\bigcup_{A\in\Cal A}\Cal B_A$ is
stationary over $I$.

\spheader 5A1Sc If $\zeta$ is an ordinal of uncountable cofinality, and
$S\subseteq\zeta$ is stationary in the ordinary sense of 4A1C, then
$S$ is stationary over $\zeta$ in the sense of 5A1R.
\prooflet{\Prf\ Let
$f:[\zeta]^{<\omega}\to[\zeta]^{\le\omega}$ be a function.
Consider $C_f=\{\xi:\xi<\zeta$, $f(J)\subseteq\xi$ for every
$J\in[\xi]^{<\omega}\}$.   Because $\cf\zeta>\omega$, $C_f$ is closed and
cofinal in $\zeta$ (cf.\ 4A1B(c-iii)), so meets $S$.   As $f$ is
arbitrary, $S$ is stationary over $\zeta$.\ \Qed}

\leader{5A1T}{}\dvAnew{2017}\cmmnt{ In \S539 %539V
we shall need an important property of $\omega_1$.

\medskip

\noindent}{\bf Theorem} (a) There is a family
$\ofamily{\xi}{\omega_1}{e_{\xi}}$ such that
$e_{\xi}:\xi\to\Bbb N$ is an injective function for each $\xi<\omega_1$ and
$e_{\eta}\symmdiff(e_{\xi}\restr\eta)$ is finite
whenever $\eta<\xi<\omega_1$.

(b) There is a sequence $\sequencen{\le_n}$ of partial orders on $\omega_1$ such
that

\inset{$(\omega_1,\le_n)$ is a tree of height at most
$n+1$ for each $n\in\Bbb N$,

$\eta\le_0\xi$ iff $\eta=\xi$,

$\le_n\,\subseteq\,\le_{n+1}$ for every $n\in\Bbb N$,

$\bigcup_{n\in\Bbb N}\le_n$ is the usual well-ordering of $\omega_1$.}

\proof{\footnote{I regret that I cannot recall
where I saw this proof.}{\bf (a)}
Choose $\ofamily{\xi}{\omega_1}{e_{\xi}}$ inductively;  as
well as the obvious inductive hypothesis that
$e_{\eta}\symmdiff(e_{\xi}\restr\eta)$ is finite
whenever $\eta<\xi$, we need to ensure that $\Bbb N\setminus e_{\xi}[\xi]$ is
always infinite.

Start the induction with $e_0$ the empty function.   For the inductive step
to $\xi+1$, where $\xi<\omega_1$,
take any $k\in\Bbb N\setminus e_{\xi}[\xi]$ and set
$e_{\xi+1}=e_{\xi}\cup\{(\xi,k)\}$.   For the inductive step to a non-zero
limit ordinal $\xi<\omega_1$, choose a strictly increasing sequence
$\sequencen{\eta_n}$ of ordinals with supremum $\xi$ and define
$\sequencen{e'_n}$, $\sequencen{k_n}$ inductively as follows.
Start with $e'_0=e_{\eta_0}$ and
$k_0\in\Bbb N\setminus e_{\eta_0}[\eta_0]$.   Given $e'_n$ and
$k_0,\ldots,k_n\in\Bbb N$ such that
$e'_n:\eta_n\to\Bbb N\setminus\{k_0,\ldots,k_n\}$ is injective and
$e'_n\symmdiff e_{\eta_n}$ is finite, we see that

\Centerline{$e'_n\symmdiff(e_{\eta_{n+1}}\restr\eta_n)
\subseteq(e'_n\symmdiff e_{\eta_n})
   \cup(e_{\eta_n}\symmdiff(e_{\eta_{n+1}}\restr\eta_n))$}

\noindent is finite while
$e_{\eta_{n+1}}$ is injective and $\Bbb N\setminus e_{\eta_{n+1}}[\eta_{n+1}]$
is infinite.   We therefore have room to adjust $e_{\eta_{n+1}}$ at finitely
many points to obtain an injective function
$e'_{n+1}:\eta_{n+1}\to\Bbb N\setminus\{k_0,\ldots,k_n\}$, extending $e'_n$,
such that $e'_{n+1}\symmdiff e_{\eta_{n+1}}$ is finite.
Because $e'_{n+1}\symmdiff e_{\eta_{n+1}}$ is finite,
$\Bbb N\setminus e'_{n+1}[\eta_{n+1}]$ is infinite and we can find
$k_{n+1}\in\Bbb N\setminus(\{k_0,\ldots,k_n\}\cup e'_{n+1}[\eta_{n+1}])$
before continuing this internal induction.
Finally set $e_{\xi}=\bigcup_{n\in\Bbb N}e'_n$;  then
$e_{\xi}:\xi\to\Bbb N\setminus\{k_i:i\in\Bbb N\}$ is injective and
$e_{\eta_n}\symmdiff(e_{\xi}\restr\eta_n)=e_{\eta_n}\symmdiff e'_n$ is finite for
every $n\in\Bbb N$.   Now if $\eta$ is any ordinal less than $\xi$, there is an
$n\in\Bbb N$ such that $\eta<\eta_n$, so that

\Centerline{$e_{\eta}\symmdiff(e_{\xi}\restr\eta)
\subseteq (e_{\eta}\symmdiff(e_{\eta_n}\restr\eta))
  \cup(e_{\eta_n}\symmdiff(e_{\xi}\restr\eta_n))$}

\noindent is finite.   Thus the outer induction continues and we get a suitable
family $\ofamily{\xi}{\omega_1}{e_{\xi}}$.

\medskip

{\bf (b)} Take a family $\ofamily{\xi}{\omega_1}{e_{\xi}}$ as in (a).   For
$n\in\Bbb N$ and $\xi$, $\eta<\omega_1$, say that $\eta\le_n\xi$ iff either
$\eta=\xi$ or

\Centerline{$\eta<\xi$,\quad $e_{\xi}(\eta)<n$, \quad
$e_{\eta}\symmdiff(e_{\xi}\restr\eta)\subseteq\omega_1\times n$.}

\noindent We need to check the following.

\medskip

\quad{\bf (i)} If $\zeta\le_n\eta\le_n\xi$ then $\zeta\le_n\xi$.   \Prf\ If
either $\zeta=\eta$ or $\eta=\xi$ this is trivial.   Otherwise,
$\zeta<\eta<\xi$ and

\Centerline{$e_{\zeta}\symmdiff(e_{\xi}\restr\zeta)
\subseteq(e_{\zeta}\symmdiff(e_{\eta}\restr\zeta))
  \cup(e_{\eta}\symmdiff(e_{\xi}\restr\eta))
\subseteq\omega_1\times n$.}

\noindent Of course $\zeta<\xi$.   If $e_{\xi}(\zeta)=e_{\eta}(\zeta)$,
then $e_{\xi}(\zeta)<n$ because $e_{\eta}(\zeta)<n$;  otherwise,
$e_{\xi}(\zeta)<n$ because
$e_{\eta}\symmdiff(e_{\xi}\restr\eta)\subseteq\omega_1\times n$.   So
$\zeta\le_n\xi$.\ \Qed

Thus $\le_n$ is transitive;  the definition ensured that it would be reflexive;
and it is antisymmetric because $\le$ is.

\medskip

\quad{\bf (ii)} If $\zeta\le_n\xi$ and $\eta\le_n\xi$ then either
$\zeta\le_n\eta$ or $\eta\le_n\zeta$.   \Prf\ We can suppose that
$\zeta\le\eta$.   If $\zeta=\eta$ or $\eta=\xi$
then of course $\zeta\le_n\eta$.   If $\zeta<\eta<\xi$, then

\Centerline{$e_{\zeta}\symmdiff(e_{\eta}\restr\zeta)
\subseteq(e_{\zeta}\symmdiff(e_{\xi}\restr\zeta))
  \cup(e_{\eta}\symmdiff(e_{\xi}\restr\eta))
\subseteq n$.}

\noindent If $e_{\eta}(\zeta)=e_{\xi}(\zeta)$ then $e_{\eta}(\zeta)<n$ because
$e_{\xi}(\zeta)<n$;  otherwise $e_{\eta}(\zeta)<n$ because
$e_{\eta}\symmdiff(e_{\xi}\restr\eta)\subseteq\omega_1\times n$.   So
$\zeta\le_n\eta$.\ \Qed

\medskip

\quad{\bf (iii)} Thus $\{\eta:\eta\le_n\xi\}$ is totally ordered by $\le_n$, for
every $\xi$.   Moreover,
$\{\eta:\eta\le_n\xi\}\subseteq\{\xi\}\cup\{\eta:\eta<\xi$, $e_{\xi}(\eta)<n\}$
has at most $n+1$ members, because $e_{\xi}$ is injective.
So $\le_n$ is well-founded and is a tree ordering with height at most $n$.

\medskip

\quad{\bf (iv)} It is also immediate from the definition that
$\sequencen{\le_n}$ is non-decreasing.   Finally, if $\eta<\xi$, there is an
$n\in\Bbb N$ such that $e_{\xi}(\eta)<n$ and
$e_{\eta}\symmdiff(e_{\xi}\restr\eta)\subseteq\omega_1\times n$,
because $e_{\eta}\symmdiff(e_{\xi}\restr\eta)$ is itself finite.   So
$\le\,=\bigcup_{n\in\Bbb N}\le_n$.
}%end of proof of 5A1T

%\exercises{\leader{5A1X}{Basic exercises (a)}

%\leader{5A1Y}{Further exercises (a)}

%}%end of exercises

%\endnotes{
%\Notesheader{5A1}

%}%end of notes

\discrpage

\frfilename{mt5a2.tex}
\versiondate{25.2.21}
\copyrightdate{2007}

\def\chaptername{Appendix}
\def\sectionname{Pcf theory}

\newsection{5A2}

In \S542 I call on some results from Shelah's pcf theory.   As
I have still not come across an elementary textbook for this material, 
I copy out
part of the appendix of {\smc Fremlin 93}, itself drawn largely from
{\smc Burke \& Magidor 90}.

\leader{5A2A}{Reduced products}\cmmnt{ We need the following
elementary generalization of the construction in 351M.}   Let
$\familyiI{P_i}$ be a family of partially ordered sets with product
$P$.

\spheader 5A2Aa Let $\Cal F$ be a filter on $I$.   We have an equivalence
relation $\equiv_{\Cal F}$ on $P$, given by saying that
$f\equiv_{\Cal F}g$ if $\{i:f(i)=g(i)\}\in \Cal F$.   I write
$P|\Cal F$
for the set of equivalence classes under this relation, the
{\bf partial order reduced product} of $\langle P_i\rangle_{i\in I}$ modulo
$\Cal F$.   Now $P|\Cal F$ is again a partially ordered set, writing

\Centerline{$f^{\bullet}\le g^{\bullet}
\iff f\le_{\Cal F}g\iff\{i:f(i)\le g(i)\}\in\Cal F$.}

\noindent Observe that if every $P_i$ is totally ordered and $\Cal F$
is an ultrafilter, then $P|\Cal F$ is totally ordered.

\spheader 5A2Ab For any filter $\Cal F$ on $I$ we have

$$\eqalign{\min_{i\in I}\add P_i
=\add P&
\le\sup_{F\in\Cal F}\add(\prod_{i\in F}P_i)
=\sup_{F\in\Cal F}\min_{i\in F}\add P_i\cr&
\le\add(P|\Cal F),\cr}$$

%\Centerline{$\min_{i\in I}\add P_i=\add P
%\le\sup_{F\in\Cal F}\add(\prod_{i\in F}P_i)
%=\sup_{F\in\Cal F}\min_{i\in F}\add P_i
%\le\add(P|\Cal F)$,}

\Centerline{$\cf(P|\Cal F)
\le\min_{F\in\Cal F}\cf(\prod_{i\in F}P_i)\le\cf P$.}

\noindent\prooflet{\Prf\ By 511Hg,
$\add(\prod_{i\in F}P_i)=\min_{i\in F}\add P_i$ for any $F\in\Cal F$,
and in
particular when $F=I$.   For $p\in P|\Cal F$ choose $f_p\in P$ such that
$f_p^{\ssbullet}=p$.   If $F\in\Cal F$ then $p\mapsto f_p\restr F$ is a
Tukey function from $P|\Cal F$ to $\prod_{i\in F}P_i$, so
$P|\Cal F\prT\prod_{i\in F}P_i$ and 513Ee tells us that
$\add(\prod_{i\in F}P_i)\le\add(P|\Cal F)$ and
$\cf(P|\Cal F)\le\cf(\prod_{i\in F}P_i)$.   Also $f\mapsto f\restr F$ is a
dual Tukey function from $P$ to $\prod_{i\in F}P_i$, so
$\prod_{i\in F}P_i\prT P$ and $\cf(\prod_{i\in F}P_i)\le\cf P$.\ \Qed}

\spheader 5A2Ac Note that if $\Cal F$, $\Cal G$ are filters on $I$
and $\Cal F\subseteq\Cal G$, then $\add(P|\Cal F)\le\add(P|\Cal G)$
and $\cf(P|\Cal F)\ge\cf(P|\Cal G)$.   \prooflet{\Prf\ If
$f\le_{\Cal F}g$ then $f\le_{\Cal G}g$.   So we have a canonical surjective
order-preserving map $\psi:P|\Cal F\to P|\Cal G$ given by saying that
$\psi(\pi_{\Cal F}(f))=\pi_{\Cal G}(f)$ for every $f\in P$, where
$\pi_{\Cal F}(f)$, $\pi_{\Cal G}(f)$ are the equivalence classes of $f$ in
$P|\Cal F$ and $P|\Cal G$ respectively.   By
513E(b-iii), $\psi$ is a dual Tukey function, so
$P|\Cal G\prT P|\Cal F$ and we can use 513Ee again.\ \Qed}
%5{}42D

\leader{5A2B}{Theorem} %RVMC A1I
Let $\lambda>0$ be a cardinal and
$\langle\theta_{\zeta}\rangle_{\zeta<\lambda}$ a family of regular
infinite cardinals, all greater than $\lambda$.   Set
$P=\prod_{\zeta<\lambda}\theta_{\zeta}$.
For any filter $\Cal F$ on $\lambda$, let $P|\Cal F$ be the corresponding
reduced product and $\pi_{\Cal F}:P\to P|\Cal F$
the canonical map.   For any cardinal $\delta$ set

\Centerline{$\frak F_{\delta}=\{\Cal F:\Cal F$ is an ultrafilter on
$\lambda$, $\cf(P|\Cal F)=\delta\},$}

\Centerline{$\frak F^*_{\delta}
=\bigcup_{\delta'\ge\delta}\frak F_{\delta'}$;}

\noindent if $\frak F^*_{\delta}\ne\emptyset$, let $\Cal G_{\delta}$
be the filter $\bigcap\frak F^*_{\delta}$.   Now

(a) if $\frak F^*_{\delta}\ne\emptyset$, then
$\add(P|\Cal G_{\delta})\ge\delta$;

(b) for every $\delta$ there is a set
$F\in[P]^{\le\delta}$ such that $\pi_{\Cal F}[F]$ is
cofinal with $P|\Cal F$ for every $F\in\frak F_{\delta}$;

(c) $\frak F_{\cf P}\ne\emptyset$.

\proof{ The case of finite $\lambda$ is trivial throughout,
as then

\Centerline{$\cf P=\max_{\zeta<\lambda}\theta_{\zeta}$,}

\Centerline{$\frak F_{\delta}
=\{\Cal F:$ there is a $\zeta<\lambda$ such that
$\{\zeta\}\in\Cal F$ and $\theta_{\zeta}=\delta\}$,}

\Centerline{$\frak F^*_{\delta}
=\{\Cal F:\{\zeta:\theta_{\zeta}\ge\delta\}\in\Cal F\}$,}

\Centerline{$\Cal G_{\delta}
=\{G:\{\zeta:\theta_{\zeta}\ge\delta\}\subseteq G\subseteq\lambda\}$.}

\noindent So henceforth let us take it that $\lambda$ is infinite.

If $\Cal F$ is an ultrafilter on $\lambda$, then
$P|\Cal F$ is a non-empty totally ordered set with no
greatest member, so its additivity and cofinality are the same;
thus

\Centerline{$\frak F_{\delta}=\{\Cal F:\Cal F$ is an ultrafilter on
$\lambda$, $\add(P|\Cal F)=\delta\}$}

\noindent for every $\delta$, and

\Centerline{$\min_{\zeta\in F}\theta_{\zeta}
\le\delta\le\cf(\prod_{\zeta\in F}\theta_{\zeta})$}

\noindent whenever $F\in\Cal F\in\frak F_{\delta}$, by 5A2Ab.

Write
$L=\{\zeta:\zeta<\lambda$, $\theta_{\zeta}=\lambda^+\}$,
$M=\lambda\setminus L$.   If $\Cal F$ is an ultrafilter on $\lambda$
and $L\in\Cal F$, then
$\cf(\prod_{\zeta\in L}\theta_{\zeta})=\lambda^+$, because the set of
constant functions is cofinal with $\prod_{\zeta\in L}\theta_{\zeta}$,
so $\cf(P|\Cal F)$ must be $\lambda^+$;  otherwise, $M\in\Cal F$ and
$\cf(P|\Cal F)>\lambda^+$.

\medskip

{\bf (a)} Set $\delta'=\add(P|\Cal G_{\delta})$.

\medskip

\quad{\bf (i)} $\delta'$ is a regular infinite cardinal (513C(a-i)) and

\Centerline{$\delta'\ge\min_{\zeta<\lambda}\theta_{\zeta}
>\lambda$}

\noindent by 5A2Ab again.
If $\delta=\lambda^+$ then of course $\delta'\ge\delta$;  so
suppose that $\delta>\lambda^+$.   In this case $L\notin\Cal F$ for
any $\Cal F\in\frak F^*_{\delta}$, so
$M\in\Cal G_{\delta}$ and
$\delta'\ge\min_{\zeta\in M}\theta_{\zeta}>\lambda^+$.

\medskip

\quad{\bf (ii)} \Quer\ If $\delta'<\delta$ then (translating 513C(a-i)
into a statement about the pre-order
$\le_{\Cal G_{\delta}}$) there is a family
$\langle f_{\alpha}\rangle_{\alpha<\delta'}$ in $P$ such that
$f_{\alpha}\le_{\Cal G_{\delta}}f_{\beta}$ whenever
$\alpha\le\beta<\delta'$ but
there is no $f\in P$ such that $f_{\alpha}\le_{\Cal G_{\delta}}f$
for every $\alpha<\delta'$.   Choose $h_{\xi}\in P$,
$\alpha_{\xi}<\delta'$
inductively, for $\xi<\lambda^+$, as follows.  $h_0=f_0$.
Given $h_{\xi}$, set

\Centerline{$B_{\xi\alpha}
=\{\zeta:\zeta\in M,\,h_{\xi}(\zeta)\ge f_{\alpha}(\zeta)\}$}

\noindent for each $\alpha<\delta'$;  let
$\alpha_{\xi}<\delta'$ be such that
$f_{\alpha_{\xi}}\not\le_{\Cal G_{\delta}}h_{\xi}$,
so that $B_{\xi\alpha}\notin\Cal G_{\delta}$ when
$\alpha_{\xi}\le\alpha<\delta'$.   Choose
$\Cal F_{\xi}\in\frak F^*_{\delta}$ such that
$B_{\xi,\alpha_{\xi}}\notin\Cal F_{\xi}$.
Now, because $\cf(P|\Cal F_{\xi})\ge\delta>\delta'$, there
is an $h_{\xi+1}\in P$ such that $f_{\alpha}\le_{\Cal F_{\xi}}h_{\xi+1}$
for every $\alpha<\delta'$;  we may take $h_{\xi+1}\ge h_{\xi}$.

For non-zero limit ordinals $\xi<\lambda^+$ take
$h_{\xi}(\zeta)=\sup_{\eta<\xi}h_{\eta}(\zeta)$ for every
$\zeta<\lambda$.

Set $\alpha=\sup_{\xi<\lambda^+}\alpha_{\xi}<\delta'$.   Then
$\langle B_{\xi\alpha}\rangle_{\xi<\lambda^+}$ is a non-decreasing
family in $\Cal P\lambda$.   So there must be a $\xi<\lambda^+$ such
that $B_{\xi\alpha}=B_{\xi+1,\alpha}$.

By the choice of $h_{\xi+1}$, $B_{\xi+1,\alpha}\in\Cal F_{\xi}$.
So $B_{\xi\alpha}\in\Cal F_{\xi}$ and
$f_{\alpha}\le_{\Cal F_{\xi}}h_{\xi}$.   Because $\alpha\ge\alpha_{\xi}$,
$f_{\alpha_{\xi}}\le_{\Cal G_{\delta}}f_{\alpha}$, so

\Centerline{$f_{\alpha_{\xi}}\le_{\Cal F_{\xi}}f_{\alpha}
\le_{\Cal F_{\xi}}h_{\xi}$}

\noindent and $B_{\xi,\alpha_{\xi}}\in\Cal F_{\xi}$;  contrary to the
choice of $\Cal F_{\xi}$.\ \Bang

\medskip

{\bf (b)(i)}
If $\frak F_{\delta}=\emptyset$, we can take $F=\emptyset$;  so
suppose that $\frak F_{\delta}$ is non-empty.
As in (a-i) above, we must have
$\delta\ge\min_{\zeta<\lambda}\theta_{\zeta}>\lambda$, and the case
$\delta=\lambda^+$ is again elementary.   \Prf\ Take $F\subseteq P$ 
to be the set
of constant functions with values less than $\lambda^+$.   If
$\Cal F\in\frak F_{\delta}$ then $M\notin\Cal F$ and $L\in\Cal F$.
So for any $h\in P$ we have $\alpha=\sup_{\zeta\in L}h(\zeta)<\lambda^+$,
and there is an $f\in F$ such that $f(\zeta)=\alpha$ for every
$\zeta$, in which case $h\le_{\Cal F}f$;  thus $\pi_{\Cal F}[F]$ is cofinal
with $P|\Cal F$, as required.\ \Qed

So suppose from now on
that $\delta>\lambda^+$, so that $M\in\Cal F$ for every
$\Cal F\in\frak F_{\delta}$.   Of course $\delta$, being the
cofinality of a non-empty totally ordered set
with no greatest member, is regular (511He-511Hf, 513C(a-i)).
For each $\Cal F\in\frak F_{\delta}$, choose a family
$\ofamily{\beta}{\delta}{g_{\Cal F\beta}}$ in $P$ such that
$\{\pi_{\Cal F}(g_{\Cal F\beta}):\beta<\delta\}$ is cofinal with $P|\Cal F$.

\medskip

\quad{\bf (ii)} \Quer\ Suppose, if possible, that there is no $F$ of the
required type.   In this case, we can find families
$\langle f_{\xi\alpha}\rangle_{\xi<\lambda^+,\alpha<\delta}$ in $P$ and
$\langle\Cal F_{\xi}\rangle_{\xi<\lambda^+}$ in $\frak F_{\delta}$ such
that

\inset{($\alpha$) $f_{\eta\alpha}\le_{\Cal F_{\xi}}f_{\xi 0}$ whenever
$\alpha<\delta$, $\eta<\xi<\lambda^+$;

($\beta$) $\{\pi_{\Cal F_{\xi}}(f_{\xi\alpha}):\alpha<\delta\}$
is cofinal with $P|\Cal F_{\xi}$ for every $\xi<\lambda^+$;

($\gamma$) $f_{\eta\alpha}\le f_{\xi\alpha}$ whenever
$\alpha<\delta$, $\eta\le\xi<\lambda^+$;

($\delta$) if $\xi<\lambda^+$, $\alpha<\delta$ and
$\cf\alpha=\lambda^+$ then

\Centerline{$f_{\xi\alpha}(\zeta)
=\min\{\sup_{\beta\in C}f_{\xi\beta}(\zeta):C
  \text{ is a closed cofinal set in }\alpha\}$}

\noindent for every $\zeta\in M$;

($\epsilon$) $f_{\xi\beta}\le_{\Cal F_{\xi}}f_{\xi\alpha}$
whenever $\xi<\lambda^+$, $\beta\le\alpha<\delta$.}

\noindent\Prf\ Construct
$\langle f_{\xi\alpha}\rangle_{\xi<\lambda^+,\alpha<\delta}$ inductively,
taking $\lambda^+\times\delta$ with its lexicographic well-ordering (that
is, $(\xi,\alpha)\le(\eta,\beta)$ if either $\xi<\eta$ or $\xi=\eta$ and
$\alpha<\beta$).   Given that
$\langle f_{\eta\beta}\rangle_{(\eta,\beta)<(\xi,\alpha)}$ satisfies the
inductive hypothesis so far, proceed according to the nature of $\alpha$,
as follows.

{\it Zero} If $\alpha=0$, then, because $\#(\xi\times\delta)\le\delta$, the
counter-hypothesis tells us that there is an
$\Cal F_{\xi}\in\frak F_{\delta}$ such that
$\{\pi_{\Cal F_{\xi}}(f_{\eta\beta}):\eta<\xi$, $\beta<\delta\}$ is
not cofinal with $P|\Cal F_{\xi}$.
Accordingly we can find $f_{\xi 0}\in P$ such that

\Centerline{$f_{\eta\alpha}\le_{\Cal F_{\xi}}f_{\xi 0}$ whenever
$\eta<\xi$ and $\alpha<\delta$,}

\noindent and because $\add P\ge\lambda^+>\#(\xi)$, we can also insist that

\Centerline{$f_{\eta 0}\le f_{\xi 0}$ whenever $\eta<\xi$.}

{\it Successor}  If $\alpha=\beta+1$ is a successor ordinal, set

\Centerline{$f_{\xi\alpha}(\zeta)
=\max(f_{\xi\beta}(\zeta),g_{\Cal F_{\xi}\beta}(\zeta),
    \sup_{\eta<\xi}f_{\eta\alpha}(\zeta))$ for every $\zeta<\lambda$;}

\noindent this is acceptable because $\cf\theta_{\zeta}>\lambda$ for every
$\zeta$.

{\it Cofinality $\lambda^+$} If
$\cf\alpha=\lambda^+$, set

$$\eqalign{f_{\xi\alpha}(\zeta)
&=\sup_{\eta<\xi}f_{\eta\alpha}(\zeta)\text{ if }\zeta\in L,\cr
&=\min\{\sup_{\beta\in C}f_{\xi\beta}(\zeta):
  C\text{ is a closed cofinal set in }\alpha\}
  \text{ if }\zeta\in M.\cr}$$

\noindent This time, note that if $\zeta\in M$, then
$f_{\xi\alpha}(\zeta)<\theta_{\zeta}$ because
there is a closed cofinal set in $\alpha$ with cardinal
$\lambda^+<\theta_{\zeta}$.

{\it Otherwise} If $\alpha$ is a non-zero
limit ordinal and $\cf\alpha\ne\lambda^+$, choose $f_{\xi\alpha}$ such that

\Centerline{$f_{\eta\alpha}\le f_{\xi\alpha}$ for every $\eta<\xi$,}

\Centerline{$f_{\xi\beta}\le_{\Cal F_{\xi}}f_{\xi\alpha}$ for every
$\beta<\alpha$;}

\noindent this is possible because $\add P$ and $\add(P|\Cal F_{\xi})$ are
both at least $\lambda^+>\max(\#(\xi),\#(\alpha))$.

Now let us work through the list of conditions to be satisfied.

($\alpha$) is written into the case $\alpha=0$ of the induction.

($\beta$) Because $g_{\Cal F_{\xi}\alpha}\le f_{\xi,\alpha+1}$ for every 
$\alpha$
and $\{\pi_{\Cal F_{\xi}}(g_{\Cal F_{\xi}\alpha}):\alpha<\delta\}$ 
is cofinal with $P|\Cal F_{\xi}$,
$\{\pi_{\Cal F_{\xi}}(f_{\xi\alpha}):\alpha<\delta\}$ is cofinal with
$P|\Cal F_{\xi}$.

($\gamma$) The construction ensures that we shall have
$f_{\eta\alpha}(\zeta)\le f_{\xi\alpha}(\zeta)$ in all the required cases
except possibly when $\cf\alpha=\lambda^+$ and $\zeta\in M$.   But in this
case, taking $\eta<\xi$ and a closed cofinal set $C\subseteq\alpha$
such that $f_{\xi\alpha}(\zeta)=\sup_{\beta\in C}f_{\xi\beta}(\zeta)$, the
inductive hypothesis will assure us that

\Centerline{$f_{\eta\alpha}(\zeta)
\le\sup_{\beta\in C}f_{\eta\beta}(\zeta)
\le\sup_{\beta\in C}f_{\xi\beta}(\zeta)=f_{\xi\alpha}(\zeta)$,}

\noindent so there is no problem.

($\delta$) is written into the formula for the inductive step when
$\cf\alpha=\lambda^+$.

($\epsilon$) We certainly have $f_{\xi\alpha}\le f_{\xi,\alpha+1}$, so
$f_{\xi\alpha}\le_{\Cal F_{\xi}}f_{\xi,\alpha+1}$, for
every $\alpha$.   If $\cf\alpha=\lambda^+$, then
because the intersection of
fewer than $\cf\alpha$ closed cofinal subsets of $\alpha$ is again a
closed cofinal set in $\alpha$ (4A1Bd), there
will be a closed cofinal set $C\subseteq\alpha$ such that
$f_{\xi\alpha}(\zeta)=\sup_{\beta\in C}f_{\xi\beta}(\zeta)$ for every
$\zeta\in M$.   So $f_{\xi\beta}\le_{\Cal F_{\xi}}f_{\xi\alpha}$ for every
$\beta\in C$;  by the inductive hypothesis,
$f_{\xi\beta}\le_{\Cal F_{\xi}}f_{\xi\alpha}$ for every
$\beta<\alpha$.   For other limit ordinals $\alpha$, we have
$f_{\xi\beta}\le_{\Cal F_{\xi}}f_{\xi\alpha}$ for every
$\beta<\alpha$ directly from the choice of $f_{\xi\alpha}$.

So the procedure works.\ \Qed

\medskip

\quad{\bf (iii)} The next step is to choose a non-decreasing family
$\langle h_{\eta}\rangle_{\eta<\lambda^+}$ in $P$ and a strictly
increasing family
$\langle\gamma(\eta)\rangle_{\eta<\lambda^+}$ in $\delta$ such that

\inset{$\gamma(\eta)=\sup_{\eta'<\eta}\gamma(\eta')$ whenever
$\eta<\lambda^+$ is a limit ordinal (in particular, $\gamma(0)=0$);

$f_{\xi,\gamma(\eta)}(\zeta)<h_{\eta}(\zeta)$ whenever
$\xi$, $\eta<\lambda^+$ and $\zeta\in M$ (choosing $h_{\eta}$);

$h_{\eta}\le_{\Cal F_{\xi}}f_{\xi,\gamma(\eta+1)}$ whenever
$\xi$, $\eta<\lambda^+$ (choosing $\gamma(\eta+1)$).}

\noindent Set $h(\zeta)=\sup_{\eta<\lambda^+}h_{\eta}(\zeta)$
for $\zeta\in M$, $h(\zeta)=0$ for $\zeta\in L$,
$\alpha=\sup_{\eta<\lambda^+}\gamma(\eta)<\delta$ (because
$\delta=\cf\delta>\lambda^+$);  then $\cf\alpha=\lambda^+$ and
$\{\gamma(\eta):\eta<\lambda^+\}$ is a closed cofinal set in
$\alpha$.   Thus

\Centerline{$f_{\xi\alpha}(\zeta)
\le\sup_{\eta<\lambda^+}f_{\xi,\gamma(\eta)}(\zeta)\le h(\zeta)$}

\noindent for every $\xi<\lambda^+$ and $\zeta\in M$, by (ii-$\delta$).
So if we set

\Centerline{$A_{\xi}=\{\zeta:\zeta\in M,\,
f_{\xi\alpha}(\zeta)=h(\zeta)\}$}

\noindent for each $\xi<\lambda^+$, we shall have
$A_{\eta}\subseteq A_{\xi}$ whenever
$\eta\le\xi<\lambda^+$, by (ii-$\gamma$).

