% 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\Sous{\Cal S\text{ous}}  %used in vol 6
\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{mt4a.tex} 
\versiondate{3.9.13} 
\copyrightdate{2000} 
      
\def\chaptername{Appendix} 
\def\sectionname{Introduction} 
      
\gdef\topparagraph{} 
\gdef\bottomparagraph{Appendix to Vol.\ 4 {\it intro.}} 
      
\centerline{\bf Appendix to Volume 4} 
      
\medskip 
      
\centerline{\bf Useful facts} 
      
\medskip 
      
As is to be expected, we are coming in this volume to depend on a wide 
variety of more or less recondite information, and only an exceptionally 
broad mathematical education will have covered it all.   While all the 
principal ideas are fully expressed in standard textbooks, there are 
many minor points where I need to develop variations on the familiar 
formulations.   A little under half the material, by word-count, is in   
general topology (\S4A2), where I begin with some pages of definitions.   
I follow this with a section on Borel and Baire  
$\sigma$-algebras, Baire-property algebras and cylindrical algebras (\S4A3),  
worked out a little more thoroughly than the rest of the material.  
The other sections are on set theory (\S4A1), linear analysis (\S4A4), 
topological groups (\S4A5) and Banach algebras (\S4A6).    
 
      
\discrpage 
      
      
\frfilename{mt4a1.tex}
\versiondate{27.1.13}
\copyrightdate{2000}

\def\refquery{\medskip reference\query \medskip}
\def\JWII{{\smc Just \& Weese 97}}
\def\Kunen{{\smc Kunen 80}}
\def\Levy{{\smc Levy 79}}
\def\JechGH{{\smc Jech 78}}
\def\JechIC{{\smc Jech 03}}

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

\newsection{4A1}

For this volume, we need fragments from four topics in set theory and
one in Boolean algebra.   The
most important are the theory of closed cofinal sets and stationary
sets (4A1B-4A1C)
and infinitary combinatorics (4A1D-4A1H).  %4A1D 4A1E 4A1F 4A1G 4A1H
Rather more specialized, we have the theory of normal (ultra)filters
(4A1J-4A1L) %4A1J 4A1K 4A1L
and a mention of Ostaszewski's $\clubsuit$ (4A1M-4A1N), used for an
example in \S439.   I conclude with a simple result on the cardinality
of $\sigma$-algebras (4A1O).

\leader{4A1A}{Cardinals again (a)} An infinite cardinal $\kappa$ is
{\bf regular} if\cmmnt{ it is not the supremum of fewer than
$\kappa$
smaller cardinals, that is, if} $\cf\kappa=\kappa$.   Any infinite
successor cardinal is regular.\cmmnt{  (\Kunen, I.10.37;
{\smc Just \& Weese 96}, 11.18;  \JechGH, p.\ 40;
\Levy, IV.3.11.)}\cmmnt{ In particular,} $\omega_1=\omega^+$ is regular.
%4{}19

\spheader 4A1Ab If $\zeta$ is an ordinal and $X$ is a set then I say that a
family $\langle x_{\xi}\rangle_{\xi<\zeta}$ in $X$ {\bf runs over $X$
with cofinal repetitions} if $\{\xi:\xi<\zeta,\,x_{\xi}=x\}$ is
cofinal with $\zeta$ for every $x\in X$.
%4{}29
Now if $X$ is any non-empty set and $\kappa$ is a cardinal greater than
or equal to $\max(\omega,\#(X))$, there
is a family $\langle x_{\xi}\rangle_{\xi<\kappa}$ running over $X$ with
cofinal repetitions.  \prooflet{\Prf\ By 3A1Ca, there is a surjection
$\xi\mapsto(x_{\xi},\alpha_{\xi}):\kappa\to X\times\kappa$.\ \Qed}

%\#([X]^{<\omega})\le\max(\omega,\#(X))  is 3A1Cd

\spheader 4A1Ac {\bf The cardinal $\frak c$} (i) Every non-trivial
interval in $\Bbb R$ has cardinal $\frak c$.
\cmmnt{({\smc Enderton 77}, p.\ 131.)}
%4{}19

\quad(ii)\dvAnew{2009} If $\#(A)\le\frak c$ and $D$ is countable, then
$\#(A^D)\le\frak c$.   
\cmmnt{($\#(A^D)\le\#(\Cal P\Bbb N)^{\Bbb N})
=\#(\Cal P(\Bbb N\times\Bbb N))=\#(\Cal P\Bbb N)$.)}
%\refquery

\quad(iii) $\cf(2^{\kappa})>\kappa$ for every infinite
cardinal $\kappa$;  in particular, $\cf\frak c>\omega$.
\cmmnt{(\Kunen, I.10.40;  {\smc Just \& Weese 96},
11.24;  \JechGH, p.\ 46;  \JechIC, 5.11;  \Levy, V.5.2.)}
%4{}39 %5{}11

\spheader 4A1Ad {\bf The Continuum Hypothesis} This is the statement
`$\frak c=\omega_1$';  it is neither provable nor disprovable from the
ordinary axioms of mathematics\cmmnt{, including the Axiom of Choice}.
\cmmnt{As such, it belongs to Volume 5 rather than to the present
volume.   But I do at one point refer to one of its immediate
consequences.}   If the continuum hypothesis is true, then there is a
well-ordering $\preccurlyeq$ of $[0,1]$ such that
$([0,1],\preccurlyeq)$
has order type $\omega_1$\prooflet{ (because there is a bijection
$f:[0,1]\to\omega_1$, and we can set $s\preccurlyeq t$ if
$f(s)\le f(t)$)}.
%4{}63

\leader{4A1B}{Closed cofinal sets} Let $\alpha$ be an ordinal.
%we need the results for ordinals in 5{}A5F

\spheader 4A1Ba Note that a subset $F$ of $\alpha$ is closed in the
order topology iff $\sup A\in F$ whenever
$A\subseteq F$ is non-empty and $\sup A<\alpha$.
\cmmnt{(4A2S(a-ii).)}

\spheader 4A1Bb If $\alpha$ has uncountable cofinality, and
$A\subseteq\alpha$ has supremum $\alpha$, then
$A'=\{\xi:0<\xi<\alpha$, $\xi=\sup(A\cap\xi)\}$ is a closed cofinal
set in $\alpha$.
\prooflet{\Prf\ ($\alpha$)
For any $\eta<\alpha$ we can choose inductively a
strictly increasing sequence $\sequencen{\xi_n}$ in $A$ starting from
$\xi_0\ge\eta$;  now $\xi=\sup_{n\in\Bbb N}\xi_n\in A'$ and
$\xi\ge\eta$;  this shows that $A'$ is cofinal with $\alpha$.   ($\beta$)
If $\emptyset\ne B\subseteq A'$ and $\sup B=\xi<\alpha$, then
$A\cap\xi\supseteq A\cap\eta$ for every $\eta\in B$, so

\Centerline{$\xi=\sup B=\sup_{\eta\in B}\sup(A\cap\eta)
=\sup(\bigcup_{\eta\in B}A\cap\eta)\le\sup A\cap\xi\le\xi$}

\noindent and $\xi\in A'$.\ \Qed
%4{}29

}In particular, \cmmnt{taking $A=\alpha=\omega_1$,} the set of
non-zero
countable limit ordinals is a closed cofinal set in $\omega_1$.
%4{}19

\spheader 4A1Bc(i) If $\ofamily{\xi}{\alpha}{F_{\xi}}$ is a family of
subsets of $\alpha$,
the {\bf diagonal intersection} of $\ofamily{\xi}{\alpha}{F_{\xi}}$ is
$\{\xi:\xi<\alpha,\,\xi\in F_{\eta}$ for every $\eta<\xi\}$.

\quad(ii) If $\kappa$ is a regular uncountable cardinal and
$\ofamily{\xi}{\kappa}{F_{\xi}}$ is any family of closed cofinal sets
in $\kappa$, its diagonal intersection $F$ is again a closed cofinal set
in $\kappa$.
\prooflet{\Prf\ $F=\bigcap_{\xi<\kappa}(F_{\xi}\cup[0,\xi])$ is
certainly closed.   To see that it is cofinal, argue as follows.
Start from any $\zeta_0<\kappa$.   Given $\zeta_n<\kappa$, set

\Centerline{$\zeta_{n+1}
=\sup_{\xi<\zeta_n}(\min(F_{\xi}\setminus\zeta_n)+1)$;}

\noindent this is defined because every $F_{\xi}$ is cofinal with
$\kappa$, and is less than $\kappa$ because $\cf\kappa=\kappa$.   At
the
end of the induction, set $\zeta^*=\sup_{n\in\Bbb N}\zeta_n$;  then
$\zeta_0\le\zeta^*$ and $\zeta^*<\kappa$ because $\cf\kappa>\omega$.
If $\xi$, $\eta<\zeta^*$, there is an $n\in\Bbb N$ such that
$\max(\xi,\eta)<\zeta_n$, in which case
$F_{\xi}\cap(\zeta^*\setminus\eta)
\supseteq F_{\xi}\cap\zeta_{n+1}\setminus\zeta_n$ is non-empty.   As
$\eta$ is arbitrary and $F_{\xi}$ is closed, $\zeta^*\in F_{\xi}$;  as
$\xi$ is arbitrary, $\zeta^*\in F$;  as $\zeta_0$ is arbitrary, $F$ is
cofinal.\ \Qed}

%(\Kunen, II.6.14;  \JechGH, p.\ 58;  \JechIC, 8.4;
%\JWII, Theorem 21.7;
%\Levy, IV.4.20.)

\quad(iii) In particular, if $f:\kappa\to\kappa$ is any function, then
$\{\xi:\xi<\kappa,\,f(\eta)<\xi$ for every $\eta<\xi\}$ is a closed
cofinal set in $\kappa$\cmmnt{, being the diagonal intersection of
$\ofamily{\xi}{\kappa}{\kappa\setminus(f(\xi)+1)}$}.
%4{}23

\spheader 4A1Bd If $\alpha$ has uncountable cofinality, $\Cal F$ is a
non-empty family of closed cofinal sets
in $\alpha$ and $\#(\Cal F)<\cf\alpha$, then $\bigcap\Cal F$ is a
closed
cofinal set in $\alpha$.   \prooflet{\Prf\ Being the intersection of
closed sets it is surely closed.   Set
$\lambda=\max(\omega,\#(\Cal F))$ and let
$\ofamily{\xi}{\lambda}{F_{\xi}}$ run over $\Cal F$ with
cofinal repetitions.   Starting from any $\zeta_0<\alpha$, we can
choose $\langle\zeta_{\xi}\rangle_{1\le\xi\le\lambda}$ such that

\inset{-- if $\xi<\lambda$ then
$\zeta_{\xi}\le\zeta_{\xi+1}\in F_{\xi}$;

-- if $\xi\le\lambda$ is a non-zero limit ordinal,
$\zeta_{\xi}=\sup_{\eta<\xi}\zeta_{\eta}$.}

\noindent (Because $\lambda<\cf\alpha$, $\zeta_{\xi}<\alpha$ for
every $\xi$.)   Now $\zeta_0\le\zeta_{\lambda}<\alpha$, and if
$F\in\Cal F$, $\zeta<\zeta_{\lambda}$ there is a $\xi<\lambda$ such
that $F=F_{\xi}$ and $\zeta\le\zeta_{\xi}$, in which case
$\zeta\le\zeta_{\xi+1}\le\zeta_{\lambda}$ and $\zeta_{\xi+1}\in F$.
This shows that either $\zeta_{\lambda}\in F$ or
$\zeta_{\lambda}=\sup(F\cap\zeta_{\lambda})$, in which case again
$\zeta_{\lambda}\in F$.   As $F$ is arbitrary,
$\zeta_{\lambda}\in\bigcap\Cal F$;
as $\zeta_0$ is arbitrary, $\bigcap\Cal F$ is cofinal.\ \Qed
%4{}19 5{}A5

}In particular, the intersection of any sequence of closed cofinal sets
in $\omega_1$ is again a closed cofinal set in $\omega_1$.
%4{}19 %4{}11

\leader{4A1C}{Stationary sets (a)} Let $\kappa$ be a cardinal.   A
subset of $\kappa$ is
{\bf stationary} in $\kappa$ if it meets every closed cofinal set in
$\kappa$;  otherwise it is {\bf non-stationary}.
%4{}19

\spheader 4A1Cb If $\kappa$ is a cardinal of uncountable cofinality,
the intersection of any stationary subset of $\kappa$ with a closed
cofinal
set in $\kappa$ is again a stationary set\prooflet{ (because the
intersection
of two closed cofinal sets is a closed cofinal set)};  the family of
non-stationary subsets of $\kappa$ is a $\sigma$-ideal, the
{\bf non-stationary ideal} of $\kappa$.
\cmmnt{(\Kunen, II.6.9;  \JWII, Lemma 21.11;  \JechIC, p.\ 93;
\Levy, IV.4.35.)}
%4{}19

\spheader 4A1Cc {\bf Pressing-Down Lemma\cmmnt{ (Fodor's theorem)}} If
$\kappa$ is a regular uncountable cardinal, $A\subseteq\kappa$ is
stationary and $f:A\to\kappa$ is such that $f(\xi)<\xi$ for every
$\xi\in A$, then there is a stationary set $B\subseteq A$ such that
$f$ is constant on $B$.
\leaveitout{\Prf\Quer\ Otherwise, we have closed cofinal sets
$F_{\eta}\subseteq\omega_1$ such
that $f(\xi)\ne\eta$ for $\xi\in A\cap F_{\eta}$.   Set
$F=\{\xi:\xi<\omega_1,\,\xi\in F_{\eta}$ for every $\eta<\xi\}$;  then
$F$ is a closed cofinal set, so meets $A$.   \Bang\Qed}
\cmmnt{(\Kunen, II.6.15;  \JWII, Theorem 21.2;
\JechGH, Theorem 22;  \JechIC, 8.7;  \Levy, IV.4.40.)}
%4{}19

\spheader 4A1Cd There are disjoint stationary sets $A$,
$B\subseteq\omega_1$.
\cmmnt{(This is easily deduced from 419G or 438Cd, and is also a
special case
of very much stronger results.   See 541Ya in Volume 5, or
\Kunen, II.6.12;
\JWII, Corollary 23.4;
\JechGH, p.\ 59;   \JechIC, 8.8;  \Levy, IV.4.48.)}
%4{}39

\leader{4A1D}{$\Delta$-systems (a)} A family
$\family{\xi}{A}{I_{\xi}}$
of sets is a {\bf $\Delta$-system} with {\bf root} $I$ if
$I_{\xi}\cap I_{\eta}=I$ for all distinct $\xi$, $\eta\in A$.

\spheader 4A1Db {\bf $\Delta$-system Lemma} If $\#(A)$ is a regular
uncountable cardinal and $\langle I_{\xi}\rangle_{\xi\in A}$ is any
family of finite sets, there is a set $D\subseteq A$ such that
$\#(D)=\#(A)$ and $\family{\xi}{D}{I_{\xi}}$ is a $\Delta$-system.
\prooflet{(\Kunen, II.1.6;  \JWII, Theorem 16.3.  For the present
volume
we need only the case $\#(A)=\omega_1$, which is treated in
\JechGH, p.\ 225 and \JechIC, 9.18.)}
%4{}17 %4{}35  %5{}13 %5{}53
%5{}14 5{}15 for cardinals > omega_1;
% check refs for this case

\leader{4A1E}{Free sets (a)} Let $A$ be a set with cardinal at least
$\omega_2$, and $\langle J_{\xi}\rangle_{\xi\in A}$ a family of
countable sets.   Then there are distinct $\xi$, $\eta\in A$ such that
$\xi\notin J_{\eta}$ and $\eta\notin J_{\xi}$.
\prooflet{\Prf\ Let $K\subseteq A$ be a set with cardinal $\omega_1$,
and set
$L=K\cup\bigcup_{i\in K}J_i$.   Then $L$ has cardinal $\omega_1$, so
there is a $\xi\in A\setminus L$.   Now there is an $\eta\in
K\setminus
J_{\xi}$, and this pair $(\xi,\eta)$ serves.\ \Qed}
%for Losert's example, 4{}53

\spheader 4A1Eb If $\family{\xi}{A}{K_{\xi}}$ is a disjoint family of
sets indexed by an uncountable subset $A$ of $\omega_1$,
and $\langle J_{\eta}\rangle_{\eta<\omega_1}$ is a family
of countable sets, there is an uncountable $B\subseteq A$ such that
$K_{\xi}\cap J_{\eta}=\emptyset$ whenever $\eta$, $\xi\in B$ and
$\eta<\xi$.
\prooflet{\Prf\ Choose $\ofamily{\xi}{\omega_1}{\zeta_{\xi}}$
inductively in such a way that $\zeta_{\xi}\in A$ and
$K_{\zeta_{\xi}}\cap J_{\zeta_{\eta}}=\emptyset$,
$\zeta_{\xi}>\zeta_{\eta}$
for every $\eta<\xi$.   Set $B=\{\zeta_{\xi}:\xi<\omega_1\}$.\ \Qed}
%4{}35

\leader{4A1F}{Selecting subsequences (a)} Let $\familyiI{K_i}$ be a
countable family of
sets such that $\bigcap_{i\in J}K_i$ is infinite for every finite
subset $J$ of $I$.   Then there is an infinite set $K$ such that
$K\setminus K_i$ is finite and $K_i\setminus K$ is infinite for every
$i\in I$.
\prooflet{\Prf\ We can suppose that $I\subseteq\Bbb N$.   Choose
$\sequencen{k_n}$ inductively such that
$k_n\in\bigcap_{i\in I,i\le n}K_i\setminus\{k_i:i<n\}$ for every
$n\in\Bbb N$, and set $K=\{k_{2n}:n\in\Bbb N\}$.\ \Qed

}Consequently there is a family
$\ofamily{\xi}{\omega_1}{K_{\xi}}$ of
infinite subsets of $\Bbb N$ such that $K_{\xi}\setminus K_{\eta}$ is
finite if $\eta\le\xi$, infinite if $\xi<\eta$.
\prooflet{ (Choose the $K_{\xi}$ inductively.)}
%4{}66

\spheader 4A1Fb Let $\familyiI{\Cal J_i}$ be a countable family of
subsets of $[\Bbb N]^{\omega}$ such that
$\Cal J_i\cap\Cal PK\ne\emptyset$ for every
$K\in[\Bbb N]^{\omega}$ and $i\in I$.   Then there is an infinite
$K\subseteq\Bbb N$ such that for every $i\in I$ there is a
$J\in\Cal J_i$ such that $K\setminus J$ is finite.   \prooflet{\Prf\
The case $I=\emptyset$ is trivial;  suppose that $\sequence{n}{i_n}$ runs
over $I$.   Choose $K_n$, $k_n$ inductively, for $n\in\Bbb N$, by
taking

\Centerline{$K_0=\Bbb N$,\quad$k_n\in K_n$,
\quad$K_{n+1}\subseteq K_n\setminus\{k_n\}$,
\quad$K_{n+1}\in\Cal J_{i_n}$}

\noindent for every $n$;  set  $K=\{k_n:n\in\Bbb N\}$.\ \Qed}
%4{}37G

\leader{4A1G}{Ramsey's theorem} If $n\in\Bbb N$, $K$ is finite and
$h:[\Bbb N]^n\to K$ is any function, there is an infinite
$I\subseteq\Bbb N$ such that $h$ is constant on $[I]^n$.
\prooflet{({\smc Bollob\'as 79}, p.\ 105, Theorem 3;
\JWII, 15.3;
\JechGH, 29.1;  \JechIC, 9.1;
\Levy, IX.3.7.   For the present volume we need only the
case $n=\#(K)=2$.)}

\leader{4A1H}{\cmmnt{The Marriage Lemma \dvrocolon{again}}}\cmmnt{ In 449L
it will be useful to have an infinitary version of the Marriage Lemma
available.

\medskip

\noindent}{\bf Proposition} Let $X$ and $Y$ be sets, and
$R\subseteq X\times Y$ a set such that $R[\{x\}]$ is finite for every
$x\in X$ and
$\#(R[I])\ge\#(I)$ for every finite set $I\subseteq X$.   Then there
is an injective function $f:X\to Y$ such that $(x,f(x))\in R$ for every
$x\in X$.

\proof{ For each finite $J\subseteq X$ there is an injective function
$f_J:J\to Y$ such that $f_J\subseteq R$ (identifying $f_J$ with its
graph), by the ordinary Marriage Lemma (3A1K) applied to
$R\cap(J\times R[J])$.   Let $\Cal F$ be any ultrafilter on
$[X]^{<\omega}$ containing $\{J:I\subseteq J\in[X]^{<\omega}\}$ for
every $I\in[X]^{<\omega}$;  then for each $x\in X$ there must be an
$f(x)\in R[\{x\}]$ such that $\{J:f_J(x)=f(x)\}\in\Cal F$, because
$R[\{x\}]$ is finite.   Now $f\subseteq R$ is a function from $X$ to
$Y$
and must be injective, because for any $x$, $x'\in X$ there is a
$J\in[X]^{<\omega}$ such that $f$ and $f_J$ agree on $\{x,x'\}$.
}%end of proof of 4A1H

\leader{4A1I}{Filters (a)} Let $X$ be a non-empty set.
If $\Cal E\subseteq\Cal PX$ is non-empty
and has the finite intersection property,

\Centerline{$\Cal F=\{A:A\subseteq X,\,A\supseteq\bigcap\Cal E'$ for
some non-empty finite $\Cal E'\subseteq\Cal E\}$}

\noindent is the smallest filter on $X$ including $\Cal E$, the filter
{\bf generated} by $\Cal E$.

If $\Cal E\subseteq\Cal PX$ is non-empty and
downwards-directed, then it has the finite intersection property iff
it does not contain $\emptyset$;  in this case we say that $\Cal E$ is a
{\bf filter base};
$\Cal F=\{A:A\subseteq X,\,A\supseteq E$ for some $E\in\Cal E\}$, and
$\Cal E$ is a base for the filter $\Cal F$.

In general, if $\Cal E$ is a family of subsets of $X$,
then there is a filter on $X$ including
$\Cal E$ iff $\Cal E$ has the finite intersection property;  in this
case, there is an ultrafilter on $X$ including
$\Cal E$\prooflet{ (2A1O)}.

\spheader 4A1Ib If $\kappa$ is a cardinal and $\Cal F$ is a filter
then $\Cal F$ is {\bf $\kappa$-complete} if $\bigcap\Cal E\in\Cal F$
whenever $\Cal E\subseteq\Cal F$ and $0<\#(\Cal E)<\kappa$.
%4{}35
Every filter is $\omega$-complete.

\spheader 4A1Ic A filter $\Cal F$ on a regular uncountable cardinal
$\kappa$ is {\bf normal} if ($\alpha$) $\kappa\setminus\xi\in\Cal F$ for
every $\xi<\kappa$ ($\beta$) whenever
$\langle F_{\xi}\rangle_{\xi<\kappa}$ is a family in $\Cal F$, its
diagonal intersection belongs to $\Cal F$.

\leader{4A1J}{Lemma} A normal filter $\Cal F$ on a regular uncountable
cardinal $\kappa$ is $\kappa$-complete.

\proof{ If $\lambda<\kappa$ and $\ofamily{\xi}{\lambda}{F_{\xi}}$ is a
family in $\Cal F$, set $F_{\xi}=\kappa$ for $\lambda\le\xi<\kappa$,
and
let $F$ be the diagonal intersection of
$\ofamily{\xi}{\kappa}{F_{\xi}}$;  then
$\bigcap_{\xi<\lambda}F_{\xi}\supseteq F\setminus\lambda$ belongs to
$\Cal F$.
}%end of proof of 4A1J

\leader{4A1K}{Theorem} Let $X$ be a set and $\Cal F$ a non-principal
$\omega_1$-complete ultrafilter on $X$.   Let $\kappa$ be the least
cardinal of any non-empty set $\Cal E\subseteq\Cal F$ such that
$\bigcap\Cal E\notin\Cal F$.   Then $\kappa$ is a regular uncountable
cardinal, $\Cal F$ is $\kappa$-complete, and
there is a function $g:X\to\kappa$ such that $g[[\Cal F]]$ is a normal
ultrafilter on $\kappa$.

\proof{{\bf (a)} By the definition of $\kappa$, $\Cal F$ is
$\kappa$-complete.   Because $\Cal F$ is $\omega_1$-complete,
$\kappa>\omega$.   Let $H$ be the set of all functions $h:X\to\kappa$
such that $h^{-1}[\kappa\setminus\xi]\in\Cal F$ for every
$\xi<\kappa$.
Then $H$ is not empty.   \Prf\ Let $\ofamily{\xi}{\kappa}{E_{\xi}}$
be a
family in $\Cal F$ such that
$E=\bigcap_{\xi<\kappa}E_{\xi}\notin\Cal F$.   Because $\Cal F$ is an
ultrafilter, $X\setminus E\in\Cal F$.   Set $h(x)=0$ if
$x\in E$, $h(x)=\min\{\xi:x\notin E_{\xi}\}$ if $x\in X\setminus E$;
then

\Centerline{$h^{-1}[\kappa\setminus\xi]
\supseteq(X\setminus E)\cap\bigcap_{\eta<\xi}E_{\eta}\in\Cal F$}

\noindent for every $\xi<\kappa$, because $\Cal F$ is
$\kappa$-complete,
so $h\in H$.\ \Qed

\medskip

{\bf (b)} For $h$, $h'\in H$, say that $h\prec h'$ if
$\{x:h(x)<h'(x)\}\in\Cal F$.   Then there is a $g\in H$ such that
$h\not\prec g$ for any $h\in H$.   \Prf\Quer\ Otherwise, there is a
sequence $\sequencen{h_n}$ in $H$ such that $h_{n+1}\prec h_n$ for
every $n\in\Bbb N$.   In this case
$E_n=\{x:h_{n+1}(x)<h_n(x)\}\in\Cal F$ for every
$n$.   Because $\Cal F$ is $\omega_1$-complete, there is an
$x\in\bigcap_{n\in\Bbb N}E_n$;  but now $\sequencen{h_n(x)}$ is a
strictly decreasing sequence of ordinals, which is impossible.\
\Bang\Qed

\medskip

{\bf (c)} I should check that $\kappa$ is regular.   \Prf\ If
$\ofamily{\xi}{\lambda}{\alpha_{\xi}}$ is any family in $\kappa$ with
$\lambda<\kappa$, then $g^{-1}[\kappa\setminus\alpha_{\xi}]\in\Cal F$
for every $\xi$, so (because $\Cal F$ is $\kappa$-complete)

\Centerline{$g^{-1}[\kappa\setminus\sup_{\xi<\lambda}\alpha_{\xi}]
=\bigcap_{\xi<\lambda}g^{-1}[\kappa\setminus\alpha_{\xi}]\in\Cal F$,}

\noindent and $\sup_{\xi<\lambda}\alpha_{\xi}\ne\kappa$.\ \Qed

\medskip

{\bf (d)} The image filter $g[[\Cal F]]$ is an ultrafilter on $\kappa$,
by 2A1N.   Because $g\in H$, $g^{-1}[\kappa\setminus\xi]\in\Cal F$
and $\kappa\setminus\xi\in g[[\Cal F]]$ for any $\xi<\kappa$.   \Quer\
Suppose, if possible, that $g[[\Cal F]]$ is not normal.   Then there
is a family $\ofamily{\xi}{\kappa}{A_{\xi}}$ in $g[[\Cal F]]$ such that
its diagonal intersection $A$ does not belong to $g[[\Cal F]]$, that is,
$g^{-1}[A]\notin\Cal F$ and $X\setminus g^{-1}[A]\in\Cal F$.   Define
$h:X\to\kappa$ by setting

$$\eqalign{h(x)&=0\text{ if }g(x)\in A,\cr
&=\min\{\eta:\eta<g(x),\,g(x)\notin A_{\eta}\}
  \text{ if }g(x)\notin A.\cr}$$

\noindent Then

\Centerline{$h^{-1}[\kappa\setminus\xi]
\supseteq(X\setminus g^{-1}[A])\cap\bigcap_{\eta<\xi}g^{-1}[A_{\eta}]
\in\Cal F$}

\noindent for every $\xi<\kappa$.   Thus $h\in H$.   But also
$h(x)<g(x)$ for every $x\in X\setminus g^{-1}[A]$, so $h\prec g$,
contrary to the choice of $g$.\ \Bang

Thus $g[[\Cal F]]$ is a normal filter, and the theorem is proved.
}%end of proof of 4A1K

%4{}35

\leader{4A1L}{Theorem} Let $\kappa$ be a regular uncountable cardinal,
and $\Cal F$
a normal ultrafilter on $\kappa$.   If $S\subseteq[\kappa]^{<\omega}$,
there is a set $F\in\Cal F$ such that, for each $n\in\Bbb N$, $[F]^n$
is either a subset of $S$ or disjoint from $S$.

\proof{{\bf (a)} For each $n\in\Bbb N$ there is an
$F_n\in\Cal F$ such that either $[F_n]^n\subseteq S$ or
$[F_n]^n\cap S=\emptyset$.   \Prf\ Induce on $n$.   If $n=0$ we can take
$F_n=\kappa$, because $[\kappa]^0=\{\emptyset\}$.   For the inductive
step to $n+1$, set
$S_{\xi}=\{I:I\in[\kappa]^{<\omega}$, $I\cup\{\xi\}\in S\}$ for each
$\xi<\kappa$.   By the inductive hypothesis, there is for
each $\xi<\kappa$ a set $E_{\xi}\in\Cal F$ such that either
$[E_{\xi}]^n\subseteq S_{\xi}$ or $[E_{\xi}]^n\cap S_{\xi}=\emptyset$.
Let $E$ be the diagonal intersection of
$\ofamily{\xi}{\kappa}{E_{\xi}}$, so that $E\in\Cal F$.

Suppose that $A=\{\xi:[E_{\xi}]^n\subseteq S_{\xi}\}$ belongs to
$\Cal F$.   Then $E\cap A\in\Cal F$.   If $I\in[E\cap A]^{n+1}$, set
$\xi=\min I$.   Then $I\setminus\{\xi\}\subseteq E_{\xi}$, so that
$I\setminus\{\xi\}\in S_{\xi}$ and $I\in S$.   Thus
$[E\cap A]^{n+1}\subseteq S$.   Similarly, if $A\notin\Cal F$, then
$E\setminus A\in\Cal F$ and $[E\setminus A]^{n+1}\cap S=\emptyset$.
Thus we can take one of $E\cap A$, $E\setminus A$ for $F_{n+1}$, and
the induction continues.\ \Qed

\medskip

{\bf (b)} At the end of the induction, take $F=\bigcap_{n\in\Bbb N}F_n$;
this serves.
}%end of proof of 4A1L
%4{}35

\vleader{72pt}{4A1M}{Ostaszewski's $\clubsuit$} This is the statement

\inset{Let $\Omega$ be the family of non-zero countable limit
ordinals.   Then there is a family
$\langle\theta_{\xi}(n)\rangle_{\xi\in\Omega,n\in\Bbb N}$ such that
($\alpha$) for each
$\xi\in\Omega$, $\sequencen{\theta_{\xi}(n)}$ is a strictly increasing
sequence with supremum $\xi$ ($\beta$) for any uncountable
$A\subseteq\omega_1$ there is a $\xi\in\Omega$ such that
$\theta_{\xi}(n)\in A$ for every $n\in\Bbb N$.}

\cmmnt{\noindent This is an immediate consequence of Jensen's
$\diamondsuit$ (\JWII, Exercise 22.9), which is itself a consequence
of G\"odel's Axiom of Constructibility (\Kunen, \S II.7;  \JWII, \S22;
\JechGH, \S22;  \JechIC, 13.21).}
%4{}39

\leader{4A1N}{Lemma} Assume $\clubsuit$.   Then there is a family
$\langle C_{\xi}\rangle_{\xi<\omega_1}$ of sets such that (i)
$C_{\xi}\subseteq\xi$ for every $\xi<\omega_1$ (ii) $C_{\xi}\cap\eta$
is finite whenever $\eta<\xi<\omega_1$ (iii) for any uncountable sets
$A$, $B\subseteq\omega_1$ there is a $\xi<\omega_1$ such that
$A\cap C_{\xi}$ and $B\cap C_{\xi}$ are both infinite.

\proof{{\bf (a)} Let
$\langle\theta_{\xi}(n)\rangle_{\xi\in\Omega,n\in\Bbb N}$ be a family
as in 4A1M.   Let $f:\omega_1\to[\omega_1]^2$ be a surjection (3A1Cd).
For $\xi\in\Omega$, set

\Centerline{$C_{\xi}=\bigcup_{i\in\Bbb N}
f(\theta_{\xi}(i+1))\cap\xi\setminus\theta_{\xi}(i)$.}

\noindent Then $C_{\xi}\subseteq\xi$, and if $\eta<\xi$ there is some
$n\in\Bbb N$ such that $\theta_{\xi}(n)\ge\eta$, so that

\Centerline{$C_{\xi}\cap\eta
\subseteq\bigcup_{i\le n}f(\theta_{\xi}(i))$}

\noindent is finite.   For $\xi\in\omega_1\setminus\Omega$ set
$C_{\xi}=\emptyset$.   Then $\langle C_{\xi}\rangle_{\xi<\omega_1}$
satisfies (i) and (ii) above.

\medskip

{\bf (b)} Now suppose that $A$, $B\subseteq\omega_1$ are uncountable.
Choose $\langle\alpha_{\xi}\rangle_{\xi<\omega_1}$,
$\langle\beta_{\xi}\rangle_{\xi<\omega_1}$,
%$\langle\gamma_{\xi}\rangle_{\xi<\omega_1}$,
$\langle I_{\xi}\rangle_{\xi<\omega_1}$ inductively, as follows.
$\beta_{\xi}$ is to be the smallest ordinal such that
$\{\alpha_{\eta}:\eta<\xi\}\cup\bigcup_{\eta<\xi}I_{\eta}
\subseteq\beta_{\xi}$;  $I_{\xi}$ is to be a doubleton subset of
$\omega_1\setminus(\beta_{\xi}\cup\bigcup_{\eta\le\beta_{\xi}}f(\eta))$
meeting both $A$ and $B$;  and $\alpha_{\xi}<\omega_1$ is to be such
that $f(\alpha_{\xi})=I_{\xi}$.   Set
$D=\{\alpha_{\xi}:\xi<\omega_1\}$.
This construction ensures that $\ofamily{\xi}{\omega_1}{\alpha_{\xi}}$
and $\ofamily{\xi}{\omega_1}{\beta_{\xi}}$ are strictly increasing,
with
$\beta_{\xi}<\alpha_{\xi}<\beta_{\xi+1}$ for every $\xi$, so that
$f(\delta)$ meets both $A$ and $B$ for
every $\delta\in D$, while $f(\delta)\subseteq\delta'$ and
$f(\delta')\cap(\delta\cup f(\delta))=\emptyset$
whenever $\delta<\delta'$ in $D$.

By the choice of
$\langle\theta_{\xi}(n)\rangle_{\xi\in\Omega,n\in\Bbb N}$, there is a
$\xi\in\Omega$ such that $\theta_{\xi}(n)\in D$ for
every $n\in\Bbb N$.   But this means that

\Centerline{$f(\theta_{\xi}(i))\subseteq\theta_{\xi}(i+1)\subseteq\xi$,
\quad$f(\theta_{\xi}(i+1))\cap(\theta_{\xi}(i)\cup f(\theta_{\xi}(i)))
=\emptyset$}

\noindent for every $i\in\Bbb N$, so
$C_{\xi}=\bigcup_{i\ge 1}f(\theta_{\xi}(i))$ meets both $A$ and
$B$
in infinite sets.
}%end of proof
%4{}29N

\leader{4A1O}{The size of $\sigma$-algebras:  Proposition} Let $\frak A$
be a Boolean algebra, $B$ a subset of $\frak A$, and $\frak B$ the
$\sigma$-subalgebra of $\frak A$ generated by $B$\cmmnt{ (331E)}.
Then $\#(\frak B)\le\max(4,\#(B^{\Bbb N}))$.   In particular, if
$\#(B)\le\frak c$ then $\#(\frak B)\le\frak c$.

\proof{{\bf (a)} If $\#(B)\le 1$, this is trivial, since then
$\#(\frak B)\le 4$.   So we need consider only the case $\#(B)\ge 2$.

\medskip

{\bf (b)} Set $\kappa=\#(B^{\Bbb N})$;  then whenever
$\#(A)\le\kappa$,
that is, there is an injection from $A$ to $B^{\Bbb N}$, then

\Centerline{$\#(A^{\Bbb N})\le\#((B^{\Bbb N})^{\Bbb N})
=\#(B^{\Bbb N\times\Bbb N})=\#(B^{\Bbb N})=\kappa$.}

\noindent As we are supposing that $B$ has more than one element,
$\kappa\ge\#(\{0,1\}^{\Bbb N})=\frak c\ge\omega_1$.

\medskip

{\bf (c)} Define $\langle B_{\xi}\rangle_{\xi<\omega_1}$ inductively,
as follows.   $B_0=B\cup\{0\}$.   Given $\ofamily{\eta}{\xi}{B_{\eta}}$,
where $0<\xi<\omega_1$, set $B'_{\xi}=\bigcup_{\eta<\xi}B_{\eta}$ and

$$\eqalign{B_{\xi}
&=\{1\Bsetminus b:b\in B'_{\xi}\}\cr
&\qquad\cup\{\sup_{n\in\Bbb N}b_n:\sequencen{b_n}
  \text{ is a sequence in }B'_{\xi}
  \text{ with a supremum in }\frak A\};\cr}$$

\noindent continue.

An easy induction on $\xi$ (relying on 3A1Cc and (b) above) shows that
every $B_{\xi}$ has cardinal at most $\kappa$.   So
$C=\bigcup_{\xi<\omega_1}B_{\xi}$ has cardinal at most $\kappa$.

\medskip

{\bf (d)} Now $\ofamily{\xi}{\omega_1}{B_{\xi}}$ is a non-decreasing
family, so if $\sequencen{c_n}$ is any sequence in $C$ there is some
$\xi<\omega_1$ such that every $c_n$ belongs to
$B_{\xi}\subseteq B'_{\xi+1}$.   But this means that if
$\sup_{n\in\Bbb N}c_n$ is defined in $\frak A$, it belongs to
$B_{\xi+1}\subseteq C$.   At the same time,

\Centerline{$1\Bsetminus c_0\in B_{\xi+1}\subseteq C$.}

\noindent This shows that $C$ is closed under complementation and
countable suprema;  since it contains $0$, it is a $\sigma$-subalgebra
of $\frak A$;  since it includes $B$, it includes $\frak B$, and
$\#(\frak B)\le\#(C)\le\kappa$, as claimed.

\medskip

{\bf (d)} Finally, if $\#(B)\le\frak c$,
$\#(\frak B)\le\max(4,\#(B^{\Bbb N}))\le\frak c$ by 4A1A(c-ii).
}%end of proof of 4A1O

\leader{4A1P}{An incidental fact} If $I$ is a countable set and
$\epsilon>0$, there is a family
$\langle\epsilon_i\rangle_{i\in I}$ of strictly positive real numbers
such that $\sum_{i\in I}\epsilon_i\le\epsilon$.  \prooflet{\Prf\ Let
$f:I\to\Bbb N$ be an injection and set
$\epsilon_i=2^{-f(i)-1}\epsilon$.\ \Qed}
%4{}12 %4{}19  used in 3{}42

\exercises{
%\leader{4A1X}{Basic exercises (a)}

%\leader{4A1Y}{Further exercises (a)}

}%end of exercises

%\cmmnt{\Notesheader{4A1}

%}%end of notes

\discrpage

\frfilename{mt4a2.tex}
\versiondate{21.4.13/28.12.18}
\copyrightdate{2002}

\def\Cech{{\smc \v{C}ech 66}}
\def\Csaszar{{\smc Cs\'asz\'ar 78}}
\def\Engelking{{\smc Engelking 89}}
\def\Gaal{{\smc Gaal 64}}
\def\Kuratowski{{\smc Kuratowski 66}}
\def\Kechris{{\smc Kechris 95}}

\def\glitem#1{{\it #1}\enskip}

\def\chaptername{Appendix}
\def\sectionname{General topology}

\newsection{4A2}

Even more than in previous volumes, naturally enough, the work of this
volume depends on results from general topology.   We have now reached
the point where some of the facts I rely on are becoming hard
to find as explicitly stated theorems in standard textbooks.   I find
myself therefore writing out rather a lot of proofs.   You should not
suppose that the results to which I attach proofs, rather than
references, are particularly deep;  on the contrary, in many cases I am
merely spelling out solutions to classic exercises.