\medskip

\quad{\bf (iv)} As $\#(M)\le\lambda$, there must be some
$\xi<\lambda^+$ such that $A_{\xi}=A_{\xi+1}$.   Let
$C\subseteq\alpha$ be a closed cofinal set such that

\Centerline{$f_{\xi+1,\alpha}(\zeta)
=\sup_{\beta\in C}f_{\xi+1,\beta}(\zeta)$}

\noindent for every $\zeta\in M$.
Set $C'=\gamma^{-1}[C]$.   Then $C'$ is a closed cofinal subset of
$\lambda^+$.   \Prf\ It is closed because $\gamma:\lambda^+\to\alpha$ is
order-continuous, therefore continuous (4A2Ro).   Next, $\gamma[\lambda^+]$
is closed and cofinal in
$\alpha$, while $\cf\alpha=\lambda^+$ is uncountable, so
$C\cap\gamma[\lambda^+]$ is cofinal with $\alpha$ and $\gamma[\lambda^+]$
(4A1Bd again), and $C'$ is cofinal with $\lambda^+$.\ \QeD\

For each $\eta\in C'$ write $\eta'$ for the next member of $C'$
greater than $\eta$;  then

\Centerline{$h_{\eta}\le_{\Cal F_{\xi+1}}f_{\xi+1,\gamma(\eta+1)}
\le_{\Cal F_{\xi+1}}f_{\xi+1,\gamma(\eta')}$,}

\Centerline{$f_{\xi\alpha}\le_{\Cal F_{\xi+1}}f_{\xi+1,0}
=f_{\xi+1,\gamma(0)}$}

\noindent so there is a $\zeta_{\eta}\in M$ such that

\Centerline{$h_{\eta}(\zeta_{\eta})
\le f_{\xi+1,\gamma(\eta')}(\zeta_{\eta})$,
\quad$f_{\xi\alpha}(\zeta_{\eta})\le f_{\xi+1,\gamma(0)}(\zeta_{\eta})
<h_0(\zeta_{\eta})\le h(\zeta_{\eta})$.}

\noindent Let $\zeta\in M$ be such that

\Centerline{$B=\{\eta:\eta\in C',\,\zeta_{\eta}=\zeta\}$}

\noindent is cofinal with $\lambda^+$.
Then $f_{\xi\alpha}(\zeta)<h(\zeta)$ so
$\zeta\notin A_{\xi}$.   On the other hand,

\Centerline{$f_{\xi+1,\alpha}(\zeta)
=\sup_{\beta\in C}f_{\xi+1,\beta}(\zeta)
\ge\sup_{\eta\in B}f_{\xi+1,\gamma(\eta')}(\zeta)
\ge\sup_{\eta\in B}h_{\eta}(\zeta)=h(\zeta)$}

\noindent because $\ofamily{\eta}{\lambda^+}{h_{\eta}}$ is non-decreasing.
So $\zeta\in A_{\xi+1}$;  which is impossible.\ \Bang

This contradiction completes the proof of (b).

\medskip

{\bf (c)(i)} Set $\Delta=\{\delta:\frak F_{\delta}\ne\emptyset\}$,
$\Cal G=\bigcup_{\delta\in\Delta}\Cal G_{\delta}$.   Since
$\Cal G_{\delta}\subseteq\Cal G_{\delta'}$ for $\delta\le\delta'$ in
$\Delta$,
$\Cal G$ is a filter on $\lambda$ and there is an ultrafilter
$\Cal H$ on $\lambda$ including $\Cal G$.   For any
$\delta\in\Delta$, $\Cal H\supseteq\Cal G_{\delta}$, so

\Centerline{$\cf(P|\Cal H)=\add(P|\Cal H)
\ge\add(P|\Cal G_{\delta})\ge\delta>\lambda$,}

\noindent using 5A2Ac and (a) above.   Consequently
$\delta^*=\cf(P|\Cal H)$ is the greatest element of $\Delta$.

\medskip

\quad{\bf (ii)} For each $\delta\le\delta^*$ choose a set
$F_{\delta}\in[P]^{\le\delta}$ such that $\pi_{\Cal F}[F_{\delta}]$ is
cofinal with
$P|\Cal F$ for every $\Cal F\in\frak F_{\delta}$ (using (b) above).
Set $F=\bigcup_{\delta\in\Delta}F_{\delta}$ and

\Centerline{$G=\{\sup I:I\in[F]^{<\omega}\}\subseteq P$.}

\noindent Then $\#(G)\le\delta^*$.   I claim that $G$ is cofinal with
$P$.   \Prf\Quer\ Suppose otherwise;  take $h\in P$ such
that $h\not\le g$ for every $g\in G$.   Write

\Centerline{$A_g=\{\zeta:h(\zeta)>g(\zeta)\}$}

\noindent for each $g\in G$.   Because $G$ is upwards-directed,
$\{A_g:g\in G\}$ is a filter base, and there is an ultrafilter $\Cal
F$
on $\lambda$ containing every $A_g$.   Now there is a
$\delta\in\Delta$
such that $\Cal F\in\frak F_{\delta}$, so that
$\pi_{\Cal F}[F_{\delta}]$
is cofinal with $P|\Cal F$, and there is an $f\in F_{\delta}$
such that $h\le_{\Cal F}f$.   But in this
case $A=\{\zeta:h(\zeta)\le f(\zeta)\}$
and $A_f=\lambda\setminus A$ both belong to $\Cal F$.\ \Bang\Qed

\medskip

\quad{\bf (iii)}
Accordingly $\cf P\le\#(G)\le\delta^*$.   But also of course
$\delta^*=\cf(P|\Cal H)\le\cf P$,
so $\delta^*=\cf P$.   Now we
have $\Cal H\in\frak F_{\delta^*}=\frak F_{\cf P}$.
}%end of proof of 5A2B

\leader{5A2C}{Theorem} Let $\lambda>0$ be a cardinal and
$\langle\theta_{\zeta}\rangle_{\zeta<\lambda}$ a family of regular
infinite cardinals, all greater than $\lambda$.   Set
$P=\prod_{\zeta<\lambda}\theta_{\zeta}$.
Let $\Cal F$ be an ultrafilter on $\lambda$
and $\kappa$ a regular
infinite cardinal with $\lambda<\kappa\le\cf(P|\Cal F)$.
\cmmnt{(In the language of 5A2B, $\Cal F\in\frak F_{\kappa}^*$.)}
Then there is a
family $\langle\theta'_{\zeta}\rangle_{\zeta<\lambda}$ of regular
infinite cardinals such that $\lambda<\theta'_{\zeta}\le\theta_{\zeta}$ for
every $\zeta<\lambda$ and $\cf(P'|\Cal F)=\kappa$, where
$P'=\prod_{\zeta<\lambda}\theta'_{\zeta}$.

\proof{{\bf (a)}
If $\lambda$ is finite, there is a $\zeta<\lambda$ such that
$\{\zeta\}\in\Cal F$, $\cf(P|\Cal F_{\zeta})=\theta_{\zeta}$ and we just
have to take $\theta'_{\zeta}=\kappa$;   
if $\kappa=\lambda^+$ we may take $\theta'_{\zeta}=\lambda^+$ for
every $\zeta$;  if $\kappa=\cf(P|\Cal F)$ we may take
$\theta'_{\zeta}=\theta_{\zeta}$;  so let us assume that
$\omega_1\le\lambda^+<\kappa<\cf(P|\Cal F)$.   In this case
$\{\zeta:\zeta<\lambda$, $\theta_{\zeta}=\lambda^+\}\notin\Cal F$, by 5A2Ab, so
$M=\{\zeta:\zeta<\lambda$, $\theta_{\zeta}>\lambda^+\}$ belongs
to $\Cal F$.

\medskip

{\bf (b)} For each ordinal $\gamma<\kappa$ choose a
relatively closed cofinal set $C_{\gamma}\subseteq\gamma$ with
$\text{otp}(C_{\gamma})=\cf\gamma$.
Choose families $\langle f_{\alpha}\rangle_{\alpha<\kappa}$,
$\langle g_{\alpha\gamma}\rangle_{\alpha,\gamma<\kappa}$ in $P$
inductively, as follows.
Given $\langle f_{\beta}\rangle_{\beta<\alpha}$, where
$\alpha<\kappa$,
and $\gamma<\kappa$, define $g_{\alpha\gamma}\in P$ by setting

$$\eqalign{g_{\alpha\gamma}(\zeta)
&=\sup\{f_{\beta}(\zeta):\beta\in C_{\gamma}\cap\alpha\}+1
\text{ if this is less than }\theta_{\zeta},\cr
&=0\text{ otherwise}.\cr}$$

\noindent Now choose $f_{\alpha}\in P$ such that

\Centerline{$f_{\beta}\le_{\Cal F}f_{\alpha}\enskip\forall\enskip
\beta<\alpha$,
\quad$g_{\alpha\gamma}\le_{\Cal F}f_{\alpha}\enskip\forall\enskip
\gamma<\kappa$;}

\noindent this is possible because $\kappa<\cf(P|\Cal F)$.
Observe that if $\alpha=\beta+1$ then $C_{\alpha}=\{\beta\}$ so that
$g_{\alpha\alpha}=f_{\beta}+1$ and
$f_{\alpha}\not\le_{\Cal F}f_{\beta}$.   Continue.

\wheader{5A2C}{6}{2}{2}{36pt}

{\bf (c)} Suppose that for each $\zeta<\lambda$ we are given a set
$S_{\zeta}\subseteq \theta_{\zeta}$ with $\#(S_{\zeta})\le\lambda$.
Then there is an $\alpha<\kappa$ such that

\Centerline{for every $h\in\prod_{\zeta<\lambda}S_{\zeta}$,
if $f_{\alpha}\le_{\Cal F}h$ then
$f_{\beta}\le_{\Cal F}h$ for every $\beta<\kappa$.}

\noindent\Prf\Quer\ If not, then (because $\kappa$ is regular) we can
find a family $\langle h_{\xi}\rangle_{\xi<\kappa}$ in
$\prod_{\zeta<\lambda}S_{\zeta}$ and a strictly increasing family
$\langle\phi(\xi)\rangle_{\xi<\kappa}$ in $\kappa$ such that

\Centerline{$f_{\phi(\xi)}\le_{\Cal F}h_{\xi}
\le_{\Cal F}f_{\phi(\xi+1)}$ for all $\xi<\kappa$,}

\Centerline{$\phi(\xi)=\sup_{\eta<\xi}\phi(\eta)$ for limit ordinals
$\xi<\kappa$.}

\noindent Set

\Centerline{$C=\{\xi:\xi<\kappa,\,\phi(\xi)=\xi\}$,}

\noindent so that $C$ is a closed cofinal set in $\kappa$.   Let
$\alpha\in C$ be such that $\alpha=\sup(C\cap\alpha)$ and
$\cf\alpha=\lambda^+$.
Then (because $\lambda^+\ge\omega_1$) $C\cap C_{\alpha}$ is cofinal
with $\alpha$.

For $\beta\in C\cap C_{\alpha}$ and $\zeta<\lambda$ we have

\Centerline{$\#(C_{\alpha}\cap\beta)
\le\text{otp}(C_{\alpha}\cap\beta)<\text{otp}(C_{\alpha})
=\lambda^+\le \theta_{\zeta}$,}

\noindent so

\Centerline{$\theta_{\zeta}
>\sup_{\xi\in C_{\alpha}\cap\beta}f_{\xi}(\zeta)+1
=g_{\beta\alpha}(\zeta)$.}

\noindent Now

\Centerline{$g_{\beta\alpha}\le_{\Cal F}f_{\beta}
=f_{\phi(\beta)}\le_{\Cal F}h_{\beta}\le_{\Cal F}f_{\phi(\beta+1)}
\le_{\Cal F}f_{\beta'}$,}

\noindent where $\beta'$ is the next member of $C\cap C_{\alpha}$
greater than $\beta$.
So there is a $\zeta_{\beta}<\lambda$ such that

\Centerline{$g_{\beta\alpha}(\zeta_{\beta})
\le h_{\beta}(\zeta_{\beta})\le f_{\beta'}(\zeta_{\beta})$.}

\noindent Because $\lambda<\cf\alpha$ there is a $\zeta<\lambda$
such that

\Centerline{$B=\{\beta:\beta\in C\cap C_{\alpha},\,
\zeta_{\beta}=\zeta\}$}

\noindent is cofinal with $\alpha$.   But now observe that if
$\beta$, $\gamma\in B$ and $\beta'<\gamma$ then
$\beta'\in C\cap C_{\alpha}\cap\gamma$ so

\Centerline{$h_{\beta}(\zeta)\le f_{\beta'}(\zeta)<g_{\gamma\alpha}
(\zeta)\le h_{\gamma}(\zeta).$}

\noindent It follows that

\Centerline{$\lambda^+=\#(B)
=\#(\{h_{\beta}(\zeta):\beta\in B\})\le\#(S_{\zeta})
\le\lambda$,}

\noindent which is absurd.\ \Bang\Qed

\medskip

{\bf (d)} Consequently
$E=\{f_{\alpha}^{\ssbullet}:\alpha<\kappa\}$
has a least upper bound in $P|\Cal F$.   \Prf\Quer\
If not, choose a family
$\langle h_{\xi}\rangle_{\xi<\lambda^+}$ in $P$ inductively,
as follows.   Because $\kappa<\cf(P|\Cal F)$, there is an
$h_0\in P$ such that
$f_{\alpha}\le_{\Cal F}h_0$ for every $\alpha<\kappa$.   Given $h_{\xi}$
such that $h_{\xi}^{\bullet}=\pi_{\Cal F}(h_{\xi})$ is an upper bound
for $E$ in $P|\Cal F$, 
then $h_{\xi}^{\bullet}$ cannot be the least upper bound of
$E$, so there is an $h_{\xi+1}\in P$ such that $h_{\xi+1}^{\bullet}$
is an upper bound of $E$ strictly less than $h_{\xi}^{\bullet}$.
For non-zero limit ordinals $\xi<\lambda^+$, set

\Centerline{$S_{\xi\zeta}=\{h_{\eta}(\zeta):\eta<\xi\}
\subseteq\theta_{\zeta}$}

\noindent for each $\zeta<\lambda$.   By (c) above, there is an
$\alpha_{\xi}<\kappa$ such that

\Centerline{for every $h\in\prod_{\zeta<\lambda}S_{\xi\zeta}$
either $f_{\alpha_{\xi}}\not\le_{\Cal F}h$ or
$f_{\alpha}\le_{\Cal F}h\enskip\forall\enskip\alpha<\kappa$.}

\noindent Set

\Centerline{$h_{\xi}(\zeta)=\min(\{\eta:\eta\in S_{\xi\zeta},\,
f_{\alpha_{\xi}}(\zeta)\le\eta\}\cup\{h_0(\zeta)\})\in S_{\xi\zeta}$}

\noindent for each $\zeta<\lambda$.   Then
$f_{\alpha_{\xi}}\le_{\Cal F}h_{\xi}$ (because
$f_{\alpha_{\xi}}(\zeta)\le h_{\xi}(\zeta)$ whenever
$f_{\alpha_{\xi}}(\zeta)\le h_0(\zeta)$) and
$h_{\xi}\in\prod_{\zeta<\lambda}S_{\xi\zeta}$, so
$f_{\alpha}\le_{\Cal F}h_{\xi}$ for every $\alpha<\kappa$ and
$h_{\xi}^{\bullet}$ is an upper bound for $E$.   Also, if
$\eta<\xi$, then $h_{\xi}(\zeta)\le h_{\eta}(\zeta)$ whenever
$f_{\alpha_{\xi}}(\zeta)\le h_{\eta}(\zeta)$, so $h_{\xi}\le_{\Cal F}
h_{\eta}$.   Continue.

Having got the family $\langle h_{\xi}\rangle_{\xi<\lambda^+}$, set

\Centerline{$S_{\zeta}=\bigcup_{\xi<\lambda^+}S_{\xi\zeta}
=\{h_{\xi}(\zeta):\xi<\lambda^+\}\subseteq\theta_{\zeta}$}

\noindent for each $\zeta<\lambda$.   For each $\alpha<\kappa$,
$\zeta<\lambda$ set

\Centerline{$g_{\alpha}(\zeta)=\min(\{\eta:f_{\alpha}(\zeta)
\le\eta\in S_{\zeta}\}\cup\{h_0(\zeta)\})\in S_{\zeta}$.}

\noindent Then, by the same arguments as above,

\Centerline{$f_{\alpha}\le_{\Cal F}g_{\alpha}
\le_{\Cal F}h_{\xi}$ for every $\alpha<\kappa$,
$\xi<\lambda^+$.}

For each $\alpha<\kappa$ there is a non-zero limit ordinal $\xi<\lambda^+$
such that $g_{\alpha}(\zeta)\in S_{\xi\zeta}$ for every
$\zeta<\lambda$, because $\ofamily{\xi}{\lambda^+}{S_{\xi\zeta}}$ is
non-decreasing for each $\zeta$.
Because $\lambda^+<\kappa$ there is a non-zero
limit ordinal $\xi<\lambda^+$ such that

\Centerline{$A
=\{\alpha:g_{\alpha}(\zeta)\in S_{\xi\zeta}
   \enskip\forall\enskip\zeta<\lambda\}$}

\noindent is cofinal with $\kappa$.   In particular, there is an
$\alpha\in A$ such that $\alpha\ge\alpha_{\xi}$.   In this case

\Centerline{$f_{\alpha_{\xi}}\le_{\Cal F}f_{\alpha}
\le_{\Cal F}g_{\alpha}\le_{\Cal F}h_{\xi+1}
\le_{\Cal F}h_{\xi}\not\le_{\Cal F}h_{\xi+1}$,}

\noindent so there is a $\zeta<\lambda$ such that

\Centerline{$f_{\alpha_{\xi}}(\zeta)\le f_{\alpha}(\zeta)
\le g_{\alpha}(\zeta)\le h_{\xi+1}(\zeta)<h_{\xi}(\zeta)$.}

\noindent But now observe that

\Centerline{$f_{\alpha_{\xi}}(\zeta)\le g_{\alpha}(\zeta)
\in S_{\xi\zeta}$}

\noindent so $h_{\xi}(\zeta)\le g_{\alpha}(\zeta)<h_{\xi}(\zeta)$,
which is absurd.\ \Bang\Qed

\medskip

{\bf (e)} Let $g\in P$ be such that $g^{\bullet}=\sup E$ in
$P|\Cal F$
and $g(\zeta)>0$ for every $\zeta<\lambda$.  For each $\zeta<\lambda$
set $\hat\theta_{\zeta}=\cf(g(\zeta))<\theta_{\zeta}$
and choose a cofinal set $D_{\zeta}\subseteq g(\zeta)$ of order type
$\hat\theta_{\zeta}$.   For $\alpha<\kappa$ and $\zeta<\lambda$ set

\Centerline{$\hat g_{\alpha}(\zeta)
=\min\{\eta:f_{\alpha}(\zeta)\le\eta\in D_{\zeta}\}$}

\noindent if $f_{\alpha}(\zeta)<g(\zeta)$,  $\min D_{\zeta}$
otherwise.   Then $\hat g_{\alpha}\le_{\Cal F}\hat g_{\beta}$
whenever $\alpha\le\beta<\kappa$.
Also if $h\in Q=\prod_{\zeta<\lambda}D_{\zeta}$ then
$h^{\bullet}<g^{\bullet}$ so there is an $\alpha<\kappa$ such that
$h^{\bullet}<f_{\alpha}^{\bullet}\le\hat g_{\alpha}^{\bullet}$.
Thus $\{\hat g_{\alpha}^{\bullet}:\alpha<\kappa\}$ is a cofinal subset of
$\{h^{\bullet}:h\in Q\}$.

\medskip

{\bf (f)}
Because each $D_{\zeta}$ is order-isomorphic to $\hat\theta_{\zeta}$,
we can identify $Q$ with
$\hat P=\prod_{\zeta<\lambda}\hat\theta_{\zeta}$.
Now if $\alpha<\kappa$, then
there is a $\beta<\kappa$ such that $\hat g_{\alpha}\le_{\Cal F}f_{\beta}$
while $f_{\beta+1}\not\le_{\Cal F}f_{\beta}$ (see (b) above), so
$\hat g_{\beta+1}\not\le_{\Cal F}g_{\alpha}$.   Because $\kappa$ is regular, 

\Centerline{$\cf(\hat P|\Cal F)=\cf(\{h^{\bullet}:h\in Q\})
=\cf(\{\hat g_{\alpha}^{\bullet}:\alpha<\kappa\})=\kappa$.}

\medskip

{\bf (g)} It may be that some of the $\hat\theta_{\zeta}$ are
less than or equal to $\lambda$.   But taking
$I=\{\zeta:\hat\theta_{\zeta}\le\lambda\}$, we have $I\notin\Cal F$.
\Prf\Quer\  If $I\in\Cal F$,
then for each $\zeta\in I$ set $\hat S_{\zeta}=D_{\zeta}$ and for
$\zeta\in\lambda\setminus I$ set $\hat S_{\zeta}=\{0\}$.   
Then $\#(\hat S_{\zeta})\le\lambda$ for every $\zeta<\lambda$.   By (c), there
is an $\alpha<\kappa$ such that

\Centerline{for every $h\in\prod_{\zeta<\lambda}\hat S_{\zeta}$,
if $f_{\alpha}\le_{\Cal F}h$ then
$f_{\beta}\le_{\Cal F}h\enskip\forall\enskip\beta<\kappa$.}

\noindent But as $f_{\alpha+1}\le_{\Cal F}g$, and $I\in\Cal F$, there
must be an $h\in\prod_{\zeta<\lambda}\hat S_{\zeta}$ such that
$f_{\alpha}\le_{\Cal F}h$, and now $g\le_{\Cal F}h$ because
$g^{\bullet}$ is the
least upper bound of $E$;  but $h(\zeta)<g(\zeta)$ for every
$\zeta\in I$, so this is impossible.\ \Bang\Qed

So $\{\zeta:\hat\theta_{\zeta}>\lambda\}\in\Cal F$.   But this means
that if we set

$$\eqalign{\theta'_{\zeta}
&=\hat\theta_{\zeta}\text{ when }\hat\theta_{\zeta}>\lambda,\cr
&=\theta_{\zeta}\text{ when }\hat\theta_{\zeta}\le\lambda\cr}$$

\noindent and $P'=\prod_{\zeta<\lambda}\theta'_{\zeta}$,
then $P'|\Cal F\cong\hat P|\Cal F$ so $\cf(P'|\Cal F)=\kappa$, as
required.
}%end of proof of 5A2C
%5{}42D

\leader{5A2D}{Definitions (a)} Let $\alpha$, $\beta$, $\gamma$ and
$\delta$ be cardinals.   \cmmnt{Following {\smc Shelah 92} and
{\smc Shelah 94},} I write

\Centerline{$\covSh(\alpha,\beta,\gamma,\delta)$}

\noindent for the least cardinal of any family
$\Cal E\subseteq[\alpha]^{<\beta}$ such that for every
$A\in[\alpha]^{<\gamma}$ there is a $\Cal D\in[\Cal E]^{<\delta}$ with
$A\subseteq\bigcup\Cal D$.
In the trivial cases in which there is no such family
$\Cal E$ I write $\covSh(\alpha,\beta,\gamma,\delta)=\infty$.

\medskip

{\bf (b)} For cardinals $\alpha$, $\gamma$
write $\Theta(\alpha,\gamma)$ for the supremum of all cofinalities

\Centerline{$\cf(\prod_{\zeta<\lambda}\theta_{\zeta})$}

\noindent for families $\langle\theta_{\zeta}\rangle_{\zeta<\lambda}$
such that $\lambda<\gamma$ is a cardinal, every
$\theta_{\zeta}$ is a regular infinite cardinal and
$\lambda<\theta_{\zeta}<\alpha$ for every $\zeta<\lambda$.
\cmmnt{(This carries some of the same information as the cardinal
pp$_{\kappa}(\alpha)$ of {\smc Shelah 94}, p.\ 41.)}
%5{}23M 5{}42D

\medskip

\wheader{5A2D}{0}{0}{0}{60pt}
\noindent{\bf Remarks\cmmnt{ (i)}}\cmmnt{ Immediately from the definitions,
we see that}

\Centerline{$\covSh(\alpha,\beta',\gamma,\delta')
\le\covSh(\alpha',\beta,\gamma',\delta)$,
\quad$\Theta(\alpha,\gamma)\le\Theta(\alpha',\gamma')$}

\noindent whenever $\alpha\le\alpha'$, $\beta\le\beta'$,
$\gamma\le\gamma'$ and $\delta\le\delta'$.

\cmmnt{\medskip

{\bf (ii)} The definition of $\Theta$ demands a moment's thought in
trivial cases.   If $\gamma=0$ there is no $\lambda<\gamma$, so we are
taking the supremum of an empty set of cofinalities, and
$\Theta(\alpha,0)=0$ for every $\alpha$.   If $\gamma>0$
then we are allowed $\lambda=0$ and
$\prod_{\zeta<\lambda}\theta_{\zeta}=\{\emptyset\}$, so
$\Theta(\alpha,\gamma)\ge 1$ for every $\alpha$.   If $\gamma>1$ we are
allowed $\lambda=1$, so $\Theta(\alpha^+,\gamma)\ge\alpha$ for every
infinite $\alpha$.
}%end of comment

%if $\gamma>1$ and $\alpha$ an uncountable limit cardinal, then
%$\Theta(\alpha,\gamma)\ge\alpha$ (5{}42Dd)

\vleader{48pt}{5A2E}{Lemma} %RVMC 7Q
Let $\alpha$, $\beta$, $\gamma$, $\gamma'$ and
$\delta$ be cardinals.

(a) If $\gamma\le\gamma'\le\beta$ and $\delta\ge 2$ then

\Centerline{$\covSh(\alpha,\beta,\gamma,\delta)
\le\cf([\alpha]^{<\gamma'})\le\#([\alpha]^{<\gamma'})$.}

(b) If either $\omega\le\gamma\le\cf\alpha$ or
$\omega\le\cf\alpha<\cf\delta$ then

\Centerline{$\covSh(\alpha,\beta,\gamma,\delta)
\le\max(\alpha,
  \sup_{\theta<\alpha}\covSh(\theta,\beta,\gamma,\delta))$.}

\proof{{\bf (a)}
If $\Cal E$ is a cofinal subset of $[\alpha]^{<\gamma'}$ of
cardinal $\cf([\alpha]^{<\gamma'})$, then $\Cal E$ witnesses that
$\covSh(\alpha,\gamma',\gamma',\delta)\le\cf([\alpha]^{<\gamma'})$.   Now

\Centerline{$\covSh(\alpha,\beta,\gamma,\delta)
\le\covSh(\alpha,\gamma',\gamma',\delta)\le\cf([\alpha]^{<\gamma'})$.}

\medskip

{\bf (b)} Set $\kappa
=\max(\alpha,\sup_{\theta<\alpha}\covSh(\theta,\beta,\gamma,\delta))$
and $\lambda=\cf\alpha$.   Let $\ofamily{\xi}{\lambda}{\zeta_{\xi}}$
enumerate a cofinal subset of $\alpha$.   For each $\xi<\lambda$, let
$\Cal E_{\xi}\subseteq[\zeta_{\xi}]^{<\beta}$ be a set with cardinal at most
$\kappa$ such that for every $A\in[\zeta_{\xi}]^{<\gamma}$ there is a
$\Cal D\in[\Cal E_{\xi}]^{<\delta}$ such that
$A\subseteq\bigcup\Cal D$.   Set
$\Cal E=\bigcup_{\xi<\lambda}\Cal E_{\xi}$, so that
$\Cal E\subseteq[\alpha]^{<\beta}$ has cardinal at most $\kappa$.
Take $A\in[\alpha]^{<\gamma}$.

If $\omega\le\gamma\le\lambda$ then $\sup A<\alpha$ and there is a
$\xi<\lambda$ such that $A\subseteq\zeta_{\xi}$.   Now there is a
$\Cal D\in[\Cal E_{\xi}]^{<\delta}\subseteq[\Cal E]^{<\delta}$ such
that $A\subseteq\bigcup\Cal D$.

If $\omega\le\lambda<\cf\delta$, then for each $\xi<\lambda$ there is
a $\Cal D_{\xi}\in[\Cal E_{\xi}]^{<\delta}$ such that
$A\cap\zeta_{\xi}\subseteq\bigcup\Cal D_{\xi}$.   Set
$\Cal D=\bigcup_{\xi<\lambda}\Cal D_{\xi}$;  because
$\lambda<\cf\delta$, $\Cal D\in[\Cal E]^{<\delta}$, while

\Centerline{$A=\bigcup_{\xi<\lambda}A\cap\zeta_{\xi}
\subseteq\bigcup\Cal D$.}

Thus in either case $\Cal E$ witnesses that
$\covSh(\alpha,\beta,\gamma,\delta)\le\kappa$.
}%end of proof of 5A2E
%5{}41S 5{}42D 5{}42E

\leader{5A2F}{Lemma} Let $\alpha$, $\gamma$ be cardinals.   If
$\alpha\le 2^{\gamma}$, then $\Theta(\alpha,\gamma)\le 2^{\gamma}$.

\proof{ If $\gamma\le\omega$ then
$\Theta(\alpha,\gamma)\le\max(1,\alpha)\le 2^{\gamma}$.
If $\gamma>\omega$,
$\lambda<\gamma$ and $\theta_{\zeta}<2^{\gamma}$ for every
$\zeta<\lambda$, then

\Centerline{$\cf(\prod_{\zeta<\lambda}\theta_{\zeta})
\le\#(\prod_{\zeta<\lambda}\theta_{\zeta})
\le(2^{\gamma})^{\gamma}=2^{\gamma}$.}
}%end of proof of 5A2F

\leader{5A2G}{Theorem} %RVCM A1K
For any cardinals $\alpha$ and $\gamma$,

\Centerline{$\covSh(\alpha,\gamma,\gamma,\omega_1)
\le\max(\omega,\alpha,\Theta(\alpha,\gamma))$.}

\proof{{\bf (a)} To begin with (down to the end of (f) below) let us
suppose that we have
$\alpha\ge\gamma=\gamma_0^+>\cf\alpha>\omega$, and set
$\kappa=\max(\alpha,\Theta(\alpha,\gamma))$.

Take a family $\Cal E\subseteq[\alpha]^{\le\gamma_0}$ such that

\inset{(i) $\Cal E$ contains all singleton subsets of $\alpha$;

(ii) $\Cal E$ contains a cofinal subset of $\alpha$;

(iii) If $E\in\Cal E$ then $\{\xi:\xi+1\in E\}\in\Cal E$;

(iv) if $E\in\Cal E$ then
there is an $F\in\Cal E$ such that $\sup(F\cap\xi)=\xi$ whenever
$\xi\in E$ and $\omega\le\cf\xi\le\gamma_0$;

(v) if $E\in\Cal E$ then $\{\xi:\xi\in E,\,
\cf\xi>\gamma_0\}\in\Cal E$;

(vi) if $E\in\Cal E$ and $\cf(\prod_{\eta\in E}\eta)\le\kappa$,
then $\{g:g\in\prod_{\eta\in E}\eta,\,g[E]\in\Cal E\}$ is cofinal with
$\prod_{\eta\in E}\eta$;

(vii) $\#(\Cal E)\le\kappa$.}

\noindent To see that this can be done, observe that whenever
$E\in[\alpha]^{\le\gamma_0}$ there is an $F\in[\alpha]^{\le\gamma_0}$
such that $\sup(F\cap\xi)=\xi$ whenever $\xi\in E$ and
$\omega\le\cf\xi\le\gamma_0$;  thus condition (iv) can be
achieved, like conditions
(iii) and (v), by ensuring that $\Cal E$ is closed under suitable
functions
from $[\alpha]^{\le\gamma_0}$ to itself;  while condition (vi)
requires that for each $E\in\Cal E$ we have  an appropriate
family with cardinal at most $\kappa$ included in $\Cal E$.

Write $\Cal J$ for the $\sigma$-ideal of $\Cal P\alpha$ generated by
$\Cal E$.   Note that if $A\in\Cal J$ then $\{\xi:\xi+1\in A\}$ belongs to
$\Cal J$, by (iii).

\medskip

{\bf (b)} \Quer\ If
$\covSh(\alpha,\gamma,\gamma,\omega_1)>\kappa$, there must be a set
in $[\alpha]^{\le\gamma_0}$
not covered by any sequence from $\Cal E$, that is, not belonging to
$\Cal J$;  that is, there is a
function $f:\gamma_0\to\alpha$ such that $f[\gamma_0]\notin\Cal J$.
Accordingly $\Cal I=\{f^{-1}[E]:E\in\Cal J\}$
is a proper $\sigma$-ideal of $\Cal P\gamma_0$.
By condition (a-i), $\Cal I$ contains all singletons in
$\Cal P\gamma_0$.