The style of `general' topology, as it has evolved over the last hundred
years, is to develop a language capable of squeezing the utmost from
every step of argument.   While this does sometimes lead to absurdly
obscure formulations, it remains a natural, and often profitable,
response to the remarkably dense network of related ideas in this area.
I therefore follow the spirit of the subject in giving the results I
need in the full generality achievable within the terminology I use.
For the convenience of anyone coming to the theory for the first time,
I repeat some of them in the forms in which they are actually applied.
I should remark, however, that in some cases materially stronger results
can be proved with little extra effort;  as always, this appendix is to
be thought of not as a substitute for a thorough study of the subject,
but as a guide connecting standard approaches to the general theory with
the special needs of this volume.

\vleader{72pt}{4A2A}{Definitions}\cmmnt{ I begin the section with a
glossary
of terms not defined elsewhere.

\medskip}

\glitem{Baire space} A topological space $X$ is a {\bf Baire space} if
$\bigcap_{n\in\Bbb N}G_n$ is dense in $X$ whenever $\sequencen{G_n}$ is a
sequence of dense open subsets of $X$.

\glitem{Base of neighbourhoods} If $X$ is a topological space and
$x\in X$,
a {\bf base of neighbourhoods} of $x$ is a family $\Cal V$ of
neighbourhoods of $x$ such that
every neighbourhood of $x$ includes some member of $\Cal V$.
%Engelking 1.1

\glitem{boundary} If $X$ is a topological space and $A\subseteq X$, the
{\bf boundary} of $A$ is
$\partial A=\overline{A}\setminus\interior A
=\overline{A}\cap\overline{X\setminus A}$.
%Engelking 1.3

\glitem{\cadlag} If $X$ is a Hausdorff space, a function
$f:\coint{0,\infty}\to X$ is {\bf c\`adl\`ag}  (`continue \`a
droit, limite \`a gauche') (or {\bf RCLL}
(`right continuous, left limits'), an {\bf $\pmb{R}$-function},)
%Rogers & Williams 94
%also  corlol  (continuous on right, limits on left)
if $\lim_{s\downarrow t}f(s)=f(t)$
for every $t\ge 0$ and $\lim_{s\uparrow t}f(s)$ is defined in
$X$ for every $t>0$.

\glitem{\callal} If $X$ is a Hausdorff space, a function
$f:\coint{0,\infty}\to X$ is {\bf\callal}  (`continue \`a l'une,
limite \`a l'autre') if $f(0)=\lim_{s\downarrow 0}f(s)$ and, for every
$t>0$, $\lim_{s\downarrow t}f(s)$ and $\lim_{s\uparrow t}f(s)$
are defined in $X$, and at least one of them is equal to $f(t)$.

\glitem{\v{C}ech-complete} A completely regular Hausdorff topological
space
$X$ is {\bf \v{C}ech-complete} if it is homeomorphic to a G$_{\delta}$
subset of a compact Hausdorff space.
%4@34 %4@35
%Engelking 3.9

   %>component

\glitem{closed interval} Let $X$ be a totally ordered set.   A {\bf
closed interval} in $X$ is an interval of one of the forms $\emptyset$,
$[x,y]$,
$\ocint{-\infty,y}$, $\coint{x,\infty}$ or $X=\ooint{-\infty,\infty}$
where $x$, $y\in X$\cmmnt{ (see the definition of `interval' below)}.

\glitem{coarser topology} If $\frak S$ and $\frak T$ are two topologies
on a set $X$, we say that $\frak S$ is {\bf coarser} than $\frak T$ if
$\frak S\subseteq\frak T$.   (Equality allowed.)

\glitem{compact support} Let $X$ be a topological space and
$f:X\to\Bbb R$
a function.   I say that $f$ has {\bf compact support} if
$\overline{\{x:x\in X,\,f(x)\ne 0\}}$ is compact in $X$.

\glitem{countably compact} A topological space $X$ is {\bf countably
compact} if every countable open cover of $X$ has a finite subcover.
\cmmnt{({\bf Warning!} some authors reserve the term for Hausdorff
spaces.)}
%Engelking 3.10
A subset of a topological space is countably compact if it is countably
compact in its subspace topology.

\glitem{countably paracompact} A topological space $X$ is {\bf countably
paracompact} if given any countable open cover $\Cal G$ of $X$ there is
a locally finite family $\Cal H$ of open sets which refines $\Cal G$ and
covers $X$.
\cmmnt{({\bf Warning!} some authors reserve the term for Hausdorff
spaces.)}
%Engelking 5.2

\glitem{countably tight} A topological space $X$ is {\bf countably
tight}
(or has {\bf countable tightness}) if whenever $A\subseteq X$ and
$x\in\overline{A}$ there is a countable set $B\subseteq A$ such that
$x\in\overline{B}$.
%4@25
%Engelking 1.7.13

\leaveitout{\glitem{development} If $X$ is a topological space, a
{\bf development} for the topology of $X$ is a sequence
$\sequencen{\Cal G_n}$ of open covers of $X$ such that whenever
$H\subseteq X$ is open and $x\in H$, there is an $n\in\Bbb N$ such that
$\bigcup\{G:x\in G\in\Cal G_n\}\subseteq H\}$.
}%end of leaveitout

\glitem{direct sum, disjoint union} Let $\familyiI{(X_i,\frak T_i)}$ be
a family of topological spaces, and set
$X=\{(x,i):i\in I$, $x\in X_i\}$.
The {\bf disjoint union topology} on $X$ is $\frak T=\{G:G\subseteq X$,
$\{x:(x,i)\in G\}\in\frak T_i$ for every $i\in I\}$;  $(X,\frak T)$ is
the {\bf (direct) sum} of $\familyiI{(X_i,\frak T_i)}$.

If $X$ is a set, $\familyiI{X_i}$ a partition of $X$, and
$\frak T_i$ a topology on $X_i$ for every $i\in I$, then the {\bf
disjoint union topology} on $X$ is
$\{G:G\subseteq X$, $G\cap X_i\in\frak T_i$ for every $i\in I\}$.
%Engelking \S2.2)
%4@14

\glitem{dyadic} A Hausdorff space is {\bf dyadic} if it is a continuous
image of $\{0,1\}^I$ for some set $I$.
%Engelking exercises

\glitem{equicontinuous} If $X$ is a topological space, $(Y,\Cal W)$ a
uniform space, and $F$ a set of functions from $X$ to $Y$, then $F$ is {\bf
equicontinuous} if for every $x\in X$ and $W\in\Cal W$ the set
$\{y:(f(x),f(y))\in W$ for every $f\in F\}$ is a neighbourhood of $x$.
%Kelley

\glitem{finer topology} If $\frak S$ and $\frak T$ are two topologies on
a set $X$, we say that $\frak S$ is {\bf finer} than $\frak T$ if
$\frak S\supseteq\frak T$.   (Equality allowed.)

\glitem{first-countable} A topological space $X$ is
{\bf first-countable} if every point has a countable base of
neighbourhoods.
%4@39
%Engelking 1.1

\glitem{half-open} Let $X$ be a totally ordered set.   A {\bf half-open
interval} in $X$ is a set of one of the forms $\coint{x,y}$,
$\ocint{x,y}$ where $x$, $y\in X$ and $x<y$\cmmnt{ (see the definition
of `interval' below)}.
%this excludes \emptyset

\glitem{hereditarily Lindel\"of} A topological space is {\bf
hereditarily Lindel\"of} if every subspace is Lindel\"of.
%4@39, %4@33, %4@25

\glitem{hereditarily metacompact} A topological space is {\bf
hereditarily metacompact} if every subspace is metacompact.
%4@38

\glitem{hereditarily separable}\dvAnew{2024} A topological space is 
{\bf hereditarily separable} if every subspace is separable.
%4@38

\glitem{indiscrete} If $X$ is any set, the {\bf indiscrete} topology on
$X$ is the topology $\{\emptyset,X\}$.
%4@77

\glitem{interval} Let $(P,\le)$ be a partially ordered set.
An {\bf interval} in $P$ is a set of one of the forms
$[p,q]=\{r:p\le r\le q\}$, $\coint{p,q}=\{r:p\le r<q\}$,
$\ocint{p,q}=\{r:p<r\le q\}$, $\ooint{p,q}=\{r:p<r<q\}$,
$\coint{p,\infty}=\{r:p\le r\}$,
$\ocint{-\infty,q}=\{r:r\le q\}$, $\ooint{p,\infty}=\{r:p<r\}$,
$\ooint{-\infty,q}=\{r:r<q\}$, $\ooint{-\infty,\infty}=P$, where $p$,
$q\in P$.   \cmmnt{Note that every interval is
order-convex, but even in a totally ordered set not every order-convex set
need be an interval in this sense;  an interval always has end-points, if
we allow $\pm\infty$.}

\glitem{irreducible} If $X$ and $Y$ are topological spaces, a continuous
surjection $f:X\to Y$ is {\bf irreducible} if $f[F]\ne Y$ for any closed
proper subset $F$ of $X$.
%Engelking 3.1.C
%what one actually seems to need is that $\overline{f[F]}\ne Y$ for
%closed $F\subset X$

\glitem{isolated} If $X$ is a topological space, a family
$\Cal A$ of subsets of $X$ is {\bf isolated}
%or `relatively open'
if $A\cap\overline{\bigcup(\Cal A\setminus\{A\})}$ is empty for every
$A\in\Cal A$\cmmnt{;  that is, if $\Cal A$ is disjoint and every
member of $\Cal A$ is a relatively open set in $\bigcup\Cal A$}.

\glitem{Lindel\"of} A topological space is {\bf Lindel\"of} if every
open
cover has a countable subcover.    \cmmnt{({\bf Warning!} some authors
reserve the term for regular spaces.)}
%4@12
%Engelking 3.8 requires `regular'

\glitem{Lipschitz}\dvArevised{2022} 
If $(X,\rho)$ and $(Y,\sigma)$ are metric spaces, a
function $f:X\to Y$ is {\bf $\gamma$-Lipschitz}, or
{\bf $(\gamma,\rho,\sigma)$-Lipschitz}, where $\gamma\ge 0$, if
$\sigma(f(x),f(y))\le\gamma\rho(x,y)$ for all $x$, $y\in X$.   $f:X\to Y$
is {\bf Lipschitz} or {\bf $(\rho,\sigma)$-Lipschitz}
if it is $(\gamma,\rho,\sigma)$-Lipschitz for some $\gamma\ge 0$.  
%4@71

\glitem{locally finite} If $X$ is a topological space, a family $\Cal A$
of
subsets of $X$ is {\bf locally finite} if for every $x\in X$ there is an
open set which contains $x$ and meets only finitely many members of
$\Cal A$.
%4@A2
%Engelking 1.1 (for indexed families)

\glitem{lower semi-continuous} If $X$ is a topological space and $T$ a
totally ordered set, a function $f:X\to T$ is
{\bf lower semi-continuous} if $\{x:f(x)>t\}$ is open for every
$t\in T$.  \cmmnt{(Cf.\ 225H, 3A3Cf.)}
%Engelking 1.7.14 (for real-valued functions)

%\glitem{manifold} If $r\ge 1$ is an integer, an {\bf $r$-dimensional
%manifold} is a Hausdorff space $X$ such that every point of $X$ belongs
%to an open set homeomorphic to the open ball
%$\{x:\|x\|<1\}\subseteq\BbbR^r$.

\glitem{metacompact} A topological space is
{\bf metacompact} if every open
cover has a point-finite refinement which is an open cover.
\cmmnt{({\bf Warning!} some authors reserve the term for Hausdorff
spaces.)}
%Engelking 5.3

\leaveitout{
\glitem{monotonically normal} Let $(X,\frak T)$ be a
topological space
and $Q$ the set of disjoint pairs $(E,F)$ of closed sets in $X$.   $X$ is
{\bf monotonically normal} if there is a function $W:Q\to\frak T$ such that
($\alpha$) $E\subseteq W(E,F)$ and $F\cap\overline{W(E,F)}=\emptyset$ for
all $(E,F)\in Q$ ($\beta$) if $(E,F)$ and $(E',F')\in Q$, $E\subseteq E'$
and $F'\subseteq F$ then $W(E,F)\subseteq W(E',F')$.
}%end of leaveitout

\leaveitout{\glitem{Moore space} A {\bf Moore space} is a regular
Hausdorff space with a development.
}

\glitem{neighbourhood} If $X$ is a topological space and $x\in X$, a
{\bf
neighbourhood} of $x$ is any subset of $X$ including an open set which
contains $x$.
%Engelking 1.1
%4@A2

\glitem{network} Let $(X,\frak T)$ be a topological space.   A
{\bf network} for $\frak T$ is a family $\Cal E\subseteq\Cal PX$ such
that whenever $x\in G\in\frak T$ there is an $E\in\Cal E$ such that
$x\in E\subseteq G$.
%Engelking 3.1

\glitem{normal} A topological space $X$ is {\bf normal} if for any
disjoint
closed sets $E$, $F\subseteq X$ there are disjoint open sets $G$, $H$
such that $E\subseteq G$ and $F\subseteq H$.   \cmmnt{({\bf Warning!}
some authors reserve the term for Hausdorff spaces.)}

\glitem{open interval} Let $X$ be a totally ordered set.   An
{\bf open interval} in $X$ is a set of one of the the forms
$\ooint{x,y}$, $\ooint{x,\infty}$, $\ooint{-\infty,x}$ or
$\ooint{-\infty,\infty}=X$ where $x$, $y\in X$\cmmnt{ (see
the definition of `interval' above)}.

\glitem{open map} If $(X,\frak T)$ and $(Y,\frak S)$ are topological
spaces, a function $f:X\to Y$ is {\bf open} if $f[G]\in\frak S$ for
every $G\in\frak T$.

\glitem{order-convex} Let $(P,\le)$ be a partially ordered set.   A
subset $C$ of $P$ is {\bf order-convex} if
$[p,q]=\{r:p\le r\le q\}$ is included in $C$ whenever $p$, $q\in C$.

\glitem{order topology}  Let $(X,\le)$ be a totally ordered set.   Its
{\bf order topology} is that generated by intervals of the form
$\ooint{x,\infty}\cmmnt{\mskip5mu =\{y:y>x\}}$,
$\ooint{-\infty,x}\cmmnt{\mskip5mu =\{y:y<x\}}$ as $x$ runs over $X$.
%Engelking 1.7.4

\glitem{paracompact} A topological space is {\bf paracompact} if every
open cover has a locally finite refinement which is an
open cover.   \cmmnt{({\bf Warning!}  some authors reserve the term
for Hausdorff spaces.)}
%Engelking 5.1

\glitem{perfect} A topological space is {\bf perfect} if
it is compact and has no isolated points.

\glitem{perfectly normal} A topological space is {\bf perfectly normal}
if it is normal and every closed set is a G$_{\delta}$ set.
\cmmnt{({\bf
Warning!} remember that some authors reserve the term `normal' for
Hausdorff spaces.)}

\glitem{point-countable, point-finite} A family $\Cal A$ of sets is
{\bf point-countable} if no
point belongs to more than countably many members of $\Cal A$.
%4@A2
Similarly, an indexed family $\familyiI{A_i}$ of sets is
{\bf point-finite} if $\{i:x\in A_i\}$ is finite for every $x$.
%4@38

\glitem{Polish} A topological space $X$ is {\bf Polish} if it is
separable
and its topology can be defined from a metric under which $X$ is
complete.

\glitem{pseudometrizable} A topological space $(X,\frak T)$ is
{\bf pseudometrizable} if $\frak T$ is defined by a single
pseudometric\cmmnt{ (2A3F)}.

\glitem{refine(ment)} If $\Cal A$ is a family of sets, a {\bf
refinement}
of $\Cal A$ is a family $\Cal B$ of sets such that every member of
$\Cal B$ is included in some member of $\Cal A$;  in this case I say
that $\Cal B$ {\bf refines} $\Cal A$.   \cmmnt{({\bf Warning!} I do
not suppose that $\bigcup\Cal B=\bigcup\Cal A$.)}
%4@A2

\glitem{relatively countably compact} If $X$ is a topological space, a
subset $A$ of $X$ is {\bf relatively countably compact} if every
sequence in $A$ has a cluster point in $X$.   \cmmnt{({\bf Warning!}
This is {\it not} the same as supposing that $A$ is included in a
countably compact subset of $X$.)}

\glitem{scattered} A topological space $X$ is {\bf scattered} if every
non-empty subset of $X$ has an isolated point (in its subspace
topology).

\glitem{second-countable} A topological space is
{\bf second-countable} if the topology has a
countable base\cmmnt{, that is, if its weight is at most $\omega$}.
%Engelking 1.1
%4@13

\glitem{semi-continuous} {\it see} lower semi-continuous, upper
semi-continuous.

\glitem{sequential} A topological space is {\bf sequential} if every
sequentially closed set in $X$ is closed.
%Engelking 1.6

\glitem{sequentially closed} If $X$ is a topological space, a subset $A$
of
$X$ is {\bf sequentially closed} if $x\in A$ whenever $\sequencen{x_n}$
is a sequence in $A$ converging to $x\in X$.

\glitem{sequentially compact} A topological space is
{\bf sequentially compact} if every sequence has a convergent sequence.
A subset of a
topological space is sequentially compact if it is sequentially compact
in its subspace topology.   \cmmnt{({\bf Warning!} some authors
reserve the term for Hausdorff spaces.)}
%4@16 %4@25

\glitem{sequentially continuous} If $X$ and $Y$ are topological spaces,
a function $f:X\to Y$ is {\bf sequentially continuous} if
$\sequencen{f(x_n)}\to f(x)$ in $Y$ whenever $\sequencen{x_n}\to x$ in
$X$.

\glitem{subbase} If $(X,\frak T)$ is a topological space, a {\bf
subbase}
for $\frak T$ is a family $\Cal U\subseteq\frak T$ which generates
$\frak T$, in the sense that $\frak T$ is the coarsest topology on $X$
including $\Cal U$.   \cmmnt{({\bf Warning!} most authors reserve the
term for families $\Cal U$ with union $X$.)}

\glitem{totally bounded} If $(X,\Cal W)$ is a uniform space, a subset
$A$ of $X$ is {\bf totally bounded} if for every $W\in\Cal W$ there is a
finite set $I\subseteq X$ such that $A\subseteq W[I]$.   If $(X,\rho)$
is a metric space, a subset of $X$ is totally bounded if it is totally
bounded for the associated uniformity\cmmnt{ (3A4B)}.

\glitem{uniform convergence} If $X$ is a set, $(Y,\sigma)$ is a metric
space and $\Cal A$ is a family
of subsets of $X$ then the {\bf topology of uniform convergence} on
members of $\Cal A$ is the topology on $Y^X$ generated by the
pseudometrics $(f,g)\mapsto\min(1,\sup_{x\in A}\sigma(f(x),g(x)))$ as
$A$ runs over
$\Cal A\setminus\{\emptyset\}$.   \cmmnt{(It is elementary to
verify that the formula here defines a pseudometric.)}

\glitem{upper semi-continuous} If $X$ is a topological space and $T$ is a
totally ordered set, a function
$f:X\to T$ is {\bf upper semi-continuous} if $\{x:f(x)<t\}$ is
open for every $t\in T$.
%Engelking 1.7.14

\glitem{weakly $\alpha$-favourable} A topological space $(X,\frak T)$ is
{\bf weakly $\alpha$-favourable} if there is a function
$\sigma:\bigcup_{n\in\Bbb N}(\frak T\setminus\{\emptyset\})^{n+1}
\to\frak T\setminus\{\emptyset\}$ such that (i)
$\sigma(G_0,\ldots,G_n)\subseteq G_n$ whenever $G_0,\ldots,G_n$ are
non-empty open sets (ii) whenever
$\sequencen{G_n}$ is a sequence in $\frak T\setminus\{\emptyset\}$ such
that $G_{n+1}\subseteq\sigma(G_0,\ldots,G_n)$ for every $n$, then
$\bigcap_{n\in\Bbb N}G_n$ is non-empty.
%Kechris uses `Choquet space'

\glitem{weight} If $X$ is a topological space, its {\bf weight} $w(X)$
is the smallest cardinal of any base for the topology.
%4@25
%Engelking 1.1

\glitem{$C_b$} If $X$ is a topological space, $C_b(X)$ is the space of
bounded continuous real-valued functions defined on $X$.

\glitem{F$_{\sigma}$} If $X$ is a topological space, an {\bf
F$_{\sigma}$}
set in $X$ is one expressible as the union of a sequence of closed sets.
%4@22
%Engelking 1.3

\glitem{G$_{\delta}$} If $X$ is a topological space, a {\bf
G$_{\delta}$}
set in $X$ is one expressible as the intersection of a
sequence of open sets.
%Engelking 1.3

\glitem{K$_{\sigma}$} If $X$ is a topological space, a
{\bf K$_{\sigma}$} set in $X$ is one expressible as the union of a
sequence of compact sets.
%4{}59

%\glitem{lim sup} If $\family{i}{I}{A_i}$ is a non-empty family of sets
%in a topological space and $\Cal F$ is a filter on $I$,
%I write $\limsup_{i\to\Cal F}A_i$ for
%$\bigcap_{J\in\Cal F}\overline{\bigcup_{i\in J}A_i}$.

$\Cal PX$ If $X$ is any set, the {\bf usual topology} on $\Cal PX$ is
that generated by the sets $\{a:a\subseteq X$, $a\cap J=K\}$ where
$J\subseteq X$ is finite and $K\subseteq J$.

\glitem{T$_0$} If $(X,\frak T)$ is a topological space, we say that it
is
{\bf T$_0$} if for any two distinct points of $X$ there is an open set
containing one but not the other.
%Engelking 1.5

\glitem{T$_1$} If $(X,\frak T)$ is a topological space, we say that it
is
{\bf T$_1$} if singleton sets are closed.

\glitem{$\pi$-base} If $(X,\frak T)$ is a topological space, a
{\bf $\pi$-base} for $\frak T$ is a set $\Cal U\subseteq\frak T$ such
that every non-empty open set includes a non-empty member of $\Cal U$.

\glitem{$\sigma$-compact} A topological space $X$ is
{\bf $\sigma$-compact}
if there is a sequence of compact subsets of $X$ covering $X$.

\glitem{$\sigma$-disjoint} A family of sets is
{\bf $\sigma$-disjoint} if it is expressible as
$\bigcup_{n\in\Bbb N}\Cal A_n$ where every $\Cal A_n$ is disjoint.

\glitem{$\sigma$-isolated} If $X$ is a topological space, a
family of subsets of $X$ is {\bf $\sigma$-isolated} if it is expressible
as $\bigcup_{n\in\Bbb N}\Cal A_n$ where every $\Cal A_n$ is an isolated
family.

\glitem{$\sigma$-metrically-discrete} If $(X,\rho)$ is a metric space, a
family of subsets of $X$ is {\bf $\sigma$-metrically-discrete}
if it is expressible as $\bigcup_{n\in\Bbb N}\Cal A_n$ where
$\rho(x,y)\ge 2^{-n}$ whenever $n\in\Bbb N$, $A$ and $B$ are distinct
members of $\Cal A_n$, $x\in A$ and $y\in B$.

\leader{4A2B}{Elementary facts about general topological spaces}
{\bf (a) Bases and networks} (i) Let $(X,\frak T)$ be a topological
space and $\Cal U$ a subbase for $\frak T$.   Then
$\{X\}\cup\{U_0\cap U_1\cap\ldots\cap U_n:U_0,\ldots,U_n\in\Cal U\}$ is
a base for $\frak T$.   \prooflet{(For this is a base for a topology, by
3A3Mc.)}

\quad (ii) Let $X$ and $Y$ be topological spaces, and $\Cal U$ a subbase
for the topology of $Y$.   Then a function $f:X\to Y$ is continuous iff
$f^{-1}[U]$ is open for every $U\in\Cal U$.   \prooflet{(\Engelking,
1.4.1(ii)).}
%4@18 %4@17 %4@23

\quad (iii) If $X$ and $Y$ are topological spaces, $\Cal E$ is a network
for the topology of $Y$, and $f:X\to Y$ is a function such that
$f^{-1}[E]$ is open for every $E\in\Cal E$, then $f$ is continuous.
\prooflet{(The topology generated by $\Cal E$ includes the given
topology on $Y$.)}

\quad (iv) If $X$ is a topological space and $\Cal U$ is a subbase for
the topology of $X$, then a filter $\Cal F$ on $X$ converges to $x\in X$
iff $\{U:x\in U\in\Cal U\}\subseteq\Cal F$.   \prooflet{(If the
condition is satisfied, $\Cal F\cup\{A:A\subseteq X,\,x\notin A\}$ is a
topology on $X$ including $\Cal U$.)}
%4@39

\quad (v) If $X$ and $Y$ are topological spaces with subbases
$\Cal U$, $\Cal V$ respectively, then
$\{U\times Y:U\in\Cal U\}\cup\{X\times V:V\in\Cal V\}$ is a subbase for
the product topology of $X\times Y$.   \prooflet{(\Kuratowski, \S15.I.)}

%A base for a topology $\frak T$ is just a network for
%$\frak T$ consisting of members of $\frak T$.

\quad (vi) If $\Cal U$ is a (sub-\nobreak)base for a topology on $X$,
and $Y\subseteq X$, then
$\{Y\cap U:U\in\Cal U\}$ is a (sub-\nobreak)base for the subspace
topology of $Y$.   \prooflet{(\Csaszar, 2.3.13(e)-(f).)}

\quad (vii) If $X$ is a topological space, $\Cal E$ is a network for the
topology of $X$, and $Y$ is a subset of $X$, then
$\{E\cap Y:E\in\Cal E\}$ is a network for the topology of $Y$.

\quad (viii) If $X$ is a topological space and $\Cal A$ is a
($\sigma$-\nobreak)isolated family of subsets of $X$, then
$\{A\cap Y:A\in\Cal A'\}$ is ($\sigma$-\nobreak)isolated whenever
$Y\subseteq X$ and $\Cal A'\subseteq\Cal A$.

\quad (ix) If a topological space $X$ has a $\sigma$-isolated
network, so has every subspace of $X$.

%partitions into open sets
\spheader 4A2Bb
If $\familyiI{H_i}$ is a partition of a topological space $X$ into open
sets and $F_i\subseteq H_i$ is closed\cmmnt{ (either in $X$ or in
$H_i$)} for each $i\in I$, then $F=\bigcup_{i\in I}F_i$ is closed in
$X$.
\prooflet{($X\setminus F=\bigcup_{i\in I}(H_i\setminus F_i)$.)}
%4@43

%filters
\spheader 4A2Bc If $X$ is a topological space, $A\subseteq X$ and
$x\in X$,
then $x\in\overline{A}$ iff there is an ultrafilter on $X$, containing
$A$, which converges to $x$.
\prooflet{($\{A\}\cup\{G:x\in G\subseteq X$, $G$ is open$\}$ has the
finite intersection property;  use 4A1Ia.)}
%4@22

%semi-continuity
\spheader 4A2Bd {\bf Semi-continuity} Let $X$ be a topological space.

\quad(i) A function $f:X\to\Bbb R$ is lower semi-continuous iff $-f$ is
upper semi-continuous.
\cmmnt{(\Cech, 18D.8.)}
A function $f:X\to\Bbb R$ is lower semi-continuous iff
$\Omega=\{(x,\alpha):x\in X$, $\alpha\ge f(x)\}$ is closed in
$X\times\Bbb R$.
%\query not \Cech not \Engelking
\prooflet{(If $f$ is lower semi-continuous and
$\alpha<\beta<f(x)$ then $\{y:f(y)>\beta\}\times\ooint{-\infty,\beta}$ is a
neighbourhood of $(x,\alpha)$;  so $\Omega$ is closed.   If $\Omega$ is
closed then for any $\gamma\in\Bbb R$ the set
$\{x:f(x)>\gamma\}=\{x:(x,\gamma)\notin\Omega\}$ is open;  so $f$ is lower
semi-continuous.)}

\quad(ii) If $T$ is a totally ordered set, $f:X\to T$ is lower
semi-continuous, $Y$ is another topological space,
and $g:Y\to X$ is continuous, then $fg:Y\to T$ is lower
semi-continuous.
\prooflet{($\{y:(fg)(y)>t\}=g^{-1}[\{x:f(x)>t\}]$.)}
%4@43, 4@44
In particular, if $f:X\to T$ is lower semi-continuous and
$Y\subseteq X$, then $f\restr Y$ is lower semi-continuous.
Similarly, if $f:X\to T$ is upper semi-continuous
and $g:Y\to X$ is continuous, then $fg:Y\to T$ is upper
semi-continuous.

\quad(iii) If $f$, $g:X\to\ocint{-\infty,\infty}$ are lower
semi-continuous so is $f+g:X\to\ocint{-\infty,\infty}$.
\cmmnt{(\Cech, 18D.8.)}
%\prooflet{($\{x:(f+g)(x)>\alpha\}
%=\bigcup_{\beta\in\Bbb R}
%\{x:f(x)>\beta\}\cap\{x:g(x)>\alpha-\beta\}$.)}

\quad(iv) If $f$, $g:X\to[0,\infty]$ are lower semi-continuous so
is $f\times g:X\to[0,\infty]$.
\cmmnt{(\Cech, 18D.8.)}
%\prooflet{($\{x:(f\times g)(x)>\alpha\}
%=\bigcup_{\beta>0}\{x:f(x)>\beta\}\cap\{x:g(x)>\Bover{\alpha}{\beta}\}$
%if $\alpha>0$.)}
%4@43

\quad(v) If $\Phi$ is any non-empty set of lower semi-continuous
functions from $X$ to $[-\infty,\infty]$, then
$x\mapsto\sup_{f\in\Phi}f(x):X\to[-\infty,\infty]$ is lower
semi-continuous.

\quad(vi) $f:X\to\Bbb R$ is continuous iff $f$ is both upper
semi-continuous and lower semi-continuous iff $f$ and $-f$ are both
lower semi-continuous.

\quad(vii)
If $f:X\to[-\infty,\infty]$ is lower semi-continuous, and
$\Cal F$ is a filter on $X$ converging to $y\in X$, then
$f(y)\le\liminf_{x\to\Cal F}f(x)$.
%\prooflet{\query.} %not Engelking
%4{}94C

\quad(viii)
If $X$ is compact and not empty, and $f:X\to[-\infty,\infty]$ is lower
semi-continuous then $K=\{x:f(x)=\inf_{y\in X}f(y)\}$ is non-empty and
compact.
\prooflet{\Prf\ Setting $\gamma=\inf_{y\in X}f(y)\in[-\infty,\infty]$,
$\{\{x:f(x)\le\alpha\}:\alpha>\gamma\}$ is a downwards-directed family of
non-empty closed sets, so its intersection $K$ is a non-empty closed set.\
\Qed}

\quad(ix)
If $f$, $g:X\to[0,\infty]$ are lower semi-continuous and
$f+g$ is continuous at $x\in X$ and finite there,
then $f$ and $g$ are continuous at $x$.
\prooflet{\Prf\ If $\epsilon>0$ there is a neighbourhood $G$ of $x$ such
that $(f+g)(y)\le(f+g)(x)+\epsilon$ for every $y\in G$ and
$g(y)\ge g(x)-\epsilon$ for every $y\in G$, so that
$f(y)\le f(x)+2\epsilon$ for every $y\in G$.\ \Qed}

% If $f:X\to[-\infty,\infty]$ is lower semi-continuous and
% $\Cal F$ is a convergent ultrafilter on $X$ then
% $f(\lim\Cal F)\le\lim f[[\Cal F]]
% If $f$, $g$ are lower semi-continuous so are $f\vee g$, $f\wedge g$

%separability
\spheader 4A2Be {\bf Separable spaces} (i)
If $\familyiI{A_i}$ is a countable family of separable subsets of a
topological space $X$ then $\bigcup_{i\in I}A_i$
and $\overline{\bigcup_{i\in I}A_i}$ are separable.
\prooflet{(If $D_i\subseteq A_i$ is countable and dense for each $i$,
$\bigcup_{i\in I}D_i$ is countable and dense in both
$\bigcup_{i\in I}A_i$ and its closure.)}
%4@18 tacitly

\quad (ii) If $\familyiI{X_i}$ is a family of separable topological
spaces and $\#(I)\le\frak c$, then $\prod_{i\in I}X_i$ is separable.
\prooflet{(\Engelking, 2.3.16.)}

\quad (iii) A continuous image of a separable topological space is
separable.   \prooflet{(\Engelking, 1.4.11.)}

\spheader 4A2Bf {\bf Open maps} (i) Let $\familyiI{X_i}$ be any family of
topological
spaces, with product $X$.   If $J\subseteq I$ is any set, and we write
$X_J$ for $\prod_{i\in J}X_i$, then the canonical map
$x\mapsto x\restr J:X\to X_J$ is open.
\prooflet{(\Engelking, p.\ 79.)}
%4@15

\quad(ii) Let $X$ and $Y$ be topological spaces and $f:X\to Y$ a
continuous open map.   Then
$\interior{f^{-1}[B]}=f^{-1}[\interior B]$ and
$\overline{f^{-1}[B]}=f^{-1}[\overline{B}]$ for every $B\subseteq Y$.
\prooflet{\Prf\ Because $f$ is continuous, $f^{-1}[\interior B]$ is an
open set included in
$f^{-1}[B]$, so is included in $\interior f^{-1}[B]$.   Because $f$ is
open, $f[\interior f^{-1}[B]]$ is an open set included in
$f[f^{-1}[B]]\subseteq B$, so
$f[\interior f^{-1}[B]]\subseteq\interior B$, that is,
$\interior f^{-1}[B]\subseteq f^{-1}[\interior B]$.
Now apply this to $Y\setminus B$ and take complements.\ \Qed}

It follows that $f^{-1}[B]$ is nowhere dense in $X$ whenever $B\subseteq Y$
is nowhere dense in $Y$.   \prooflet{($\interior\overline{f^{-1}[B]}
=\interior f^{-1}[\overline{B}]=f^{-1}[\interior\overline{B}]=\emptyset$.)}
If $f$ is surjective and $B\subseteq Y$, then
$B$ is nowhere dense in $Y$ iff $f^{-1}[B]$ is nowhere dense in $X$.
\prooflet{(For $\interior\overline{f^{-1}[B]}
=f^{-1}[\interior\overline{B}]$ is empty iff $\interior\overline{B}$ is
empty.)}
%4@43

\quad(iii)
Let $X$ and $Y$ be topological spaces and $f:X\to Y$ a
continuous open map.   Then $H\mapsto f^{-1}[H]$ is an order-continuous
Boolean homomorphism from the regular open algebra of $Y$ to the regular
open algebra of $X$.   \prooflet{\Prf\ If $H\subseteq Y$ is a regular
open set,

\Centerline{$\interior\overline{f^{-1}[H]}=\interior f^{-1}[\overline{H}]
=f^{-1}[\interior\overline{H}]=f^{-1}[H]$}

\noindent by (ii), so $f^{-1}[H]$ is a regular open set in $X$.   If
$F\subseteq Y$ is nowhere dense, then
$f^{-1}[F]$ is nowhere dense in $X$, as noted in (ii) above.   By 314Ra,
$H\mapsto f^{-1}[H]=\interior\overline{f^{-1}[H]}$ is an order-continuous
Boolean
homomorphism from $\RO(Y)$ to $\RO(X)$.\ \QeD}   If $f$ is surjective, then
the homomorphism is injective\prooflet{ (because
$f^{-1}[H]\ne\emptyset$ whenever
$H\ne\emptyset$)}, and for $H\subseteq Y$, $H$ is a regular
open set in $Y$ iff $f^{-1}[H]$ is a regular open set in $X$\prooflet{
(because in this case $f^{-1}[H]=f^{-1}[\interior\overline{H}]$)}.
% 5{}27

\quad(iv)
If $X_0$, $Y_0$, $X_1$, $Y_1$ are topological spaces,
and $f_i:X_i\to Y_i$ is an open map for each $i$, then
$(x_0,x_1)\mapsto(f_0(x_0),f_1(x_1)):X_0\times X_1\to Y_0\times Y_1$ is
open.   \prooflet{(\Engelking, 2.3.29.)}

%product spaces
\spheader 4A2Bg Let $\familyiI{X_i}$ be a family of topological spaces
with product $X$.

\quad(i) If $A\subseteq X$ is determined by coordinates in
$J\subseteq I$\cmmnt{ in the sense of 254M}, then $\overline{A}$ and
$\interior A$ are also determined by coordinates in $J$.
\prooflet{\Prf\ Let
$\pi:X\to\prod_{i\in J}X_i$ be the canonical map.   Then
$A=\pi^{-1}[\pi[A]]$, so (f) tells us that
$\interior A=\pi^{-1}[\interior\pi[A]]$ and
$\overline{A}=\pi^{-1}[\overline{\pi[A]}]$;  but these are both
determined by coordinates in $J$.\ \Qed}

\quad(ii) If $F\subseteq X$ is closed, there is a smallest set
$J^*\subseteq I$ such that $F$ is determined by coordinates in
$J^*$.

\prooflet{\Prf\ Let $\Cal J$ be the family of all those sets
$J\subseteq I$ such that $F$ is determined by coordinates in $\Cal J$.
If $J_1$, $J_2\in\Cal J$, then $J_1\cap J_2\in\Cal J$ (254Ta).
Set $J^*=\bigcap\Cal J$.   \Quer\ Suppose, if possible, that $F$ is not
determined by coordinates in $J^*$.   Then there are $x\in F$,
$y\in X\setminus F$ such that $x\restr J^*=y\restr J^*$.   Because
$X\setminus F$ is open, there is a finite set $K\subseteq I$ such that
$z\notin F$ whenever $z\in X$ and $z\restr K=y\restr K$.   Because
$\Cal J$ is closed under finite intersections, there is a $J\in\Cal J$
such that $K\cap J=K\cap J^*$.   Define $z\in X$ by setting $z(i)=x(i)$
for $i\in J$, $z(i)=y(i)$ for $i\in I\setminus J$.   Then
$z\restr J=x\restr J$, so $z\in F$, but $z\restr K=y\restr K$, so
$z\notin F$.\ \Bang

Thus $J^*\in\Cal J$ and is the required smallest member of $\Cal J$.\
\Qed}%end of prooflet

%locally finite families
\spheader 4A2Bh Let $X$ be a topological space.

\quad(i) If $\Cal E$ is a locally finite family of closed subsets of
$X$, then $\bigcup\Cal E'$ is closed for every $\Cal E'\subseteq\Cal E$.
\prooflet{(\Engelking, 1.1.11.)}

\quad(ii) If $\familyiI{f_i}$ is a
family in $C(X)$ such that $\familyiI{\{x:f_i(x)\ne 0\}}$ is locally
finite, then we have a continuous function $f:X\to\Bbb R$ defined by
setting $f(x)=\sum_{i\in I}f_i(x)$ for every $x\in X$.
\prooflet{\Prf\ For any $x$, $\{i:f_i(x)\ne 0\}$ is finite, so $f$
is well-defined.   If $x_0\in X$ and $\epsilon>0$, there is a
neighbourhood $V$ of $x_0$ such that
$J=\{i:i\in I,\,f_i(x)\ne 0$ for some $x\in V\}$ is finite;  now there
is a neighbourhood $W$ of $x_0$,
included in $V$, such that
$\sum_{i\in J}|f_i(x)-\sum_{i\in J}f_i(x_0)|<\epsilon$ for every
$x\in W$, so that $|f(x)-f(x_0)|<\epsilon$ for every $x\in W$.   As
$x_0$ and $\epsilon$ are arbitrary, $f$ is continuous.\ \Qed}

%boundary
\spheader 4A2Bi Let $X$ be a topological space and $A$, $B$
two subsets of $X$.   Then the boundary $\partial(A*B)$ is included in
$\partial A\cup\partial B$, where $*$ is
any of $\cup$, $\cap$, $\setminus$, $\symmdiff$.  \prooflet{(Generally,
if $F\subseteq X$, $\{A:\partial A\subseteq F\}
=\{A:\overline{A}\setminus F\subseteq\interior A\}$ is a subalgebra of
$\Cal PX$.)}

%\leaveitout{\spheader 4A2Bi
%\quad(i) If $G\subseteq X$ is open then
%$G\cap\partial B\subseteq\partial(G\cap B)$.
%\prooflet{($G\cap\overline{B}\subseteq\overline{G\cap B}$,
%$G\cap\overline{X\setminus B}\subseteq\overline{G\setminus B}$.)}

%\quad(ii) \quad(iii) $A\cap\partial B\subseteq\partial
%A\cup\partial(A\cap B)$.
%4@74?
%}%end of leaveitout

%dense subspaces
\spheader 4A2Bj Let $X$ be a topological space and $D$ a dense subset of
$X$, endowed with its subspace topology.