Let $H$ be the set of all functions $h:\gamma_0\to\alpha$ such that
$f(\xi)\le h(\xi)$ for every $\xi<\gamma_0$ and
$h[\gamma_0]\in\Cal J$.   Because
$\Cal E$ contains a cofinal set $C\subseteq\alpha$ (condition (a-ii)),
we can find an
$h\in H$;  just take $h:\gamma_0\to C$ such that $f(\xi)\le h(\xi)$
for every $\xi$.

\medskip

{\bf (c)} Because $\Cal I$ is a proper
$\sigma$-ideal, there cannot be any sequence
$\langle h_n\rangle_{n\in\Bbb N}$ in $H$ such that
$\{\xi:h_{n+1}(\xi)\ge h_n(\xi)\}\in\Cal I$ for every $n\in\Bbb N$.
Consequently there is an $h^*\in H$ such that

\Centerline{$\{\xi:h(\xi)\ge h^*(\xi)\}\notin\Cal I$ for every $h\in H$.}

\noindent We know that $h^*[\gamma_0]\in\Cal J$;  let
$\langle E_n\rangle_{n\in\Bbb N}$ be a sequence in $\Cal E$ covering
$h^*[\gamma_0]$.
For $\xi<\gamma_0$ write $\theta_{\xi}=\cf(h^*(\xi))$, so that
each $\theta_{\xi}$ is 0 or
1 or a regular infinite cardinal less than $\alpha$.   Set

\Centerline{$I=\{\xi:\xi<\gamma_0$, $f(\xi)=h^*(\xi)\}$,}

\Centerline{$I'=\{\xi:\xi<\gamma_0,\,f(\xi)<h^*(\xi),\,
\theta_{\xi}=1\}$,}

\Centerline{$I_n=\{\xi:\xi<\gamma_0$,
$f(\xi)<h^*(\xi)$, $\omega\le\theta_{\xi}\le\gamma_0$,
$h^*(\xi)\in E_n\}$ for $n\in\Bbb N$,}

\Centerline{$J_n=\{\xi:\xi<\gamma_0$, $f(\xi)<h^*(\xi)$,
$\gamma_0<\theta_{\xi}$, $h^*(\xi)\in E_n\}$ for $n\in\Bbb N$.}

\noindent Note that if $\theta_{\xi}=0$ then $h^*(\xi)=0=f(\xi)$, so
$I$, $I'$, $\sequencen{I_n}$ and $\sequencen{J_n}$ constitute a cover
of $\gamma_0$.

\medskip

{\bf (d)} For each $n\in\Bbb N$ set
$G_n=\{\eta:\eta\in E_n$, $\cf\eta>\gamma_0\}\in\Cal E$;  note that
$h^*(\xi)\in G_n$ for $\xi\in J_n$.   Then
$\cf(\prod_{\eta\in G_n}\eta)\le\Theta(\alpha,\gamma)$.   \Prf\
For $\eta\in G_n$ set $\theta'_{\eta}=\cf\eta$;  then $\theta'_{\eta}$
is a regular cardinal and $\#(G_n)\le\gamma_0<\theta'_{\eta}<\alpha$
for each $\eta\in G_n$.   If for each $\eta\in G_n$ we choose a
cofinal set $C_{\eta}\subseteq\eta$ of order type $\theta'_{\eta}$,
then

\Centerline{$\cf(\prod_{\eta\in G_n}\eta)
=\cf(\prod_{\eta\in G_n}C_{\eta})
=\cf(\prod_{\eta\in G_n}\theta'_{\eta})
\le\Theta(\alpha,\gamma)$}

\noindent by the definition of $\Theta(\alpha,\gamma)$.\ \Qed

Consequently, by (a-vi),

\Centerline{$\{g:g\in\prod_{\eta\in G_n}\eta,\,g[G_n]\in\Cal E\}$}

\noindent is cofinal with $\prod_{\eta\in G_n}\eta$.

\medskip

{\bf (e)} Define $h:\gamma_0\to\alpha$ as follows.

\quad(i) If $\xi\in I$ set $h(\xi)=h^*(\xi)$.

\quad(ii) If $\xi\in I'$ let $h(\xi)$ be the predecessor of
$h^*(\xi)$.

\quad(iii) For each $n\in\Bbb N$ take $F_n\in\Cal E$
such that $\eta=\sup(F_n\cap\eta)$ whenever $\eta\in E_n$ and
$\omega\le\cf\eta\le\gamma_0$.   If
$\xi\in I_n\setminus\bigcup_{m<n}I_m$, take $h(\xi)\in F_n$ such that
$f(\xi)\le h(\xi)<h^*(\xi)$.

\quad(iv) For each $n\in\Bbb N$ and $\eta\in G_n$ set

\Centerline{$g^*(\eta)=\sup\{f(\xi):\xi<\gamma_0,\,f(\xi)<h^*(\xi)=\eta\}$.}

\noindent Then $g^*(\eta)<\eta$, because $\gamma_0<\cf\eta$.   By
(d), there is a $g_n\in\prod_{\eta\in G_n}\eta$ such that
$g_n[G_n]\in\Cal E$
and $g^*(\eta)\le g_n(\eta)$ for every $\eta\in G_n$.   So for
$\xi\in J_n\setminus\bigcup_{m<n}J_m$ we may set
$h(\xi)=g_n(h^*(\xi))$ and see that $h^*(\xi)\in G_n$ and

\Centerline{$f(\xi)\le g^*h^*(\xi)\le g_nh^*(\xi)=h(\xi)<h^*(\xi)$,}

\noindent while $h(\xi)\in g_n[G_n]$.

\medskip

{\bf (f)} Now we see that

\Centerline{$h[\gamma_0]\subseteq h^*[\gamma_0]\cup\{\eta:\eta+1\in
h^*[\gamma_0]\}\cup\bigcup_{n\in\Bbb N}F_n\cup\bigcup_{n\in\Bbb N}
g_n[G_n]\in\Cal J$,}

\noindent while $f(\xi)\le h(\xi)$ for every $\xi<\gamma_0$, so
$h\in H$.   Consequently

\Centerline{$I=\{\xi:h(\xi)\ge h^*(\xi)\}\notin\Cal I$.}

\noindent But also

\Centerline{$f[I]\subseteq h^*[\gamma_0]\in\Cal J$,}

\noindent so $I\in\Cal I$, which is absurd.  \Bang

\medskip

{\bf (g)} Thus the special case described in (a) is dealt with, and we
may return to the general case.   I proceed by induction on $\alpha$
for fixed $\gamma$.

\medskip

\quad{\bf (i)} To start the induction, observe that
if either $\alpha\le\omega$ or $\gamma\le\omega$ or $\alpha<\gamma$,
then

\Centerline{$\covSh(\alpha,\gamma,\gamma,\omega_1)
\le\cf([\alpha]^{<\gamma})\le\max(\alpha,\omega)$.}

\medskip

\quad{\bf (ii)} For the inductive step to $\alpha$ when {\it either}
$\cf\alpha\ge\gamma\ge\omega$ {\it or} $\cf\alpha=\omega$,
5A2Eb tells us that

$$\eqalign{\covSh(\alpha,\gamma,\gamma,\omega_1)
&\le\max(\alpha,
  \sup_{\alpha'<\alpha}\covSh(\alpha',\gamma,\gamma,\omega_1))\cr
&\le\max(\omega,\alpha,\sup_{\alpha'<\alpha}\Theta(\alpha',\gamma))
\le\max(\omega,\alpha,\Theta(\alpha,\gamma))\cr}$$

\noindent by the inductive hypothesis.

\medskip

\quad{\bf (iii)} For the inductive step to $\alpha$ when
$\omega<\cf\alpha<\gamma\le\alpha$, observe that

\Centerline{$[\alpha]^{<\gamma}
=\bigcup_{\delta<\gamma}[\alpha]^{\le\delta}$.}

\noindent For each cardinal $\delta<\gamma$ we have a set
$\Cal E_{\delta}\subseteq [\alpha]^{\le\delta}$ such that
$\#(\Cal E_{\delta})
\le\covSh(\alpha,\delta^+,\delta^+,\omega_1)$ and every member of
$[\alpha]^{\le\delta}$
can be covered by a sequence from $\Cal E_{\delta}$.   Set
$\Cal E=\bigcup_{\cf\alpha\le\delta<\gamma}\Cal E_{\delta}$;  then
$\Cal E\subseteq[\alpha]^{<\gamma}$ and every member of
$[\alpha]^{<\gamma}$ can be covered by a sequence from $\Cal E$.
So

$$\eqalignno{\covSh(\alpha,\gamma,\gamma,\omega_1)
&\le\#(\Cal E)
\le\max(\gamma,\sup_{\cf\alpha\le\delta<\gamma}
  \covSh(\alpha,\delta^+,\delta^+,\omega_1))\cr
&\le\max(\gamma,\alpha,
  \sup_{\cf\alpha\le\delta<\gamma}\Theta(\alpha,\delta^+))\cr
\displaycause{by (a)-(f) above}
&\le\max(\alpha,\Theta(\alpha,\gamma)).\cr}$$

This completes the proof.

\medskip

\noindent{\bf Remark} This is taken from {\smc Shelah 94}, Theorem
II.5.4,
where a stronger result is proved, giving an exact description of
many of the numbers
$\covSh(\alpha,\beta,\gamma,\delta)$ in terms of cofinalities of
reduced products $\prod_{\zeta<\lambda}\theta_{\zeta}|\Cal F$.
}%end of proof of 5A2G
%5{}42D 5{}42E

\leader{5A2H}{Lemma} %RVMC A1L
Let $\gamma$ be an infinite regular cardinal and
$\alpha\ge\Theta(\gamma,\gamma)$ a cardinal.   Then
$\Theta(\Theta(\alpha,\gamma),\gamma)\le\Theta(\alpha,\gamma)$.

\proof{{\bf (a)} The case $\gamma=\omega$ is elementary, since
$\Theta(\alpha,\omega)\le\alpha$ for every cardinal $\alpha$.   So we may
suppose that $\gamma$ is uncountable.   

\medskip

{\bf (b)} \Quer\ Suppose, if possible, that
$\Theta(\alpha,\gamma)<\Theta(\Theta(\alpha,\gamma),\gamma)$.

\medskip

\quad{\bf (i)} $\Theta(\gamma,\gamma)\le\alpha<\Theta(\alpha,\gamma)$ so
$\gamma<\alpha$ and $\Theta(\alpha,\gamma)$ is infinite.

\medskip

\quad{\bf (ii)} There is a family 
$\langle\theta_{\zeta}\rangle_{\zeta<\lambda}$
of regular infinite cardinals such that $\lambda<\gamma$,
$\lambda<\theta_{\zeta}<\Theta(\alpha,\gamma)$ for every
$\zeta<\lambda$
and $\cf(\prod_{\zeta<\lambda}\theta_{\zeta})>\Theta(\alpha,\gamma)$.
As $\Theta(\alpha,\gamma)\ne 0$, $\lambda\ne 0$.
By 5A2Bc, there is an ultrafilter $\Cal F$ on $\lambda$ such that
$\cf(\prod_{\zeta<\lambda}\theta_{\zeta}|\Cal F)>\Theta(\alpha,\gamma)$.

\medskip

\quad{\bf (iii)} Set $L=\{\zeta:\zeta<\lambda,\,\theta_{\zeta}<\alpha\}$;  as

\Centerline{$\#(L)\le\lambda<\min(\gamma,\theta_{\zeta})$,
\quad$\theta_{\zeta}<\alpha$}

\noindent for every $\zeta\in L$,

\Centerline{$\cf(\prod_{\zeta\in L}\theta_{\zeta})\le\Theta(\alpha,\gamma)
<\cf(\prod_{\zeta<\lambda}\theta_{\zeta}|\Cal F)$}  

\noindent and $L\notin\Cal F$ (5A2Ab).
Set $M=\lambda\setminus L\in\Cal F$.   Let $\Cal F'=\Cal F\cap\Cal PM$ 
be the induced ultrafilter on $M$, so that
$\prod_{\zeta<\lambda}\theta_{\zeta}|\Cal F
\cong\prod_{\zeta\in M}\theta_{\zeta}|\Cal F'$, and
$\cf(\prod_{\zeta\in M}\theta_{\zeta}|\Cal F')>\Theta(\alpha,\gamma)$.   

\medskip

\quad{\bf (iv)} For each $\zeta\in M$, we have
$\theta_{\zeta}<\Theta(\alpha,\gamma)$, so there must be a family
$\langle\theta_{\zeta\eta}\rangle_{\eta<\lambda_{\zeta}}$ of regular
infinite cardinals with $\lambda_{\zeta}<\gamma$,
$\lambda_{\zeta}<\theta_{\zeta\eta}<\alpha$ for every
$\eta<\lambda_{\zeta}$ and
$\theta_{\zeta}\le\cf(\prod_{\eta<\lambda_{\zeta}}\theta_{\zeta\eta})$.
Again by 5A2Bc, %RVMC A1Ic
there is an ultrafilter $\Cal F_{\zeta}$ on $\lambda_{\zeta}$ such
that $\theta_{\zeta}
\le\cf(\prod_{\eta<\lambda_{\zeta}}\theta_{\zeta\eta}|\Cal F_{\zeta})$.
Because

\Centerline{$\lambda_{\zeta}<\gamma\le\alpha\le\theta_{\zeta}$,}

\noindent 5A2C tells us that there
is a family $\langle\theta'_{\zeta\eta}\rangle_{\eta<\lambda_{\zeta}}$
of regular infinite cardinals such that
$\lambda_{\zeta}<\theta'_{\zeta\eta}\le \theta_{\zeta\eta}$ for every
$\eta$ and $\theta_{\zeta}
=\cf(\prod_{\eta<\lambda_{\zeta}}\theta'_{\zeta\eta}|\Cal F_{\zeta})$.

\woddheader{5A2H}{4}{2}{2}{66pt}
{\bf (c)(i)} Set

\Centerline{$I=\{(\zeta,\eta):\zeta\in M,\,\eta<\lambda_{\zeta}\}$,}

\Centerline{$\Cal H
=\{H:H\subseteq I,\,
  \{\zeta:\{\eta:(\zeta,\eta)\in H\}\in\Cal F_{\zeta}\}
\in\Cal F'\}$,}

\Centerline{$P=\prod_{(\zeta,\eta)\in I}\theta'_{\zeta\eta}$.}

\noindent Then $\Cal H$ is an ultrafilter on $I$, and
$\cf(P|\Cal H)\ge\cf(\prod_{\zeta\in M}\theta_{\zeta}|\Cal F')$.
\Prf\ Let $F\subseteq P$ be a set with cardinal $\cf(P|\Cal H)$ such
that
$\{f^{\bullet}:f\in F\}$ is cofinal with $P|\Cal H$.   For
$f\in P$ and $\zeta\in M$, define
$f_{\zeta}\in\prod_{\eta<\lambda_{\zeta}}\theta'_{\zeta\eta}$
by setting $f_{\zeta}(\eta)=f(\zeta,\eta)$ for each
$\eta<\lambda_{\zeta}$, and let $f_{\zeta}^{\bullet}$ be the image of
$f_{\zeta}$ in
$\prod_{\eta<\lambda_{\zeta}}\theta'_{\zeta\eta}|\Cal F_{\zeta}$.
For each $\zeta\in M$ let
$\langle u_{\zeta\xi}\rangle_{\xi<\theta_{\zeta}}$ be a strictly
increasing cofinal family in the totally ordered set
$\prod_{\eta<\lambda_{\zeta}}\theta'_{\zeta\eta}|\Cal F_{\zeta}$.
Now, for $f\in F$, take a function
$g_{f}\in\prod_{\zeta\in M}\theta_{\zeta}$
such that $f_{\zeta}^{\bullet}\le u_{\zeta,g_f(\zeta)}$ for every
$\zeta\in M$.

If $g\in\prod_{\zeta\in M}\theta_{\zeta}$, then we can find an $h\in P$
such that $h_{\zeta}^{\bullet}=u_{\zeta,g(\zeta)}$ for each
$\zeta\in M$.   Let $f\in F$ be such that $h\le_{\Cal H}f$. Then

\Centerline{$\{\zeta:g(\zeta)\le g_f(\zeta)\}
\supseteq\{\zeta:h_{\zeta}^{\bullet}\le f_{\zeta}^{\bullet}\}
\in\Cal F'$,}

\noindent so $g\le_{\Cal F'}g_f$.
Accordingly $\{g_f:f\in F\}$ is cofinal with
$\prod_{\zeta\in M}\theta_{\zeta}|\Cal F'$ and 
$\cf(\prod_{\zeta\in M}\theta_{\zeta}|\Cal F')\le\#(F)=\cf(P|\Cal H)$, as
claimed.\ \Qed

\medskip

\quad{\bf (ii)} Note that as $\#(M)\le\lambda<\gamma$, $\lambda_{\zeta}<\gamma$
for every $\zeta\in M$ and $\gamma$ is uncountable and regular,

\Centerline{$\#(I)\le\max(\omega,\#(M),\sup_{\zeta\in M}\lambda_{\zeta})
<\gamma$.}

\medskip

{\bf (d)(i)} Putting (c-i) and (b-iii) together,
$\cf(P|\Cal H)>\Theta(\alpha,\gamma)$.   Set
$J=\{(\zeta,\eta):(\zeta,\eta)\in I,\,\theta'_{\zeta\eta}\ge\gamma\}$.
Since 

$$\eqalignno{\#(J)&
\le\#(I)<\gamma
\Displaycause{(c-ii) just above}
\le\theta'_{\zeta\eta}
\le\theta_{\zeta\eta}
<\alpha
\cr}$$

\noindent whenever $(\zeta,\eta)\in J$, 

\Centerline{$\cf(\prod_{(\zeta,\eta)\in J}\theta'_{\zeta\eta})
\le\Theta(\alpha,\gamma)<\cf(P|\Cal H)$}

\noindent and $J\notin\Cal H$ by 5A2Ab once more.   It follows that
$K=I\setminus J\in\Cal H$.   Set
$M'=\{\zeta:\zeta\in M,\,\{\eta:(\zeta,\eta)\in K\}\in\Cal F_{\zeta}\}
\in\Cal F'$.

\medskip

\quad{\bf (ii)} If $\zeta\in M'$, then $F=\{\eta:\eta<\lambda_{\zeta}$,
$\theta'_{\zeta\eta}<\gamma\}$ belongs to $\Cal F_{\zeta}$.   Now

$$\eqalignno{\theta_{\zeta}&
=\cf(\prod_{\eta<\lambda_{\zeta}}\theta'_{\zeta\eta}|\Cal F')
\le\cf(\prod_{\eta\in F}\theta'_{\zeta\eta})
\le\Theta(\gamma,\gamma)
\Displaycause{because $\#(F)\le\lambda_{\zeta}<\theta'_{\zeta\eta}<\gamma$
for every $\eta\in F$}
\le\alpha
\le\theta_{\zeta}\cr}$$

\noindent because $\zeta\in M$.   .
So in fact $\theta_{\zeta}=\alpha$ for $\zeta\in M'$ and we have

\Centerline{$\Theta(\alpha,\gamma)
<\cf(\prod_{\zeta\in M}\theta_{\zeta}|\Cal F')
\le\cf(\prod_{\zeta\in M'}\theta_{\zeta})
=\cf(\prod_{\zeta<\delta}\alpha)$,}

\noindent where $\delta=\#(M')$, while at the same time $\alpha$
is infinite and regular.

\medskip

\quad{\bf (iii)}
But if $\alpha$ is infinite and regular and $1\le\delta<\alpha$,
$\cf(\prod_{\zeta<\delta}\alpha)=\alpha$.
Accordingly $\Theta(\alpha,\gamma)<\alpha$;  
which contradicts (b-i) above.\ \Bang

This contradiction completes the proof.
}%end of proof of 5A2H

\leader{5A2I}{Lemma} %RVMC A1M
Let $\alpha$ and $\gamma$ be cardinals.   Set
$\delta=\sup_{\alpha'<\alpha}\Theta(\alpha',\gamma)$.

(a) If $\cf\alpha\ge\gamma$ then
$\Theta(\alpha,\gamma)\le\max(\alpha,\delta)$.

(b) If $\cf\alpha<\gamma$ then
$\Theta(\alpha,\gamma)\le\max(\alpha,\delta^{\cf\alpha})$, where
$\delta^{\cf\alpha}$ is the cardinal power.
%5{}42

\proof{ Let $\langle\theta_{\zeta}\rangle_{\zeta<\lambda}$
be a family of regular infinite
cardinals with $\lambda<\theta_{\zeta}<\alpha$
for each $\zeta$ and $\lambda<\gamma$.

\medskip

{\bf case 1} If $\alpha'=\sup_{\zeta<\lambda}\theta_{\zeta}$ is less than
$\alpha$, set

\Centerline{$I=\{\zeta:\zeta<\lambda,\,\theta_{\zeta}<\alpha'\}$,}

\Centerline{$J=\{\zeta:\zeta<\lambda,\,\theta_{\zeta}=\alpha'\}$.}

\noindent Then

\Centerline{$\cf(\prod_{\zeta\in I}\theta_{\zeta})
\le\Theta(\alpha',\gamma)\le\delta$,
\quad$\cf(\prod_{\zeta\in J}\theta_{\zeta})
\le\max(1,\alpha')\le\alpha$.}

\noindent If $\lambda=0$ then
$\cf(\prod_{\zeta<\lambda}\theta_{\zeta})=1\le\alpha$;  if $\lambda>0$ then
$\alpha$ is infinite and

\Centerline{$\cf(\prod_{\zeta<\lambda}\theta_{\zeta})
\le\max(\omega,\cf(\prod_{\zeta\in I}\theta_{\zeta}),
   \cf(\prod_{\zeta\in J}\theta_{\zeta}))
\le\max(\alpha,\delta)$.}

This is enough to deal with (a).

\medskip

{\bf case 2} If $\alpha'=\alpha=0$ then $\lambda=0=\delta$, so

\Centerline{$\cf(\prod_{\zeta<\lambda}\theta_{\zeta})=1
=\delta^{\cf\alpha}$.}

\noindent So (b) is true if $\alpha=0$.

\medskip

{\bf case 3} If $\alpha'=\alpha>0$ and $\cf\alpha<\gamma$, then
$\lambda>0$ and $\alpha$ is a supremum of strictly smaller infinite
cardinals, so must be uncountable.   Let
$\langle\alpha_{\xi}\rangle_{\xi<\cf\alpha}$ be a strictly increasing
family of cardinals with supremum $\alpha$, starting from
$\alpha_0=0$ and $\alpha_1=\omega$ and with
$\alpha_{\xi}=\sup_{\eta<\xi}\alpha_{\eta}$ for non-zero limit ordinals
$\xi<\cf\alpha$.   Set

\Centerline{$P_{\xi}
=\prod_{\zeta<\lambda,\alpha_{\xi}\le\theta_{\zeta}<\alpha_{\xi+1}}
   \theta_{\zeta}$}

\noindent for each $\xi<\cf\alpha$.   Then

\Centerline{$\cf P_{\xi}\le\Theta(\alpha_{\xi+1},\gamma)\le\delta$}

\noindent for each $\xi<\cf\alpha$, so

\Centerline{$\cf(\prod_{\zeta<\lambda}\theta_{\zeta})
=\cf(\prod_{\xi<\cf\alpha}P_{\xi})\le\delta^{\cf\alpha}$.}

\noindent Putting this together with case 1, we have a proof of (b) when
$\alpha>0$.
}%end of proof of 5A2I

\leaveitout{\exercises{\leader{5A2X}{Basic exercises (a)}

\leader{5A2Y}{Further exercises (a)}

}%end of exercises

\endnotes{
\Notesheader{5A2}

}%end of notes
}%end of leaveitout

\discrpage


\frfilename{mt5a3.tex}
\versiondate{20.5.23}
\copyrightdate{2007}

\Loadfourteens

\def\chaptername{Appendix}
\def\sectionname{Forcing}

%\def\uBp{\Cal{UB}\eurm{p}}
\def\uBp{\Cal{U}\mskip-1mu\widehat{\Cal B}}

\newsection{5A3}

My discussion of forcing is based on {\smc Kunen 80};  in particular,
I start from pre-ordered sets rather than Boolean algebras, and the class
$V^{\Bbb P}$ of terms in a forcing language will consist of subsets of
$V^{\Bbb P}\times P$.   I find however that I wish to diverge almost
immediately from standard formulations in a technical respect, which I
describe in 5A3A, introducing what I call `forcing notions'.
I do not refer to generic filters or models of ZFC, preferring to express
all results in terms of the forcing relation (5A3C).   I give some space to
the interpretation of names (5A3E, 5A3F) and, in particular, to names for
real numbers derived from elements of $L^0(\RO(\Bbb P))$ (5A3L).

\leader{5A3A}{Forcing notions (a)}
A {\bf forcing notion} is a quadruple
$\Bbb P=(P,\le,\Bbbone,\uparrow)$ or $\Bbb P=(P,\le,\Bbbone,\downarrow)$
where $(P,\le)$ is a pre-ordered set (that is, $\le$ is a transitive
reflexive relation on $P$), $\Bbbone\in P$, and

\inset{if $\Bbb P=(P,\le,\Bbbone,\uparrow)$ then $\Bbbone\le p$ for every
$p\in P$,

if $\Bbb P=(P,\le,\Bbbone,\downarrow)$ then $p\le\Bbbone$ for every
$p\in P$.}

\noindent In this context members of $P$ are\cmmnt{ commonly} called
{\bf conditions}.

\spheader 5A3Ab\cmmnt{ I had better try to explain what I am doing here.
The problem is the following.
Consider two of the standard examples of pre-ordered set in this
context.   For a set $I$,
$\Fn_{<\omega}(I;\{0,1\})$ is the set of functions from finite subsets of
$I$ to $\{0,1\}$;  for a non-trivial Boolean algebra $\frak A$,
$\frak A^+$ is the set of non-zero elements of $\frak A$.
In each case, we have a relevant direction.   In
$\Fn_{<\omega}(I;\{0,1\})$, a condition $p$ is stronger than a condition
$q$ if $p$ extends $q$, that is, if $p\supseteq q$;
in $\frak A^+$, $p$ is stronger than $q$ if $p\Bsubseteq q$.   So the
forcing notions, in the terminology I have chosen, are

\Centerline{$(\Fn_{<\omega}(I;\{0,1\}),\subseteq,\emptyset,\uparrow)$,
\quad$(\frak A^+,\Bsubseteqshort,1_{\frak A},\downarrow)$.}

\noindent Generally,} I will say that a forcing notion
$(P,\le,\Bbbone,\uparrow)$ is {\bf active upwards}, while
$(P,\le,\Bbbone,\downarrow)$ is {\bf active downwards}.

\cmmnt{\spheader 5A3Ac Of course this is unconventional.
It is much more usual
to take all forcing notions to be active in the same direction (usually
downwards) and to use local definitions (e.g., saying that `$p\le q$ if $p$
extends $q$') to ensure that this will be appropriate.

However the great majority of forcing notions, like the two examples in (b)
above, come with structures which strongly suggest a natural interpretation
of `$\le$';  and these structures are not arbitrary, but are essential to
our intuitive conception of the pre- or partial order we are studying.
I prefer, therefore, to maintain the notation I would use for the
same objects
in any other context, and to indicate separately the orientation which is
relevant when using them to build a forcing language.
}

\spheader 5A3Ad\cmmnt{ This approach demands further changes in the
language.}   It
will no longer be helpful to talk about conditions in $P$ being `larger' or
`less than' others.   Instead, I will use the word `{\bf stronger}':  if
$\Bbb P=(P,\le,\Bbbone,\uparrow)$, then $p\in P$ will be stronger than
$q\in P$ if $p\ge q$;  if
$\Bbb P=(P,\le,\Bbbone,\downarrow)$, then $p\in P$ will be stronger than
$q\in P$ if $p\le q$.
\cmmnt{(So $p$ will be stronger than $\Bbbone$ for every $p\in P$.)}

Similarly,\cmmnt{ the words `cofinal' and `coinitial' are now
inappropriate, and I
will turn to the word `dense', as favoured by most authors
discussing forcing;}  if $\Bbb P=(P,\le,\Bbbone,\updownarrows)$
is a forcing notion, a subset $Q$ of
$P$ is {\bf dense} if for every $p\in P$ there is a $q\in Q$ such that $q$
is stronger than $p$.   In the same way,\cmmnt{ I can say that} two
conditions $p$,
$q$ in $P$ are `compatible' if there is an $r\in P$ stronger than both.
We\cmmnt{ shall} have a standard topology on $P$ generated by sets of
the form
$\{q:q$ is stronger than $p\}$, and a corresponding regular open algebra
$\RO(\Bbb P)$\cmmnt{, as in 514M}.   An antichain for $\Bbb P$
will be a set $A\subseteq P$ such that any
two distinct conditions in $A$ are incompatible, and
$\Bbb P$ will be ccc if every antichain for $\Bbb P$ is countable.
The `saturation' $\sat\Bbb P$ of $\Bbb P$ will be the least cardinal
$\kappa$ such that there is no antichain with cardinal $\kappa$.

\leader{5A3B}{Forcing languages} Let
$\Bbb P=(P,\le,\Bbbone,\updownarrows)$
be a forcing notion.

\spheader 5A3Ba The class of {\bf $\Bbb P$-names}, that is,
terms of the forcing language defined by $\Bbb P$, is

\Centerline{$V^{\Bbb P}=\{A:A$ is a set and
$A\subseteq V^{\Bbb P}\times P\}$}

\noindent\cmmnt{({\smc Kunen 80}, VII.2.5)\footnote{In this section, we 
need the Axiom of
Foundation (`for any non-empty set $A$, there is an $a\in A$ such that
$a\cap A=\emptyset$');  here, to determine whether a set $A$ belongs to 
$V^{\Bbb P}$, we need to induce on the rank of $A$.}.   In this context,}
I will say that the {\bf domain} of a name $A\in V^{\Bbb P}$ is the set
$\dom A\subseteq V^{\Bbb P}$ of first members of elements of $A$.

\spheader 5A3Bb For any set $X$, $\check X$ will be the $\Bbb P$-name
$\{(\check x,\Bbbone):x\in X\}\in V^{\Bbb P}$\cmmnt{ ({\smc Kunen 80},
VII.2.10)}.

\leader{5A3C}{The Forcing Relation}\cmmnt{ ({\smc Kunen 80}, VII.3.3)}
Suppose that $\Bbb P=(P,\le,\Bbbone,\updownarrows)$
is a forcing notion, $p\in P$, $\phi$, $\psi$ are formulae of set theory,
and $\dot x_0,\ldots,\dot x_n\in V^{\Bbb P}$.

\spheader 5A3Ca $p\VVdP\,\dot x_0=\dot x_1$ iff

\inset{whenever $(\dot y,q)\in\dot x_0$ and $r\in P$ is
stronger than both $p$ and $q$, there are a $(\dot y',q')\in\dot x_1$ and
an $r'$ stronger than both $r$ and $q'$ such that
$r'\VVdP\,\dot y=\dot y'$,

whenever $(\dot y,q)\in\dot x_1$ and $r\in P$ is
stronger than both $p$ and $q$, there are a $(\dot y',q')\in\dot x_0$ and
an $r'$ stronger than both $r$ and $q'$ such that
$r'\VVdP\,\dot y=\dot y'$.}

\cmmnt{\noindent Note that $p\VVdP\,\dot x=\dot x$ for every $\Bbb P$-name
$\dot x$ and every $p\in P$ (induce on the rank of $\dot x$).}

\spheader 5A3Cb $p\VVdP\,\dot x_0\in\dot x_1$ iff

\inset{whenever $q\in P$ is stronger than $p$ there are a
$(\dot y,q')\in\dot x_1$ and an $r$ stronger than both $q$ and $q'$ such
that $r\VVdP\,\dot x_0=\dot y$.}

\spheader 5A3Cc
$p\VVdP\,\phi(\dot x_0,\ldots,\dot x_n)\,
\&\,\psi(\dot x_0,\ldots,\dot x_n)$ iff

\Centerline{$p\VVdP\,\phi(\dot x_0,\ldots,\dot x_n)$ and
  $p\VVdP\,\psi(\dot x_0,\ldots,\dot x_n)$.}

\spheader 5A3Cd $p\VVdP\,\neg\phi(\dot x_0,\ldots,\dot x_n)$ iff

\Centerline{there is no $q$ stronger than $p$ such that
$q\Vdash_{\Bbb P}\,\phi(\dot x_0,\ldots,\dot x_n)$.}

\spheader 5A3Ce $p\VVdP\,\Exists x,\,\phi(x,\dot x_0,\ldots,\dot x_n)$
iff

\inset{for
every $q$ stronger than $p$ there are an $r$ stronger than $q$ and a
$\dot y\in V^{\Bbb P}$ 

\hfill such that
$r\VVdP\,\phi(\dot y,\dot x_0,\ldots,
\dot x_n)$.\cmmnt{\footnote{\smallerfonts This formulation
is appropriate if we wish to explore forcing without using the axiom of
choice.   Subject to AC, we have an alternative condition:
$p\eightVVdash_{\Bbb P}\,\Exists x,\,\phi(x,\dot x_0,\ldots,\dot x_n)$
iff there is a $\dot y\in V^{\Bbb P}$ such that
$p\eightVVdash_{\Bbb P}\,\phi(\dot y,\dot x_0,\ldots,\dot x_n)$.}}}

\spheader 5A3Cf You will see that if $p\VVdP\,\phi$ and $q$ is stronger than
$p$ then $q\VVdP\,\phi$.

\spheader 5A3Cg\cmmnt{ In this context} I will write $\VVdP$ for
$\Bbbone\VVdP$.

\leader{5A3D}{The Forcing Theorem} If $\phi$ is any theorem of ZFC, and
$\Bbb P$ is any forcing notion, then $\VVdP\,\phi$.
\prooflet{({\smc Kunen 80}, VII.4.2.)}

\leader{5A3E}{Names for functions}\dvAformerly{5{}A3H} Let $\Bbb P$ be a 
forcing notion, $P$ its set of conditions, and
$R\subseteq V^{\Bbb P}\times V^{\Bbb P}\times P$ a set.
Consider the $\Bbb P$-names

\Centerline{$\dot f=\{((\dot x,\dot y),p):(\dot x,\dot y,p)\in R\}$,}

\Centerline{$\dot A=\{(\dot x,p):(\dot x,\dot y,p)\in R\}$,
\quad$\dot B=\{(\dot y,p):(\dot x,\dot y,p)\in R\}$.}

(a) The following are equiveridical:

\quad(i) $\VVdP\,\dot f$ is a function;

\quad(ii) whenever $(\dot x_0,\dot y_0,p_0)$,
$(\dot x_1,\dot y_1,p_1)$ belong to $R$,
$p\in\Bbb P$ is stronger than both $p_0$ and $p_1$
and $p\VVdP\,\dot x_0=\dot x_1$, then $p\VVdP\,\dot y_0=\dot y_1$.

(b) In this case,

\Centerline{$p\VVdP\,\dot f(\dot x)=\dot y$}

\noindent whenever $(\dot x,\dot y,p)\in R$,

\Centerline{$\VVdP\,\dom\dot f=\dot A$ and $\dot f[\dot A]=\dot B$,}

\noindent and the following are equiveridical:

\quad(i) $\VVdP\,\dot f$ is injective;

\quad(ii) whenever $(\dot x_0,\dot y_0,p_0)$,
$(\dot x_1,\dot y_1,p_1)$ belong to $R$,
$p\in\Bbb P$ is stronger than both $p_0$ and $p_1$
and $p\VVdP\,\dot y_0=\dot y_1$, then $p\VVdP\,\dot x_0=\dot x_1$.


\cmmnt{\medskip

\noindent{\bf Remark} In the formula for $\dot f$ here, brackets take
different meanings at different points.   In the expression
$((\dot x,\dot y),p)$, the
inner brackets must be interpreted in the forcing language, while the outer
brackets, like the brackets in the expression $(\dot x,\dot y,p)\in R$, are
interpreted in the ordinary universe\dvro{.}{;}\cmmnt{ {\smc Kunen 80}
might write

\Centerline{$\dot f=\{($op$(\dot x,\dot y),p):(\dot x,\dot y,p)\in R\}$.}}
}

\proof{ Elementary.
}%end of proof of 5A3E

\leader{5A3F}{More notation}\dvAformerly{5{}A3E}\cmmnt{ In 5A3C I took it 
for granted that every
formula of set theory would have a version in $V^{\Bbb P}$.   I should
perhaps explain some of the versions I have in mind.}
Let $\Bbb P=(P,\le,\Bbbone,\updownarrows)$ be a forcing notion.

\spheader 5A3Fa If $\dot y_0$, $\dot y_1\in V^{\Bbb P}$ then
$\dot x=\{(\dot y_0,\Bbbone),(\dot y_1,\Bbbone)\}\in V^{\Bbb P}$, and

\Centerline{$\VVdP\,\dot x=\{\dot y_0,\dot y_1\}$\dvro{.}{;}}

\noindent\cmmnt{so we have a suitable formal expression for pair sets in
$V^{\Bbb P}$. }Similarly, if we think of the formula $(x,y)$ as being an
abbreviation for $\{\{x\},\{x,y\}\}$, we get a $\Bbb P$-name

\Centerline{$\dot z
=\{(\{(\dot y_0,\Bbbone)\},\Bbbone),
  (\{(\dot y_0,\Bbbone),(\dot y_1,\Bbbone)\},\Bbbone)\}$}

\noindent such that

\Centerline{$\VVdP\,\dot z=(\dot y_0,\dot y_1)$.}

\spheader 5A3Fb Now let
$\familyiI{\dot x_i}$ be a family of $\Bbb P$-names, and set

\Centerline{$\dot f=\{((\varchecki,\dot x_i),\Bbbone):i\in I\}$.}

\noindent As in 5A3E,

\Centerline{$\VVdP\,\dot f$ is a function with domain $\check I$,}

\noindent and

\Centerline{$\VVdP\,\dot f(\varchecki)=\dot x_i$}

\noindent for every $i\in I$.\cmmnt{ For obvious reasons I do not wish
to spell this procedure out every time, and} I will use the\cmmnt{ rather
elliptic} formula

\Centerline{$\family{i}{\check I}{\dot x_i}$}

\noindent to signify the $\Bbb P$-name $\dot f$.

\spheader 5A3Fc\cmmnt{ Similarly,}
$\dot T=\{(\dot x_i,\Bbbone):i\in I\}$ is a $\Bbb P$-name such that
$\VVdP\,\dot x_i\in\dot T$
for every $i\in I$, and whenever $p\in\Bbb P$ and $\dot x$ is a
$\Bbb P$-name such that $p\VVdP\,\dot x\in\dot T$, there are an $i\in I$
and a $q$ stronger than $p$ such that $q\VVdP\,\dot x=\dot x_i$;  I will
write $\{\dot x_i:i\in\check I\}$ for $\dot T$.

\spheader 5A3Fd
In the same spirit, if I have a family
$\familyiI{\dot x_i}$ of $\Bbb P$-names for real numbers between $0$ and $1$,
I will allow myself to write
`$\sup_{i\in\check I}\dot x_i$' to signify a $\Bbb P$-name such that

\Centerline{$\VVdP\,\sup_{i\in\check I}\dot x_i
=\sup\{\dot x_i:i\in\check I\}$\dvro{.}{,}}

\noindent\cmmnt{without taking the trouble to spell out any exact
formula to
represent the supremum.  }I will do the same for limits of sequences;
if $\sequencen{\dot x_n}$ is a sequence of $\Bbb P$-names for real numbers,
and

\Centerline{$\VVdP\,\sequencen{\dot x_n}$ is convergent,}

\noindent then I will write `$\lim_{n\to\infty}\dot x_n$' to mean a
$\Bbb P$-name $\dot x$ such that

\Centerline{$\VVdP\,\sequencen{\dot x_n}\to\dot x\in\Bbb R$.}

\cmmnt{\noindent Of course this is tolerable only because it is
possible to set
out a general rule for devising a suitable name $\dot x\in V^{\Bbb P}$
from the given sequence $\sequencen{\dot x_n}$.   See 5A3L below.}

\vleader{72pt}{5A3G}{Boolean truth values}\dvAformerly{5{}A3F} Let
$\Bbb P$ be a forcing notion and $P$ its set of conditions.

\spheader 5A3Ga If $\phi$ is a
formula of set theory, and $\dot x_0,\ldots,\dot x_n\in V^{\Bbb P}$, then

\Centerline{$\{p:p\in P$, $p\VVdP\,\phi(\dot x_0,\ldots,\dot x_n)\}$}

\noindent is a regular open set in $P$\cmmnt{ (use 514Md)};
I will denote it $\Bvalue{\phi(\dot x_0,\ldots,\dot x_n)}$.

\spheader 5A3Gb If $\phi$ and $\psi$ are formulae of set theory
and $\dot x_0,\ldots,\dot x_n\in V^{\Bbb P}$, then

\Centerline{
$\Bvalue{\phi(\dot x_0,\ldots,\dot x_n)\,\&\,\psi(\dot x_0,\ldots,\dot x_n)}
=\Bvalue{\phi(\dot x_0,\ldots,\dot x_n)}
\cap\Bvalue{\psi(\dot x_0,\ldots,\dot x_n)}$}

\noindent and

\Centerline{$\Bvalue{\neg\phi(\dot x_0,\ldots,\dot x_n)}
=P\setminus\overline{\Bvalue{\phi(\dot x_0,\ldots,\dot x_n)}}$\dvro{.}{,}}

\cmmnt{\noindent the complement of
$\Bvalue{\phi(\dot x_0,\ldots,\dot x_n)}$ in $\RO(\Bbb P)$.}

\spheader 5A3Gc If $\phi$ is a formula of set theory, $A$ is a set,
and $\dot x_0,\ldots,\dot x_n$ are $\Bbb P$-names, then

\Centerline{$\Bvalue{\exists\,x\in\check A,\,\phi(x,\dot x_0,\ldots,\dot x_n)}
=\interior\overline{\bigcupop_{a\in A}
   \Bvalue{\phi(\check a,\dot x_0,\ldots,\dot x_n)}}$,}

\noindent the supremum of
$\{{\Bvalue{\phi(\check a,\dot x_0,\ldots,\dot x_n}}:a\in A\}$ in
$\RO(\Bbb P)$.  \prooflet{(Use 5A3Ce.)}

\leader{5A3H}{Concerning $\pmb{\varBspcheck}$s\dvAformerly{5{}A3G} (a)}
\cmmnt{The reader is entitled to an attempt at
consistency on the following point of notation, among others.}
For any set $X$ and any forcing notion $\Bbb P$ there is a
corresponding $\Bbb P$-name $\check X$\cmmnt{ (5A3Bb)}.   We start with
$\check\emptyset=\emptyset$.   If $1=\{\emptyset\}$ is the next von Neumann
ordinal, we get a name

\Centerline{$\check 1
=\{(\check\emptyset,\Bbbone)\}=\{(\emptyset,\Bbbone)\}$;}

\noindent and\cmmnt{ we can check directly from 5A3C that}

\Centerline{$\VVdP\,\check 1=\{\emptyset\}$,}

\noindent that is\cmmnt{, if you like},

\Centerline{$\VVdP\,\check 1=1$,}

\noindent where in this formula the first $1$ is interpreted in the
ordinary universe and the second is interpreted in the forcing language.
Similarly, if we take `$2$' to be an abbreviation for
`$\{\emptyset,\{\emptyset\}\}$',\cmmnt{ we have}

\Centerline{$\VVdP\,\check 2=2$,}

\noindent and so on.   Indeed\cmmnt{ we get}

\Centerline{$\VVdP\,\check\omega$ is the first infinite ordinal,}

\Centerline{$\VVdP\,\check\Bbb Q$ is the set of rational numbers,}

\noindent so the same convention would lead to

\Centerline{$\VVdP\,\check\omega=\omega$,\, 
$\check\Bbb N=\Bbb N$,\, $\check\Bbb Q=\Bbb Q$.}

\noindent\cmmnt{(This formula does not depend on which construction of
the set of
rational numbers we use, provided that we use the same method both in the
ordinary universe and in the forcing language.)  Of course it is
{\it not} the case (except for forcing notions of particular types) that

\Centerline{$\VVdP\,\check\omega_1$ is the first uncountable ordinal,
\quad$\VVdP\,\check\Bbb R$ is the set of real numbers.}}

\cmmnt{\spheader 5A3Hb For `absolute' objects, therefore,
like $\omega+7$ or $\bover{22}7$,
appearing in sentences of a forcing language, I shall have a
choice between formulations

\Centerline{$(\omega+7)\varBspcheck$,\quad$(\bover{22}7)\varBspcheck$}

\noindent (working directly from 5A3Bb),

\Centerline{$\omega+7$,\quad$\bover{22}7$}

\noindent (regarding the phrases `$\omega+7$' and `$\bover{22}7$' as 
abbreviations for
expressions in set theory which can be evaluated either in the ordinary
universe or in the forcing language), or

\Centerline{$\check\omega\mskip5mu\check+\mskip5mu\check 7$,
\quad$22\varBspcheck\check\div\check 7$}

\noindent (regarding $\omega$, $+$, $7$, $22$ and $\div$ as sets to which the 
rule of
5A3Bb can be applied, and then interpreting the combination in the forcing
language).   The least cluttered versions, $\omega+7$ and $\bover{22}7$, 
look better, and
these will ordinarily be my choice.   But it means that when you see the
symbol $\Bbb Q$ in a sentence of the forcing language, it is likely to mean
two things at once, a superposition of `the set of rational numbers' and
`the $\Bbb P$-name $\check{\Bbb Q}$', with algebraic operations and
relations attached correspondingly.
}

\spheader 5A3Hc\cmmnt{ `Absoluteness' is treated properly in {\smc Kunen 80},
\S IV.3.   I shall not attempt to even sketch the concept here.   But we shall
need a couple of basic examples.}    Let $\Bbb P$ be a forcing notion and $P$
its set of conditions.   

\medskip

\quad{\bf (i)} If $A$ and $B$ are sets, $p\in P$ and 
$p\VVdP\,\check A\subseteq\check B$ then $A\subseteq B$.   
\prooflet{\Prf\ Induce on the rank of $B$.   If $a\in A$
then $(\check a,\Bbbone)\in\check A$ so $\VVdP\,\check a\in\check A$ and
$p\VVdP\,\check a\in\check B$.   Now there must be a 
$\Bbb P$-name $\dot y$, a $q\in P$ and an $r$ stronger than both $p$ and
$q$ such that $(\dot y,q)\in\check B$ and $r\VVdP\,\check a=\dot y$;  in which
case $\dot y$ must be of the form $\check b$ where $b\in B$, and
$r\VVdP\,\check a=\check b$.   By the inductive hypothesis $a=b$ and 
$a\in B$.\ \Qed}

\medskip

\quad{\bf (ii)} If $\alpha$, $\beta\in\Bbb Q$ and 
$p\VVdP\,\check\alpha\le\check\beta$ then $\alpha\le\beta$.
\prooflet{\Prf\ I leave the proof as an exercise, since the details necessarily
depend on the precise construction you use for $\Bbb Q$.   But they resolve
quickly into a handful of similar statements concerning arithmetic in $\Bbb N$, 
and (i) here, together with 5A3Eb, should be enough to deal with these.\ \Qed}

\leader{5A3I}{Regular open algebras} If
$\Bbb P=(P,\le,\Bbbone,\updownarrows)$ is a forcing notion with regular
open algebra $\RO(\Bbb P)$, then we have a natural map
$\iota:P\to\RO(\Bbb P)^+$ defined by saying that

\Centerline{$\iota(p)
=\interior\overline{\{q:q\text{ is stronger than }p\}}$}

\noindent for $p\in P$\cmmnt{ ({\smc Kunen 80}, II.3.3)};
and (allowing for the possible reversal of the direction of $\Bbb P$)
$\iota$ is a dense
embedding of the pre-ordered set $(P,\le)$ in the partially ordered set
$(\RO(\Bbb P)^+,\subseteq)$\cmmnt{,
in the sense of {\smc Kunen 80}, \S VII.7}.
Consequently, taking $\widehat{\Bbb P}$ to be the forcing notion
$(\RO(\Bbb P)^+,\subseteq,P,\downarrow)$,\cmmnt{ we shall have}

\Centerline{$\VVdP\,\phi$ if and only if
$\VVdash_{\widehat{\Bbb P}}\,\phi$}

\noindent for every sentence $\phi$ of
set theory\cmmnt{ ({\smc Kunen 80},
VII.7.11)}.   It follows that if two forcing notions
have isomorphic regular open algebras, then they force exactly the same
theorems of set theory.

\leader{5A3J}{}\cmmnt{ The following technical device will be useful at
one point.

\medskip

\noindent}{\bf Definition} Let $\Bbb P$ be a forcing notion.
I will say that a
$\Bbb P$-name $\dot X$ is {\bf discriminating} if whenever $(\dot x,p)$ and
$(\dot y,q)$ are distinct members of $\dot X$, and $r$ is stronger than
both $p$ and $q$, then $r\VVdP\,\dot x\ne\dot y$.

\leader{5A3K}{Lemma} Let $\Bbb P$ be a forcing notion, and $P$ its
set of conditions.

(a) For any $\Bbb P$-name $\dot X$,
there is a discriminating $\Bbb P$-name $\dot X_1$
such that $\VVdP\,\dot X_1=\dot X$.

(b) Let $\dot X$ be a
discriminating $\Bbb P$-name, and $f:\dot X\to V^{\Bbb P}$ a function.
Let $\dot g$ be the $\Bbb P$-name

\Centerline{$\{((\dot x,f(\dot x,p)),p):
(\dot x,p)\in\dot X\}$.\cmmnt{\footnote{Once again I present a formula
in which some ordered pairs
are to be interpreted in the ordinary universe, but another is to be
interpreted in the forcing language.}}}

\noindent Then

\Centerline{$\VVdP\,\dot g$ is a function with domain $\dot X$.}

\proof{{\bf (a)(i)} Set

\Centerline{$\dot X_0=\{(\dot x,q):$ there is some
$p\in P$ such that $(\dot x,p)\in\dot X$ and $q\in P$
is stronger than $p\}$.}

\noindent Then $\VVdP\,\dot X=\dot X_0$.   \Prf\ Because
$\dot X\subseteq\dot X_0$, $\VVdP\,\dot X\subseteq\dot X_0$.   In the other
direction, if $\dot x$ is a $\Bbb P$-name and $p$ is such that
$p\VVdP\,\dot x\in\dot X_0$, there are an $(\dot x_1,p_1)\in\dot X_0$ and a
$q$, stronger than both $p$ and $p_1$, such that $q\VVdP\,\dot x=\dot x_1$.
Now there is a $p_2$ such that $(\dot x_1,p_2)\in\dot X$ and
$p_1$ is stronger than $p_2$.   In this case, $q$ is stronger than $p_2$
and $p_2\VVdP\,\dot x_1\in\dot X$, so $q\VVdP\,\dot x=\dot x_1\in\dot X$,
while $q$ is stronger than $p$.   As $\dot x$ and $p$ are arbitrary,
$\VVdP\,\dot X_0\subseteq\dot X$.\ \Qed

Let $\dot X_1\subseteq\dot X_0$ be a maximal discriminating name.

\medskip

\quad{\bf (ii)} Because $\dot X_1\subseteq\dot X_0$,
$\VVdP\,\dot X_1\subseteq\dot X_0=\dot X$.   But we also have
$\VVdP\,\dot X\subseteq\dot X_1$.   \Prf\ Suppose that $\dot x$
is a $\Bbb P$-name and $p\in P$ is such that
$p\VVdP\,\dot x\in\dot X$.    Then there must be an
$(\dot x_1,p_1)\in\dot X$ and a $q$ stronger than both $p$ and $p_1$
such that $q\VVdP\,\dot x=\dot x_1$.   In this case,
$(\dot x_1,q)\in\dot X_0$.   By the maximality of $\dot X_1$, there are a
$(\dot y,q')\in\dot X_1$ and an $r$ stronger than both $q$ and $q'$
such that $r\VVdP\,\dot x_1=\dot y$.   Now $r$ is stronger than $p$ and

\Centerline{$r\VVdP\,\dot x=\dot x_1=\dot y\in\dot X_1$.}

\noindent As $\dot x$ and $p$ are arbitrary,
$\VVdP\,\dot X\subseteq\dot X_1$.\ \Qed

So $\VVdP\,\dot X_1=\dot X$, as required.

\medskip

{\bf (b)} Consider

\Centerline{$\{(\dot x,f(\dot x,p),p):(\dot x,p)\in\dot X\}
\subseteq V^{\Bbb P}\times V^{\Bbb P}\times P$.}

\noindent If $(\dot x_0,p_0)$ and $(\dot x_1,p_1)$ belong to $\dot X$,
$p$ is stronger than both $p_0$ and $p_1$, and $p\VVdP\,\dot x_0=\dot x_1$,
then $(\dot x_0,p_0)=(\dot x_1,p_1)$, because $\dot X$ is a discriminating
name;  so $p\VVdP\,f(\dot x_0,p_0)=f(\dot x_1,p_1)$.   By
5A3E,

\Centerline{$\VVdP\,\dot g$ is a function and $\dom\dot g=\dot X$.}
}%end of proof of 5A3K

\leader{5A3L}{Real numbers in forcing languages}
Let $\Bbb P$ be any forcing notion, and $P$ its set of conditions.

\spheader 5A3La\cmmnt{ I have tried to avoid committing myself to any
declaration
of what a real number actually `is';  in fact I believe that at the deepest
level this should be regarded as
an undefined concept, and that the descriptions offered by
Weierstrass and Dedekind are essentially artificial.   But if we are to
make sense of real analysis in forcing models we must fix on some
formulation, so} I will say that a real number is the set of strictly
smaller rational numbers.\cmmnt{ (I leave it to you to decide whether a
rational number is an equivalence class of pairs of integers, or a
coprime pair $(m,n)$
where $m\in\Bbb Z$ and $n\in\Bbb N\setminus\{0\}$, or something else
altogether, provided only that you fix on a construction expressible by a
formula of set theory.)}   \cmmnt{Observe that this has the desirable
effect that}

\Centerline{$\VVdP\,\check\alpha$ is a real number}

\noindent for every real number $\alpha$.

\spheader 5A3Lb Consider the Dedekind complete Boolean algebra
$\RO(\Bbb P)$ and the corresponding space
$L^0=L^0(\RO(\Bbb P))$\cmmnt{ as defined in 364A}.

\medskip

\quad{\bf (i)} For every $u\in L^0$, set

\Centerline{$\vec u
=\{(\check\alpha,p):\alpha\in\Bbb Q$, $p\in\Bvalue{u>\alpha}\}$.}

\noindent Then

\Centerline{$\VVdP\,\vec u$ is a real number.}

\prooflet{\noindent\Prf\ Asw $\VVdP\,\check\Bbb Q=\Bbb Q$,
$\VVdP\,\vec u\subseteq\Bbb Q$.   If $p\in P$, there are an
$n\in\Bbb Z$ and a $q$ stronger than $p$ such that $q\in\Bvalue{u>n}$,
in which case $(\check n,q)\in\vec u$ and $q\VVdP\,\vec u\ne\emptyset$;
accordingly $\VVdP\,\vec u\ne\emptyset$.   Again, if $p\in P$, there
are an $n\in\Bbb N$ and a $q$ stronger than $p$ such that
$q\in\Bvalue{u\le n}$, in which case $\alpha\le n$ whenever
$(\check\alpha,r)\in\vec u$ and $r$ is stronger than $q$, so that

\Centerline{$q\VVdP\,\check n$ is an upper bound for $\vec u$;}

\noindent accordingly $\VVdP\,\vec u$ is bounded above.

If $p\in P$ and $\dot\alpha$ is a $\Bbb P$-name such that

\Centerline{$p\VVdP\,\dot\alpha\in\vec u$,}

\noindent then for any $q$ stronger than $p$ there are an $r$ stronger than
$q$ and an $\alpha\in\Bbb Q$ such that $r\VVdP\,\dot\alpha=\check\alpha$
and $r\in\Bvalue{u>\alpha}$.   Now there are a $\beta\in\Bbb Q$ and an $r'$
stronger than $r$ such that $\beta>\alpha$ and $r'\in\Bvalue{u>\beta}$;  in
which case $r'\VVdP\,\dot\alpha<\check\beta\in\vec u$.   As $q$ is
arbitrary,

\Centerline{$p\VVdP\,\dot\alpha$ is not the greatest member of $\vec u$;}

\noindent as $p$ and $\dot\alpha$ are arbitrary,

\Centerline{$\VVdP\,\vec u$ has no greatest member.}

If $p\in P$ and $\dot\alpha$, $\dot\beta$ are $\Bbb P$-names such that

\Centerline{$p\VVdP\,\dot\alpha\in\Bbb Q$,
$\dot\alpha\le\dot\beta\in\vec u$,}

\noindent then for any $q$ stronger than $p$ there are an $r$ stronger than
$q$ and $\alpha$, $\beta\in\Bbb Q$ such that $(\check\beta,r)\in\vec u$ and

\Centerline{$r\VVdP\,\dot\alpha=\check\alpha$, $\dot\beta=\check\beta$,
$\check\alpha\le\check\beta$.}

\noindent In this case, $\alpha\le\beta$ (5A3H(c-ii)), 
$r\in\Bvalue{u>\beta}\subseteq\Bvalue{u>\alpha}$,
$(\check\alpha,r)\in\vec u$ and $r\VVdP\,\dot\alpha=\check\alpha\in\vec u$.
As $q$ is arbitrary,

\Centerline{$p\VVdP\,\dot\alpha\in\vec u$;}

\noindent as $p$, $\dot\alpha$ and $\dot\beta$ are arbitrary,

\Centerline{$\VVdP\,\alpha\in\vec u$ whenever $\alpha\in\Bbb Q$ and
$\alpha\le\beta\in\vec u$, so $\vec u$ is a real number.  \Qed}
}%end of prooflet

\medskip

\quad{\bf (ii)}\cmmnt{ Observe next
that} $\Bvalue{\vec u>\check\alpha}=\Bvalue{u>\alpha}$ for every
$\alpha\in\Bbb Q$.   \prooflet{\Prf\ For $p\in P$,

$$\eqalign{p\in\Bvalue{\vec u>\check\alpha}
&\iff p\VVdP\,\check\alpha<\vec u\cr
&\iff p\VVdP\,\check\alpha\in\vec u\cr
&\iff\text{ for every }q\text{ stronger than }p
\text{ there are }q'\in P,\,\beta\in\Bbb Q\cr
&\mskip50mu\text{ and an }r\text{ stronger than both }q\text{ and }q'\cr
&\mskip50mu
\text{ such that }(\check\beta,q')\in\vec u
\text{ and }r\VVdP\,\check\beta=\check\alpha\cr
&\iff\text{ for every }q\text{ stronger than }p
\text{ there is an }r\text{ stronger than }q\cr
&\mskip50mu
\text{ such that }(\check\alpha,r)\in\vec u\cr
&\iff\text{ for every }q\text{ stronger than }p
\text{ there is an }r\text{ stronger than }q\cr
&\mskip50mu
\text{ such that }r\in\Bvalue{u>\alpha}\cr
&\iff p\in\Bvalue{u>\alpha}\cr}$$

\noindent (514Md, because $\Bvalue{u>\alpha}$ is a regular open subset of
$P$).\ \Qed}

\medskip

\quad{\bf (iii)} In the other direction,
if we have a $\Bbb P$-name $\dot x$ for a real number (that is, a
$\Bbb P$-name such that
$\VVdP\,\dot x$ is a real number), then there is a unique
$u\in L^0$ such that $\VVdP\,\dot x=\vec u$.   \prooflet{\Prf\
For every $\alpha\in\Bbb Q$ we have a
Boolean value $\Bvalue{\dot x>\check\alpha}$ belonging to
$\RO(\Bbb P)$ (5A3G).   It is easy to see that

\Centerline{$\Bvalue{\dot x>\check\alpha}
=\sup_{\beta\in\Bbb Q,\beta>\alpha}\Bvalue{\dot x>\check\beta}$}

\noindent for every $\alpha\in\Bbb Q$,

\Centerline{$\inf_{n\in\Bbb Z}\Bvalue{\dot x>\check n}=0$,
\quad$\sup_{n\in\Bbb Z}\Bvalue{\dot x>\check n}=1$.}

\noindent We therefore have a unique $u\in L^0$ such that

\Centerline{$\Bvalue{u>\alpha}
=\Bvalue{\dot x>\check\alpha}$}

\noindent for every $\alpha\in\Bbb R$ (364Ae).   Now
$\Bvalue{\vec u>\check\alpha}=\Bvalue{\dot x>\check\alpha}$ for every
$\alpha\in\Bbb Q$, that is,

\Centerline{$p\VVdP\,\vec u>\check\alpha$ iff
$p\VVdP\,\dot x>\check\alpha$}

\noindent for every $\alpha\in\Bbb Q$ and $p\in P$, that is,

\Centerline{$p\VVdP\,\check\alpha\in\vec u$ iff
$p\VVdP\,\check\alpha\in\dot x$}

\noindent for every $\alpha\in\Bbb Q$ and $p\in P$, that is (since both
$\vec u$ and $\dot x$ are $\Bbb P$-names for subsets of $\Bbb Q$),

\Centerline{$\VVdP\,\vec u=\dot x$.  \Qed}}

\medskip

\quad{\bf (iv)} It follows that if $\dot x$ is a $\Bbb P$-name and
$p\in P$ is such that $p\VVdP\,\dot x\in\Bbb R$, then there is a
$u\in L^0$ such that $p\VVdP\,\dot x=\vec u$.   \prooflet{(For there is a
$\Bbb P$-name $\dot y$ such that $\VVdP\,\dot y\in\Bbb R$ and
$p\VVdP\,\dot x=\dot y$.)}

\medskip

\quad{\bf (v)} If $\alpha\in\Bbb R$, then
$(\alpha\chi 1)\sspvec=\check\alpha$.
\prooflet{\Prf\ For any $\beta\in\Bbb Q$ and
$p\in P$,

$$\eqalign{\Bvalue{(\alpha\chi 1)\sspvec>\check\beta}
=\Bvalue{\alpha\chi 1>\beta}
&=1\text{ if }\beta<\alpha,\,0\text{ otherwise}\cr
&=\Bvalue{\check\alpha>\check\beta}\text{ in either case}.
  \text{ \Qed}\cr}$$
}

\spheader 5A3Lc Suppose that $u$, $v\in L^0$.

\medskip

\quad{\bf (i)} $\Bvalue{\vec u<\vec v}=\Bvalue{v-u>0}$.   \prooflet{\Prf

$$\eqalignno{\Bvalue{\vec u<\vec v}
&=\Bvalue{\Exists\alpha\in\Bbb Q,\,\vec u\le\alpha<\vec v}\cr
&=\sup_{\alpha\in\Bbb Q}\Bvalue{\vec u\le\check\alpha<\vec v}\cr
\displaycause{taking the supremum in $\RO(\Bbb P)$, 5A3Gc}
&=\sup_{\alpha\in\Bbb Q}(\Bvalue{\vec v>\check\alpha}
  \Bsetminus\Bvalue{\vec u>\check\alpha})\cr
\displaycause{taking the relative complements in $\RO(\Bbb P)$}
&=\sup_{\alpha\in\Bbb Q}(\Bvalue{v>\alpha}
  \Bsetminus\Bvalue{u>\alpha})
=\Bvalue{v-u>0}.  \text{ \Qed}\cr}$$
}%end of prooflet

\medskip

\quad{\bf (ii)} In particular, if $u\le v$ in $L^0$, then

\Centerline{$\VVdP\,\vec u\le\vec v$ in $\Bbb R$\dvro{,}{}}

\noindent\prooflet{since $\Bvalue{\vec v<\vec u}=0$,  }and
$\Bvalue{\vec u=\vec v}=\Bvalue{u-v=0}$ for any $u$, $v\in L^0$.