\quad(i) A set $A\subseteq D$ is nowhere dense in $D$ iff it is nowhere
dense in $X$.   \prooflet{\Prf\

$$\eqalignno{A\text{ is nowhere dense in }X
&\iff X\setminus\overline{A}\text{ is dense in }X\cr
\displaycause{writing $\overline{A}$ for the closure of $A$ in $X$}
&\iff D\setminus\overline{A}\text{ is dense in }X\cr
\displaycause{3A3Ea}
&\iff D\setminus\overline{A}\text{ is dense in }D\cr
&\iff D\setminus(D\cap\overline{A})\text{ is dense in }D\cr
&\iff A\text{ is nowhere dense in }D\cr}$$

\noindent because $D\cap\overline{A}=\overline{A}^{(D)}$
is the closure of $A$ in $D$.\ \Qed}

\quad(ii) A set $G\subseteq D$ is a regular open set in $D$ iff it is
expressible as $D\cap H$ for some regular open set $H\subseteq X$.
\prooflet{\Prf\ ($\alpha$) If $G$ is a regular open subset of $D$, set
$H=\interior\overline{G}$, taking both the closure and the interior in $X$.
Then $H$ is a regular open set in $X$.   Now $D\cap H$ is a relatively open
subset of $D$ included in $D\cap\overline{G}=\overline{G}^{(D)}$, so
$D\cap H\subseteq\interior_D\overline{G}^{(D)}=G$.   In the other
direction, $\overline{G}\cup\overline{D\setminus G}=\overline{D}=X$,
so $\overline{G}\supseteq X\setminus\overline{D\setminus G}$ and
$H\supseteq X\setminus\overline{D\setminus G}\supseteq G$.   So
$G=H\cap D$ is of the required form.   ($\beta$) If $H\subseteq X$ is a
regular open set such that $G=D\cap H$,
set $V=X\setminus\overline{H}$;  then $H=X\setminus\overline{V}$.
Now

\Centerline{$\overline{V\cap D}^{(D)}
=D\cap\overline{V\cap D}=D\cap\overline{V}=D\setminus H=D\setminus G$,}

\noindent so
$G=D\setminus\overline{V\cap D}^{(D)}$ is the complement of the closure of
an open set in $D$, and is a regular open set in $D$.\ \Qed}

%Moore spaces
%\leaveitout{\spheader 4A2B? If $X$ has a development, it has a
%$\sigma$-isolated network.   \prooflet{({\smc Fremlin 84}, A4O.)}
%}%end of leaveitout

\leader{4A2C}{G$_{\delta}$, F$_{\sigma}$, zero and cozero sets} Let
$X$ be a topological space.

\spheader 4A2Ca(i) The union of two G$_{\delta}$ sets in $X$ is a
G$_{\delta}$ set.
\prooflet{(\Engelking, p.\ 26;  \Kuratowski, \S5.V.)}
%4@41

\quad (ii) The intersection of countably many G$_{\delta}$ sets is a
G$_{\delta}$ set.
\prooflet{(\Engelking, p.\ 26;  \Kuratowski, \S5.V.)}
%4@41 %4@24

\quad (iii) If $Y$ is another topological space, $f:X\to Y$ is
continuous and $E\subseteq Y$ is G$_{\delta}$ in $Y$,
then $f^{-1}[E]$ is G$_{\delta}$ in $X$.
\prooflet{($f^{-1}[\bigcap_{n\in\Bbb N}H_n]
=\bigcap_{n\in\Bbb N}f^{-1}[H_n]$.)}
%4@39

\quad (iv) If $Y$ is a G$_{\delta}$ set in $X$ and $Z\subseteq Y$ is a
G$_{\delta}$ set for the subspace topology of $Y$, then $Z$ is a
G$_{\delta}$ set in $X$.
\prooflet{(\Kuratowski, \S5.V.)}
%4@19 %4@39

\quad (v) A set $E\subseteq X$ is an F$_{\sigma}$ set iff $X\setminus E$
is a G$_{\delta}$ set.
\cmmnt{(\Kuratowski, \S5.V.)}
%\prooflet{($\bigcup_{n\in\Bbb N}F_n
%=X\setminus\bigcap_{n\in\Bbb N}X\setminus F_n$.)}

\spheader 4A2Cb(i) A zero set is closed.   A cozero set is open.

\quad (ii) The union of two zero sets is a zero set.
\prooflet{(\Csaszar, 4.2.36.)}
%4@12 %4@19
The intersection of two cozero sets is a cozero set.
%4@22

\quad (iii) The intersection of a sequence of zero sets is a zero set.
\prooflet{(If $f_n:X\to\Bbb R$ is continuous for each $n$,
$x\mapsto\sum_{n=0}^{\infty}\min(2^{-n},|f_n(x)|)$ is continuous.)}
%4@12
The union of a sequence of cozero sets is a cozero set.
%4@22

\quad (iv) If $Y$ is another topological space, $f:X\to Y$ is continuous
and $L\subseteq Y$ is a zero set, then $f^{-1}[L]$ is a zero set.
If $f:X\to Y$ is continuous and $H\subseteq Y$ is a cozero set, then
$f^{-1}[H]$ is a cozero set.
\prooflet{(\Cech, 28B.3.)}
%4@43
If $K\subseteq X$ and $L\subseteq Y$ are zero sets then $K\times L$ is a
zero set in $X\times Y$.
\prooflet{($K\times L=\pi_1^{-1}[K]\cap\pi_2^{-1}[L]$.)}

\quad (v) If $H\subseteq X$ is a (co-\nobreak)zero set and
$Y\subseteq X$, then $H\cap Y$ is a (co-\nobreak)zero set in $Y$.
\prooflet{(Use (iv).)}
%4@12 4@43

\quad (vi) A cozero set is the union of
a non-decreasing sequence of zero sets.
\prooflet{(If $f:X\to\Bbb R$ is continuous,
$X\setminus f^{-1}[\{0\}])=\bigcup_{n\in\Bbb N}g_n^{-1}[\{0\}]$, where
$g_n(x)=\max(0,2^{-n}-|f(x)|)$.)}
%4@12 %4@21
In particular, a cozero set is an F$_{\sigma}$ set;  \cmmnt{taking
complements,} a zero set is a G$_{\delta}$ set.
%4@19

\quad (vii) If $\Cal G$ is a partition of $X$ into open sets, and
$H\subseteq X$ is such that $H\cap G$ is a cozero set in $G$ for every
$G\in\Cal G$, then $H$ is a cozero set in $X$.
\prooflet{(If $f_G:G\to\Bbb R$ is continuous for every $G\in\Cal G$,
then $f:X\to\Bbb R$ is continuous, where $f(x)=f_G(x)$ for
$x\in G\in\Cal G$.)}
Similarly, if $F\subseteq X$ is such that $F\cap G$ is a zero set in $G$
for every $G\in\Cal G$, then $F$ is a zero set in $X$.
%4@43

\vleader{72pt}{4A2D}{Weight} Let $X$ be a topological space.

\spheader 4A2Da(i)
$w(Y)\le w(X)$ for every subspace $Y$ of $X$\prooflet{ (4A2B(a-vi))}.
%4@25

\quad(ii) If $X=\prod_{i\in I}X_i$ then
$w(X)\le\max(\omega,\#(I),\sup_{i\in I}w(X_i))$.
\prooflet{(\Engelking, 2.3.13.)}

\spheader 4A2Db A disjoint family\cmmnt{ $\Cal G$} of non-empty open
sets in $X$ has cardinal at most $w(X)$.
\prooflet{(If $\Cal U$ is a base for the topology of $X$, then every
non-empty member of $\Cal G$ includes a
non-empty member of $\Cal U$, so we have an injective function from
$\Cal G$ to $\Cal U$.)}
%4@25 %4@38

\spheader 4A2Dc A point-countable family\cmmnt{ $\Cal G$} of open sets
in $X$ has cardinal at most $\max(\omega,w(X))$.
\prooflet{\Prf\ If $X=\emptyset$, this is trivial.   Otherwise, let
$\Cal U$ be a base for the topology of $X$ with $\#(\Cal U)=w(X)>0$.
Choose a function $f:\Cal G\to\Cal U$ such that
$\emptyset\ne f(G)\subseteq G$ whenever
$G\in\Cal G\setminus\{\emptyset\}$.   Then $\Cal G_U=\{G:f(G)=U\}$ is
countable for every $U\in\Cal U$, so there is an injection
$h_U:\Cal G_U\to\Bbb N$;  now
$G\mapsto(f(G),h_{f(G)}(G)):\Cal G\to\Cal U\times\Bbb N$ is injective,
so $\#(\Cal G)\le\#(\Cal U\times\Bbb N)=\max(\omega,w(X))$.\ \Qed}
%4@25

\spheader 4A2Dd If $X$ is a dyadic Hausdorff
space then $X$ is a continuous image of $\{0,1\}^{w(X)}$.
\prooflet{\Prf\ There are a set $I$
and a continuous surjection $f:\{0,1\}^I\to X$;
because any power of $\{0,1\}$ is compact, so is $X$.   If $w(X)$ is
finite, $\#(X)=w(X)\le\#(\{0,1\}^{w(X)})$ and the result is trivial;  so
we may suppose that $w(X)$ is infinite.    Let $\Cal U$ be
a base for the topology of $X$ with cardinality
$w(X)$.   Set $Z=\{0,1\}^I$ and let $\Cal E$ be the
algebra of subsets of $Z$ determined by coordinates in finite sets, so
that $\Cal E$ is an algebra of subsets of $Z$ and is a base for the
topology of $Z$.
For each pair $U$, $V$ of members of $\Cal U$ such that
$\overline{U}\subseteq V$, $f^{-1}[V]\subseteq Z$ is open;  the
set $\{E:E\in\Cal E$, $E\subseteq f^{-1}[V]\}$ is upwards-directed and
covers the compact set $f^{-1}[\overline{U}]$, so there
is an $E_{UV}\in\Cal E$ such that
$f^{-1}[\overline{U}]\subseteq E_{UV}\subseteq f^{-1}[V]$.   Let
$J\subseteq I$ be a set with cardinal at most $\max(\omega,w(X))$ such
that every $E_{UV}$ is determined by coordinates in $J$.   Fix any
$w\in\{0,1\}^{I\setminus J}$ and define $g:\{0,1\}^J\to X$ by setting
$g(z)=f(z,w)$ for every $z\in\{0,1\}^J$, identifying $Z$ with
$\{0,1\}^J\times\{0,1\}^{I\setminus J}$.   Then $g$ is continuous.
\Quer\ If $g$ is not surjective, set
$H=X\setminus g[\{0,1\}^J]$.   Take $x\in H$;  take $V\in\Cal U$ such
that $x\in V\subseteq H$;  take an open set $G$ such that
$x\in G\subseteq\overline{G}\subseteq V$
(this must be possible because $X$, being compact and Hausdorff, is
regular -- see 3A3Bb);  take $U\in\Cal U$ such that $x\in U\subseteq G$, so that
$x\in U\subseteq\overline{U}\subseteq V$.
Because $f$ is surjective, there is a
$(u,v)\in\{0,1\}^J\times\{0,1\}^{I\setminus J}$
such that $f(u,v)=x$.   Now $(u,v)\in f^{-1}[U]\subseteq E_{UV}$;  as
$E_{UV}$ is determined by coordinates in $J$,
$(u,w)\in E_{UV}\subseteq f^{-1}[V]$ and $g(u)=f(u,w)\in V$;  but $V$ is
supposed to be
disjoint from $g[\{0,1\}^J]$.\ \BanG\  So $g$ is surjective, and $X$ is
a continuous image of $\{0,1\}^J$.   Since
$\#(J)\le\max(\omega,w(X))=w(X)$, $\{0,1\}^J$ and
$X$ are continuous images of $\{0,1\}^{w(X)}$.\ \Qed}

\spheader 4A2De If $X$ is a dyadic
Hausdorff space then $X$ is separable iff it is
a continuous image of $\{0,1\}^{\frak c}$.
\prooflet{\Prf\ $\{0,1\}^{\frak c}$ is separable (4A2B(e-ii)), so any
continuous image of it is separable.   If $X$ is a separable dyadic
Hausdorff space, let $A\subseteq X$ be a countable
dense set.   If $G$, $G'\subseteq X$ are distinct regular open sets,
then $G\cap A\ne G\cap A'$.   Thus $X$ has at most $\frak c$ regular
open sets;  since $X$ is compact
and Hausdorff, therefore regular, its regular open sets form a base
(4A2F(b-ii)), and $w(X)\le\frak c$.   By (d), $X$ is a continuous image of
$\{0,1\}^{\max(\omega,w(X))}$ which is
in turn a continuous image of $\{0,1\}^{\frak c}$.\ \Qed}

\leader{4A2E}{The countable chain condition (a)}(i) Let $\familyiI{X_i}$
be a family of topological spaces.   If $\prod_{i\in J}X_i$ is ccc
for every finite $J\subseteq I$, then $\prod_{i\in I}X_i$ is ccc.
\prooflet{({\smc Kunen 80}, II.1.9;  {\smc Fremlin 84}, 12I.)}

\quad(ii) A separable topological space is ccc.   \prooflet{(If $D$ is a
countable dense set and $\Cal G$ is a disjoint family of non-empty open
sets, we have a surjection from a subset of $D$ onto $\Cal G$.)}

\quad(iii) The product of any family of separable topological spaces is
ccc.   \prooflet{\Prf\ By 4A2B(e-ii) and (ii) here, the product of
finitely many separable spaces is separable, therefore ccc;  so we can
apply (i).\ \Qed}

\quad(iv)\dvAnew{2014}
Any continuous image of a ccc topological space is ccc.
\prooflet{(If $f:X\to Y$ is a continuous surjection and $\Cal H$ is an
uncountable disjoint family of open subsets of $Y$, then
$\{f^{-1}[H]:H\in\Cal H\}$ is an uncountable disjoint family of
open subsets of $X$.)}

\spheader 4A2Eb Let $\familyiI{X_i}$ be a family of topological
spaces, and suppose that $X=\prod_{i\in I}X_i$ is ccc.   For
$J\subseteq I$ and $x\in X$ set
$X_J=\prod_{i\in J}X_i$, $\pi_J(x)=x\restr J$.

\quad(i) If $G\subseteq X$ is open, there is an open set $W\subseteq G$
determined by coordinates in a countable subset of $I$ such that
$G\subseteq\overline{W}$.
\prooflet{\Prf\ Let $\Cal W$ be the family of subsets of $X$ determined
by coordinates in countable sets.   Then $\Cal W$ is a $\sigma$-algebra
(254Mb) including the standard base
$\Cal U$ for the topology of $X$.   Let $\Cal U_0$ be a maximal disjoint
family in $\{U:U\in\Cal U,\,U\subseteq G\}$.   Then $\Cal U_0$ is
countable, so $W=\bigcup\Cal U_0$ belongs to $\Cal W$.   No member of
$\Cal U$ can be included in $G\setminus W$, so $G\setminus\overline{W}$
must be empty, and we have a suitable set.\ \QeD}   So
$\overline{G}=\overline{W}$
and $\interior\overline{G}$ are
determined by coordinates in a countable set\prooflet{ (4A2B(g-i))};
in particular, if $G$ is a regular open set, then it is determined by
coordinates in a countable set.

\quad(ii) If $f:X\to\Bbb R$ is continuous, there are a countable set
$J\subseteq I$ and a continuous function $g:X_J\to\Bbb R$ such that
$f=g\pi_J$.   \prooflet{\Prf\ For each
$q\in\Bbb Q$, set $F_q=\overline{\{x:f(x)<q\}}$.   By (i), $F_q$ is
determined by coordinates in a countable set.   Because
$\Bbb Q$ is countable, there is a countable $J\subseteq I$ such that
every $F_q$ is determined by coordinates in $J$.   Also
$\{x:f(x)<\alpha\}=\bigcup_{q\in\Bbb Q,q<\alpha}F_q$ is determined by
coordinates in $J$ for every $\alpha\in\Bbb R$, so $f(x)=f(y)$ whenever
$x\restr J=y\restr J$, and there is a $g:X_J\to\Bbb R$ such that
$f=g\pi_J$.   Now if $H\subseteq\Bbb R$ is
open, $g^{-1}[H]=\pi_J[f^{-1}[H]]$ is open (4A2B(f-i)),
so $g$ is continuous.\ \Qed}

\quad(iii) If $A\subseteq X$ is nowhere dense there is a countable set
$J\subseteq I$ such that $\pi_J^{-1}[\pi_J[A]]$ is nowhere dense.
\prooflet{\Prf\
By (ii), there are a countable set $J$ and an open set
$W\subseteq X\setminus\overline{A}$ such that $W$ is determined by
coordinates in $J$ and $X\setminus\overline{A}\subseteq\overline{W}$;
now $W$ is dense in $X$ and $\pi_J^{-1}[\pi_J[A]]\subseteq X\setminus W$ is
nowhere dense.\ \Qed}

\leader{4A2F}{Separation axioms (a) Hausdorff spaces}
(i) A Hausdorff space is T$_1$.
\cmmnt{(\Cech, 27A.1.)}
%4@15 %4@16 %4@23

\quad (ii) If $X$ is a Hausdorff space and $\sequencen{x_n}$ is a
sequence in $X$,
then a point $x$ of $X$ is a cluster point of $\sequencen{x_n}$ iff
there is a non-principal ultrafilter $\Cal F$ on $\Bbb N$ such that
$x=\lim_{n\to\Cal F}x_n$.   \prooflet{(If $x$ is a cluster point of
$\sequencen{x_n}$, apply 4A1Ia to
$\{\{n:n\ge n_0,\,x_n\in G\}:n_0\in\Bbb N$, $G\subseteq X$ is open,
$x\in G\}$.)}


\quad(iii) A topological space $X$ is Hausdorff iff $\{(x,x):x\in X\}$ is
closed in $X\times X$.
%Engelking ex 2.3.C
\prooflet{(\Cech\ 27A.7;  \Kuratowski, I.15.IV.)}
%If $I$ is a set, $\Cal F$ is a filter on $I$, $X$ and $Y$ are Hausdorff
%spaces, $f:I\to X$ is a function, $g:X\to Y$ is a continuous function,
%and $\lim_{i\to\Cal F}f(i)=x$, then $\lim_{i\to\Cal F}gf(i)=g(x)$.