\medskip

\quad{\bf (iii)} $\VVdP\,(u+v)\sspvec=\vec u+\vec v$.
\prooflet{\Prf\ For any $\alpha\in\Bbb Q$,

$$\eqalignno{\Bvalue{\vec u+\vec v>\check\alpha}
&=\Bvalue{\Exists\,\beta\in\Bbb Q,\,\vec u>\beta,\,
  \vec v>\check\alpha-\beta}\cr
&=\sup_{\beta\in\Bbb Q}\Bvalue{\vec u>\check\beta,\,
  \vec v>\check\alpha-\check\beta}\cr
&=\sup_{\beta\in\Bbb Q}(\Bvalue{\vec u>\check\beta}\Bcap
  \Bvalue{\vec v>(\alpha-\beta)\varBspcheck})\cr
&=\sup_{\beta\in\Bbb Q}
  (\Bvalue{u>\beta}\Bcap\Bvalue{v>\alpha-\beta})\cr
&=\Bvalue{u+v>\alpha}\cr
\displaycause{364D}
&=\Bvalue{(u+v)\sspvec>\check\alpha}.  \text{ \Qed}\cr}$$
}

\medskip

\quad{\bf (iv)} $\VVdP\,(u\times v)\sspvec=\vec u\vec v$.
\prooflet{\Prf\
If $u$, $v\ge 0$ in $L^0$ and $\alpha\ge 0$ in $\Bbb Q$,

$$\eqalignno{\Bvalue{\vec u\vec v>\check\alpha}
&=\Bvalue{\Exists\,\beta\in\Bbb Q,\,\beta>0,\,\vec u>\beta,\,
  \vec v>\Bover{\check\alpha}{\beta}}\cr
&=\sup_{\beta\in\Bbb Q,\beta>0}\Bvalue{\vec u>\check\beta,\,
  \vec v>\Bover{\check\alpha}{\check\beta}}\cr
&=\sup_{\beta\in\Bbb Q,\beta>0}\Bvalue{\vec u>\check\beta}\cap
  \Bvalue{\vec v>(\Bover{\alpha}{\beta})\varBspcheck}\cr
&=\sup_{\beta\in\Bbb Q,\beta>0}\Bvalue{u>\beta}\cap
  \Bvalue{v>\Bover{\alpha}{\beta}}\cr
&=\Bvalue{u\times v>\alpha}\cr
&=\Bvalue{(u\times v)\sspvec>\check\alpha}.\cr}$$

\noindent So in this case

\Centerline{$\VVdP\,(u\times v)\sspvec=\vec u\vec v$.}

\noindent Since we have an appropriate distributive law in $L^0$ (364D), it
follows from (iii) that the same is true for general $u$, $v\in L^0$.\
\Qed}

\medskip

\quad{\bf (v)} If $\alpha\in\Bbb R$, then
$\VVdP\,(\alpha u)\sspvec=\check\alpha\vec u$.  \prooflet{(Put (iv) and
(b-v) together.)}

\spheader 5A3Ld{\bf (i)} Suppose that $\familyiI{u_i}$ is a non-empty family in
$L^0$ with supremum $u\in L^0$.   Then

\Centerline{$\VVdP\,\vec u=\sup_{i\in\check I}\vec u_i$ in $\Bbb R$.}

\prooflet{\noindent\Prf\ By (c-ii),

\Centerline{$\VVdP\,\vec u_i\le\vec u$}

\noindent for every $i\in I$, so

\Centerline{$\VVdP\,\sup_{i\in\check I}\vec u_i\le\vec u$.}

\noindent In the other direction, \Quer\ suppose, if possible, that

\Centerline{$\notVVdash_{\Bbb P}\,\vec u$ is the least upper bound of
$\{\vec u_i:i\in\check I\}$.}

\noindent Then there are a $p\in P$ and an $\alpha\in\Bbb Q$ such that

\Centerline{$p\VVdP\,\check\alpha<\vec u$ is an upper bound for
$\{\vec u_i:i\in\check I\}$.}

\noindent In this case,

\Centerline{$p\in\Bvalue{\vec u>\check\alpha}
=\Bvalue{u>\alpha}=\sup_{i\in I}\Bvalue{u_i>\alpha}
=\interior\overline{\bigcupop_{i\in I}\Bvalue{u_i>\alpha}}$}

\noindent (364L(a-ii), 314P).
There are therefore a $q$ stronger than $p$ and an
$i\in I$ such that $q\in\Bvalue{u_i>\alpha}$;  but in this case
$q\VVdP\,\vec u_i>\check\alpha$, which is impossible, because
$p\VVdP\,\vec u_i\le\check\alpha$.\ \BanG\  So

\Centerline{$\VVdP\,\vec u=\sup_{i\in\check I}\vec u_i$.  \Qed}
}

\medskip

\quad{\bf (ii)} And if $\familyiI{u_i}$ is a non-empty family in
$L^0$ with infimum $u\in L^0$, then

\Centerline{$\VVdP\,\vec u=\inf_{i\in\check I}\vec u_i$ in $\Bbb R$}

\noindent because $\VVdP\,(-u)\sspvec=-\vec u$ (using (c)).

\medskip

\spheader 5A3Le Suppose that $\sequencen{u_n}$ is a sequence in
$L^0$, order*-convergent\cmmnt{ (in the sense of \S367)} to $u\in L^0$.
Then

\Centerline{$\VVdP\,\vec u=\lim_{n\to\infty}\vec u_n$.}

\prooflet{\noindent\Prf\ By 367Gb, 

\Centerline{$u=\sup_{n\in\Bbb N}\inf_{m\ge n}u_m
=\inf_{n\in\Bbb N}\sup_{m\ge n}u_m$}

\noindent so (d) tells us that

\Centerline{$\VVdP\,\vec u=\sup_{n\in\check\Bbb N}\inf_{m\ge n}\vec u_m
=\inf_{n\in\check\Bbb N}\sup_{m\ge n}\vec u_m$,}

\noindent that is,

\Centerline{$\VVdP\,\vec u=\lim_{n\to\infty}\vec u_n$.  \Qed}
}

\leader{5A3M}{Forcing with Boolean algebras}
Suppose that $\frak A$ is a Dedekind complete
Boolean algebra, not $\{0\}$.   \cmmnt{As noted in 5A3Ab,}
$\Bbb P=(\frak A^+,\Bsubseteqshort,1_{\frak A},\downarrow)$ is a forcing
notion.   We have a natural
isomorphism between $\RO(\Bbb P)$ and $\frak A$, matching
each $G\in\RO(\Bbb P)$ with $\sup G\in\frak A$\cmmnt{ (514Sb)};
\cmmnt{by 514M(d-ii),} $\sup G$, taken in $(\frak A^+,\Bsubseteq)$, 
will belong to $G$ unless
$G=\emptyset$.   \cmmnt{In this context,} I
will usually identify the two algebras, so that
$\Bvalue{\phi}$ becomes
$\sup\{a:a\in\frak A^+$, $a\VVdP\,\phi\}$,
and\cmmnt{ we shall have} $\Bvalue{\phi}\VVdP\,\phi$ except when
$\VVdP\,\neg\phi$.   Note that $\Bvalue{\neg\phi}=1\Bsetminus\Bvalue{\phi}$
(5A3Gb).

The identification of $\RO(\Bbb P)$ with $\frak A$ itself
simplifies\cmmnt{ some
of} the discussion in 5A3L.   We have a $\Bbb P$-name $\vec u$ associated
with each $u\in L^0(\frak A)$, and the formula

\Centerline{$\Bvalue{\vec u=\vec v}=\Bvalue{u-v=0}$ in $\RO(\Bbb P)$ 
when $u$, $v\in L^0(\RO(\Bbb P))$}

\noindent of 5A3L(c-ii) turns into

\Centerline{whenever $u$, $v\in L^0(\frak A)$ and $a\in\frak A^+$,
$u\times\chi a=v\times\chi a\iff a\VVdP\,\vec u=\vec v$.}

\prooflet{\noindent\Prf\

$$\eqalignno{a\VVdP\,\vec u=\vec v&
\iff a\in\Bvalue{\vec u=\vec v}\text{ interpreted in }\RO(\Bbb P)\cr&
\iff a\in\Bvalue{u-v=0}\text{ interpreted in }\RO(\Bbb P)
\Displaycause{here thinking of $u$ and $v$ as members of $L^0(\RO(\Bbb P))$}
\iff a\Bsubseteq\Bvalue{u-v=0}\text{ interpreted in }\frak A
\Displaycause{now thinking of $u$ and $v$ as members of $L^0(\frak A)$}
\iff (u-v)\times\chi a=0\text{ in }L^0(\frak A)\cr&
\iff u\times\chi a=v\times\chi a.  \text{ \Qed}\cr}$$
}

\leader{5A3N}{Ordinals and cardinals} Let $\Bbb P$ be a forcing notion, and
$P$ its set of conditions.

\spheader 5A3Na For any ordinal $\alpha$,

\Centerline{$\VVdP\,\check\alpha$ is an ordinal;}

\noindent\cmmnt{moreover, }if $p\in P$ and $\dot x$ is a
$\Bbb P$-name such that

\Centerline{$\VVdP\,\dot x$ is an ordinal,}

\noindent there are a $q$ stronger than $p$ and an ordinal $\alpha$ such
that

\Centerline{$q\VVdP\,\dot x=\check\alpha$}

\cmmnt{\noindent ({\smc Jech 03}, 14.23;  see {\smc Kunen 80}, IV.3.14).}  

\spheader 5A3Nb If $\Bbb P$ is ccc, then

\Centerline{$\VVdP\,\check\kappa$ is a cardinal}

\noindent for every cardinal\cmmnt{ (that is, initial
ordinal)} $\kappa$\cmmnt{ ({\smc Kunen 80}, VII.5.6;  {\smc Jech 03},
14.34)}.   In particular,

\Centerline{$\VVdP\,\check\omega_1$ is a cardinal, so is the first
uncountable cardinal,}

\noindent and we can write

\Centerline{$\VVdP\,\omega_1=\check\omega_1$, $\omega_2=\check\omega_2$}

\noindent etc.\cmmnt{, if we are sure of being understood.}

\spheader 5A3Nc Again suppose that $\Bbb P$ is ccc, and that we have a set $A$,
a $\Bbb P$-name $\dot X$ and a cardinal $\kappa$ such that

\Centerline{$\VVdP\,\dot X\subseteq\check A$ and $\#(\dot X)\le\check\kappa$.}

\noindent Then
there is a set $B\subseteq A$ such that $\#(B)\le\max(\omega,\kappa)$ and

\Centerline{$\VVdP\,\dot X\subseteq\check B$.}

\prooflet{\noindent\Prf\ Let $\dot f$ be a $\Bbb P$-name such that

\Centerline{$\VVdP\,\dot f$ is an injective function with domain $\dot X$ and
$\dot f[\dot X]\subseteq\check\kappa$.}

\noindent For $\xi<\kappa$ set

\Centerline{$B_{\xi}=\{a:a\in A$ and there is a $p\in P$
such that $p\VVdP\,\check a\in\dot X$ \& $\dot f(\check a)=\check\xi\}$.}

\noindent For each $a\in B_{\xi}$, choose $p_{\xi a}\in P$ such that

\Centerline{$p_{\xi a}\VVdP\,\check a\in\dot X$ and
$\dot f(\check a)=\check\xi$;}

\noindent then if $a$, $b\in B_{\xi}$ and $q$ is stronger than both
$p_{\xi a}$ and $p_{\xi b}$, we have

\Centerline{$q\VVdP\,
\dot f(\check a)=\dot f(\check b)$ so $\check a=\check b$}

\noindent and $a=b$ (5A3Hc).   Thus $\family{a}{B_{\xi}}{p_{\xi a}}$ is an
antichain in $\Bbb P$ and $B_{\xi}$ must be countable;  setting
$B=\bigcup_{\xi<\kappa}B_{\xi}$, $B\subseteq A$ and
$\#(B)\le\max(\omega,\kappa)$.

Now suppose that $p\in P$ and that $\dot x$ is a $\Bbb P$-name such that
$p\VVdP\,\dot x\in\dot X$.   Then

\Centerline{$p\VVdP\,\dot x\in\check A$ and
$\dot f(\dot x)\in\check\kappa$,}

\noindent so
there are a $q$ stronger than $p$, an $a\in A$ and a $\xi<\kappa$ such that

\Centerline{$q\VVdP\,\dot x=\check a$ and $\dot f(\dot x)=\check\xi$.}

\noindent Now $a\in B_{\xi}\subseteq B$, so $q\VVdP\,\dot x\in\check B$.
As $p$ and $\dot x$ are arbitrary,

\Centerline{$\VVdP\,\dot X\subseteq\check B$,}

\noindent as required.\ \Qed}

\spheader 5A3Nd If $\Bbb P$ is ccc, then

\Centerline{$\VVdP\,\cff[\check I]^{\le\omega}
=(\cff[I]^{\le\omega})\varBspcheck$}

\noindent for every set $I$.   \prooflet{\Prf\ Write $\kappa$ for
$\cff[I]^{\le\omega}$.   (i) Let
$\Cal K\subseteq[I]^{\le\omega}$ be a cofinal family with
$\#(\Cal K)=\kappa$.   Then
$\VVdP\,\check\kappa$ is a cardinal, so

\Centerline{$\VVdP\,\check{\Cal K}\subseteq[\check I]^{\le\omega}$
and $\#(\check{\Cal K})=\check\kappa$.}

\noindent If $\dot J$ is a $\Bbb P$-name such that

\Centerline{$\VVdP\,\dot J\in[\check I]^{\le\omega}$,}

\noindent then by (c) there is a countable set $K\subseteq I$ such that
$\VVdP\,\dot J\subseteq\check K$;  now there is an $L\in\Cal K$ such that
$K\subseteq L$, and

\Centerline{$\VVdP\,
\dot J\subseteq\check K\subseteq\check L\in\check{\Cal K}$.}

\noindent As $\dot J$ is arbitrary,

\Centerline{$\VVdP\,\check{\Cal K}$ is cofinal with
$[\check I]^{\le\omega}$ and $\cff[\check I]^{\le\omega}\le\check\kappa$.}

\noindent(ii) \Quer\ If

\Centerline{$\neg\VVdP\,\check\kappa\le\cff[\check I]^{\le\omega}$,}

\noindent then there are a $p\in P$ and an ordinal $\delta$ such
that

\Centerline{$p\VVdP\,
\cff[\check I]^{\le\omega}=\check\delta<\check\kappa$.}

\noindent Now there must be a family $\ofamily{\xi}{\delta}{\dot J_{\xi}}$
of $\Bbb P$-names such that

\Centerline{$p\VVdP\,
\{\dot J_{\xi}:\xi<\check\delta\}$
is cofinal with $[\check I]^{\le\omega}$.}

\noindent By (c) again, there must be for each $\xi<\kappa$ a countable
$K_{\xi}\subseteq I$ such that
$p\VVdP\,\dot J_{\xi}\subseteq\check K_{\xi}$.   Because
$\delta<\cff[I]^{\le\omega}$, there is a $K\in[I]^{\le\omega}$ such that
$K\not\subseteq K_{\xi}$ for every $\xi<\delta$.   In this case,

\Centerline{$p\VVdP\,\check K\in[\check I]^{\le\omega}$ so there is a
$\xi<\check\delta$ such that $\check K\subseteq\dot J_{\xi}$,}

\noindent and there must be a $\xi<\delta$ and a $q$ stronger than $p$ such
that

\Centerline{$q\VVdP\,\check K\subseteq\dot J_{\xi}
\subseteq\check K_{\xi}$.}

\noindent But this implies that $K\subseteq K_{\xi}$, which isn't so.\
\Bang

We conclude that

\Centerline{$\VVdP\,\check\kappa\le\cff[\check I]^{\le\omega}$
and $\check\kappa=\cff[\check I]^{\le\omega}$.
\Qed}}

\vleader{72pt}{5A3O}{Iterated forcing}\cmmnt{ ({\smc Kunen 80}, VIII.5.2)}
If $\Bbb P$ is a forcing notion and $P$ its set of conditions,
and we have a quadruple
$\dot{\Bbb Q}=(\dot Q,\dot\le,\dot 1,\dot\epsilon)$
of $\Bbb P$-names such that $(\dot 1,\Bbbone_{\Bbb P})\in\dot Q$ and

\doubleinset{$\VVdP\,\dot\le$ is a pre-order on $\dot Q$,
$\dot\epsilon$ is a direction of activity
and every member of $\dot Q$ is stronger than $\dot 1$,}

\noindent then $\Bbb P*\dot{\Bbb Q}$ is the forcing notion defined by
saying that its conditions are objects of the form $(p,\dot q)$ where

\Centerline{$p\in P$,
\quad$\dot q\in\dom\dot Q$,
\quad$p\VVdP\,\dot q\in\dot Q$,}

\noindent and that $(p,\dot q)$ is stronger than $(p',\dot q')$ if
$p$ is stronger than $p'$ and

\Centerline{$p\VVdP\,\dot q$ is stronger than $\dot q'$.}

\noindent(\cmmnt{Strictly speaking, I should add that}
$\Bbbone_{\Bbb P*\dot\Bbb Q}
=(\Bbbone_{\Bbb P},\dot 1)$.)\cmmnt{\footnote{This
formulation gives us the freedom to take $\dot\epsilon$ to be
non-trivial.
I do not mean to suggest that it would be reasonable to take
advantage of this.}}

\leader{5A3P}{Martin's axiom} Let $\kappa$ be a regular uncountable
cardinal such that $2^{\lambda}\le\kappa$ for every $\lambda<\kappa$.
Then there is a ccc forcing notion $\Bbb P$ such that

\Centerline{$\VVdP\,\frak m=\frak c=\check\kappa$.}

\prooflet{\noindent ({\smc Kunen 80}, VIII.6.3;  {\smc Jech 03}, 16.13).}

\leader{5A3Q}{Countably closed forcings (a)} Let $\Bbb P$ be a forcing
notion, and $P$ its set of conditions.
$\Bbb P$ is {\bf countably closed} if
whenever $\sequencen{p_n}$ is a sequence in $P$ such that $p_{n+1}$ is
stronger than $p_n$ for every $n$, there is a $p\in P$ which is stronger
than every $p_n$.

\spheader 5A3Qb If $\Bbb P$ is a countably closed forcing notion, then
$\VVdP\,\Cal P\Bbb N=(\Cal P\Bbb N)\varBspcheck$.
\prooflet{\Prf\ Let $P$ be the set of
conditions of $\Bbb P$.   If $p\in P$
and $\dot x$ is a $\Bbb P$-name such that
$p\VVdP\,\dot x\subseteq\Bbb N$, choose $\sequencen{p_n}$ inductively in
$P$ such that $p_0=p$ and, for each $n\in\Bbb N$, $p_{n+1}$ is stronger
than $p_n$ and either $p_{n+1}\VVdP\,\check n\in\dot x$ or
$p_{n+1}\VVdP\,\check n\notin\dot x$.   Let $q\in P$ be stronger than every
$p_n$, and set $A=\{n:q\VVdP\,\check n\in\dot x\}$.   Then
$q\VVdP\,\check n\notin\dot x$ for every $n\in\Bbb N\setminus A$, so
$q\VVdP\,\dot x=\check A\in(\Cal P\Bbb N)\varBspcheck$.   As $p$ and $\dot x$
are arbitrary,
$\VVdP\,\Cal P\Bbb N\subseteq(\Cal P\Bbb N)\varBspcheck$.   For the reverse 
inequality, we have $\VVdash\,(\Cal PX)\varBspcheck\subseteq\Cal P\check X$
for any forcing notion and set $X$, so here we have 
$\VVdP\,(\Cal P\Bbb N)\varBspcheck\subseteq\Cal P\check\Bbb N=\Cal P\Bbb N$.\ 
\Qed}

Consequently $\VVdP\,\Bbb R=\check\Bbb R$.   \prooflet{\Prf\ The argument just
above shows that
$\VVdP\,\Cal P\Bbb Q=(\Cal P\Bbb Q)\varBspcheck$,
and now it is easy to see that

$$\eqalign{\VVdP\,\Bbb R
&=\{A:\emptyset\ne A\subseteq\Bbb Q,
  \,A\text{ is bounded above and has no greatest element},\cr
&\mskip 200mu
  q\in A\text{ whenever }q\le q'\in A\}\cr
&=\check\Bbb R.  \text{ \Qed}\cr}$$

\noindent}Similarly, $\VVdP\,[0,1]=[0,1]\varBspcheck$.

%\exercises{\leader{5A3X}{Basic exercises (a)}

%\leader{5A3Y}{Further exercises (a)}

%}%end of exercises


\endnotes{
\Notesheader{5A3} In terms of the discussion in {\smc Kunen 80}, \S VII.9,
you will see that I follow an extreme version of the `syntactical'
approach to forcing.
In the first place, this is due to a philosophical prejudice;  I do not
believe in models of ZF.   But it seems to me that quite apart from this
there is a fundamental difference between the sentences

\Centerline{$\frak m=\frak c$}

\noindent and

\Centerline{$\VVdP\,\,\frak m=\frak c$}

\noindent associated with the fact that the symbols $\frak m$, $\frak c$
and even $=$ must be reinterpreted in the second version.   I have tried in
this section to develop a language which can express and accommodate the
difference.   It puts a substantial burden on the reader, especially in
such formulae as $\sup_{i\in\check I}\dot x_i$ (5A3F) and
$((\dot y,f(\dot y,p)),p)$ (5A3K), where you may have to read quite
carefully to determine which parts of the formulae are supposed to be in
the forcing language, and which are in the ordinary language of set theory.
There is an additional complication in 5A3L, where I use the same symbol
$\Bvalue{\mskip15mu}$ for two quite different functions;
but here at least the
objects $\Bvalue{u>\alpha}$, $\Bvalue{\vec u>\check\alpha}$ belong to the
same set $\RO(\Bbb P)$, even if the formulae inside the brackets have to be
parsed by very different rules.
I hope that the clue of a superscripted letter $\dot x$ or
$\check I$ or $\vec u$ will alert you to the need for thought.
Once we have grasped this nettle, however, we are in a position to move
between the two languages, as in 5A3K;  and statements of results such as
5A3P can be shortened by taking it for granted that the preamble
`$2^{\lambda}\le\kappa$ for every $\lambda<\kappa$' refers to the
ground universe, while the conclusion `$\frak m=\frak c=\check\kappa$' is
to be interpreted in the forcing universe.

Of course a large number of different types of forcing notion have been
described and investigated.   In 5A3N I mention some basic facts about ccc
forcings.   Another important class is that of countably closed forcings
(5A3Q).
}%end of notes

\discrpage



\frfilename{mt5a4.tex}
\versiondate{15.8.13/4.4.22}
\copyrightdate{2007}

\def\chaptername{Appendix}
\def\sectionname{General topology}

\def\Engelking{{\smc Engelking 89}}

\newsection{5A4}

The principal new topological concepts required in this volume are
some of the standard cardinal functions of topology (5A4A-5A4B).
As usual, there are some particularly interesting phenomena involving
compact spaces (5A4C).   For special purposes in \S513, we need to
know some non-trivial facts about metrizable spaces (5A4D).
The rest of the section is made up of scraps which are either elementary or
standard.

\leader{5A4A}{Definitions} Let $(X,\frak T)$ be a topological space.

\spheader 5A4Aa The {\bf weight} of $X$, $w(X)$, is the least cardinal
of any base for $\frak T$.

\spheader 5A4Ab The {\bf $\pi$-weight} of $X$ is
$\pi(X)=\ci(\frak T\setminus\{\emptyset\})$, the smallest cardinal of any
$\pi$-base for $\frak T$.

\spheader 5A4Ac The {\bf density} $d(X)$ of
$X$ is the smallest cardinal of any dense subset of $X$.

\spheader 5A4Ad The {\bf cellularity} of $X$ is

\Centerline{$c(X)
\cmmnt{\mskip5mu=\cdownarrow(\frak T\setminus\{\emptyset\})}
=\sup\{\#(\Cal G):\Cal G\subseteq\frak T\setminus\{\emptyset\}$ is
disjoint$\}$.}

\noindent The {\bf saturation} of $X$ is

\Centerline{$\sat(X)
\cmmnt{\mskip5mu=\sat^{\downarrow}(\frak T\setminus\{\emptyset\})}
=\sup\{\#(\Cal G)^+:\Cal G\subseteq\frak T\setminus\{\emptyset\}$ is
disjoint$\}$\dvro{.}{,}}

\cmmnt{\noindent that is, the smallest cardinal $\kappa$ such that
there is no disjoint family of $\kappa$ non-empty open sets.}

\spheader 5A4Ae The {\bf tightness} of $X$, $t(X)$, is the smallest
cardinal $\kappa$ such that whenever $A\subseteq X$ and
$x\in\overline{A}$ there is a $B\in[A]^{\le\kappa}$ such that
$x\in\overline{B}$.   \cmmnt{(Recall that
$[A]^{\le\kappa}=\{B:B\subseteq A$, $\#(B)\le\kappa\}$.)}
%can drop I think

\spheader 5A4Af The {\bf Nov\'ak number}
$n(X)$ is the smallest cardinal of any family of nowhere dense subsets
of $X$ covering $X$;  or $\infty$ if there is no such family.

\spheader 5A4Ag(i) The {\bf Lindel\"of number} $L(X)$ is the least cardinal $\kappa$
such that every open cover of $X$ has a subcover with cardinal at most
$\kappa$.

\quad(ii) The {\bf hereditary Lindel\"of number} $\hL(X)$ is
$\sup_{Y\subseteq X}L(Y)$.
%5{}31A

\spheader 5A4Ah(i) If $x\in X$, the {\bf character of $x$ in
$X$}, $\chi(x,X)$, is the smallest cardinal of any
base of neighbourhoods of $x$.

\quad(ii) The {\bf character} of $X$ is $\chi(X)=\sup_{x\in X}\chi(x,X)$.

\spheader 5A4Ai The {\bf network weight} of $X$, $\nw(X)$,
is the smallest cardinal of any network for $\frak T$.

%weight:  w(X) Engelking p 12
%character \chi(x,X) \chi(X)  Engelking p 12
%density  d(X)  Engelking p 25
%tightness  t(X)  Engelking p 60
%cellularity=Souslin number  c(X)  Engelking p 59
%network weight  nw(X)  Engelking p 127
%L(X)  Engelking p 127  hL(X) Engelking p 222

\cmmnt{\medskip

\noindent{\bf Remark} Recall that $X$ is called `second-countable'
iff $w(X)\le\omega$, `separable' iff $d(X)\le\omega$, `ccc'
iff $c(X)\le\omega$ (that is, $\sat(X)\le\omega_1$), `Lindel\"of' if
$L(X)\le\omega$, `hereditarily Lindel\"of' if $\hL(X)\le\omega$,
`first-countable' if $\chi(X)\le\omega$
and `countably tight' iff $t(X)\le\omega$.
}%end of comment

\leader{5A4B}{Proposition} Let $(X,\frak T)$ be a topological space.

(a)\dvArevised{2023}

\Centerline{$c(X)\le d(X)
%5{}21Oa
%\le\min(\#(X),\pi(X))
\le\pi(X)\le w(X)\le\#(\frak T)\le 2^{\nw(X)}$,}
%5{}31A

\Centerline{$t(X)\le\chi(X)\le w(X)\le\max(\#(X),\chi(X))$.}
%5{}31Xk

\noindent $\sat(X)=c(X)^+$ unless $\sat(X)$ is weakly inaccessible,
in which case $\sat(X)=c(X)$.
%5{}14Da
%If $X$ is regular then $w(X)\le 2^{d(X)}$.

(b)\dvArevised{2023} If $Y$ is a subspace of $X$,
then $w(Y)\le w(X)$,
%5{}A4Bh
$t(Y)\le t(X)$,
%531U
$\nw(Y)\le\nw(X)$
%5{}31A
and $\chi(y,Y)\le\chi(y,X)$ for every $y\in Y$.
%5{}31P
%$\chi(Y)\le\chi(X)$
%If $Y$ is open, then $\sat(Y)\le\sat(X)$,
%$c(Y)\le c(X)$ and $\pi(Y)\le\pi(X)$.

(c)\dvArevised{2023} If a topological space $Y$ is a continuous
image of $X$, then $d(Y)\le d(X)$,
%5{}36Cf
$c(Y)\le c(X)$,
%5{}A4Be
$t(Y)\le t(X)$,
%5{}31T
$L(Y)\le L(X)$
%5{}34K
and $\nw(Y)\le\nw(X)$.
%5{}21 notes
%$\sat(Y)\le\sat(X)$

(d) If $\Cal G$ is a family of open subsets
of $X$, then there is a subfamily $\Cal H\subseteq\Cal G$ such that
$\#(\Cal H)<\sat(X)$ and
$\overline{\bigcup\Cal H}=\overline{\bigcup\Cal G}$.
%5{}14Db 5{}16Qb

(e) Let $\familyiI{X_i}$ be a family of non-empty topological spaces
with product $X$, and $\lambda$ a cardinal such that
$\#(I)\le 2^{\lambda}$.   Then

%\Centerline{$w(X)\le\max(\omega,\#(I),\sup_{i\in I}w(X_i))$,}

\Centerline{$d(X)\le\max(\omega,\lambda,\sup_{i\in I}d(X_i))$}
%5{}15H %5{}21N %5{}31Xd  cf. 4A2B(e-ii)

%\Centerline{$\pi(X)\le\max(\omega,\#(I),\sup_{i\in I}\pi(X_i))$}

\noindent and

\Centerline{$c(X)
=\sup_{J\subseteq I\text{ is finite}}c(\prod_{i\in J}X_i)$.}
%5{}14Ud
%\le\max(\omega,\sup_{i\in I}\min(d(X_i),2^{c(X_i)}))$.}

(f) If $\Cal G$ is any family of open subsets of $X$, there is an
$\Cal H\subseteq\Cal G$ such that $\#(\Cal H)\le\hL(X)$ and
$\bigcup\Cal H=\bigcup\Cal G$.
%5{}31A

(g) If $X$ is Hausdorff then $\#(X)\le 2^{\max(c(X),\chi(X))}$.
%5{}31Xk

(h) Suppose that $X$ is metrizable.

\quad(i) $d(X)=w(X)$.

\quad(ii) $d(Y)\le d(X)$ for every $Y\subseteq X$.
%5{}29B    %5{}31 5{}28   %5{}34 %5{}28
So any discrete subset of $X$ has cardinal at most $d(X)$.
%5{}31M

%\quad(ii) $c(X)=d(X)$.

\quad(iii) Let $\rho$ be a metric on $X$ defining its topology.   Then $X$
is separable iff there is no uncountable $A\subseteq X$ such that
$\inf_{x,y\in A,x\ne y}\rho(x,y)>0$.
%5{}34D

\proof{{\bf (a)} Let $D\subseteq X$
be a dense set with cardinal $d(X)$.   If
$\Cal G\subseteq\frak T\setminus\{\emptyset\}$ is disjoint, we have a
surjection from $D\cap\bigcup\Cal G$ to $\Cal G$, so
$\#(\Cal G)\le d(X)$;  as $\Cal G$ is arbitrary, $c(X)\le d(X)$.

Let $\Cal U$ be a $\pi$-base for $\frak T$ with cardinal $\pi(X)$.
Then there is a set $D\subseteq X$,
with cardinal at most $\#(\Cal U)$, meeting every non-empty
member of $\Cal U$;  now
$D$ is dense, so $d(X)\le\#(D)\le\pi(X)$.

Any base for $\frak T$ is a $\pi$-base for $\frak T$, so
$\pi(X)\le w(X)$.   Of course $w(X)\le\#(\frak T)$.

Let $\Cal E$ be a network for $\frak T$ with cardinal $\nw(X)$;  then
$\frak T\subseteq\{\bigcup\Cal E':\Cal E'\subseteq\Cal E\}$, so
$\#(\frak T)\le 2^{\#(\Cal E)}=2^{\nw(X)}$.

If $\Cal U$ is a base for $\frak T$ with cardinal $w(X)$, and $x\in X$, then
$\Cal U_x=\{U:x\in U\in\Cal U\}$ is a base of neighbourhoods of $x$, so
$\chi(x,X)\le\#(\Cal U_x)\le w(X)$;  as $x$ is arbitrary,
$\chi(X)\le w(X)$.

If $A\subseteq X$ and $x\in\overline{A}$, there are a base $\Cal V$ of
neighbourhoods of $X$ with $\#(\Cal V)\le\chi(X)$ and a set
$B\in[A]^{\le\chi(X)}$ meeting every member of $\Cal V$, so that
$x\in\overline B$.

If $X$ is finite, every point $x$ of $X$ has a smallest neighbourhood
$V_x$, and $\{V_x:x\in X\}$ is a base for $\frak T$, so $w(X)\le\#(X)$.
If $X$ is infinite, then for each $x\in X$ choose a base $\Cal U_x$ of
neighbourhoods of $x$ with $\#(\Cal U_x)=\chi(x,X)\le\chi(X)$.   Set
$\Cal U=\{\interior U:U\in\bigcup_{x\in X}\Cal U_x\}$;
then $\Cal U$ is a base for $\frak T$ so

\Centerline{$w(X)\le\#(\Cal U)\le\max(\omega,\#(X),\chi(X))
=\max(\#(X),\chi(X))$.}

Taking $P$ to be the partially ordered set
$(\frak T\setminus\{\emptyset\},\subseteq)$, $c(X)=c^{\downarrow}(P)$ and
$\sat(X)=\sat^{\downarrow}(P)$, so 513B, inverted, tells us that
$\sat(X)=c(X)^+$ unless $\sat(X)$ is weakly inaccessible,
in which case $\sat(X)=c(X)$.

\medskip

{\bf (b)} If $\Cal U$ is a base for $\frak T$, then
$\{U\cap Y:U\in\Cal U\}$ is a base for the topology of $Y$,
so $w(Y)\le w(X)$.
If $A\subseteq Y$ and $y\in Y$ belongs to the closure of
$A$ in $Y$ it belongs to the closure of $A$ in $X$ so there is a
$B\in[A]^{\le t(X)}$ such that $y\in\overline B$ taken either in $X$ or in $Y$.
If $\Cal E$ is a network for $\frak T$, then
$\{E\cap Y:E\in\Cal E\}$ is a network for the topology of $Y$,
so $\nw(Y)\le\nw(X)$.
If $y\in Y$ and $\Cal V$ is a base of
neighbourhoods of $y$ in $X$, then $\{V\cap Y:V\in\Cal V\}$ is a base of
neighbourhoods of $y$ in $Y$, so $\chi(y,Y)\le\chi(y,X)$.

\medskip

{\bf (c)} Let $f:X\to Y$ be a continuous surjection.   If $D\subseteq X$ is
dense, then $f[D]$ is
dense in $Y$ (3A3Eb), and $d(Y)\le\#(f[D])\le\#(D)$;  as $D$ is arbitrary,
$d(Y)\le d(X)$.

If $\Cal H$ is a disjoint family of non-empty open sets in $Y$, then
$\Cal G=\{f^{-1}[H]:H\in\Cal H\}$ is a disjoint family of non-empty open
sets in $X$, so $\#(\Cal H)=\#(\Cal G)\le c(X)$;  as $\Cal H$ is arbitrary,
$c(Y)\le c(X)$.

If $B\subseteq Y$ and $y\in\overline{B}$ there is an $x\in X$ such that
$f(x)=y$.  Now $x\in f^{-1}[\overline B]\subseteq\overline{f^{-1}[B]}$ so there
is a $C\subseteq[f^{-1}[B]]^{\le t(X)}$ such that $x\in\overline C$
and $y\in\overline{f[C]}$ while $f[C]\in[B]^{\le t(X)}$.

If $\Cal H$ is an open cover of $Y$, then
$\Cal G=\{f^{-1}[H]:H\in\Cal H\}$ is an open cover of $X$;  let
$\Cal G_0\in[\Cal G]^{\le L(X)}$ be a subcover;  choose
$\Cal H_0\subseteq\Cal H$ such that $\#(\Cal H_0)=\Cal G_0$ and
$\Cal G_0=\{f^{-1}[H]:H\in\Cal H_0\}$;  then $\Cal H_0$ covers $Y$.
As $\Cal H$ is arbitrary, $L(Y)\le L(X)$.

If $\Cal A$ is a network for $\frak T$, then $\{f[A]:A\in\Cal A\}$ is a
network for the topology of $Y$, so $\nw(Y)\le\nw(X)$.

\medskip

{\bf (d)} Let $\Cal V$ be a maximal disjoint family of non-empty open sets
included in members of $\Cal G$.   Then $\#(\Cal V)<\sat(X)$.   Let
$\Cal H\subseteq\Cal G$ be such that $\#(\Cal H)\le\#(\Cal V)$ and every
member of $\Cal V$ is included in a member of $\Cal H$.   If $G\in\Cal G$
then $G\setminus\overline{\bigcup\Cal H}$ meets no member of $\Cal V$, so
must be empty;  so this $\Cal H$ serves.

\medskip

{\bf (e)(i)} By \Engelking, 2.3.15,
$d(X)\le\max(\omega,\lambda,\sup_{i\in I}d(X_i))$.

\medskip

\quad{\bf (ii)} Set
$\kappa=\sup_{J\subseteq I\text{ is finite}}c(\prod_{i\in J}X_i)$.
All the
finite products $\prod_{i\in J}X_i$ are continuous images of $X$, so
$c(X)\ge\kappa$, by (c).   \Quer\ Suppose, if possible, that
$c(X)>\kappa$.   Let $\Cal V$ be the usual base for the topology of $X$,
consisting of sets of the form $\prod_{i\in I}G_i$ where $G_i\subseteq X_i$
is open for every $i$ and $\{i:G_i\ne X_i\}$ is finite.
Let $\ofamily{\xi}{\kappa^+}{W_{\xi}}$ be a disjoint family of non-empty
open sets in $X$.   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 for distinct
$\xi$, $\eta\in A$,

\Centerline{$\emptyset=W'_{\xi}\cap W'_{\eta}
=U_{\xi}\cap U_{\eta}\cap V_{\xi}\cap V_{\eta}$.}

\noindent Since $V_{\xi}$ and $V_{\eta}$ and $U_{\xi}\cap U_{\eta}$ are
determined by coordinates in the disjoint sets $I_{\xi}\setminus J$,
$I_{\eta}\setminus J$ and $J$ respectively, one of them must be empty,
and this can only be $U_{\xi}\cap U_{\eta}$.   Thus
$\family{\xi}{A}{U_{\xi}}$ is disjoint.   But now observe that each
$U_{\xi}$ 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
$\pi_J(x)=x\restr J$ for every $x\in X$.   So $\family{\xi}{A}{H_{\xi}}$
witnesses that $c(\prod_{i\in J}X_i)\ge\kappa^+$, which contradicts the
definition of $\kappa$.\ \Bang

Thus $c(X)=\sup_{J\subseteq I\text{ is finite}}c(\prod_{i\in J}X_i)$.

\medskip

{\bf (f)} This is just because $L(\bigcup\Cal G)\le\hL(X)$.

\medskip

{\bf (g)} If $X$ is finite, $c(X)=\#(X)$ and the result is trivial.
Otherwise, set $\kappa=\max(c(X),\chi(X))$ and for each $x\in X$ let
$\ofamily{\xi}{\kappa}{U_{\xi}(x)}$ run over a base of neighbourhoods of
$x$ consisting of open sets.   Let $f:[X]^2\to[\kappa]^2$ be such that
whenever $x$, $y\in X$ are distinct then there are $\xi$,
$\eta\in f(\{x,y\})$ such that $U_{\xi}(x)$ and $U_{\eta}(y)$ are disjoint.
\Quer\ If $\#(X)>2^{\kappa}$ then by the Erd\H{o}s-Rado theorem
(5A1Ha) there is a $C\subseteq X$ such that $\#(C)>\kappa$ and $f$ is
constant on $[C]^2$;  let $\{\xi,\eta\}$ be the constant value.
For $x\in C$ set $G_x=U_{\xi}(x)\cap U_{\eta}(x)$;  then
$\family{x}{C}{G_x}$ is a disjoint family of non-empty open sets, so
$c(X)\ge\#(C)>\kappa$.\ \Bang

\medskip

{\bf (h)} Fix a metric $\rho$ on $X$ defining its topology.

\medskip

\quad{\bf (i)} If $d(X)<\omega$ then $X$ is finite and the result is
trivial.
Otherwise, let $D$ be a dense subset of $X$ with cardinal $d(X)$;
setting $U(x,\epsilon)=\{y:\rho(y,x)<\epsilon\}$,
$\{U(x,2^{-n}):x\in D$, $n\in\Bbb N\}$ is a base for $\frak T$, so
$w(X)\le\max(\#(D),\omega)=d(X)$.   Since we know from (a) that
$d(X)\le w(X)$, we have equality.

\medskip

\quad{\bf (ii)} Put (i) together with (b) above to see that $d(Y)\le d(X)$.
If $Y$ is discrete, then $\#(Y)=d(Y)\le d(X)$.

\medskip

\quad{\bf (iii)}($\alpha$) If there is an uncountable $A\subseteq X$ such
that $\inf_{x,y\in A,x\ne y}\rho(x,y)>0$, then $A$ is not separable in its
subspace topology, so $X$ is not separable, by (ii).
($\beta$) If there is no such $A$, then
for each $n\in\Bbb N$ let $A_n$ be a maximal subset of
$X$ such that $\rho(x,y)\ge 2^{-n}$ for all distinct $x$, $y\in A_n$.
In this case $\bigcup_{n\in\Bbb N}A_n$ is dense, so
$d(X)\le\max(\omega,\sup_{n\in\Bbb N}\#(A_n))=\omega$ and $X$ is separable.
}%end of proof of 5A4B

\leader{5A4C}{Compactness} Let $X$ be a compact Hausdorff space.

\spheader 5A4Ca{\bf (i)} $\nw(X)=w(X)$.
%5{}31A
\prooflet{(\Engelking, 3.1.19.)}

\medskip

\quad{\bf (ii)} There is a set
$Y\subseteq X$, with cardinal at most the cardinal power
$d(X)^{\omega}$, which meets every
non-empty G$_{\delta}$ subset of $X$.
%5{}21N
\prooflet{\Prf\ Let $D\subseteq X$ be a dense set with cardinal $d(X)$.
For each sequence $\omega\in D^{\Bbb N}$ choose a cluster point
$x_{\omega}$ of $\sequencen{\omega(n)}$;  set
$Y=\{x_{\omega}:\omega\in D^{\Bbb N}\}$.   Then
$\#(Y)\le\#(D^{\Bbb N})=d(X)^{\omega}$.   If $\sequencen{G_n}$ is a
sequence of open sets in $X$ with non-empty intersection, take
$x\in\bigcap_{n\in\Bbb N}G_n$ and choose inductively a sequence
$\sequencen{H_n}$ of open sets such that $x\in H_n$ and
$\overline{H}_{n+1}\subseteq H_n\cap G_n$ for every $n$.   Let
$\omega\in D^{\Bbb N}$ be such that $\omega(n)\in H_n$ for every $n$;
then

\Centerline{$x_{\omega}\in Y\cap\bigcap_{n\in\Bbb N}\overline{H}_n
\subseteq\bigcap_{n\in\Bbb N}G_n$.}

\noindent As $\sequencen{G_n}$ is arbitrary, $Y$ is a suitable set.\ \Qed}
%works for countably compact X, and we could have Y countably compact too

\spheader 5A4Cb
If $X$ is perfectly normal it is first-countable.
%5{}31Q
\prooflet{(Every singleton set in $X$ is a zero set (4A2Fi), so is
a G$_{\delta}$ set;  by 4A2Kf, $X$ is first-countable.)}

\spheader 5A4Cc
If $w(X)\le\kappa$, $X$ is
homeomorphic to a closed subspace of $[0,1]^{\kappa}$.
%5{}33D %5{}36C
\prooflet{(\Engelking, 3.2.5.)}

\spheader 5A4Cd{\bf (i)} If $Y$ is a Hausdorff space and $f:X\to Y$ is a
continuous irreducible surjection, then $d(X)=d(Y)$.
%5{}31Q
\prooflet{\Prf\ We know that $d(Y)\le d(X)$ (5A4Bc).   In the other
direction, let $D\subseteq Y$ be a dense set with cardinal $d(Y)$, and
$C\subseteq X$ a set with cardinal $\#(D)$ such that $f[C]=D$.   If
$G\subseteq X$ is open and not empty, $f[X\setminus G]$ is a closed proper
subset of $Y$ (because $f$ is irreducible),
so $D\not\subseteq f[X\setminus G]$ and
$C\not\subseteq X\setminus G$.   As $G$ is arbitrary, $C$ is dense, and
witnesses that $d(X)\le d(Y)$.\ \Qed}
%$\pi(X)=\pi(Y)$

\medskip

\quad{\bf (ii)} If $f:X\to\{0,1\}^{\kappa}$ is
a continuous irreducible surjection, where $\kappa\ge\omega$, then
$\chi(x,X)\ge\kappa$ for every $x\in X$.
\prooflet{\Prf\ Let $\Cal V$ be a base of neighbourhoods of $x$ with cardinal
$\chi(x,X)$.   For each $\xi<\kappa$, set
$G_{\xi}=\{y:y\in X$, $f(y)(\xi)=f(x)(\xi)\}$.   For $V\in\Cal V$, set
$I_V=\{\xi:\xi<\kappa$, $V\subseteq G_{\xi}\}$;  then
$f[V]\subseteq\{z:z\in\{0,1\}^{\kappa}$, $z\restr I_V=f(x)\restr I_V\}$;
but $f[X\setminus V]$ is a closed proper subset of $\{0,1\}^{\kappa}$, so
$\interior f[V]$ is non-empty and $I_V$ is finite.   As $\Cal V$ is a base
of neighbourhoods of $x$, $\kappa=\bigcup_{V\in\Cal V}I_V$.
As $\kappa$ is infinite, $\Cal V$ is infinite, and
$\kappa\le\#(\Cal V)=\chi(x,X)$.\ \Qed}

\medskip

\quad{\bf (iii)}\dvArevised{2015} So if there is a continuous
surjection from a closed subset of $X$ onto $\{0,1\}^{\kappa}$,
there is a non-empty closed
$K\subseteq X$ such that $\chi(x,K)\ge\kappa$ for every $x\in K$.
%5{}31M %5{}31N %5{}31P
\prooflet{\Prf\ Let $f:F\to\{0,1\}^{\kappa}$ be a continuous surjection,
where $F\subseteq X$ is closed.
By 4A2G(i-i), there is a closed $K\subseteq F$ such that
$f\restr K$ is an irreducible surjection onto $\{0,1\}^{\kappa}$, and we
can use (ii).\ \Qed}

\medskip

\quad{\bf (iv)} If $Y$ and $Z$ are Hausdorff spaces and $f:X\to Y$,
$g:Y\to Z$ are continuous irreducible surjections then
$gf:X\to Z$ is irreducible.
%5{}31Q
\prooflet{(If $F\subseteq X$ is a closed proper subset, then $f[F]$ is
a closed proper subset of $Y$ and $g[f[F]]\ne Z$.)}

\spheader 5A4Ce If $\sequencen{x_n}$
is a sequence in $X$ with at most one cluster point in
$X$, then $\sequencen{x_n}$ is convergent.
%5{}36C
\prooflet{\Prf\ Because $X$ is compact, $\sequencen{x_n}$ has at least one
cluster point;  let $x$ be such a point.   \Quer\ If $\sequencen{x_n}$ does
not converge to $x$, let $G$ be an open set containing $x$ such that
$J=\{n:n\in\Bbb N$, $x_n\notin G\}$ is infinite.   Then there must be a
point $y$ in $\bigcap_{n\in\Bbb N}\overline{\{x_i:i\in J\setminus n\}}$;
and now $y$ is a cluster point of $\sequencen{x_n}$ in $X\setminus G$, so
cannot be equal to $x$.\ \Bang\Qed}

\spheader 5A4Cf Let $Y$ be a Hausdorff space and
$f:X\to Y$ a continuous function.   If $\Cal E$ is a non-empty
downwards-directed family of closed subsets of $X$, then
$f[\bigcap\Cal E]=\bigcap_{F\in\Cal E}f[F]$.
%5{}36C
\prooflet{\Prf\ Of course
$f[\bigcap\Cal E]\subseteq\bigcap_{F\in\Cal E}f[F]$.
If $y\in\bigcap_{F\in\Cal E}f[F]$, then
$\{F\cap f^{-1}[\{y\}]:F\in\Cal E\}$ is a downwards-directed family of
closed subsets of $X$, so has non-empty intersection;  and any point of the
intersection witnesses that $y\in f[\bigcap\Cal E]$.\ \Qed}

\leader{5A4D}{Vietoris topologies:  Proposition}
Let $X$ be a separable metrizable space and $\Cal K$
the set of compact subsets of $X$ with the topology induced by the Vietoris
topology on the set of closed subsets of $X$\cmmnt{ (4A2T)}.
%Note that this is the topology defined by a Hausdorff metric.

(a) $\Cal K$ is second-countable.
%actually it is separable and metrizable.

(b) If $Y$ is a topological space and $R\subseteq Y\times X$ is
usco-compact, then $y\mapsto R[\{y\}]:Y\to\Cal K$ is Borel measurable.
%5{}13N

(c) There is a sequence $\sequencen{f_n}$ of Borel measurable
functions from $\Cal K\setminus\{\emptyset\}$ to $X$ such that
$\{f_n(K):n\in\Bbb N\}$ is a dense subset of $K$ for every
$K\in\Cal K\setminus\{\emptyset\}$.
%5{}13L
%maybe bits of (b) and (c) work without "separable"

\proof{{\bf (a)} Let $\Cal U$ be a
countable base for the topology of $X$.
Let $\Cal V$ be the family of sets of the form

\Centerline{$\{K:K\in\Cal K$, $K\cap U_i\ne\emptyset$ for $i<n$,
$K\subseteq\bigcup_{i<n}U_i\}$}

\noindent where $n\in\Bbb N$ and $U_i\in\Cal U$ for $i<n$;  then
$\Cal V$ is a countable family of
open sets in $\Cal K$ and is a base for the topology of $\Cal K$.

\medskip

{\bf (b)} If $G\subseteq X$ is open, then

\Centerline{$\{y:y\in Y$, $R[\{y\}]\subseteq G\}
=Y\setminus R^{-1}[X\setminus G]$}

\noindent is open.   Also $G$ can be expressed as
$\bigcup_{n\in\Bbb N}F_n$ where every $F_n$ is closed, so

\Centerline{$\{y:R[\{y\}]\cap G\ne\emptyset\}
=\bigcup_{n\in\Bbb N}R^{-1}[F_n]$}

\noindent is F$_{\sigma}$, therefore Borel.   Thus

\Centerline{$\Cal W=\{W:W\subseteq\Cal K$, $\{y:R[\{y\}]\in W\}$ is
Borel$\}$}

\noindent includes a
subbase for the topology of $\Cal K$.   It therefore includes a base;
because $\Cal K$ is second-countable, therefore hereditarily Lindel\"of,
every open set is a countable union of members of
$\Cal W$ and belongs to $\Cal W$, that is,
$y\mapsto R[\{y\}]$ is Borel measurable.

\medskip

{\bf (c)(i)} Note first that if $G\subseteq X$ is open, then
$K\mapsto\overline{K\cap G}:\Cal K\to\Cal K$ is Borel measurable.   \Prf\
If $H\subseteq X$ is open, then

\Centerline{$\{K:\overline{K\cap G}\cap H\ne\emptyset\}
=\{K:K\cap(G\cap H)\ne\emptyset\}$}

\noindent is open.   Next, we can express $H$ as the union
$\bigcup_{n\in\Bbb N}H_n$ of a non-decreasing
sequence of open sets such that
$\overline{H}_n\subseteq H$ for every $n$, so

\Centerline{$\{K:\overline{K\cap G}\subseteq H\}
=\bigcup_{n\in\Bbb N}\{K:K\cap G\subseteq\overline{H}_n\}
=\bigcup_{n\in\Bbb N}\{K:K\cap(G\setminus\overline{H}_n)=\emptyset\}$}

\noindent is F$_{\sigma}$, therefore Borel.   As in (b), this is enough.\
\Qed

\medskip

\quad{\bf (ii)} Let $\sequencen{U_n}$ run over a base for the topology of
$X$.   For each $n\in\Bbb N$ define $g_n:\Cal K\to\Cal K$ by setting

$$\eqalign{g_n(K)
&=\overline{K\cap U_n}\text{ if }K\cap U_n\ne\emptyset,\cr
&=K\text{ otherwise}.\cr}$$

\noindent Since $\{K:K\cap U_n\ne\emptyset\}$ is open,
(i) tells us that $g_n$ is Borel measurable.   Set $h_n=g_n\ldots g_1g_0$;
then $h_n$ also is Borel measurable, for each $n$.   Now, for each
$K\in\Cal K\setminus\{\emptyset\}$, $\sequencen{h_n(K)}$ is a
non-increasing sequence of non-empty compact sets, so has non-empty
intersection.   Morover, for each $n$, $h_n(K)$ is either disjoint from
$U_n$ or included in $\overline{U}_n$;  so
$\bigcap_{n\in\Bbb N}h_n(K)$ has exactly one point;  call this point
$f(K)$.   Of course $f(K)\in h_0(K)\subseteq K$.

Now $f:\Cal K\setminus\{\emptyset\}\to X$ is Borel measurable.   \Prf\
If $F\subseteq X$ is closed, then

\Centerline{$f^{-1}[F]
=\bigcap_{n\in\Bbb N}\{K:F\cap h_n(K)\ne\emptyset\}$}

\noindent is a Borel set because every $h_n$ is Borel measurable and
$\{K:F\cap K=\emptyset\}$ is open.\ \Qed

\medskip

\quad{\bf (iii)} Set $f_n=fg_n$ for each $n$.   Then $f_n(K)\in K$ for
every $n\in\Bbb N$ and $K\in\Cal K\setminus\{\emptyset\}$,
$f_n:\Cal K\setminus\{\emptyset\}\to X$ is Borel measurable for each $n$,
and $f_n(K)\in\overline{K\cap U_n}$ whenever $K\cap U_n\ne\emptyset$;
so $\{f_n(K):n\in\Bbb N\}$ is dense in $K$ for every
$K\in\Cal K\setminus\{\emptyset\}$.
}%end of proof of 5A4D

\leader{5A4E}{Category and the Baire property}
Let $X$ be a topological space;  write $\widehat{\Cal B}(X)$ for its
Baire-property algebra\cmmnt{ (4A3R\formerly{4{}A3Q})}.

\spheader 5A4Ea Suppose that
$\familyiI{G_i}$ is a disjoint family of open sets and
$\familyiI{E_i}$ is a family of nowhere dense sets.   Then
$\bigcup_{i\in I}G_i\cap E_i$ is nowhere dense.
%5{}17M
\prooflet{(Elementary;  see (a-i) of the proof of 4A3S\Latereditions.)}

\spheader 5A4Ec Let $Y$ be another topological space.

\medskip

\quad{\bf (i)}
If $A\subseteq X$ is nowhere dense in $X$, then $A\times Y$ is
nowhere dense in $X\times Y$.
\prooflet{($\overline{A\times Y}=\overline{A}\times Y$.)}
So if $A\subseteq X$ is meager in $X$, then $A\times Y$ is meager in
$X\times Y$.

\medskip

\quad{\bf (ii)} $\widehat{\Cal B}(X)\tensorhat\widehat{\Cal B}(Y)
\subseteq\widehat{\Cal B}(X\times Y)$.
%5{}54E
\prooflet{\Prf\ If $E\in\widehat{\Cal B}(X)$, let $G\subseteq X$ be such
that $E\symmdiff G$ is meager;  then

\Centerline{$E\times Y=(G\times Y)\symmdiff((E\symmdiff G)\times Y)
\in\widehat{\Cal B}(X\times Y)$.}

\noindent
Similarly, $X\times F\in\widehat{\Cal B}(X\times Y)$ for every
$F\in\widehat{\Cal B}(Y)$.   Because $\widehat{\Cal B}(X\times Y)$ is a
$\sigma$-algebra of sets, it includes
$\widehat{\Cal B}(X)\tensorhat\widehat{\Cal B}(Y)$.\ \Qed}

\medskip

\quad{\bf (iii)} If $Y$ is compact, Hausdorff
and not empty, then a set $A\subseteq X$ is
meager in $X$ iff $A\times Y$ is meager in $X\times Y$.   \prooflet{\Prf\
We saw in (i) that if $A$ is meager then $A\times Y$ is meager.   In the
other direction, if $A\times Y$ is meager in $X\times Y$,
let $\sequencen{W_n}$ be a sequence of dense open subsets of $X\times Y$
such that $\bigcap_{n\in\Bbb N}W_n$ is disjoint from $A\times Y$.   Choose
$\sequencen{V_n}$ inductively, as follows.   $V_0=X\times Y$.   Given that
$V_n$ is an open subset of $X\times Y$ such that $\pi_1[V_n]$ is dense in
$X$, where $\pi_1$ is the projection from $X\times Y$ onto $X$, then
$\pi_1[V_n\cap W_n]$ is dense in $X$.   Set

\Centerline{$\Cal V_n=\{G\times H:G\subseteq X$ is open, $H\subseteq Y$ is
open, $G\times\overline{H}\subseteq V_n\cap W_n\}$.}

\noindent Because $Y$ is regular, $\bigcup\Cal V_n$ is dense in
$V_n\cap W_n$ and $\pi_1[\bigcup\Cal V_n]$ is dense in $X$.
Let $\Cal V'_n\subseteq\Cal V_n$ be a maximal family such that
$\pi_1[V]\cap\pi_1[V']$ is empty whenever $V$, $V'\in\Cal V'_n$ are
disjoint;  because $G'\times H\in\Cal V_n$ whenever $G\times H\in\Cal V_n$
and $G'$ is an open subset of $G$, $\pi_1[\bigcup\Cal V'_n]$ is dense in
$X$.   Set $V_{n+1}=\bigcup\Cal V'_n$, and continue.

If
$x\in\bigcap_{n\in\Bbb N}\pi_1[V_n]$, $\sequencen{\overline{V_n[\{x\}]}}$
is a non-increasing sequence of non-empty closed subsets of $Y$, so there
is a
$y\in\bigcap_{n\in\Bbb N}\overline{V_n[\{x\}]}$, because $Y$ is compact.
For each $n$, there is a $V\in\Cal V_n$ such that $x\in\pi_1[V]$,
so $V_{n+1}[\{x\}]=V[\{x\}]$ and $(x,y)\in W_n$.   Thus
$x\in\pi_1[\bigcap_{n\in\Bbb N}W_n]$ and $x\notin A$.   As $x$ is
arbitrary, $A$ is disjoint from $\bigcap_{n\in\Bbb N}\pi_1[V_n]$ and is
meager.\ \Qed}

%works for weakly \alpha-favourable Y

\leaveitout{If $\add\Cal M(X)<\sat(Y)$, we have nwd $A_{\xi}\subseteq X$,
disjoint open $H_{\xi}\subseteq Y$, $\bigcup_{\xi}A_{\xi}\times H_{\xi}$
nwd.}

\spheader 5A4Ed Suppose that $X$ is completely regular and ccc.

\medskip

\quad{\bf (i)}
Every nowhere dense subset of $X$ is included in a nowhere dense
zero set.   \prooflet{\Prf\ If $E\subseteq X$ is nowhere dense, let
$\Cal G$ be a maximal disjoint family of cozero sets included in
$X\setminus E$.   Because $X$ is ccc, $\Cal G$ is countable, and
$G=\bigcup\Cal G$ is a cozero set.   Because $X$ is completely regular,
$X\setminus(G\cup E)$ is nowhere dense and $X\setminus G$ is a nowhere
dense zero set including $E$.\ \Qed}

\medskip

\quad{\bf (ii)}
Every meager subset of $X$ is included in a meager Baire set.
%5{}54E
\prooflet{(By (i), it is included in the union of a sequence of nowhere
dense zero sets.)}

\medskip

\quad{\bf (iii)}\dvAnew{2015}
Every subset of $X$ with the Baire property is expressible
as $G\symmdiff M$ where $G$ is a cozero set and $M$ is meager.
\prooflet{\Prf\ If $E\in\widehat{\Cal B}(X)$, express it as
$G_0\symmdiff M_0$ where $G_0$ is open and $M_0$ is meager.   Let
$\Cal H$ be a maximal disjoint family of cozero subsets of $G_0$;  as
$X$ is ccc, $\Cal H$ is countable and $G=\bigcup\Cal H$ is a cozero set;
as $X$ is completely regular, $G_0\subseteq\overline{G}$ and
$G_0\setminus G$ is nowhere dense.   So $M=(G_0\setminus G)\symmdiff M_0$
is meager and $E=G\symmdiff M$ is in the required form.\ \Qed}

\leader{5A4F}{Normal and paracompact spaces (a)}
For a normal space $X$ and an infinite
set $I$, the following are equiveridical: (i) there is a
continuous surjection from $X$ onto $[0,1]^I$;  (ii) there is a
continuous surjection from a closed subset of $X$ onto
$\{0,1\}^I$.
%5{}31L
\prooflet{\Prf\ (i)$\Rightarrow$(ii) is elementary, as $\{0,1\}^I$ is a
closed subset of $[0,1]^I$.   So suppose that (ii) is true.
The map $x\mapsto\sum_{n=0}^{\infty}2^{-n-1}$ is a
continuous surjection from $\{0,1\}^{\Bbb N}$ onto $[0,1]$;  there is
therefore a continuous surjection from $\{0,1\}^{I\times\Bbb N}$ onto
$[0,1]^I$;  but $I$ is infinite, so $\{0,1\}^I$ is homeomorphic to
$\{0,1\}^{I\times\Bbb N}$.   We therefore have a continuous surjection from
$\{0,1\}^I$ onto $[0,1]^I$.   Accordingly there is a continuous surjection
$f$ from a closed subset $F$ of $X$ onto $[0,1]^I$.
Set $f_i(x)=f(x)(i)$ for $x\in F$ and
$i\in I$;  by Tietze's theorem (4A2F(d-ix)),
there is a continuous $g_i:X\to[0,1]$ extending $f_i$;  now
$x\mapsto\familyiI{g_i(x)}:X\to[0,1]^I$ is a continuous
surjection, and (i) is true.\ \Qed}

\spheader 5A4Fb\dvArevised{2022}
Suppose that $X$ is a paracompact normal space and
$\Cal G$ is an open cover of $X$.   Then there is a continuous
pseudometric
$\rho:X\times X\to\coint{0,\infty}$ such that whenever
$\emptyset\ne A\subseteq X$
and $\sup_{x,y\in A}\rho(x,y)\le 1$ there is a $G\in\Cal G$ such
that $A\subseteq G$.
\prooflet{\Prf\ If $X$ is empty this is trivial;  suppose otherwise.
There is a locally finite open cover $\Cal H$ of $X$ refining
$\Cal G$.   Enumerate $\Cal H$ as $\ofamily{\xi}{\kappa}{H_{\xi}}$.
By 4A2F(d-vi), there is a family $\ofamily{\xi}{\kappa}{U_{\xi}}$ of open
sets, covering $X$, such that $\overline{U}_{\xi}\subseteq H_{\xi}$
for every $\xi<\kappa$.