%If $\family{j}{J}{X_j}$ is a family of Hausdorff spaces, $I$ is a set,
%$\Cal F$ is a filter on $I$, and $f:I\to X$ is a function, then
%$\lim_{i\to\Cal F}f(i)=x$ iff $\lim_{i\to\Cal F}f(i)(j)=x(j)$ for every
%$j\in J$.   \cmmnt{(Cf.\ 3{}A3Ic.)}

\spheader 4A2Fb {\bf Regular spaces}
(i) A regular T$_1$ space is Hausdorff.
\prooflet{(\Cech, 27B.7;  \Gaal, p.\ 81.)}
%4@39
Any subspace of a regular space is regular.
\cmmnt{(\Engelking, 2.1.6;  \Kuratowski, \S14.I.)}

\quad(ii) If $X$ is a regular topological space, the regular open subsets
of $X$ form a base for the topology.   \prooflet{\Prf\ If $G$ is open and
$x\in G$, there is an open set $H$ such that
$x\in H\subseteq\overline{H}\subseteq G$;  now $\interior\overline{H}$ is a
regular open set containing $x$ and included in $G$.\ \Qed}

\spheader 4A2Fc {\bf Completely regular spaces}
In a completely regular space, the cozero sets form a base
for the topology.
\prooflet{(\Cech, 28B.5.)}
%4@22

\spheader 4A2Fd {\bf Normal spaces} (i) {\bf Urysohn's Lemma}  If
$X$ is normal and $E$, $F$ are disjoint closed
subsets of $X$, then there is a continuous function $f:X\to[0,1]$ such
that $f(x)=0$ for $x\in E$ and $f(x)=1$ for $x\in F$.
\prooflet{(\Engelking, 1.5.11;  \Kuratowski, \S14.IV.)}
%4@33

\quad (ii) A regular normal space is completely regular.
%4@A2Jb

\quad (iii) A normal T$_1$ space is Hausdorff
\prooflet{(\Gaal, p.\ 86)}
%4@39
and completely regular
\prooflet{(\Csaszar, 4.2.5;  \Gaal, p.\ 110)}.
%4@26

\quad (iv) If $X$ is normal and $E$, $F$ are disjoint closed sets in $X$
there is a zero set including $E$ and disjoint from $F$.
\prooflet{(Take a continuous function $f$ which is zero on $E$ and $1$
on $F$, and set $Z=\{x:f(x)=0\}$.)}
%4@36

\quad (v) In a normal space a closed G$_{\delta}$ set is a zero set.
\prooflet{(\Engelking, 1.5.12.)}
%4@33

\quad (vi) If $X$ is a normal space and $\familyiI{G_i}$ is a
point-finite cover of $X$ by open sets, there is a family
$\familyiI{H_i}$ of open
sets, still covering $X$, such that $\overline{H}_i\subseteq G_i$ for
every $i$.
\prooflet{(\Engelking, 1.5.18;  \Cech, 29C.1;  \Gaal, p.\ 89.)}
%for (vii)

\quad (vii) If $X$ is a normal space and $\familyiI{G_i}$ is a
point-finite cover of $X$ by open sets, there is a family
$\familyiI{H'_i}$ of cozero
sets, still covering $X$, such that $H'_i\subseteq G_i$ for every $i$.
\prooflet{(Take $\familyiI{H_i}$ from (vi), and apply (iv) to the
disjoint closed sets $X\setminus G_i$, $\overline{H}_i$ to find a
suitable cozero set $H'_i$ for each $i$.)}
%4@26

\quad (viii) If $X$ is a normal space and $\familyiI{G_i}$ is a
locally finite cover of $X$ by open sets, there is a family
$\familyiI{g_i}$ of continuous functions from $X$ to $[0,1]$ such that
$g_i\le\chi G_i$ for every $i\in I$ and $\sum_{i\in I}g_i(x)=1$ for
every $x\in X$.   \prooflet{(\Engelking, proof of 5.1.9.)}
%4@15

\quad (ix) {\bf Tietze's theorem}\dvAformerly{4{}A2F(d-viii)} Let
$X$ be a normal space, $F$ a
closed subset of $X$ and $f:F\to\Bbb R$ a continuous function.   Then
there is a continuous function $g:X\to\Bbb R$ extending $f$.
\prooflet{(\Engelking, 2.1.8;  \Kuratowski, \S14.IV;  \Gaal, p.\ 203.)}
It follows that if $F\subseteq X$ is closed and $f:F\to[0,1]^I$ is a
continuous function from $F$ to any power of the unit interval, there is
a continuous function from $X$ to $[0,1]^I$ extending $f$.
\prooflet{ (Extend each of the functionals $x\mapsto f(x)(i)$ for
$i\in I$.)}

\spheader 4A2Fe {\bf Paracompact spaces} A Hausdorff paracompact space
is regular.
\prooflet{(\Engelking, 5.1.5.)}
A regular paracompact space is normal.
\prooflet{(\Engelking, 5.1.5;  \Gaal, p.\ 160.)}

%4@36
\spheader 4A2Ff {\bf Countably paracompact spaces} A normal space $X$ is
countably paracompact iff whenever $\sequencen{F_n}$
is a non-increasing
sequence of closed subsets of $X$ with empty intersection, there is
a sequence $\sequencen{G_n}$ of open sets, also with empty intersection,
such that $F_n\subseteq G_n$ for every $n\in\Bbb N$.
\prooflet{(\Engelking, 5.2.2;  \Csaszar, 8.3.56(f).)}
%4@26

\spheader 4A2Fg {\bf Metacompact spaces} (i) A paracompact space is
metacompact.

\quad (ii) A closed subspace of a metacompact space is metacompact.
%4@26

\quad (iii) A normal metacompact space is countably paracompact.
\prooflet{(\Engelking, 5.2.6;  \Csaszar, 8.3.56(c).)}
%4@26

\spheader 4A2Fh {\bf Separating compact sets} (i) If $X$ is a Hausdorff
space and $K$ and $L$ are disjoint compact subsets of $X$, there are
disjoint open sets $G$, $H\subseteq X$ such that $K\subseteq G$ and
$L\subseteq H$.
\prooflet{(\Csaszar, 5.3.18.)}
%4@22 %4@16
If $\Tau$ is an algebra of subsets of $X$ including a subbase for the
topology of $X$, there is an open $V\in\Tau$ such that
$K\subseteq V\subseteq X\setminus L$.   \prooflet{\Prf\ By 4A2B(a-i),
$\Tau$ includes a base for the topology of $X$.   So
$\Cal E=\{U:U\in\Tau$ is open, $U\subseteq G\}$ has union $G$ and there
must be a finite $\Cal E_0\subseteq\Cal E$ covering $K$;  set
$V=\bigcup\Cal E_0$.\ \Qed}

\quad(ii) If $X$ is a regular space, $F\subseteq X$ is closed, and
$K\subseteq X\setminus F$ is compact, there are disjoint open sets $G$,
$H\subseteq X$ such that $K\subseteq G$ and $F\subseteq H$.
\prooflet{(\Engelking, 3.1.6.)}
%4@22

\quad(iii) If $X$ is a completely regular space,
$G\subseteq X$ is open and $K\subseteq G$ is compact, there is a
continuous function $f:X\to[0,1]$ such that $f(x)=1$ for $x\in K$ and
$f(x)=0$ for $x\in X\setminus G$.
\prooflet{\Prf\ For each $x\in K$ there is a continuous function
$f_x:X\to[0,1]$ such that $f_x(x)=1$ and $f_x(y)=0$ for
$y\in X\setminus G$.   Set $H_x=\{y:f_x(y)>\bover12\}$.   Then
$\bigcup_{x\in K}H_x\supseteq K$, so there is a finite set
$I\subseteq K$ such that $K\subseteq\bigcup_{x\in I}H_x$.   Set
$f(y)=\min(1,2\sum_{x\in I}f_x(y))$ for $y\in X$.\ \Qed}
%for (iv), (v)

\quad (iv) If $X$ is a completely regular Hausdorff space and $K$ and
$L$ are disjoint compact subsets of $X$, there are disjoint cozero sets
$G$, $H\subseteq X$ such that $K\subseteq G$ and $L\subseteq H$.
\prooflet{\Prf\ By (i), there are disjoint open sets $G'$, $H'$ such
that $K\subseteq G'$ and $L\subseteq H'$.   By (iii), there is a
continuous function $f:X\to[0,1]$ such that $f(x)=1$ for $x\in K$ and
$f(x)=0$ for $x\in X\setminus G'$;  set $G=\{x:f(x)\ne 0\}$, so that $G$
is a cozero set and $K\subseteq G\subseteq G'$.   Similarly there is a
cozero set $H$ including $L$ and included in $H'$.\ \Qed}
%4@22

\quad (v) If $X$ is a completely regular space and $K\subseteq X$ is a
compact G$_{\delta}$ set, then $K$ is a zero set.
\prooflet{\Prf\ Let $\sequencen{G_n}$ be a sequence of open sets with
intersection $K$.   For each $n\in\Bbb N$ there is a continuous function
$f_n:X\to[0,1]$ such that $f_n(x)=1$ for $x\in K$ and $f_n(x)=0$ for
$x\in X\setminus G_n$, by (iii).   Now
$K=\bigcap_{n\in\Bbb N}\{x:1-f_n(x)=0\}$ is a zero set, by 4A2C(b-iii).\
\Qed}
%4@19 %4@43

\quad (vi) If $\sequencen{X_n}$ is a sequence of topological spaces with
product $X$, $K\subseteq X$ is compact, $F\subseteq X$ is closed and
$K\cap F=\emptyset$, there is some $n\in\Bbb N$ such that
$x\restr n\ne y\restr n$ for any $x\in F$ and $y\in K$.

\prooflet{\Prf\ For $n\in\Bbb N$ and $x\in X$ set $\pi_n(x)=x\restr n$;
set $F_n=\overline{\pi_n^{-1}[\pi_n[F]]}$.   Since
$\sequencen{\pi_n^{-1}[\pi_n[F]]}$ is non-increasing, so is
$\sequencen{F_n}$.   If $x\in K$, there is an open set $G\subseteq X$,
determined by coordinates in a finite set, such that
$x\in G\subseteq X\setminus F$;  in this case there is an $n\in\Bbb N$ such
that $\pi_n^{-1}[\pi_n[G]]=G$ is disjoint from $F$, so that
$\pi_n[G]\cap\pi_n[F]=\emptyset$, $G$ does not meet
$\pi_n^{-1}[\pi_n[F]]$ and $x\notin F_n$.   As $x$ is arbitrary,
$\sequencen{K\cap F_n}$ is a non-increasing sequence of relatively closed
subsets of $K$ with empty intersection;  as $K$ is compact, there is an
$n$ such that $K\cap F_n=\emptyset$, so that
$K\cap\pi_n^{-1}[\pi_n[F]]=\emptyset$ and $x\restr n\ne y\restr n$
whenever $x\in F$ and $y\in K$.\ \Qed}

\quad(vii) If $X$ is a compact Hausdorff space, $f:X\to\Bbb R$ is
continuous, and $\Cal U$ is a subbase for $\frak T$, then there is a
countable set
$\Cal U_0\subseteq\Cal U$ such that $f(x)=f(y)$ whenever
$\{U:x\in U\in\Cal U_0\}=\{U:y\in U\in\Cal U_0\}$.   \prooflet{(Apply (i) to
sets of the form $K=\{x:f(x)\le\alpha\}$, $L=\{x:f(x)\ge\beta\}$.)}
%4@91Xi

\spheader 4A2Fi {\bf Perfectly normal spaces} A topological space $X$ is
perfectly normal iff every closed set is a zero set.
\prooflet{(\Engelking, 1.4.9.)}

Consequently, every open set in a perfectly normal space is a cozero set
(and\cmmnt{, of course,} an F$_{\sigma}$ set).

\spheader 4A2Fj {\bf Covers of compact sets} Let $X$ be a Hausdorff space,
$K$ a compact subset of $X$, and $\familyiI{G_i}$ a family of open subsets
of $X$ covering $K$.   Then there are a finite set $J\subseteq I$ and a
family $\family{i}{J}{K_i}$ of compact sets such that
$K=\bigcup_{i\in J}K_i$ and $K_i\subseteq G_i$ for every $i\in J$.
\prooflet{\Prf\ (i) Suppose first that $I=\{i,j\}$ has just two members.
Then $K\setminus G_j$ and $K\setminus G_i$ are disjoint compact sets.
By (h-i), there are disjoint open sets $H_i$, $H_j$ such that
$K\setminus G_j\subseteq H_i$ and $K\setminus G_i\subseteq H_j$;  setting
$K_i=K\setminus H_j$ and $K_j=K\setminus H_i$ we have a suitable pair
$K_i$, $K_j$.   (ii) Inducing on $\#(I)$ we get the result for finite $I$.
(iii) In general, there is certainly a finite $J\subseteq I$ such that
$K\subseteq\bigcup_{i\in J}G_i$, and we can apply the result to
$\family{i}{J}{G_i}$.\ \Qed}

\leaveitout{
\spheader 4A2F? {\bf Monotonically normal spaces} (i)
A monotonically normal space is normal, so a
T$_1$ monotonically normal space is regular and Hausdorff.

\quad(ii) Let $(X,\frak T)$ be a regular space and $R$ the set of pairs
$(x,G)$ such that $x\in G\in\frak T$.   $(X,\frak T)$ is monotonically
normal iff there is a function $W:R\to\frak T$ such that ($\dagger$)
$x\in W(x,G)\subseteq G$ whenever $(x,G)\in R$ ($\ddagger$) if $(x,G)\in R$
and $G\subseteq H\in\frak T$ then $W(x,G)\subseteq W(x,H)$
(*) if $x$,
$y\in X$ and $\overline{\{x\}}\ne\overline{\{y\}}$ then
$W(x,X\setminus\overline{\{y\}})$ and $W(y,X\setminus\overline{\{x\}})$ are
disjoint.   \Prf\ Let $Q$ be the set of disjoint pairs of closed subsets of
$X$.   ($\alpha$) Suppose that $(X,\frak T)$ is monotonically normal.   Let
$W_1:Q\to\frak T$ be a function as in the definition in 4A2A.   If
$(x,G)\in R$, then (because $X$ is regular) $\overline{\{x\}}$ is disjoint
from $X\setminus G$;  set

\Centerline{$W(x,G)
=W_1(\overline{\{x\}},X\setminus G)
 \setminus\overline{W(X\setminus G,\overline{\{x\}})}$.}

\noindent It is easy to check that this function $W$ satisfies the
conditions ($\dagger$), ($\ddagger$) and (*) here.   ($\beta$) Suppose that
$W:R\to\frak T$ satisfies the conditions here.   For disjoint closed sets
$E$, $F\subseteq X$ set $W_1(E,F)=\bigcup_{x\in E}W(x,X\setminus F)$.
Then $W_1$ witnesses that $X$ is monotonically normal, because
$W_1(E,F)$ will be disjoint from $W(y,E)$ for every $y\in F$.\ \Qed

Note that if $W$ is any function satisfying ($\dagger$), ($\ddagger$) and
(*), and $(x,G)$, $(y,H)\in R$ are such that
$W(x,G)\cap W(y,H)\ne\emptyset$, then either $x\in H$ or $y\in G$.

\quad(iii) If $X$ is a regular monotonically normal space, every subspace
of $X$ is monotonically normal.
\Prf\ Let $W$ be a function as in (ii), and
$Y$ a subspace of $X$.   Then $Y$ is regular.
If $y\in Y$ and $H$ is a relatively open subset of $Y$, write $\tilde H$
for $\bigcup\{G:G\subseteq X$ is open, $G\cap Y\subseteq H\}$ and
$W'(y,H)=Y\cap W(y,\tilde H)$.   Then $W'$ satisfies the conditions
($\dagger$), ($\ddagger$) and (*) of (ii), so $Y$ is monotonically
normal.\ \Qed

\quad(iv) A regular monotonically normal space is countably paracompact.
\Prf\ (See {\smc Rudin 84}.) Let $X$ be a regular monotonically normal
space and $W$ a function as in (ii) above.   Let $\sequencen{F_n}$ be a
non-decreasing sequence of closed sets in $X$ with empty intersection and
$F_0=X$.   For $x\in F_n\setminus F_{n+1}$ define $\sequence{k}{V_k(x)}$ by
setting $V_0(x)=W(x,X\setminus F_{n+1})$ and $V_{k+1}(x)=W(x,V_k(x))$ for
every $k$.   Let $P$ be the set of all families $\family{n}{I}{p_n}$ where
$I\subseteq\Bbb N$ is infinite and
$p_m\in F_m\cap V_{n-m}(p_n)\setminus F_{n+1}$ whenever $m$, $n\in K$ and
$m<n$;  for $\tbf{p}=\family{n}{I}{p_n}\in P$ set

\Centerline{$S_{\tbf{p}}
=\bigcup_{m\in\Bbb N}F_m\cap\bigcap_{n\in I,n>m}V_{n-m}(p_n)
  \setminus F_{m+1}$.}

\noindent Let $\preccurlyeq$ be a well-ordering of $P$.   For
$x\in F_m\setminus F_{m+1}$, set

$$\eqalignno{U(x)&=V_m(x)\cr
\noalign{\noindent if $x\notin\bigcup_{\tbf{p}\in P}S_{\tbf{p}}$}
&=V_m(x)\setminus\bigcup_{n\in I\cap m}\overline{V_0(p_n)}\cr}$$

\noindent if $\tbf{p}=\family{n}{I}{p_n}$ is the $\preccurlyeq$-first
member of $P$ such that $x\in S_{\tbf{p}}$.   Note that in this case
$\overline{V_0(p_n)}$ is disjoint from $F_m$ and does not contain $x$ for
every $n\in I\cap m$, so
$U(x)$ contains $x$.   Set $U'(x)=W(x,U(x))$ for every $x\in X$, and
$G_n=\bigcup_{x\in F_n}U'(x)$;  then
$G_n$ is an open set including $F_n$ for each $n$.

\Quer\ Suppose, if possible, that $\bigcap_{n\in\Bbb N}G_n$ is not empty;
take $z$ in the intersection.   For each $n$, there is an $x\in F_n$ such
that $z\in U'(x)$, so
$J=\{n:z\in\bigcup_{x\in F_n\setminus F_{n+1}}U'(x)\}$
is infinite;  for each $n\in J$, choose $q_n\in F_n\setminus F_{n+1}$ such
that $z\in U'(q_n)$.

If $m$, $n\in J$ and $m<n$, then $U'(q_m)\cap U'(q_n)$ is not empty, so
$W(q_m,U(q_m))\subseteq W(q_m,V_0(q_m))$ meets
$W(q_n,U(q_n))$.   But $q_n\in F_n$ so
$q_n\notin V_0(q_m)$;  it follows that $q_m$ must belong to
$U(q_n)\subseteq V_n(q_n)$.
(See the last sentence in (ii) above.)     As $m$ and $n$ are arbitrary,
$\tbf{q}=\family{n}{J}{q_n}$ belongs to $P$, and of course every $q_n$
belongs to $S_{\tbf{q}}$.

Let $\tbf{p}=\family{n}{I}{p_n}$ be the
$\preccurlyeq$-first member of $P$ such that $S_{\tbf{p}}$ contains any
$q_n$;   let $j\in J$ be such that $q_j\in S_{\tbf{p}}$.   If $j<m<n$,
$m\in J$ and $n\in I$, then $V_{m-j}(q_m)\cap V_{n-j}(p_n)$ contains $q_j$
so is not empty, and either $p_n\in V_{m-j-1}(q_m)$ or
$q_m\in V_{n-j-1}(p_n)$;  the former is impossible because
$p_n\in F_{m+1}$, so $q_m\in V_{n-m}(p_n)$.   As $n$ is arbitrary,
$q_m\in S_{\tbf{p}}$, and $\tbf{p}$ must be
the $\preccurlyeq$-least member of
$P$ such that $q_m\in S_{\tbf{p}}$.   And this is true for every $m\in J$
such that $m>j$.

Now take $n\in I$ and $m\in J$ such that $j<n<m$.   We have
$q_j\in V_{n-j}(p_n)\cap U(q_m)$.   But $U(q_m)$ is
disjoint from $V_0(p_n)$, so this is impossible.\ \Bang

Thus $\bigcap_{n\in\Bbb N}G_n$ is empty.   As $\sequencen{F_n}$ is
arbitrary, $X$ is countably paracompact.\ \Qed
}%end of leaveitout

\leader{4A2G}{Compact and locally compact spaces (a)} In any topological
space, the union of two compact subsets is compact.

\spheader 4A2Gb A compact Hausdorff space is normal.
\prooflet{(\Engelking, 3.1.9;  \Csaszar, 5.3.23;  \Gaal, p.\ 139.)}
%4@33 %4@16 %4@46

\spheader 4A2Gc(i) If $X$ is a compact Hausdorff space,
%normal will do
$Y\subseteq X$
is a zero set and $Z\subseteq Y$ is a zero set in $Y$, then $Z$ is a
zero set in $X$.
\prooflet{(By 4A2C(b-vi) and
4A2C(a-iv), $Z$ is a G$_{\delta}$ set in $X$;  now use 4A2F(d-v).)}
%4@19

\quad(ii) Let $X$ and $Y$ be compact Hausdorff spaces,
$f:X\to Y$ a
continuous open map and $Z\subseteq X$ a zero set in $X$.   Then $f[Z]$
is a zero set in $Y$.   \prooflet{\Prf\ Let $g:X\to\Bbb R$ be a
continuous function such that $Z=g^{-1}[\{0\}]$.   Set
$G_n=\{x:x\in X$, $|g(x)|<2^{-n}\}$ for each $n\in\Bbb N$.   If
$y\in\bigcap_{n\in\Bbb N}f[G_n]$, then $f^{-1}[\{y\}]$ is a compact set
meeting all the closed sets $\overline{G}_n$, so meets their
intersection, which is $Z$.   Thus $f[Z]=\bigcap_{n\in\Bbb N}f[G_n]$ is
a G$_{\delta}$ set.   By 4A2F(d-v), it is a zero set.\ \Qed}

\spheader 4A2Gd If $X$ is a Hausdorff space, $\Cal V$ is a
downwards-directed family
of compact neighbourhoods of a point $x$ of $X$ and
$\bigcap\Cal V=\{x\}$, then $\Cal V$ is a base of neighbourhoods of $x$.
\prooflet{\Prf\ Let $G$ be any open set containing $x$.   Fix any
$V_0\in\Cal V$.   Note that because $X$ is Hausdorff, every member of
$\Cal V$ is closed (3A3Dc).   So $\{V_0\cap V\setminus G:V\in\Cal V\}$
is a family of (relatively) closed subsets of $V_0$ with empty
intersection, cannot have the finite intersection property (3A3Da),
and there is a $V\in\Cal V$ such that $V_0\cap V\setminus G=\emptyset$.
Now there is a $V'\in\Cal V$ such that $V'\subseteq V_0\cap V$ and
$V'\subseteq G$.   As $G$ is arbitrary, $\Cal V$ is a base of
neighbourhoods of $x$.\ \Qed}
%4@47? %4@46

%continuous r-v functions
\spheader 4A2Ge Let $(X,\frak T)$ be a locally compact Hausdorff space.

\quad(i) If $K\subseteq X$ is a compact set and
$G\supseteq K$ is open, then there is a continuous $f:X\to[0,1]$ with
compact support such that $\chi K\le f\le\chi G$.
\prooflet{(Let $\Cal V$ be the family of relatively compact open subsets
of $X$.   Then $\Cal V$ is upwards-directed and covers $X$, so there is
a $V\in\Cal V$ including $K$.   By 3A3Bb, $\frak T$ is completely
regular;  now 4A2F(h-iii) tells us that there is a continuous
$f:X\to[0,1]$ such that $\chi K\le f\le\chi(G\cap V)$, so that $f$ has
compact support.)}
%4@34

\quad (ii) $\frak T$ is the coarsest topology on $X$ such that every
$\frak T$-continuous real-valued function with compact support is
continuous.
\prooflet{\Prf\ Let $\Phi$ be the set of continuous functions of compact
support for $\frak T$.   If $\frak S$ is a topology on $X$ such that
every member of $\Phi$ is continuous, and $x\in G\in\frak T$, then there
is an $f\in\Phi$ such that $f(x)=1$ and $f(y)=0$ for
$y\in X\setminus G$, by (i).   Now $f$ is $\frak S$-continuous, by
hypothesis, so
$H=\{y:f(y)>\bover12\}$ belongs to $\frak S$ and $x\in H\subseteq G$.
As $x$ is arbitrary, $G=\interior_{\frak S}G$ belongs to $\frak S$;  as
$G$ is arbitrary, $\frak T\subseteq\frak S$.\ \Qed}
%4@34

%countably compact spaces
\spheader 4A2Gf(i) A topological space $X$ is countably compact iff
every sequence in $X$ has a cluster point in $X$\cmmnt{, that is, $X$
is relatively countably compact in itself.}
\prooflet{(\Engelking, 3.10.3;  \Csaszar, 5.3.31(e);  \Gaal, p.\ 129.)}
%4@16 %4@25

\quad (ii) If $X$ is a countably compact topological space and
$\sequencen{F_n}$ is a sequence of
closed sets such that $\bigcap_{i\le n}F_i\ne\emptyset$ for every
$n\in\Bbb N$, then $\bigcap_{n\in\Bbb N}F_n\ne\emptyset$.
\prooflet{(\Engelking, 3.10.3;  \Csaszar, 5.3.31(c).)}
%4@35

\quad (iii) In any topological space, a relatively compact set is
relatively countably compact\prooflet{ (2A3Ob)}.

\quad (iv) Let $X$ and $Y$ be topological spaces and $f:X\to Y$ a
continuous function.   If $A\subseteq X$ is relatively countably compact
in $X$, then $f[A]$ is relatively countably compact in $Y$.
\prooflet{\Prf\ Let $\sequencen{y_n}$ be a sequence in $f[A]$.   Then
there is a sequence $\sequencen{x_n}$ in $A$ such that $f(x_n)=y_n$ for
every $n\in\Bbb N$.   Because $A$ is relatively countably compact,
$\sequencen{x_n}$ has a cluster point $x\in X$.   If $n_0\in\Bbb N$ and
$H$ is an open set containing $f(x)$, there is an $n\ge n_0$ such that
$x_n\in f^{-1}[H]$, so that $y_n\in H$.   Thus $f(x)$ is a cluster point
of $\sequencen{y_n}$;  as $\sequencen{y_n}$ is arbitrary, $f[A]$ is
relatively countably compact.\ \Qed}

\quad (v) A relatively countably compact set in $\Bbb R$ must be
bounded.
\prooflet{(If $A\subseteq\Bbb R$ is unbounded there is a sequence
$\sequencen{x_n}$ in $A$ such that $|x_n|\ge n$ for every $n$.)}
So if $X$ is a topological space, $A\subseteq X$ is relatively countably
compact and $f:X\to\Bbb R$ is continuous, then $f[A]$ is bounded.

\quad (vi) If $X$ and $Y$ are topological spaces and $f:X\to Y$ is
continuous, then $f[A]$ is countably compact whenever $A\subseteq X$ is
countably compact.
\prooflet{(\Engelking, 3.10.5.)}

%uniform convergence
\spheader 4A2Gg(i) Let $X$ and $Y$ be topological spaces and
$\phi:X\times Y\to\Bbb R$ a continuous function.   Define
$\theta:X\to C(Y)$ by setting $\theta(x)(y)=\phi(x,y)$ for $x\in X$,
$y\in Y$.   Then $\theta$ is continuous if we give $C(Y)$ the topology
of uniform convergence on compact subsets of $Y$.
\prooflet{(As noted in \Engelking, pp.\ 157-158, the topology of uniform
convergence on compact sets is the `compact-open' topology of $C(Y)$,
as defined in 441Yi, so
the result here is covered by \Engelking, 3.4.1.)}

\quad (ii) In particular, if $Y$ is compact then $\theta$ is continuous
if we give $C(Y)$ its usual norm topology.
%4@44

\quad (iii)\dvAnew{2022} Let $X$ be a locally compact topological space, and
give $C(X)$ the topology of uniform convergence on compact subsets of $X$.
Then the function $(f,x)\mapsto f(x):C(X)\times X\to\Bbb R$ is continuous.
\prooflet{\Prf\ Take $g\in C(X)$, $y\in X$ and $\epsilon>0$.   Let $K\subseteq X$ be
a compact set such that $y\in\interior K$.   Then $V=\{f:f\in C(X)$,
$|f(x)-g(x)|\le\bover12\epsilon$ for every $x\in K\}$ is a neighbourhood of $g$, and
$U=\{x:x\in K$, $|g(x)-g(y)|\le\bover12\epsilon\}$ is a neighbourhood of $y$.   
If $f\in V$ and $x\in U$, then

\Centerline{$|f(x)-g(y)|\le|f(x)-g(x)|+|g(x)-g(y)|\le\epsilon$.  \Qed}
}

%convergent sequences
\spheader 4A2Gh(i) Suppose that $X$ is a compact space such that there
are no non-trivial convergent sequences in $X$\cmmnt{, that is, no
convergent sequences which are not eventually constant}.   If
$\sequencen{F_n}$ is a non-increasing sequence of infinite closed
subsets of $X$, then $F=\bigcap_{n\in\Bbb N}F_n$ is infinite.
\prooflet{\Prf\ Because $X$ is compact, $F$ cannot be empty (3A3Da).
Choose a sequence $\sequencen{x_n}$ such that
$x_n\in F_n\setminus\{x_i:i<n\}$ for each $n\in\Bbb N$.    If
$G\supseteq F$ is an open set, then
$\bigcap_{n\in\Bbb N}F_n\setminus G=\emptyset$, so there must be some
$n\in\Bbb N$ such that $F_n\subseteq G$, and $x_i\in G$ for $i\ge n$.
\Quer\ If $F=\{y_0,\ldots,y_k\}$, let $l\le k$ be the first point such
that whenever $G\supseteq\{y_0,\ldots,y_l\}$ is open, then
$\{i:x_i\notin G\}$ is finite.   Then there is an open set
$G'\supseteq\{y_j:j<l\}$ such that $I=\{i:x_i\notin G'\}$ is infinite.
But if $H$ is any open set containing $y_l$, then $\{i:x_i\notin G'\cup
H\}$ is finite, so $\{i:i\in I,\,x_i\notin H\}$ is finite.   Thus if we
re-enumerate $\familyiI{x_i}$ as $\sequencen{x'_n}$, $\sequencen{x'_n}$
converges to $y_l$ and is a non-trivial convergent sequence.\ \BanG\
Thus $F$ is infinite, as claimed.\ \Qed}


\quad (ii) If $X$ is an infinite scattered compact Hausdorff space it
has a non-trivial convergent sequence.
\prooflet{\Prf\ Let $\sequencen{x_n}$ be any sequence of distinct points
in $X$.   Set $F_n=\overline{\{x_i:i\ge n\}}$ for each $n$, so that
$F=\bigcap_{n\in\Bbb N}F_n$ is a non-empty set.   Because $X$ is
scattered, $F$ has an isolated point $z$ say;  let $G$ be an open set
such that $F\cap G=\{z\}$, and $H$ an open set such that
$z\in H\subseteq\overline{H}\subseteq G$ (3A3Bb).   In this case,
$I=\{i:x_i\in H\}$ must be infinite;
re-enumerate $\familyiI{x_i}$ as $\sequencen{x'_n}$.   \Quer\ If
$\sequencen{x'_n}$ does not converge to $z$, there is an open set $H'$
containing $z$ such that $\{n:x'_n\notin H'\}$ is infinite, that is,
$\{i:i\in I,\,x_i\notin H'\}$ is infinite.   In this case,
$F_n\cap\overline{H}\setminus H'$ is non-empty for every $n\in\Bbb N$,
but $F\cap\overline{H}\setminus H'=\emptyset$, which is impossible.\
\BanG\   Thus $\sequencen{x'_n}$ is a non-trivial convergent sequence in
$X$.\ \Qed}


\quad (iii) If $X$ is an extremally disconnected Hausdorff
space\cmmnt{ (definition: 3A3Af)}, it has no non-trivial convergent
sequence.   \prooflet{\Prf\Quer\ Suppose, if possible, that there is a
sequence $\sequencen{x_n}$ converging to $x\in X$ such that
$\{n:x_n\ne x\}$ is infinite.   Choose
$\sequence{i}{n_i}$ and $\sequence{i}{G_i}$ inductively, as follows.
Given that $x\notin\overline{G}_j$ for $j<i$, there is an $n_i$ such
that $x_{n_i}\ne x$ and $x_{n_i}\notin\overline{G}_j$ for every $j<i$;
now let $G_i$ be an open set such that $x_{n_i}\in G_i$ and
$x\notin\overline{G}_i$, and continue.

Since all the $n_i$ must be distinct, $\sequence{i}{x_{n_i}}\to x$.
But consider

\Centerline{$G
=\bigcup_{i\in\Bbb N}G_{2i}\setminus\bigcup_{j<2i}\overline{G}_j$,
\quad$H
=\bigcup_{i\in\Bbb N}G_{2i+1}\setminus\bigcup_{j\le 2i}\overline{G}_j$.}

\noindent Then $G$ and $H$ are disjoint open sets and $x_{n_{2i}}\in G$,
$x_{n_{2i+1}}\in H$ for every $i$.   So
$x\in\overline{G}\cap\overline{H}$.   But $\overline{G}$ is open
(because $X$ is extremally disconnected), and is disjoint from $H$, and
now $\overline{H}$ is disjoint from $\overline{G}$;  so they cannot both
contain $x$.\ \Bang\Qed}

\spheader 4A2Gi(i) If $X$ and $Y$ are compact Hausdorff spaces and
$f:X\to Y$ is a continuous surjection then there is a closed set
$K\subseteq X$ such that $f[K]=Y$ and $f\restr K$ is irreducible.
\prooflet{\Prf\
Let $\Cal E$ be the family of closed sets $F\subseteq X$
such that $f[F]=Y$.   If $\Cal F\subseteq\Cal E$ is non-empty and
downwards-directed, then for any $y\in Y$ the family
$\{F\cap f^{-1}[\{y\}]:F\in\Cal F\}$ is a downwards-directed family of
non-empty closed sets, so (because $X$ is compact) has non-empty
intersection;  this shows that $\bigcap\Cal F\in\Cal E$.   By Zorn's
Lemma, $\Cal E$ has a minimal element $K$ say.   Now $f[K]=Y$ but
$f[F]\ne Y$ for any closed proper subset of $K$,
so $f\restr K$ is irreducible.\ \Qed}

\quad(ii) If $X$ and $Y$ are compact Hausdorff spaces and $f:X\to Y$ is an
irreducible continuous surjection, then ($\alpha$) if $\Cal U$ is a
$\pi$-base for the topology of $Y$ then $\{f^{-1}[U]:U\in\Cal U\}$ is a
$\pi$-base for the topology of $X$ ($\beta$) if $Y$ has a countable
$\pi$-base so does $X$ ($\gamma$) if $x$ is an isolated point in $X$ then
$f(x)$ is an isolated point in $Y$ ($\delta$) if $Y$ has no isolated
points, nor does $X$.
\prooflet{\Prf\ ($\alpha$) If $G\subseteq X$ is a non-empty open set then
$f[X\setminus G]\ne Y$.   As $f[X\setminus G]$ is closed, there is a
non-empty $U\in\Cal U$ disjoint from $f[X\setminus G]$.   Now $f^{-1}[U]$
is a non-empty subset of $G$.   ($\beta$) Follows at once from
($\alpha$).   ($\gamma$) By ($\alpha$), with $\Cal U$ the family of all
open subsets of $Y$, there is a non-empty open set $U\subseteq Y$ such that
$f^{-1}[U]\subseteq\{x\}$, that is, $U=\{f(x)\}$.   ($\delta$) Follows at
once from ($\gamma$).\ \Qed}

\spheader 4A2Gj(i) Let $X$ be a non-empty compact Hausdorff space
without isolated points.   Then there are a closed set $F\subseteq X$
and a continuous surjection $f:F\to\{0,1\}^{\Bbb N}$.
\prooflet{\Prf\ For $\sigma\in\bigcup_{n\in\Bbb N}\{0,1\}^n$ choose
closed sets $V_{\sigma}\subseteq X$ inductively, as follows.
$V_{\emptyset}=X$.   Given that $V_{\sigma}$ is a closed set with
non-empty interior, there are distinct points $x$,
$y\in\interior V_{\sigma}$ (because $X$ has no isolated points);  let
$G$, $H$ be disjoint open subsets of $X$ such that $x\in G$ and
$y\in H$;  and let $V_{\sigma^{\smallfrown}\fraction{0}}$ and
$V_{\sigma^{\smallfrown}\fraction{1}}$ be closed sets such that

\Centerline{$x\in\interior V_{\sigma^{\smallfrown}\fraction{0}}
\subseteq G\cap\interior V_{\sigma}$,
\quad$y\in\interior V_{\sigma^{\smallfrown}\fraction{1}}
\subseteq H\cap\interior V_{\sigma}$.}

\noindent (This is possible because $X$ is regular.)   The construction
ensures that $V_{\tau}\subseteq V_{\sigma}$ whenever $\tau\in\{0,1\}^n$
extends $\sigma\in\{0,1\}^m$, and that
$V_{\tau}\cap V_{\sigma}=\emptyset$ whenever $\tau$,
$\sigma\in\{0,1\}^n$ are different.   Set
$F=\bigcap_{n\in\Bbb N}\bigcup_{\sigma\in\{0,1\}^n}V_{\sigma}$;  then
$F$ is a closed subset of $X$ and we have a continuous function
$f:F\to\{0,1\}^{\Bbb N}$ defined by saying that $f(x)(i)=\sigma(i)$
whenever $n\in\Bbb N$, $\sigma\in\{0,1\}^n$, $i<n$ and
$x\in V_{\sigma}$.   Finally, $f$ is surjective, because if
$z\in\{0,1\}^{\Bbb N}$ then $\sequencen{V_{z\restr n}}$ is a
non-increasing sequence of closed sets in the compact space $X$, so has
non-empty intersection $V$ say, and $f(x)=z$ for any $x\in V$.\ \Qed}
%for (ii) and (iii)

\quad (ii) If $X$ is a non-empty compact Hausdorff space without
isolated points, then $\#(X)\ge\frak c$.
\prooflet{(Use (i).)}
%4@19

\quad (iii) If $X$ is a compact Hausdorff space which is not scattered,
it has an infinite closed subset with a
countable $\pi$-base and no isolated points.
\prooflet{\Prf\ Because $X$ is not scattered, it has a non-empty subset
$A$ without isolated points.   Then $\overline{A}$ is compact and has no
isolated points;  by (i), there are a closed set
$F_0\subseteq\overline{A}$ and a continuous surjection
$f:F_0\to\{0,1\}^{\Bbb N}$.   By (i-i) above, there is a closed
$F\subseteq F_0$ such that $f[F]=\{0,1\}^{\Bbb N}$ and $f\restr F$ is
irreducible.   Of course $F$ is infinite;  by (i-ii), it has a countable
$\pi$-base and no isolated points.\ \Qed}


\quad (iv) Let $X$ be a compact Hausdorff space.   Then there is a
continuous surjection from $X$ onto $[0,1]$ iff $X$ is not scattered.
\prooflet{\Prf\ ($\alpha$) Suppose that $f:X\to[0,1]$ is a continuous
surjection.   By (i-i) again, there is a closed set $F\subseteq X$ such
that $f[F]=[0,1]$ and $f\restr F$ is irreducible;  by (i-ii) $F$ has no
isolated points.   So $X$ is not scattered.   ($\beta$)
If $X$ is not scattered, let $A\subseteq X$ be a
non-empty set with no isolated points.   Then $\overline{A}$ is a
non-empty compact subset of $X$ with no isolated points, so there is a
continuous surjection $g:\overline{A}\to\{0,1\}^{\Bbb N}$ ((i) of this
subparagraph).
Now there is a continuous surjection $h:\{0,1\}^{\Bbb N}\to[0,1]$ (e.g.,
set $h(y)=\sum_{n=0}^{\infty}2^{-n-1}y(n)$ for $y\in\{0,1\}^{\Bbb N}$),
so we have a continuous surjection $hg:\overline{A}\to[0,1]$.   By
Tietze's theorem (4A2F(d-ix)), there is a continuous function
$f_0:X\to\Bbb R$ extending $hg$;  setting $f(x)=\med(0,f_0(x),1)$
for $x\in X$, we have a continuous surjection $f:X\to[0,1]$.\ \Qed
}%end of prooflet

\quad(v) A Hausdorff continuous image of a scattered compact Hausdorff
space is scattered.   \prooflet{(Immediate from (iv).)}

\quad(vi) If $X$ is an uncountable first-countable compact Hausdorff
space, it is not scattered.   \prooflet{\Prf\ Let $\Cal G$ be the family
of countable open subsets of $X$, and $G^*$ its union.  No finite subset
of $\Cal G$ can cover $X$, so $X\setminus G^*$ is
non-empty.   \Quer\ If $x$ is an isolated point of $X\setminus G^*$,
then $\{x\}\cup G^*$ is a neighbourhood of $x$;  let $\sequencen{U_n}$
run over a base of open neighbourhoods of $x$ with
$\overline{U}_0\subseteq\{x\}\cup G^*$.   For each $n\in\Bbb N$,
$F_n=\overline{U}_0\setminus U_n$ is a compact set included in $G^*$, so
is covered by finitely many members of $\Cal G$, and is countable.   But
this means that $U_0=\{x\}\cup\bigcup_{n\in\Bbb N}U_0\setminus U_n$ is
countable, and $x\in G^*$.\ \BanG\  Thus $X\setminus G^*$ is a non-empty
set with no isolated points, and $X$ is not scattered.\ \Qed

}%end of prooflet
It follows that there is a continuous surjection from $X$ onto
$[0,1]$\prooflet{, by (iv)}.
%4@39

\spheader 4A2Gk A locally compact Hausdorff space is \v{C}ech-complete.
\prooflet{(\Engelking, p.\ 196.)}

\spheader 4A2Gl If $X$ is a topological space, $f:X\to\Bbb R$ is lower
semi-continuous, and $K\subseteq X$ is compact and not empty, then there
is an $x_0\in K$ such that $f(x_0)=\inf_{x\in K}f(x)$.
\prooflet{(\Gaal, p.\ 209 Theorem 3.)}
Similarly, if $g:X\to\Bbb R$ is upper
semi-continuous, there is an $x_1\in K$ such that
$g(x_1)=\sup_{x\in K}g(x)$.
%4@84 %4@75

\spheader 4A2Gm If $X$ is a Hausdorff space, $Y$ is a compact space and
$F\subseteq X\times Y$ is closed, then its projection
$\{x:(x,y)\in F\}$ is a closed subset of $X$.   \prooflet{(\Engelking,
3.1.16.)}

\spheader 4A2Gn If $X$ is a locally compact topological space,
$Y$ is a topological space and $f:X\to Y$ is a continuous open surjection,
then $Y$ is locally compact.   \prooflet{(\Engelking, 3.3.15.)}

\leader{4A2H}{Lindel\"of spaces (a)}
If $X$ is a topological space, then a subset $Y$ of $X$ is
Lindel\"of\cmmnt{ (in its subspace topology)} iff for
every family $\Cal G$ of open subsets of $X$ covering $Y$ there is a
countable subfamily of $\Cal G$ still covering $Y$.
%4@22

\spheader 4A2Hb(i) A regular Lindel\"of space $X$ is normal 
(therefore completely regular) and paracompact.
\prooflet{(\Engelking, 3.8.11 \& 5.1.2.)}
%4@15 %4@22 %4@26 %4@39?