Now for each $\xi<\kappa$ there is a continuous function
$f_{\xi}:X\to[0,1]$ such that $f_{\xi}(x)=1$ for $x\in\overline{U}_{\xi}$ and
$f_{\xi}(x)=0$ for $x\in X\setminus H_{\xi}$ (4A2F(d-i)).
Set $\rho(x,y)=2\sum_{\xi<\kappa}|f_{\xi}(x)-f_{\xi}(y)|$ for $x$, $y\in X$.
Then $\rho$ is a pseudometric on $X$, and is continuous
because $\ofamily{\xi}{\kappa}{f_{\xi}^{-1}[\,\ocint{0,1}\,]}$ is locally finite.
If $A\subseteq X$ is a non-empty set such that $\rho(x,y)\le 1$ for all
$x$, $y\in A$, take any $x\in A$.  There is a $\xi<\kappa$ such that
$x\in U_{\xi}$.   If $y\in A$,
$1-f_{\xi}(y)=|f_{\xi}(x)-f_{\xi}(y)|\le\bover12$ so $y\in H_{\xi}$.
Let $G\in\Cal G$ be such that $H_{\xi}\subseteq G$;  then  $A\subseteq G$, as
required.\ \Qed}
%5{}34Xj

\leader{5A4G}{Baire $\sigma$-algebras (a)} Let $X$ be a topological space.
Write $\CalBa_0(X)$ for the set of
cozero sets in $X$ and for ordinals $\alpha>0$ set

\Centerline{$\CalBa_{\alpha}(X)=\{\bigcup_{n\in\Bbb N}(X\setminus E_n):
\sequencen{E_n}$ is a sequence in
$\bigcup_{\beta<\alpha}\CalBa_{\beta}(X)\}$.}

\noindent Then the Baire $\sigma$-algebra $\CalBa(X)$ of $X$ is
$\bigcup_{\alpha<\omega_1}\CalBa_{\alpha}(X)$.
%5{}51F
\prooflet{\Prf\ Inducing on $\alpha$, we see that $\CalBa_{\alpha}(X)$
is included in the Baire $\sigma$-algebra of $X$ for every $\alpha$;
and $\bigcup_{\alpha<\omega_1}\CalBa_{\alpha}(X)$ is a $\sigma$-algebra of
sets containing every cozero set, so includes the Baire $\sigma$-algebra.\
\Qed}

\spheader 5A4Gb\dvAnew{2014}{\bf (i)}
If $\familyiI{X_i}$ is a family of separable
metrizable spaces with product $X$, then
$\#(\CalBa(X))\le\max(\frak c,\#(I)^{\omega})$.   \prooflet{\Prf\
By 4A3Na, $\CalBa(X)=\Tensorhat_{i\in I}\Cal B(X_i)$, where $\Cal B(X_i)$
is the Borel $\sigma$-algebra of $X_i$ for each $i$.   By 4A3Fa,
$\#(\Cal B(X_i))\le\frak c$ for each $i$, so

\Centerline{$\Cal E=\{\{x:x\in X$, $x(i)\in E\}:
i\in E$, $E\in\Cal B(X_i)\}$}

\noindent has cardinal at most $\max(\frak c,\#(I))$ and the
$\sigma$-algebra $\CalBa(X)$ it generates has cardinal at most
$\max(\frak c,\#(I))^{\omega}=\max(\frak c,\#(I)^{\omega})$ (4A1O).\ \Qed}

\medskip

\quad{\bf (ii)} If $\kappa\ge 2$ is a cardinal,
then the set\cmmnt{ $F$} of
Baire measurable functions from $\{0,1\}^{\kappa}$ to $\{0,1\}^{\omega}$
has cardinal $\kappa^{\omega}$.   \prooflet{\Prf\ The map

\Centerline{$f\mapsto\sequence{i}{\{x:f(x)(i)=1\}}:
F\to\CalBa(\{0,1\}^{\kappa})^{\Bbb N}$}

\noindent is bijective, so

\Centerline{$\#(F)=\#(\CalBa(\{0,1\}^{\kappa}))^{\omega}
\le(\max(\frak c,\kappa)^{\omega})^{\omega}=\kappa^{\omega}$.}

\noindent Of course $\#(\CalBa(\{0,1\})^{\kappa})\ge\kappa$ so
$\#(F)=\kappa^{\omega}$.\ \Qed}

\leader{5A4H}{Proposition} If $X$ is a compact metrizable space and
$(Y,\rho)$ a complete separable metric space, then
$C(X;Y)$, with the topology of uniform convergence, is Polish.
\prooflet{(\Engelking, 4.3.13 and 4.2.18.)}

\leader{5A4I}{Old friends (a)} The weight of the Stone-\v{C}ech
compactification $\beta\Bbb N$ is $\frak c$.
%5{}36C 5{}31Xd
\prooflet{(\Engelking, 3.6.11.)}

\spheader 5A4Ib{(i)} For any infinite $I$, there is a continuous surjection
from $\{0,1\}^I$ onto $[0,1]^I$.
%5{}31L
\prooflet{(Immediate from 5A4Fa, or otherwise.)}

\quad{(ii)} There is a continuous surjection from $[0,1]$ onto
$[0,1]^{\Bbb N}$.
%5{}31L
\prooflet{(The Cantor set $C\subseteq[0,1]$ is homeomorphic to
$\{0,1\}^{\Bbb N}$ (4A2Uc), so again 5A4Fa gives the result.   Or see
416Yi.)}

\quad{(iii)}\dvAnew{2023}
For any infinite $\kappa$, $t(\{0,1\}^{\kappa})=\kappa$.
\prooflet{\Prf\ $t(\{0,1\}^{\kappa})\le w(\{0,1\}^{\kappa})=\kappa$.
In the other directon, set $A=\{\chi I:I\in[\kappa]^{<\omega}\}$;  then
$\chi\kappa\in\overline A\setminus\overline B$ for every $B\in[A]^{<\kappa}$,
so $\kappa\le t(\{0,1\}^{\kappa})$.\ \Qed}

\spheader 5A4Ic If $X$ is a non-empty zero-dimensional
compact metrizable space without isolated points, it is homeomorphic to
$\{0,1\}^{\Bbb N}$.
%5{}{}38H
\prooflet{\Prf\ Let $\frak B$ be the algebra of open-and-closed subsets
of $X$.   Because $X$ has no isolated points, $\frak B$ is atomless
(316Lb).   We know that $X$ is second-countable (4A2P(a-ii));  let $\Cal U$
be a countable base for its topology;  then every member of $\frak B$ is
open, so expressible as a union of members of $\Cal U$, and compact, so
expressible as the union of a finite subset of $\Cal U$.   Accordingly
$\frak B$ is countable;  and as $X\ne\emptyset$, $\frak B\ne\{0\}$.
By 316M, $\frak B$ is isomorphic to the algebra
of open-and-closed subsets of $\{0,1\}^{\Bbb N}$;  by 311J, $X$ and
$\{0,1\}^{\Bbb N}$ are homeomorphic.\ \Qed}

\spheader 5A4Id Let $X$ be a non-empty zero-dimensional Polish space in
which no non-empty open set is compact.   Then $X$ is homeomorphic to $\NN$
with its usual topology.  \prooflet{\Prf\ Let $\rho$ be a complete metric
on $X$ defining its topology.   (i) If $U\subseteq X$ is a non-empty open
set and $\epsilon>0$, there is a partition $\sequencen{U_n}$ of $U$ into
non-empty open-and-closed sets of diameter at most $\epsilon$.
To see this, note that as $U$ is not compact,
there is a sequence $\sequencen{x_n}$ in $U$ with no cluster point in $U$
(4A2Le).   Let $\Cal V$ be the family of subsets of $U$, of diameter at
most $\epsilon$, which are open-and-closed in $X$ and contain $x_i$ for at
most finitely many $i$.   Because $X$ is zero-dimensional, $\Cal V$ is a
base for the subspace topology of $U$.
Because $U$ is Lindel\"of (4A2P(a-iii)), there is
a sequence $\sequencen{V_n}$ in $\Cal V$ covering $U$;  set
$V'_n=V_n\setminus\bigcup_{i<n}V_i$ for each $n$, so that
$\sequencen{V'_n}$ is a disjoint sequence in $\Cal V$ covering $U$.
Because no $V'_n$ can contain infinitely many of the $x_i$,
$I=\{n:V'_n\ne\emptyset\}$ is infinite, and we can re-enumerate
$\family{n}{I}{V'_n}$ as $\sequencen{U_n}$ to get an appropriate sequence.
(ii) Now set $S=\bigcup_{n\in\Bbb N}\BbbN^n$ and define
$\family{\sigma}{S}{U_{\sigma}}$ inductively in such a way that
$U_{\emptyset}=X$ and

\inset{$\sequencen{U_{\sigma^{\smallfrown}\fraction{n}}}$
is a partition of $U_{\sigma}$
into non-empty open-and-closed sets of diameter at most $2^{-k}$ whenever
$k\in\Bbb N$ and $\sigma\in\BbbN^k$.}

\noindent For $\alpha\in\NN$, $\sequence{k}{U_{\alpha\restr k}}$ is a
non-increasing sequence of non-empty closed sets and
$\diam U_{\alpha\restr k}\le 2^{-k+1}$ for every $k\ge 1$, so there is
exactly one point in $\bigcap_{k\in\Bbb N}U_{\alpha\restr k}$;  let
$f(\alpha)$ be this point.   This defines a function $f:\NN\to X$.
(iii) Because $\sequencen{U_{\sigma^{\smallfrown}\fraction{n}}}$
is a partition of $U_{\sigma}$ for every $\sigma$, $f$ is a bijection.
(iv) If $G\subseteq X$ is open, then

\Centerline{$f^{-1}[G]=\{\alpha:$ there is some $k\in\Bbb N$ such that
$U_{\alpha\restr k}\subseteq G\}$}

\noindent is open, so $f$ is continuous.   (v) If $H\subseteq\NN$ is open,
then

\Centerline{$f[H]=\bigcup\{U_{\sigma}:\sigma\in S$,
$\{\alpha:\sigma\subseteq\alpha\in\NN\}\subseteq H\}$}

\noindent is open, so $f$ is a homeomorphism.\ \Qed}
%5{}A4Je

\spheader 5A4Ie If $X$ is a non-empty Polish space without
isolated points, then it has a dense G$_{\delta}$ set which is
homeomorphic to $\NN$ with its usual topology.
%5{}22Vb
\prooflet{\Prf\ Let $\Cal U$ be a countable base for the topology of $X$,
and $D$ a countable dense subset of $X$;   set

\Centerline{$Y=X\setminus(D\cup\bigcup_{U\in\Cal U}\partial U)$.}

\noindent
Then $Y$ is a G$_{\delta}$ set in $X$, so is Polish (4A2Qd).
Because $X$ has no isolated points, $Y$ is comeager in $X$ and is dense and
not empty.
Because $\{U\cap Y:U\in\Cal U\}$ is a base for the topology of $Y$
consisting of relatively open-and-closed sets, $Y$ is zero-dimensional.
If $V\subseteq Y$ is a non-empty relatively open set, let $G\subseteq X$ be
an open set such that $G\cap Y=V$;  then $D\cap G$ is non-empty, so there
is a sequence in $V$ converging (in $X$) to a point in
$D\cap G\subseteq X\setminus V$, and $V$
cannot be compact.   By (d), $Y$ is homeomorphic to $\NN$.\ \Qed}

\spheader 5A4If\dvAnew{2022} If $X$ is any zero-dimensional Polish space there
is a closed
subspace of $\NN$ homeomorphic to $X$.  \prooflet{ Let $\rho$ be a complete
metric on $X$ defining its topology.   For each $n\in\Bbb N$,
$\Cal V_n=\{V:V\subseteq X$ is open-and-closed, $\diam(V)\le 2^{-n}\}$ is an
open cover of $X$;  let $\sequence{i}{V_{ni}}$ be a sequence in $\Cal V_n$
covering $X$;  set $U_{ni}=V_{ni}\setminus\bigcup_{j<i}V_{nj}$ for $i\in\Bbb N$,
so that $\sequence{i}{U_{ni}}$ is a partition of $X$ into open-and-closed sets
of diameter at most $2^{-n}$.   For $n\in\Bbb N$ and $x\in U_{ni}$, set
$f(x)(n)=i$;  then $f$ is a continuous function from $X$ to $\NN$.   If $x$,
$y\in X$, $n\in\Bbb N$ and $f(x)(n)=f(y)(n)=i$, then
$\rho(x,y)\le\diam U_{ni}\le 2^{-n}$, so $f$ is injective.   Because
$\{f[U_{ni}]:n$, $i\in\Bbb N\}
=\{f[X]\cap\{\alpha:\alpha(n)=i\}:n$, $i\in\Bbb N\}$
is a subbase for the topology of $\NN$, while
$\{U_{ni}:n$, $i\in\Bbb N\}$ is a subbase for the topology of $X$, $f$ is a
homeomorphism between $X$ and $f[X]$.   Finally, to see that $f[X]$ is a closed
subset of $\NN$, take a sequence $\sequence{m}{x_m}$ in $X$ such that
$\sequence{m}{f(x(m))}\to\alpha$ in $\NN$.   For any $n\in\Bbb N$ there is a
$k\in\Bbb N$ such that $f(x(m))(n)=\alpha(n)$ for every $m\ge k$, so that
$x(m)\in U_{n,\alpha(n)}$ for $m\ge k$, and $\rho(x(m),x(k))\le 2^{-n}$ for
$m\ge k$.   Thus $\sequence{m}{x(m)}$ is a Cauchy sequence in $X$ with a limit
$x\in X$, and $\alpha=f(x)$ belongs to $f[X]$.\ \Qed}



\leader{5A4J}{Compact-open topologies}\dvAnew{2015} We shall need a
couple of elementary
facts concerning some spaces of continuous functions.

\spheader 5A4Ja Let $X$ and $Y$ be topological spaces and $F$ a set of
functions from $X$ to $Y$.   The {\bf compact-open} topology on $F$
is the topology generated by sets of the form
$\{f:f\in F$, $f[K]\subseteq H\}$ where $K\subseteq X$ is compact and
$H\subseteq Y$ is open.   \cmmnt{(Cf.\ \Engelking, \S3.4.)}

\spheader 5A4Jb Let $X$ be a topological space and $\familyiI{Y_i}$ a
family of regular spaces, with product $Y$.   Set $\pi_iy=y(i)$ for
$i\in I$ and $y\in Y$.   Then
$g\mapsto\familyiI{\pi_ig}:C(X;Y)\to\prod_{i\in I}C(X;Y_i)$ is a
homeomorphism for the compact open topologies on $C(X;Y)$ and the
$C(X;Y_i)$.   \prooflet{\Prf\ ($\alpha$)
A function $g:X\to Y$ is continuous iff
$\pi_ig$ is continuous for every $i\in I$ (3A3Ib), so
$g\mapsto\familyiI{\pi_ig}:C(X;Y)\to\prod_{i\in I}C(X;Y_i)$ is a bijection.
($\beta$) If $j\in I$, $K\subseteq X$ is compact and $H\subseteq Y_j$ is
open, then $\{g:(\pi_jg)[K]\subseteq H\}=\{g:g[K]\subseteq\pi_j^{-1}[H]\}$
is open in $C(X;Y)$.   By 4A2B(a-ii),
$g\mapsto\pi_jg:C(X;Y)\to C(X;Y_j)$ is continuous.   As $j$ is arbitrary,
$g\mapsto\familyiI{\pi_ig}$ is continuous.   ($\gamma$) Now suppose that
$K\subseteq X$ is compact and $W\subseteq Y$ is open, and consider
$U=\{\pmb g:\pmb g\in\prod_{i\in I}C(X;Y_i)$,
$\phi(\pmb g)[K]\subseteq W\}$, where
$\phi(\familyiI{g_i})(x)=\familyiI{g_i(x)}$ for $x\in X$ and
$\familyiI{g_i}\in\prod_{i\in I}C(X;Y_i)$.   Take any
$\pmb h=\familyiI{h_i}$ in $U$.   Set

$$\eqalign{\Cal H
&=\{\prod_{i\in I}H_i:H_i\subseteq Y_i\text{ is open for every }i\in I,\,
\{i:H_i\ne Y_i\}\text{ is finite},\cr&
\mskip100mu
\text{there is a family }\familyiI{G_i}\text{ such that }G_i\text{ is an
open subset of }Y_i\cr&
\mskip100mu
\text{including }\overline{H}_i
\text{ for every }i\in I\text{ and }\prod_{i\in I}G_i\subseteq W\}.
\cr}$$

\noindent Then $\Cal H$ is a family of open subsets of $Y$ and because
every $Y_i$ is regular we have $W=\bigcup\Cal H$.   Now
$\phi(\pmb h)[K]$ is a compact subset of $W$ so there are
$H^{(0)},\ldots,H^{(n)}\in\Cal H$ covering $\phi(\pmb h)[K]$.
Set $K_k=K\cap\phi(\pmb h)^{-1}[\overline{H^{(k)}}]$ for $k\le n$;
then $K_0,\ldots,K_n$ are compact subsets of $X$ with union $K$.

For each $k\in K$ take $\familyiI{H_{ki}}$, $\familyiI{G_{ki}}$ such that

\Centerline{$H^{(k)}=\prod_{i\in I}H_{ki}$,}

\Centerline{$H_{ki}$, $G_{ki}$ are open subsets of $Y_i$ and
$\overline{H}_{ki}\subseteq G_{ki}$ for each $i\in I$,}

\Centerline{$J_k=\{i:H_{ki}\ne Y_i\}$ is finite,
\quad$\prod_{i\in I}G_{ki}\subseteq W$.}

\noindent We have

\Centerline{$\phi(\pmb h)[K_k]\subseteq\overline{H^{(k)}}
\subseteq\prod_{i\in I}G_{ki}$,}

\noindent so

\Centerline{$h_i[K_k]=\pi_i[\phi(\pmb h)[K_k]]\subseteq G_{ki}$}

\noindent for every $i\in I$.   Consider the set

\Centerline{$V=\{\familyiI{g_i}:\familyiI{g_i}\in\prod_{i\in I}C(X;Y_i)$,
$g_i[K_k]\subseteq G_{ki}$ whenever $k\le n$ and $i\in J_k\}$.}

\noindent Then $V$ is an open subset of $\prod_{i\in I}C(X;Y_i)$ containing
$\pmb h$.   If $\pmb g=\familyiI{g_i}$ belongs to $V$ and $k\le n$, then

\Centerline{$\phi(\pmb g)[K_k]\subseteq\prod_{i\in I}g_i[K_k]
\subseteq\prod_{i\in I}G_{ki}\subseteq W$}

\noindent because $G_{ki}=Y_i$ for $i\in I\setminus J_k$.   So
$\phi(\pmb g)[K]=\bigcup_{k\le n}\phi(\pmb g)[K_k]$ is included in $W$ and
$\pmb g\in U$.   Thus we have $\pmb h\in V\subseteq U$ and $U$ is a
neighbourhood of $\pmb h$.   As $\pmb h$ is arbitrary, $U$ is open.
As $K$ and $W$ are arbitrary, $\phi$ is continuous and
$g\mapsto\familyiI{\pi_ig}$ is a homeomorphism.\ \Qed}

\leaveitout{\Engelking, 3.4.5, claims that  $C(X;\prod_{i\in I}Y_i)$ is
always homeomorphic to $\prod_{i\in I}C(X;Y_i)$.   This is false;
take $X=Y_0=Y_1$ with the topology $\{\emptyset,\{1\},\{0,1\}\}$.
However, all is well if $X$ is Hausdorff.
}

\leaveitout{\Engelking, 3.4.5, claims that  $C(X;\prod_{i\in I}Y_i)$ is
always homeomorphic to $\prod_{i\in I}C(X;Y_i)$.   This is false;
take $Y_0=Y_1$ with the topology $\{\emptyset,\{1\},\{0,1\}\}$ and
$X=\Bbb N\cup\{\infty_1,\infty_2\}$ with the topology generated by
$\{\{n\}:n\in\Bbb N\}
  \cup\{X\setminus\{\infty_1\}\},X\setminus\{\infty_2\}\}$.
However, all is well if $X$ is Hausdorff.
}

\spheader 5A4Jc Let $X$ be a compact space, and write $\Cal E$ for the
algebra of open-and-closed subsets of $X$.   Then $f\mapsto f^{-1}[\{1\}]$
is a bijection between $C(X;\{0,1\})$ and $\Cal E$, and the compact-open
topology on $C(X;\{0,1\})$ is discrete.   \prooflet{\Prf\ ($\alpha$)
$f\mapsto f^{-1}[\{1\}]:\{0,1\}^X\to\Cal PX$ is a bijection, and a function
$f:X\to\{0,1\}$ is continuous iff $f^{-1}[\{0\}]$ and $f^{-1}[\{1\}]$ are
open, that is, iff $f^{-1}[\{1\}]\in\Cal E$.   So we have a bijection
$f\mapsto f^{-1}[\{1\}]:C(X;\{0,1\})\to\Cal E$.   ($\beta$) If
$g\in C(X;\{0,1\})$ then $K_0=g^{-1}[\{0\}]$ and $K_1=g^{-1}[\{1\}]$ are
closed, therefore compact, and

\Centerline{$\{g\}=\{f:f\in C(X;\{0,1\})$, $f[K_0]\subseteq\{0\}$,
$f[K_1]\subseteq\{1\}\}$}

\noindent is open;  accordingly the compact-open topology on
$C(X;\{0,1\})$ is discrete.\ \Qed}


\leader{5A4K}{}\dvAnew{2015}\cmmnt{ In \S547 we shall need to
calculate the uniformities of certain meager ideals.
%547R prf g

\medskip

\noindent}{\bf Proposition} Let $X$ be a set and $\Cal A$ a family of
countable sets which is stationary over $X$\cmmnt{ (definition:  5A1R)}.   Then
$\non\Cal M(X^{\Bbb N})\le\max(\#(\Cal A),\non\Cal M)$.

\medskip

\noindent{\bf Notation} Here $X$ is given its discrete topology and
$X^{\Bbb N}$ the associated product topology.   $\Cal M(X^{\Bbb N})$ is its
meager ideal and $\non\Cal M(X^{\Bbb N})$ the corresponding uniformity, the
smallest cardinal of any non-meager subset of $X^{\Bbb N}$.   $\Cal M$ is
$\Cal M(\Bbb R)$.

\proof{{\bf (a)} We can suppose that $A\subseteq X$ for every $A\in\Cal A$.
For each $A\in\Cal A$, let
$F_A\subseteq A^{\Bbb N}$ be
a non-meager set with cardinal at most $\non\Cal M$.   (If $A$ is finite, take
$F_A=A^{\Bbb N}$;  otherwise,
$\non\Cal M(A^{\Bbb N})=\non\Cal M(\Bbb R)=\non\Cal M$ by
522Wb.)  Set $F=\bigcup_{A\in\Cal A}F_A$;  then
$\#(F)\le\max(\#(\Cal A),\non\Cal M)$.

\medskip

{\bf (b)} $F$ is non-meager in $X^{\Bbb N}$.   \Prf\ Let $\sequencen{G_n}$
be a sequence of dense open subsets of $X^{\Bbb N}$.    For each
$z\in\bigcup_{k\in\Bbb N}X^k$ and $n\in\Bbb N$ choose
$w_{zn}\in\bigcup_{k\in\Bbb N}X^k$ such that
$z\subseteq w_{zn}$ and
$\{x:w_{zn}\subseteq x\in X^{\Bbb N}\}\subseteq G_n$.
For each $I\in[X]^{<\omega}$ set
$f(I)=\bigcup\{w_{zn}[\Bbb N]:z\in\bigcup_{k\in\Bbb N}I^k$, $n\in\Bbb N\}$.
Let $A\in\Cal A$ be such that $f(I)\subseteq A$ for every finite
$I\subseteq A$.

Take any $n\in\Bbb N$.
If $z\in\bigcup_{k\in\Bbb N}A^k$, set
$I=z[\Bbb N]\in[A]^{<\omega}$;  then
$w_{zn}\in\bigcup_{k\in\Bbb N}f(I)^k\subseteq\bigcup_{k\in\Bbb N}A^k$
and $x\in G_n$ whenever $w_{zn}\subseteq x\in A^{\Bbb N}$.
So $G_n\cap A^{\Bbb N}$ is dense in $A^{\Bbb N}$.   Also, of course,
$G_n\cap A^{\Bbb N}$ is open in $A^{\Bbb N}$.   As this is true for
every $n\in\Bbb N$,

\Centerline{$\emptyset\ne F_A\cap\bigcap_{n\in\Bbb N}G_n
\subseteq F\cap\bigcap_{n\in\Bbb N}G_n$.}

\noindent As $\sequencen{G_n}$ is arbitrary, $F$ is non-meager.\ \Qed

So $F$ witnesses that
$\non\Cal M(X^{\Bbb N})\le\max(\#(\Cal A),\non\Cal M)$.
}%end of proof of 5A4K

\leader{5A4L}{Irreducible surjections:  Lemma}\dvAnew{2024}
(a) Let $Q$ be a topological space
and $K$, $L$ closed subsets of $Q$ such that $K\cup L=Q$.   Set
$Z=\{(x,1):x\in K\}\cup\{(x,0):x\in L\}\subseteq Q\times\{0,1\}$, and write
$\phi:Z\to Q$ for the first-coordinate map.   Then $\phi$ is an irreducible
continuous surjection.

(b) Let $\zeta$ be an ordinal, $\langle Q_{\xi}\rangle_{\xi<\zeta}$ a family of
compact Hausdorff spaces, and $\langle\phi_{\eta\xi}\rangle_{\eta\le\xi<\zeta}$ a
family such that $\phi_{\eta\xi}:Q_{\xi}\to Q_{\eta}$ is a continuous surjection
whenever $\eta\le\xi<\zeta$.   Suppose that

\inset{$\phi_{\eta\xi}=\phi_{\eta\delta}\phi_{\delta\xi}$ whenever
$\eta\le\delta\le\xi<\omega$,

the topology of $Q_{\xi}$ is generated by
$\{\phi_{\eta\xi}^{-1}[U]:\eta<\xi$, $U\subseteq Q_{\eta}$ is open$\}$ for every
non-zero limit ordinal $\xi<\zeta$,

$\phi_{\xi,\xi+1}$ is irreducible whenever $\xi+1<\zeta$.\hfill(*)}

\noindent Then $\phi_{\eta\xi}$ is irreducible whenever $\eta\le\xi<\zeta$.

\proof{{\bf (a)} $\phi$ is a surjection because $K\cup L=Q$, and is continuous by the
definition of the product topology on $Q\times\{0,1\}$.   Suppose that
$W\subseteq Z$ is open and not empty.

If $x\in K$ and $(x,1)\in W$, then there is an open set $U\subseteq Y$
such that $(x,1)\in U\times\{1\}\subseteq W$.
In this case, $x\in U\cap K\subseteq\overline{Q\setminus L}$ and
$U\setminus L$ is not empty.   But $\phi[Z\setminus W]$ does not
meet $U\setminus L$.   \Prf\Quer\ Otherwise, take
$z\in Z\setminus W$ such that $\phi(z)\in U\setminus L$.   Then
$z$ must be equal to $(\phi(z),1)$ and $z\in U\times\{1\}\subseteq W$, which is
absurd.\ \Bang\QeD\   So $\phi[Z\setminus W]\ne Q$.

Similarly, $\phi[Z\setminus W]\ne Q$
if $W\cap\{(y,0):y\in L\}\ne\emptyset$.   But as $K\cup L=Q$, this means that
$\phi[Z\setminus W]$ can never be $Q$.

Thus $\phi[F]\ne Q$ for any closed proper subset $F$ of $Z$, and
$\phi:Z\to Q$ is irreducible.

\medskip

{\bf (b)} Induce on $\xi$.   As
$\phi_{\eta\eta}=\phi_{\eta\eta}^2:Q_{\eta}\to Q_{\eta}$ is surjective,
$\phi_{\eta\eta}$ is the identity map on $Q_{\eta}$ and the case
$\xi=\eta$ is trivial.   For the inductive step to a successor ordinal
$\xi+1$, $\phi_{\eta,\xi+1}=\phi_{\eta\xi}\phi_{\xi,\xi+1}$ is the composition of
two irreducible continuous surjections, by the inductive hypothesis and (*),
so is irreducible by 5A4C(d-iv).   For the inductive step to a limit ordinal
$\xi\in\ooint{\eta,\zeta}$, take a non-empty open set $G\subseteq Q_{\xi}$;
then there must be $\eta_0,\ldots,\eta_n\in\coint{\eta,\xi}$ and open sets
$G_i\subseteq Q_{\eta_i}$, for $i\le n$, such that
$\emptyset\ne\bigcap_{i\le n}\phi_{\eta_i\xi}^{-1}[G_i]\subseteq G$;
setting $\delta=\sup_{i\le n}\eta_i$ and
$H=\bigcap_{i\le n}\phi_{\eta_i\delta}^{-1}[G_i]$,

\Centerline{$\phi_{\delta\xi}^{-1}[H]
=\bigcap_{i\le n}\phi_{\delta\xi}^{-1}[\phi_{\eta_i\delta}^{-1}[G_i]]
=\bigcap_{i\le n}\phi_{\eta_i\xi}^{-1}[G_i]$}

\noindent is a non-empty open subset of $G$.   So

\Centerline{$\phi_{\eta\xi}[Q_{\xi}\setminus G]
=\phi_{\eta\delta}[\phi_{\delta\xi}[Q_{\xi}\setminus G]]
\subseteq\phi_{\eta\delta}[Q_{\delta}\setminus H]
\ne Q_{\eta}$}

\noindent because $\phi_{\eta\delta}$ is irreducible.
As $G$ is arbitrary, $\phi_{\eta\xi}$
is irreducible and the induction continues.
}%end of proof of 5A4L

%\endnotes{
%\Notesheader{5A4}

%}%end of notes

\discrpage
\frfilename{mt5a5.tex} 
\versiondate{3.10.13} 
\copyrightdate{2007} 
 
\def\chaptername{Appendix} 
\def\sectionname{Real analysis} 
 
\newsection{5A5} 
 
For the sake of an argument in \S534 I sketch a fragment of theory. 
 
\leader{5A5A}{Entire functions} A real function $f$ is {\bf real-analytic}  
if its domain is an open subset $G$ of $\Bbb R$ and for every $a\in G$  
there are a $\delta>0$ and a real sequence $\sequencen{c_n}$ such that  
$f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$ whenever $|x-a|<\delta$.   It is  
{\bf real-entire} if in addition its domain is the whole of $\Bbb R$. 
 
We need the following facts:  (i) if $f$ and $g$ are real-entire  
functions so is $f-g$;  (ii) if $\sequencen{c_n}$ is a real sequence such  
that $f(x)=\sum_{n=0}^{\infty}c_nx^n$ is defined in  
$\Bbb R$ for every $x\in\Bbb R$, then $f$ is real-entire;   
(iii) if in this expression not every $c_n$ is zero, then every point of  
$F=\{x:x\in\Bbb R$, $f(x)=0\}$ is isolated in $F$, so that $F$ is  
countable.    
If you have done a basic course in complex functions you should recognise  
this.    
If either you missed this out, or you are not sure you understood the  
proof of Cauchy's theorem,  
the following is a sketch of a real-variable argument. 
 
(i) is elementary.    
For (ii), observe first that if $\sequencen{c_nx^n}$ is summable then  
$\lim_{n\to\infty}c_nx^n=0$ so $\sum_{n=0}^{\infty}|c_n|t^n$ is finite  
whenever $0\le t<|x|$.    
In the present case, $\sum_{n=0}^{\infty}|c_n|t^n<\infty$ for every  
$t\ge 0$.   So if $a$, $x\in\Bbb R$,  
 
$$\eqalignno{\sum_{n=0}^{\infty}\sum_{k=0}^n 
  |\Bover{n!}{k!(n-k)!}c_nx^ka^{n-k}| 
&\le\sum_{n=0}^{\infty}\sum_{k=0}^n\Bover{n!}{k!(n-k)!}|c_n|R^n\cr 
\displaycause{where $R=\max(|x|,|a|)$} 
&=\sum_{n=0}^{\infty}|c_n|(2R)^n<\infty.\cr}$$ 
 
\noindent We therefore have 
 
\Centerline{$f(x+a) 
=\sum_{n=0}^{\infty}c_n(x+a)^n
=\sum_{n=0}^{\infty}\sum_{k=0}^n\Bover{n!}{k!(n-k)!}c_nx^ka^{n-k} 
=\sum_{k=0}^{\infty}c_{ak}x^k$}
 
\noindent where $c_{ak}=\sum_{n=k}^{\infty}\Bover{n!}{k!(n-k)!}c_na^{n-k}$  
for each $k$.   Turning this round, 
$f(x)=\sum_{k=0}^{\infty}c_{ak}(x-a)^k$ for every $x$.   This shows that  
$f$ is real-entire.   As for (iii), 
if not every $c_n$ is zero, there must be  
some neighbourhood of $0$ in which the first non-zero term $c_nx^n$  
dominates, so $f$ is not identically zero.   (The point is that 
$\sum_{k=0}^{\infty}|c_k|<\infty$, so there is some $\delta>0$ such that 
$\sum_{k=n+1}^{\infty}|c_k\delta^{k+1}|<|c_n\delta^n|$.)  
In this case, if $a\in\Bbb R$, not every  
$c_{ak}$ can be zero, and there must be some neighbourhood of $a$ in which  
the first non-zero term $c_{ak}(x-a)^k$ dominates, so that there can be no  
zeroes of $f$ in that neighbourhood except perhaps $a$ itself. 
 
\discrpage 
 
 
 
\frfilename{mt5a6.tex}
\versiondate{13.2.17}
\copyrightdate{2009}

\def\chaptername{Appendix}
\def\sectionname{Special axioms}

\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\Glsquare{{Gl}$\Square$}
\def\Sing{\mathop{\text{Sing}}}
\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\JechM{{\smc Jech 03}} %........
\def\Jech{{\smc Jech 78}} %.....
%{\smc Devlin 84}} %..... serious problems according to ARDM
%{\smc Kanamori 03} %....
%\def\JWI{{\smc Just \& Weese 96}} %.
%\def\JWII{{\smc Just \& Weese 97}} %.
%\def\DS{{\smc Drake \& Singh 96}}

\newsection{5A6}

This section contains very brief accounts of
some of the undecidable propositions
and special axioms which are used in this volume, with a few of their most
basic consequences:  the generalized
continuum hypothesis, the axiom of constructibility,
Jensen's Covering Lemma, square principles, Chang's transfer
principle, Todor\v{c}evi\'c's $p$-ideal dichotomy and the filter dichotomy.

\leader{5A6A}{The generalized continuum hypothesis (a)} The generalized
continuum hypothesis is the assertion

\Centerline{(GCH)\quad $2^{\kappa}=\kappa^+$ for every infinite cardinal
$\kappa$.}

\spheader 5A6Ab If GCH is true, then for infinite cardinals $\kappa$,
$\lambda$

$$\eqalign{\cff[\kappa]^{\le\lambda}
&=1\text{ if }\kappa\le\lambda,\cr
&=\kappa\text{ if }\lambda<\cf\kappa,\cr
&=\kappa^+\text{ otherwise}.\cr}$$

\noindent\prooflet{\Prf\ If $\kappa\le\lambda$, use 5A1F(e-i).   If
$\lambda<\kappa$ then

\Centerline{$\cff[\kappa]^{\le\lambda}\le\#([\kappa]^{\le\lambda})
\le\#(\Cal P\kappa)=2^{\kappa}=\kappa^+$.}

\noindent If $\lambda<\theta=\cf\kappa$, then
$[\kappa]^{\le\lambda}=\bigcup_{\xi<\kappa}[\xi]^{\le\lambda}$ so

\Centerline{$\cff[\kappa]^{\le\lambda}
\le\max(\kappa,\sup_{\xi<\kappa}\cff[\xi]^{\le\lambda})
\le\max(\kappa,\sup_{\xi<\kappa}\#(\xi)^+)
=\kappa$}

\noindent and we have equality (using the other part of 5A1F(e-i)).
If $\lambda=\theta$ then 5A1F(e-v) tells us that
$\cff[\kappa]^{\lambda}$ is greater than $\kappa$, so must be $\kappa^+$.
If $\theta<\lambda<\kappa$ then, by 5A1F(e-ii),

\Centerline{$\kappa<\cff[\kappa]^{\le\theta}
\le\max(\cff[\kappa]^{\le\lambda},\cff[\lambda]^{\le\theta})
\le\max(\cff[\kappa]^{\le\lambda},\lambda^+)
\le\max(\cff[\kappa]^{\le\lambda},\kappa)$,}

\noindent so again $\cff[\kappa]^{\le\lambda}=\kappa^+$.\ \Qed}

\spheader 5A6Ac If GCH is true, then for infinite cardinals $\kappa$ and
$\lambda$, the cardinal power $\kappa^{\lambda}$ is
$2^{\lambda}$ if $\kappa\le\lambda$,
$\kappa$ if $\lambda<\cf\kappa$,
$\kappa^+$ otherwise.
\prooflet{(Put (b) and 5A1F(e-iii) together.)}

\leader{5A6B}{$L$, $0^{\sharp}$ and Jensen's Covering Lemma (a)(i)}
Let $L$ be the class of {\bf
constructible} sets\prooflet{ (\JechM, \S13;
\Jech, \S12;
{\smc Kanamori 03}, \S3;  {\smc Kunen 80}, chap.\ VI)}. %;  \DS, chap.\ 8
The {\bf axiom of constructibility} is

\Centerline{(\VeqL)\quad Every set is constructible.}

\noindent \VeqL\ implies GCH
\prooflet{(\JechM, 13.20;
\Jech, Theorem 34;  {\smc Kunen 80}, \S VI.4)}. %;  \DS, \S8.6

\medskip

\quad{\bf (ii)}\cmmnt{ I will call on the following
three properties of $L$ in the
remarks below.   To make sense of them you will of course need to look at
the proper definition.   Only the third has any real content.}
Every ordinal belongs to $L$;
if $A$, $B\in L$ then $A\cap B\in L$;
if $\kappa$ is a cardinal, then $\#(L\cap\Cal P\kappa)\le\kappa^+$.

\spheader 5A6Bb $0^{\sharp}$, if it exists, is a set of sentences in a
countable formal language\prooflet{ (\JechM, \S18;
{\smc Kanamori 03}, \S9)}.   \cmmnt{I will not
attempt to explain further;  I mention $0^{\sharp}$ only so that you will
be able to explore the literature for proofs of the assertions below.}
I will write `$\exists 0^{\sharp}$' for the assertion
`$0^{\sharp}$ exists'.

{\bf Jensen's Covering Lemma} is the assertion

\doubleinset{(CL)\quad for every uncountable set $A$ of ordinals,
there is a
constructible set of the same cardinality including $A$.}

\noindent\cmmnt{Now }Jensen's Covering Theorem is

\Centerline{CL iff not-$\exists 0^{\sharp}$\dvro{.}{}}

\noindent\prooflet{({\smc Jech 03}, Theorem 18.30.)}

\spheader 5A6Bc\cmmnt{ The importance to us of $0^{\sharp}$ is that
there are
relatively direct proofs that \VeqL\ implies not-$\exists 0^{\sharp}$
(\JechM, \S18), and
that not-$\exists 0^{\sharp}$ is true
in any set forcing extension of a model of not-$\exists 0^{\sharp}$;  see
\JechM, Exercise 18.2 or \Jech, Exercise 30.2.   So} CL
implies that $\VVdP\,$CL for every forcing notion $\Bbb P$ of the kind
considered in \S5A3.

\leader{5A6C}{Theorem} Assume that CL is true.

(a) For infinite cardinals $\kappa$ and $\lambda$,

$$\eqalign{\cff[\kappa]^{\le\lambda}
&=1\text{ if }\kappa\le\lambda,\cr
&=\kappa\text{ if }\lambda<\cf\kappa,\cr
&=\kappa^+\text{ otherwise}.\cr}$$

(b) If $\kappa$ and $\lambda$ are infinite cardinals, then the cardinal
power $\kappa^{\lambda}$ is
$2^{\lambda}$ if $\kappa\le 2^{\lambda}$,
$\kappa$ if $\lambda<\cf\kappa$ and $2^{\lambda}\le\kappa$,
and $\kappa^+$ otherwise.

%(c) If $X$ is a set and $\Cal I$ is a proper ideal of subsets of $X$
%containing singletons, then $\Cal I$ is not $(\add\Cal I)^+$-saturated.

\proof{{\bf (a)(i)} If $\omega_1\le\lambda<\kappa$ then
$\cff[\kappa]^{\le\lambda}\le\kappa^+$.   \Prf\ By CL, every
$A\in[\kappa]^{\lambda}$
is included in a $B\in[L]^{\lambda}$;  now $\kappa\in L$ so
$B\cap\kappa\in L$ and $A\subseteq B\cap\kappa\in[\kappa]^{\lambda}$.
Thus $L\cap[\kappa]^{\lambda}$ is cofinal with $[\kappa]^{\lambda}$ and
$[\kappa]^{\le\lambda}$.   But $\#(L\cap\Cal P\kappa)\le\kappa^+$
(5A6B(a-ii)), so $\cf[\kappa]^{\le\lambda}\le\kappa^+$.\ \Qed

\medskip

\quad{\bf (ii)} It follows that $\cff[\kappa]^{\le\lambda}\le\kappa^+$ for all
infinite cardinals $\kappa$ and $\lambda$.   \Prf\ The case
$\lambda\ge\kappa$ is trivial, so only the case
$\lambda=\omega<\kappa$ remains.   But

\Centerline{$\cff[\kappa]^{\le\omega}
\le\max(\cff[\kappa]^{\omega_1},\cff[\omega_1]^{\le\omega})
\le\max(\kappa^+,\omega_1)=\kappa^+$}

\noindent by 5A1F(e-ii) and (i).\ \Qed

\medskip

\quad{\bf (iii)} If $\lambda<\cf\kappa$ then
$\cff[\kappa]^{\le\lambda}\le\kappa$.   \Prf\
$[\kappa]^{\le\lambda}=\bigcup_{\xi<\kappa}[\xi]^{\le\lambda}$, so

\Centerline{$\cff[\kappa]^{\le\lambda}
\le\max(\kappa,\sup_{\xi<\kappa}\cff[\xi]^{\le\lambda})
\le\max(\kappa,\sup_{\xi<\kappa}\#(\xi)^+)
=\kappa$.  \Qed}

\medskip

\quad{\bf (iv)} If $\cf\kappa\le\lambda<\kappa$ then
$\cff[\kappa]^{\le\lambda}>\kappa$.   \Prf\ Set $\theta=\cf\kappa$.
Then

$$\eqalign{\kappa
&<\cff[\kappa]^{\le\theta}
\le\max(\cff[\kappa]^{\le\lambda},\cff[\lambda]^{\le\theta})\cr
&\le\max(\cff[\kappa]^{\le\lambda},\lambda^+)
\le\max(\cff[\kappa]^{\le\lambda},\kappa)\cr}$$

\noindent(5A1F(e-v), 5A1F(e-ii), (ii) above), so
$\cff[\kappa]^{\le\lambda}>\kappa$.\ \Qed

Putting this together with 5A1F(e-i), (ii) and (iii) we have the result.

\medskip

{\bf (b)} As 5A6Ac.

%\medskip

%{\bf (c)} \JechM, 22.22
%\Cal I would have to be precipitous
}%end of proof of 5A6C

\leader{5A6D}{Square principles} {\bf (a)(i)}
Let $\Sing$ be the class of non-zero limit
ordinals which are not regular cardinals.   Global Square
%$\Glsquare$,
is the statement

\inset{there is a family
$\family{\xi}{\Sing}{C_{\xi}}$ such that

\quad for every $\xi\in\Sing$, $C_{\xi}$ is a closed cofinal set
in $\xi$;

\quad$\otp C_{\xi}<\xi$ for every $\xi\in\Sing$;

\quad if $\xi\in\Sing$ and $\zeta>0$ is such that
$\zeta=\sup(\zeta\cap C_{\xi})$, then $\zeta\in\Sing$ and
$C_{\zeta}=\zeta\cap C_{\xi}$.}

\medskip

\quad{\bf (ii)} For an infinite cardinal $\kappa$,
let $\square_{\kappa}$ be the statement

\inset{there is a family $\ofamily{\xi}{\kappa^+}{C_{\xi}}$ of
sets such that

\quad for every $\xi<\kappa^+$, $C_{\xi}\subseteq\xi$ is a
closed cofinal set in $\xi$;

\quad if $\cf\xi<\kappa$ then $\#(C_{\xi})<\kappa$;

\quad whenever $\xi<\kappa^+$ and $\zeta<\xi$ is such that
$\zeta=\sup(\zeta\cap C_{\xi})$, then
$C_{\zeta}=\zeta\cap C_{\xi}$.
}%end of inset
%5{}18I

\spheader 5A6Db \VeqL\ implies Global
Square\prooflet{ ({\smc Friedman \& Koepke 97})}.
%forthcoming article by P Welch, Sigma^* Fine Structure, Handbook of
%Set Theory, 2009?
Global Square implies that
$\square_{\kappa}$ is true for every infinite cardinal
$\kappa$\prooflet{ ({\smc Devlin 84}, VI.6.2)}.
CL implies that $\square_{\kappa}$ is true for every singular
infinite cardinal $\kappa$\prooflet{ ({\smc Devlin 84}, V.5.6)}.

\spheader 5A6Dc If $\kappa$ is an uncountable cardinal and
$\ofamily{\xi}{\kappa^+}{C_{\xi}}$ is a family as in
(a-ii), then $\otp C_{\xi}\le\kappa$ for every $\xi<\kappa^+$.
\prooflet{\Prf\ If $\cf\xi<\kappa$ this is immediate from the second clause
of $\square_{\kappa}$.
Otherwise, $\kappa$ is regular, and $\cf C_{\xi}=\kappa$.
\Quer\ If $\otp C_{\xi}>\kappa$ then $\otp C_{\xi}>\kappa+\omega$ so
there is a $\zeta\in C_{\xi}$ such that
$\otp(\zeta\cap C_{\xi})=\kappa+\omega$;  but now $\cf\zeta=\omega<\kappa$
and $C_{\zeta}=\zeta\cap C_{\xi}$ has cardinal $\kappa$, which is not
allowed.\ \Bang\Qed}

%Jech 78 speaks of `box principle', \square_{\omega_2}
%Jech 03, 23.4 has a slightly stronger version, requiring also
%  \#(C_{\alpha})<\kappa  if  \alpha<\kappa^+  a limit and
%  \cf\alpha<\kappa
%5{}18 notes

\vleader{48pt}{5A6E}{Lemma} Suppose that $\kappa$ is
an uncountable cardinal
with countable cofinality such that $\square_{\kappa}$ is true.   Then
there is a family $\ofamily{\xi}{\kappa^+}{I_{\xi}}$ of countably
infinite subsets of $\kappa$ such that

\inset{$I_{\xi}\cap I_{\eta}$ is finite whenever $\eta<\xi<\kappa^+$,

$\{\xi:\xi<\kappa^+$, $I\cap I_{\xi}$ is infinite$\}$ is countable for
every countable $I\subseteq\kappa$.}

\proof{ Let $\ofamily{\xi}{\kappa^+}{C_{\xi}}$ be a family as in
5A6D(a-ii).   Let $\sequencen{\kappa_n}$ be a non-decreasing
sequence of infinite
cardinals less than $\kappa$ with supremum $\kappa$.   Define
$\ofamily{\xi}{\kappa^+}{f_{\xi}}$ in $\prod_{n\in\Bbb N}\kappa_n^+$
as follows.   $f_0(n)=0$ for every $n$.   Given $f_{\xi}$, set
$f_{\xi+1}(n)=f_{\xi}(n)+1$ for every $n$.   Given
$\ofamily{\eta}{\xi}{f_{\eta}}$, where $\xi<\kappa^+$ is a non-zero
limit ordinal, set

\Centerline{$f_{\xi}(n)
=\sup\{f_{\eta}(n):\eta\in C_{\xi}$, $\#(\eta\cap C_{\xi})<\kappa_n\}$}

\noindent for each $n$;  because
$\{\eta:\eta\in C_{\xi}$, $\#(\eta\cap C_{\xi})<\kappa_n\}$ has
cardinal at most $\kappa_n$, $f_{\xi}(n)<\kappa_n^+$.   Continue.

We find that if $\eta<\xi<\kappa^+$ then
$\{n:f_{\xi}(n)\le f_{\eta}(n)\}$ is finite.   \Prf\ Induce on $\xi$.
If $\xi=0$ there is nothing to prove.   If $\xi=\zeta+1$ then
$\{n:f_{\xi}(n)\le f_{\eta}(n)\}=\{n:f_{\zeta}(n)<f_{\eta}(n)\}$
is finite.   If $\xi$ is a non-zero limit ordinal, let
$\zeta\in C_{\xi}$ be such that $\eta<\zeta$.   Because
$\otp C_{\xi}\le\kappa$ (5A6Dc),
$\#(\zeta\cap C_{\xi})<\kappa_m$ for some
$m$.   Now $f_{\xi}(n)\ge f_{\zeta}(n)$ for every $n\ge m$, so
$\{n:f_{\xi}(n)\le f_{\eta}(n)\}
\subseteq m\cup\{n:f_{\zeta}(n)\le f_{\eta}(n)\}$ is finite.\ \Qed

If $I\subseteq\Bbb N\times\kappa$ is countable, then
$B=\{\xi:\xi<\kappa^+$, $I\cap f_{\xi}$ is infinite$\}$ is countable,
where in this formula I am identifying $f_{\xi}$ with its graph, as usual.
\Prf\Quer\ Otherwise, let $B'\subseteq B$ be a set with order type
$\omega_1$, and set $\xi=\sup B'<\kappa^+$.   Set

\Centerline{$I'=\{(n,\alpha):(n,\alpha)\in I$,
$\alpha\le f_{\eta}(n)$ for some $\eta\in C_{\xi}\}$.}

\noindent Because $I$ and $I'$ are countable, while
$\cf C_{\xi}=\cf\xi=\omega_1$, there is a $\zeta\in C_{\xi}$ such that
$\zeta=\sup(\zeta\cap C_{\xi})$ and

\Centerline{$I'=\{(n,\alpha):(n,\alpha)\in I$,
$\alpha\le f_{\eta}(n)$ for some $\eta\in\zeta\cap C_{\xi}\}$.}

\noindent Take $\eta\in B'$ such that $\eta>\zeta$, and
$\zeta'\in C_{\xi}$ such that $\zeta'>\eta$.   Then there is an
$m\in\Bbb N$ such that $\#(\zeta\cap C_{\xi})<\kappa_m$ and
$f_{\zeta}(n)<f_{\eta}(n)<f_{\zeta'}(n)$ for every $n\ge m$.   As
$\eta\in B$, there is an $n\ge m$ such that $(n,f_{\eta}(n))\in I$;  as
$f_{\eta}(n)<f_{\zeta'}(n)$, $(n,f_{\eta}(n))\in I'$ and there is an
$\eta'\in\zeta\cap C_{\xi}$ such that $f_{\eta}(n)\le f_{\eta'}(n)$.
But now we have $\eta'\in C_{\zeta}$ and
$\#(\eta'\cap C_{\zeta})\le\#(C_{\zeta})<\kappa_n$ and
$f_{\zeta}(n)<f_{\eta'}(n)$, contrary to the choice of $f_{\zeta}$.\
\Bang\Qed

Thus if we set $I_{\xi}=f_{\xi}$ for $\xi<\kappa^+$ we have an
appropriate family of sets in $\Bbb N\times\kappa^+$ which can be
transferred to $\kappa^+$ by any bijection.
}%end of proof of 5A6E

\leader{5A6F}{Chang's transfer principle (a)}
If $\lambda_0$, $\lambda_1$, $\kappa_0$ and $\kappa_1$ are cardinals, then
$(\kappa_1,\lambda_1)\doubleheadrightarrow(\kappa_0,\lambda_0)$
means

\inset{whenever $f:[\kappa_1]^{<\omega}\to\lambda_1$ is a function, there
is an $A\in[\kappa_1]^{\kappa_0}$ such that
$\#(f[\,[A]^{<\omega}])\le\lambda_0$.}

\cmmnt{\noindent
For the original model-theoretic version of this principle, and
the proof that it comes to the same thing, see
{\smc Kanamori 03}, 8.1.   For
various combinatorial consequences, see `Chang's conjecture' in
{\smc Erd\H{o}s Hajnal M\'at\'e \& Rado 84}.}

\cmmnt{In this book, }I write CTP$(\kappa,\lambda)$ for the statement

\Centerline{$(\kappa,\lambda)\doubleheadrightarrow(\omega_1,\omega)$.}
\cmmnt{

\noindent What is commonly called `Chang's conjecture' is
CTP$(\omega_2,\omega_1)$.   For a model of
GCH + CTP$(\omega_{\omega+1},\omega_{\omega})$, see
{\smc Levinski Magidor \& Shelah 90}.}

\spheader 5A6Fb Suppose that CTP$(\kappa,\lambda)$ is true.

\medskip

\quad{\bf (i)} If $f:[\kappa]^{<\omega}\to[\lambda]^{\le\omega}$ is a
function, then there is an uncountable $A\subseteq\kappa$ such that
$\bigcup\{f(I):I\in[A]^{<\omega}\}$ is countable.   \prooflet{\Prf\
Enumerate $\Bbb N\times\Bbb N$ as $\sequencen{(k_n,m_n)}$ in
such a way that $m_n\le n$ for every $n\in\Bbb N$.   For
$I\in[\kappa]^{<\omega}$ let $\sequence{k}{f_k(I)}$ be a sequence running
over $f(I)\cup\{0\}$.   (I am passing over the trivial case $\lambda=0$.)
Now, for $n\in\Bbb N$ and
$I\in[\kappa]^n$, enumerate $I$ in ascending order as
$\ofamily{i}{n}{\xi_i}$ and set $g(I)=f_{k_n}(\{\xi_i:i<m_n\})$.
There is an uncountable $A\subseteq\kappa$ such that
$B=\{g(I):I\in[A]^{<\omega}\}$ is countable;  we may suppose that $A$ has
order type $\omega_1$.   If $J\in[A]^{<\omega}$ and $k\in\Bbb N$, let
$n\in\Bbb N$ be such that $k_n=k$ and $m_n=\#(J)$;  let $I\in[A]^n$ be such
that $J$ consists of the first $m_n$ elements of $I$;  then $f_k(J)=g(I)$
belongs to $B$.   As $J$ and $k$ are arbitrary,
$\bigcup\{f(I):I\in[A]^{<\omega}\}\subseteq B$ is countable.\ \Qed}

\medskip

\quad{\bf (ii)} If
$\ofamily{\xi}{\kappa}{A_{\xi}}$ is any family of countable subsets of
$\lambda$, then there is a countable $A\subseteq\lambda$ such that
$\{\xi:A_{\xi}\subseteq A\}$ is uncountable.   \prooflet{\Prf\ In (i), take
$f(I)=\bigcup_{\xi\in I}A_{\xi}$ for $I\in[\kappa]^{<\omega}$.\ \Qed}

\spheader 5A6Fc CL implies that
CTP$(\kappa,\lambda)$ is false except when
$\lambda\le\omega$\prooflet{ ({\smc Kanamori 03}, 8.3, 21.1 and 21.4)}.

\leader{5A6G}{Todor\v{c}evi\'c's $p$-ideal dichotomy (a)} Let $X$ be a set
and $\Cal I$ an ideal of subsets of $X$.   Then $\Cal I$ is a
{\bf $p$-ideal} if for every sequence $\sequencen{I_n}$ in $\Cal I$ there
is an $I\in\Cal I$ such that $I_n\setminus I$ is finite for every
$n\in\Bbb N$.   \cmmnt{(Compare 538Ab.)}

\spheader 5A6Gb Now Todor\v{c}evi\'c's $p$-ideal dichotomy is the statement

\doubleinset{(TPID)\quad
whenever $X$ is a set and $\Cal I\subseteq[X]^{\le\omega}$
is a $p$-ideal of countable subsets of $X$, then
{\it either} there is a $B\in[X]^{\omega_1}$ such that
$[B]^{\le\omega}\subseteq\Cal I$
{\it or} $X$ is expressible as $\bigcup_{n\in\Bbb N}X_n$ where
$\Cal I\cap\Cal PX_n\subseteq[X_n]^{<\omega}$ for every $n\in\Bbb N$.}

\noindent This is a consequence of the Proper Forcing Axiom, and
implies that $\square_{\kappa}$ is false for every
$\kappa\ge\omega_1$\cmmnt{ ({\smc Todor\v{c}evi\'c 00})}.
%mt5abits so also contradicts CL

\leader{*5A6H}{Analytic $p$-ideals:  Theorem}
Suppose that the Proper Forcing Axiom is true.
Take a non-empty set $D\subseteq\coint{0,\infty}^{\Bbb N}$ and set

\Centerline{$\Cal I=\{I:I\subseteq\Bbb N$,
  $\lim_{n\to\infty}\sup_{z\in D}\sum_{i\in I\setminus n}z(i)=0\}$,}

\noindent so that $\Cal I$ is an ideal of subsets of $\Bbb N$.
Let $\frak A$ be the quotient Boolean algebra $\Cal P\Bbb N/\Cal I$.
Then for every $\pi\in\Aut\frak A$ there are sets $I$, $J\in\Cal I$ and a
bijection
$h:\Bbb N\setminus I\to\Bbb N\setminus J$ representing $\pi$ in the sense
that $\pi(A^{\ssbullet})=(h^{-1}[A])^{\ssbullet}$ for every
$A\subseteq\Bbb N$.
\prooflet{({\smc Farah 00}, 3.4.6.)}
%5{}56

\leader{5A6I}{}{\bf $\frak u$, $\frak g$ and the filter
dichotomy: Definitions (a)} The {\bf ultrafilter number} $\frak u$ is
the least cardinal of any filter base
generating a free ultrafilter on $\Bbb N$\cmmnt{, that is,
$\min\{\ci\Cal F:\Cal F$ is a free ultrafilter on $\Bbb N\}$}.

\spheader 5A6Ib{\bf (i)} A family $A$ of infinite subsets of $\Bbb N$
is {\bf groupwise dense} if

\inset{($\alpha$) whenever $a\in A$, $a'\in[\Bbb N]^{\omega}$ and
$a'\setminus a$ is finite, then $a'\in A$,

($\beta$) whenever $\phi:\Bbb N\to\Bbb N$ is finite-to-one,
there is an infinite $c\subseteq\Bbb N$ such that
$\phi^{-1}[c]\in A$.}

\noindent (A function $f:X\to Y$ is `finite-to-one' if $f^{-1}[\{y\}]$ is
finite for every $y\in Y$.)

\medskip

\quad{\bf (ii)} The {\bf groupwise density number} $\frak g$ is
the least cardinal of any collection $\Bbb A$
of groupwise dense subsets of $[\Bbb N]^{\omega}$ such that
$\bigcap\Bbb A=\emptyset$.

%\frak g\le\cf(\frak d) ??
%\frak g is regular and uncountable (Mildenberger)

\cmmnt{\medskip

\quad{\bf (iii)} For a model in which $\omega_1=\frak u<\frak g$ see
{\smc Blass \& Laflamme 89}.}

\spheader 5A6Ic
For filters $\Cal F$ on $X$ and $\Cal G$ on $Y$,
say that $\Cal F\leRB\Cal G$ if there is a finite-to-one $\phi:Y\to X$ such
that $\Cal F=\phi[[\Cal G]]$.    (This is the {\bf Rudin-Blass ordering} of
filters.)   \cmmnt{Note that }$\Cal F\leRB\Cal F$ for every
filter $\Cal F$,
and if $\Cal F\leRB\Cal G$ and $\Cal G\leRB\Cal H$ then
$\Cal F\leRB\Cal H$ (and $\Cal F\leRK\Cal G$\cmmnt{, of course}).

\spheader 5A6Id The {\bf filter dichotomy} is the statement

\doubleinset{(FD)\quad
For every free filter $\Cal F$ on $\Bbb N$ either $\CalFr\leRB\Cal F$,
where $\CalFr$ is the Fr\'echet filter, or there is an ultrafilter $\Cal G$
on $\Bbb N$ such that $\Cal G\leRB\Cal F$.}
%5{}38S

\noindent

\leader{*5A6J}{Proposition}\cmmnt{ ({\smc Blass \& Laflamme 89})}
If $\frak u<\frak g$ then the filter dichotomy is true.

\proof{ Let $\Cal F$ be a free filter on $\Bbb N$ such that
$\CalFr\not\leRB\Cal F$, where $\CalFr$ is the Fr\'echet filter.

\medskip

{\bf (a)}   For subsets $a$, $b$, $c$ of $\Bbb N$ I will say that
{\bf $b$ interpolates between $a$ and $c$}, and write $(a(b)c)$,
if whenever $i\in a$, $k\in c$ and $i\le k$ then there is a $j\in b$ such
that $i\le j\le k$.   Now if $b\subseteq\Bbb N$ is infinite,

\Centerline{$A_b=\{a:a\in[\Bbb N]^{\omega}$, $(a(b)c)$ for some
$c\in\Cal F\}$}

\noindent is groupwise dense.   \Prf\ ($\alpha)$ If $a\in A_b$,
$a'\in[\Bbb N]^{\omega}$ and
$a'\setminus a$ is finite, let $c\in\Cal F$ be such that $(a(b)c)$.
Let $j_0\in b$ be such that
$a'\setminus a\subseteq j_0$, and set $c'=c\setminus j_0\in\Cal F$.
If $i\in a'$, $k\in c'$ and $i\le k$, either $i\le j_0$ and
$i\le j_0\le k$, or $i\in a$ and there is a $j\in b$ such that
$i\le j\le k$;  thus $(a'(b)c')$ and
$a'\in A_b$.   ($\beta$) If $\phi:\Bbb N\to\Bbb N$ is finite-to-one,
we can choose strictly increasing sequences
$\sequence{r}{l_r}$, $\sequence{r}{m_r}$ in $\Bbb N$ such that

\inset{$l_0\in\phi[\Bbb N]$,

given $l_r$, $m_r\in b$ and $i<m_r$ whenever $\phi(i)=l_r$,

given $m_r$, $l_{r+1}\in\phi[\Bbb N]$ and $i\ge m_r$ whenever
$\phi(i)=l_{r+1}$.}

\noindent Set $\psi(i)=\#(\{r:m_r\le i\})$ for $i\in\Bbb N$;  then
$\psi:\Bbb N\to\Bbb N$ is finite-to-one.   As $\Cal F$ is free,
$\CalFr\subseteq\psi[[\Cal F]]$;  as $\CalFr\not\leRB\Cal F$,
$\psi[[\Cal F]]\ne\CalFr$ and there is an infinite set
$d\subseteq\Bbb N$ such that
$d'=\Bbb N\setminus d\in\psi[[\Cal F]]$.   Of course we can suppose that
$0\notin d$.

If $i\in\psi^{-1}[d]$, $k\in\psi^{-1}[d']$ and $i\le k$, then
$\psi(i)\in d$, $\psi(k)\in d'$ and $\psi(i)\le\psi(k)$, so
$\psi(i)<\psi(k)$;  setting $r=\psi(i)$, we have $i<m_r\le k$, while
$m_r\in b$.   This shows that $(\psi^{-1}[d](b)\psi^{-1}[d'])$, so that
$\psi^{-1}[d]\in A_b$.
Next, setting $c=\{l_r:r\in d\}$, $c$ is infinite;  and if
$i\in\phi^{-1}[c]$, we have an $r\in d$ such that $\phi(i)=l_r$, in which
case (as $r>0$) $m_{r-1}\le i<m_r$ and $\psi(i)=r$.   So
$\phi^{-1}[c]\subseteq\psi^{-1}[d]$;  also $\phi^{-1}[c]$ is infinite
(because $l_r\in\phi[\Bbb N]$ for every $r$), so $\phi^{-1}[c]\in A_b$.
As $\phi$ is arbitrary, $A_b$ is groupwise dense.\ \Qed

\medskip

{\bf (b)(i)}
Now let $\Cal G$ be a non-principal ultrafilter on $\Bbb N$ which
has a filter base $B$ with cardinal $\frak u$.   Because
$\frak u<\frak g$, there is an $a\in\bigcap_{b\in B}A_b$.
This time, set $\phi(i)=\#(a\cap(i+1))$ for $i\in\Bbb N$;
as $a$ is infinite, $\phi$ is finite-to-one.   Set $m=\min a$.
If $b\in B$, $c\subseteq\Bbb N$ and $(a(b)c)$, then
$\phi[c\setminus m]\subseteq\phi[b]$.
\Prf\ If $k\in c\setminus m$, set $i=\max(a\cap(k+1))$;  then
$i\le k$ so there is a $j\in b$ such that $i\le j\le k$, and now
$a\cap(j+1)=a\cap(k+1)=a\cap(i+1)$ so $\phi(k)=\phi(j)\in\phi[b]$.\ \Qed\

\medskip

\quad{\bf (ii)}
It follows that $\phi[[\Cal G]]\subseteq\phi[[\Cal F]]$.   \Prf\ If
$G\in\phi[[\Cal G]]$ there is a $b\in B$ such that
$b\subseteq\phi^{-1}[G]$.   Now $a\in A_b$ so there is a $c\in\Cal F$ such
that $(a(b)c)$;  in this case, setting $m=\min a$,
$c\setminus m\in\Cal F$ and $\phi[c\setminus m]\in\phi[[\Cal F]]$.
By (i) just above, $\phi[c\setminus m]\subseteq\phi[b]$, while
$\phi[b]\subseteq G$.   So $G\in\phi[[\Cal F]]$.   As $G$ is arbitrary,
$\phi[[\Cal G]]\subseteq\phi[[\Cal F]]$.\ \Qed

\medskip

\quad{\bf (iii)} We supposed that
$\Cal G$ was an ultrafilter, so $\phi[[\Cal G]]$ is an ultrafilter (2A1N)
and must be equal to $\phi[[\Cal F]]\leRB\Cal F$.   Thus
we have the second alternative in the statement of FD.   As $\Cal F$ is
arbitrary, FD is true.
}%end of proof of 5A6J

\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
axioms', Comm.\ Math.\ Univ.\ Carolinae 33 (1992) 291-297.
\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

%---------------------------------------------------------------%

\end