\quad(ii) If $X$ is a Lindel\"of space and $\Cal A$ is a locally finite
family of subsets of $X$ then $\Cal A$ is countable.   \prooflet{\Prf\
The family $\Cal G$ of open sets meeting only finitely many members of
$\Cal A$ is an open cover of $X$.   If $\Cal G_0\subseteq\Cal G$ is a
countable cover of $X$ then $\{A:A\in\Cal A,\,A$ meets some member of
$\Cal G_0\}=\Cal A\setminus\{\emptyset\}$ is countable.\ \Qed}

%hereditarily Lindelof spaces
\spheader 4A2Hc(i) A topological space $X$ is hereditarily Lindel\"of
iff for any family $\Cal G$ of open
subsets of $X$ there is a countable family $\Cal G_0\subseteq\Cal G$
such that $\bigcup\Cal G_0=\bigcup\Cal G$.
\prooflet{\Prf\ ($\alpha$) If $X$ is hereditarily Lindel\"of and
$\Cal G$
is a family of open subsets of $X$, then $\Cal G$ is an open cover of
$\bigcup\Cal G$, so has a countable subcover.   ($\beta$) If $X$ is not
hereditarily Lindel\"of, let $Y\subseteq X$ be a non-Lindel\"of
subspace, and $\Cal H$ a cover of $Y$ by relatively open sets which has
no countable subcover;  setting
$\Cal G=\{G:G\subseteq X$ is open, $G\cap Y\in\Cal H\}$, there can be no
countable $\Cal G_0\subseteq\Cal G$ with union $\bigcup\Cal G$.\ \Qed}
%4@39

\quad(ii) Let $X$ be a regular hereditarily Lindel\"of space.   Then
$X$ is perfectly normal.
\prooflet{\Prf\ Let $F\subseteq X$ be closed.   Let $\Cal G$ be the
family of open sets $G\subseteq X$ such that
$\overline{G}\cap F=\emptyset$;  because $X$ is regular,
$\bigcup\Cal G=X\setminus F$;  because $X$ is hereditarily Lindel\"of,
there is a sequence $\sequencen{G_n}$ in $G$ such that
$X\setminus F=\bigcup_{n\in\Bbb N}G_n$.   This means that
$F=\bigcap_{n\in\Bbb N}X\setminus\overline{G_n}$ is a G$_{\delta}$ set.
But $X$ is normal ((b) above), so is perfectly normal.\ \Qed}
%4@35

\spheader 4A2Hd Any $\sigma$-compact topological space is Lindel\"of.
\prooflet{(\Engelking, 3.8.5.)}

\leader{4A2I}{Stone-\v{C}ech compactifications (a)}
Let $X$ be a completely regular Hausdorff space.
Then there is a compact Hausdorff space $\beta X$, the
{\bf Stone-\v{C}ech compactification} of $X$, in which $X$ can be
embedded as a dense subspace.
%4@16 %4@22
If $Y$ is another compact Hausdorff space, then every continuous
function from $X$ to $Y$ has a unique continuous extension to a
continuous function from $\beta X$ to $Y$.
\prooflet{(\Engelking, 3.6.1;  \Csaszar, 6.4d;  \Cech, 41D.5.)}
%4@18

\spheader 4A2Ib Let $I$ be any set, and write $\beta I$ for its
Stone-\v{C}ech compactification when $I$ is given its discrete topology.
Let $Z$ be the Stone space of the Boolean algebra $\Cal PI$.

\quad (i) There is a canonical homeomorphism $\phi:\beta I\to Z$ defined by
saying that $\phi(i)(a)=\chi a(i)$ for every $i\in I$ and $a\subseteq I$.
\prooflet{\Prf\ Recall that $Z$ is the set of ring homomorphisms from
$\Cal PI$ onto $\Bbb Z_2$ (311E).   If $i\in I$, let $\hat{i}$ be the
corresponding member of $Z$ defined by setting $\hat{i}(a)=\chi a(i)$
for every $a\subseteq I$.   Then $Z$ is compact and Hausdorff (311I), and
$i\mapsto\hat{i}:I\to Z$ is continuous, so has a unique extension to
a continuous function $\phi:\beta I\to Z$.

If $G\subseteq Z$ is open and not empty, it includes a set of the form
$\widehat{a}=\{\theta:\theta\in Z,\,\theta(a)=1\}$ where $a\subseteq I$
is not empty;  if $i$ is any member of $a$,
$\hat{i}\in\widehat{a}\subseteq G$ so
$G\cap\phi[\beta I]\ne\emptyset$.   This shows that $\phi[\beta I]$ is
dense in $Z$;  as $\beta I$ is compact, $\phi[\beta I]$ is compact,
therefore closed, and is equal to $Z$.   Thus $\phi$ is surjective.

If $t$, $u$ are distinct points of $\beta I$, there is an open subset
$H$ of $\beta I$ such that $t\in H$ and $u\notin\overline{H}$.   Set
$a=H\cap I$.   Then $t\in\overline{a}$, the closure of $a$ regarded as a
subset of $\beta I$, so $\phi(t)\in\overline{\phi[a]}$ (3A3Cd).   But
$\phi[a]=\{\hat{i}:i\in a\}\subseteq\widehat{a}$, which is
open-and-closed, so $\phi(t)\in\widehat{a}$.   Similarly, setting
$b=I\setminus\overline{H}$, $\phi(u)\in\widehat{b}$;  since
$\widehat{a}\cap\widehat{b}=\widehat{a\cap b}$ is empty,
$\phi(t)\ne\phi(u)$.   This shows that $\phi$ is injective, therefore a
homeomorphism between $\beta I$ and $Z$ (3A3Dd).\ \Qed}\cmmnt{

Note that if $z:\Cal PI\to\Bbb Z_2$ is a Boolean homomorphism, then
$\{J:z(J)=1\}$ is an ultrafilter on $I$;  and conversely, if $\Cal F$ is an
ultrafilter on $I$, we have a Boolean homomorphism $z:\Cal PI\to\Bbb Z_2$
such that $\Cal F=z^{-1}[\{1\}]$.}
\dvro{We}{So we} can identify $\beta I$ with the
set of ultrafilters on $I$.   Under this identification, the canonical
embedding of $I$ in $\beta I$ corresponds to matching each member of $I$
with the corresponding principal ultrafilter on $I$.

\quad (ii) $C(\beta I)$ is isomorphic, as Banach lattice, to
$\ell^{\infty}(I)$.
\prooflet{\Prf\ By 363Ha, we
can identify $\ell^{\infty}(I)$, as Banach lattice, with
$L^{\infty}(\Cal PI)=C(Z)$.   But (i) tells us that we have a canonical
identification between $C(Z)$ and $C(\beta I)$.\ \Qed}

\quad (iii) We have a one-to-one correspondence between filters $\Cal F$
on $I$ and non-empty closed sets $F\subseteq\beta I$, got by matching
$\Cal F$ with $\bigcap\{\widehat{a}:a\in\Cal F\}$, or $F$ with
$\{a:a\subseteq I,\,F\subseteq\widehat{a}\}$, where
$\widehat{a}\subseteq\beta I$ is the open-and-closed set corresponding
to $a\subseteq I$.
\prooflet{\Prf\ The identification of $\beta I$ with $Z$
means that we can regard the map
$a\mapsto\widehat{a}$ as a Boolean isomorphism between $\Cal PI$ and the
algebra of open-and-closed subsets of $\beta I$ (311I).   For any filter
$\Cal F$ on $I$, set $H(\Cal F)=\bigcap\{\widehat{a}:a\in\Cal F\}$;
because $\{\widehat{a}:a\in\Cal F\}$ is a downwards-directed family of
non-empty closed sets in the compact Hausdorff space $\beta I$,
$H(\Cal F)$ is a non-empty closed set.   If $F\subseteq\beta I$ is a
non-empty closed set, then it is elementary to check that
$\Cal H(F)=\{a:F\subseteq\widehat{a}\}$ is a filter on $I$, and
evidently $H(\Cal H(F))\supseteq F$.   But if $t\in\beta I\setminus F$,
then (because $\{\widehat{a}:a\subseteq I\}$ is a base for the topology
of $\beta I$, see 311I again) there is an $a\subseteq I$ such that
$t\in\widehat{a}$ and $F\cap\widehat{a}=\emptyset$, that is,
$F\subseteq\widehat{I\setminus a}$;  so
$\widehat{I\setminus a}\in\Cal H(F)$ and
$H(\Cal H(F))\subseteq\widehat{I\setminus a}$ and
$t\notin H(\Cal H(F))$.   Thus $H(\Cal H(F))=F$ for every non-empty
closed set $F\subseteq\beta I$.

If $\Cal F_1$ and $\Cal F_2$ are filters on $I$ and
$a\in\Cal F_1\setminus\Cal F_2$, then
$\{\widehat{b\setminus a}:b\in\Cal F_2\}$ is a downwards-directed family
of non-empty closed sets in $\beta I$, so has non-empty intersection;
if $t\in\widehat{b\setminus a}=\widehat{b}\setminus\widehat{a}$ for
every $b\in\Cal F_2$, then $t\in H(\Cal F_2)\setminus H(\Cal F_1)$.
This shows that $\Cal F\mapsto H(\Cal F)$ is injective.   It follows
that $\Cal F\mapsto H(\Cal F)$, $F\mapsto\Cal H(F)$ are the two halves
of a bijection, as claimed.\ \Qed}

\quad (iv) $\beta I$ is extremally disconnected.
\prooflet{(Because $\Cal PI$ is Dedekind complete, $Z$
is extremally disconnected (314S).)}

\quad(v) There are no non-trivial convergent sequences in
$\beta I$.
\prooflet{(4A2G(h-iii).   Compare \Engelking, 3.6.15.)}


\vleader{48pt}{4A2J}{Uniform spaces}\cmmnt{ (See \S3A4.)}  Let
$(X,\Cal W)$ be a uniform space;  give $X$ the induced topology
$\frak T$\cmmnt{ (3A4Ab)}.

\spheader 4A2Ja $\Cal W$ is generated by a family of pseudometrics.
\prooflet{(\Engelking, 8.1.10;  {\smc Bourbaki 66}, IX.1.4;  \Csaszar,
4.2.32.)}   More precisely:  if $\sequencen{W_n}$ is any sequence in
$\Cal W$, there is a pseudometric $\rho$ on $X$ such that ($\alpha$)
$\{(x,y):\rho(x,y)\le\epsilon\}\in\Cal W$ for every $\epsilon>0$
($\beta$) whenever $n\in\Bbb N$ and $\rho(x,y)<2^{-n}$ then
$(x,y)\in W_n$\prooflet{ (\Engelking, 8.1.10)}.

It follows that $\frak T$ is completely regular, therefore
regular\prooflet{ (3A3Be)}.
$\frak T$ is defined by the bounded uniformly continuous 
real-valued functions on $X$, in
the sense that it is the coarsest topology\cmmnt{ $\frak S$} on $X$
such that these are all continuous.   \prooflet{\Prf\ Let $\Rho$ be the
family of
pseudometrics compatible with $\Cal W$ in the sense of ($\alpha$) just
above.   If $x\in G\in\frak T$, there is a $\rho\in\Rho$ such that
$\{y:\rho(x,y)<1\}\subseteq G$;  setting $f(y)=\rho(x,y)$, we see that
$f$ is uniformly continuous, therefore $\frak S$-continuous, and that
$x\in\interior_{\frak S}G$.   As $x$ is arbitrary, $G\in\frak S$;  as
$G$ is arbitrary, $\frak T\subseteq\frak S$;  but of course
$\frak S\subseteq\frak T$ just because uniformly continuous functions
are continuous.\ \Qed}
%4@49

\spheader 4A2Jb If $\Cal W$ is countably generated and $\frak T$ is
Hausdorff, there is a metric $\rho$ on $X$ defining $\Cal W$ and
$\frak T$.  \prooflet{({\smc Engelking 89}, 8.1.21.)}
%Bourbaki Gen Top \query

\spheader 4A2Jc
%If $W\in\Cal W$, the interior of $W$\cmmnt{ (for the product topology
%on $X\times X$)} belongs to $\Cal W$.
%\prooflet{(\Engelking, 8.1.12.)}
If $W\in\Cal W$ and $x\in X$ then $x\in\interior W[\{x\}]$.
\prooflet{(\Engelking, 8.1.3.)}   If $A\subseteq X$ then
$\overline{A}=\bigcap_{W\in\Cal W}W[A]$.   \prooflet{(\Engelking,
8.1.4.)}
%4@A2Lc

\spheader 4A2Jd Any subset of a totally bounded set in $X$ is totally
bounded.
\prooflet{(\Engelking, 8.3.2;  \Csaszar, 3.2.70.)}
The closure of a totally bounded set is totally bounded.
\prooflet{\Prf\ If $A$ is totally bounded and $W\in\Cal W$, take
$W'\in\Cal W$ such that $W'\frsmallcirc W'\subseteq W$.   Then there is
a finite set
$I\subseteq X$ such that $A\subseteq W'[I]$.   In this case

\Centerline{$\overline{A}\subseteq W'[A]
\subseteq W'[W'[I]]=(W'\frsmallcirc W')[I]\subseteq W[I]$}

\noindent by (b).   As $W$ is arbitrary, $\overline{A}$ is totally
bounded.\ \Qed}

\spheader 4A2Je A subset of $X$ is compact iff it is
complete\cmmnt{ (definition: 3A4F)} (for its subspace uniformity) and
totally bounded.
\prooflet{(\Engelking, 8.3.16;  \Cech, 41A.8;  \Csaszar, 5.2.22;  \Gaal,
pp.\ 278-279.)}
%4@43  %4@33 %4@25
So if $X$ is complete, every closed totally bounded subset of $X$ is
compact, and the totally bounded sets are just the relatively compact
sets.
\prooflet{(A closed subspace of a complete space is complete.)}
%4@41 %4@48 %4@A4

\spheader 4A2Jf If $f:X\to\Bbb R$ is a continuous function with compact
support, it is uniformly continuous.
\prooflet{\Prf\ Set $K=\overline{\{x:f(x)\ne 0\}}$.
Let $\epsilon>0$.   For each $x\in X$, there is a $W_x\in\Cal W$ such
that $|f(y)-f(x)|\le\bover12\epsilon$ whenever $y\in W_x[\{x\}]$.   Let
$W'_x\in\Cal W$ be such that $W'_x\frsmallcirc W'_x\subseteq W_x$.   Set
$G_x=\interior W'_x[\{x\}]$;  then $x\in G_x$, by (b).   Because $K$ is
compact, there is a finite set $I\subseteq K$ such that
$K\subseteq\bigcup_{x\in I}G_x$.   Set
$W=(X\times X)\cap\bigcap_{x\in I}W'_x\in\Cal W$.   Take any
$(y,z)\in W\cap W^{-1}$.   If neither $y$ nor $z$ belongs to $K$, then
of course $|f(y)-f(z)|\le\epsilon$.   If $y\in K$, let $x\in I$ be such
that $y\in G_x$.   Then

\Centerline{$y\in W'_x[\{x\}]\subseteq W_x[\{x\}]$,
\quad$z\in W[W'_x[\{x\}]]\subseteq W'_x[W'_x[\{x\}]]
\subseteq W_x[\{x\}]$,}

\noindent so

\Centerline{$|f(y)-f(z)|\le|f(y)-f(x)|+|f(z)-f(x)|\le\epsilon$.}

\noindent The same idea works if $z\in K$.   So $|f(y)-f(z)|\le\epsilon$
for all $y$, $z\in W\cap W^{-1}$;  as $\epsilon$ is arbitrary, $f$ is
uniformly continuous.\ \Qed}
%4@A4

\spheader 4A2Jg(i) If $(Y,\frak S)$ is a completely regular space,
there is a uniformity on $Y$ which induces $\frak S$.
\prooflet{(\Engelking, 8.1.20.)}

\quad(ii)\dvArevised{2022} If $(Y,\frak S)$ is a compact completely regular
topological space, there is exactly one
uniformity on $Y$ which induces $\frak S$;  it is defined by the set
of all those pseudometrics on $Y$ which are continuous as functions from
$Y\times Y$ to $\Bbb R$.
\prooflet{(\Engelking, 8.3.13\query;  \Gaal, p.\ 304\query.)}

\quad(iii)\dvArevised{2022} 
If $(Y,\frak S)$ is a compact completely regular space and $\Cal V$ is
the uniformity on $Y$ inducing $\frak S$, then any continuous
function from $Y$ to $X$ is uniformly continuous.
%not Engelking, Kuratowski
\prooflet{(\Gaal, p.\ 305 Theorem 8\query.)}

\spheader 4A2Jh The set $U$ of uniformly continuous real-valued
functions on $X$ is a Riesz subspace of $\BbbR^X$ containing the
constant functions.   If a sequence in $U$ converges uniformly, the
limit function again belongs to $U$.
%not Kuratowski, Engelking
\prooflet{(\Csaszar, 3.2.64;  \Gaal, p.\ 237 Lemma 4.)}

\spheader 4A2Ji Let $(Y,\Cal V)$ be another uniform space.   If $\Cal F$
is a Cauchy filter on $X$ and $f:X\to Y$ is a uniformly continuous
function, then $f[[\Cal F]]$ is a Cauchy filter on $Y$.
\prooflet{(\Csaszar, 5.1.2.)}
%not Engelking ;  not \Gaal

\vleader{48pt}{4A2K}{First-countable, sequential and countably tight
spaces (a)}
%\spheader 4A2Ka
Let $X$ be a countably tight topological space.   If
$\ofamily{\xi}{\zeta}{F_{\xi}}$ is a non-decreasing family of closed
subsets of $X$ indexed by an ordinal $\zeta$, then
$E=\bigcup_{\xi<\zeta}F_{\xi}$ is an F$_{\sigma}$ set, and is
closed unless $\cf\zeta=\omega$.
\prooflet{\Prf\ If $\cf\zeta=0$, that is, $\zeta=0$, then $E=\emptyset$
is closed.   If $\cf\zeta=1$, that is, $\zeta=\xi+1$ for some ordinal
$\xi$, then $E=F_{\xi}$ is closed.   If $\cf\zeta=\omega$, there is a
sequence $\sequencen{\xi_n}$ in $\zeta$ with supremum $\zeta$, so that
$E=\bigcup_{n\in\Bbb N}F_{\xi_n}$ is F$_{\sigma}$.   If
$\cf\zeta>\omega$, take $x\in\overline{E}$.   Then there is a sequence
$\sequencen{x_n}$ in $E$ such that $x\in\overline{\{x_n:n\in\Bbb N\}}$.
For each $n$ there is a $\xi_n<\zeta$ such that $x_n\in F_{\xi_n}$, and
now $\xi=\sup_{n\in\Bbb N}\xi_n<\zeta$ and
$x\in\overline{F}_{\xi}=F_{\xi}\subseteq E$.   As
$x$ is arbitrary, $E$ is closed.\ \Qed}

\spheader 4A2Kb If $X$ is countably tight, any subspace of $X$ is
countably tight\prooflet{, just because if $A\subseteq Y\subseteq X$
then the closure of $A$ in $Y$ is the intersection of $Y$ with the
closure of $A$ in $X$}.   If $X$ is compact and countably tight, then
any Hausdorff continuous image of $X$ is countably tight.
\prooflet{ \Prf\ Let $f:X\to Y$ be a continuous surjection, where $Y$ is
Hausdorff, $B$ a subset of $Y$ and $y\in\overline{B}$.   Set
$A=f^{-1}[B]$.   Then $\overline{A}$ is compact, so $f[\overline{A}]$ is
compact, therefore closed;  because $f$ is surjective,
$y\in\overline{f[A]}\subseteq f[\overline{A}]$, and there is an
$x\in\overline{A}$ such that $f(x)=y$.   Now there is a countable set
$A_0\subseteq A$ such that $x\in\overline{A}_0$, in which case

\Centerline{$y=f(x)\in f[\overline{A}_0]\subseteq\overline{f[A_0]}$,}

\noindent while $f[A_0]$ is a countable subset of $B$.\ \Qed}

\spheader 4A2Kc If $X$ is a sequential space, it is countably tight.
\prooflet{\Prf\ Suppose that $A\subseteq X$ and $x\in\overline{A}$.
Set $B=\bigcup\{\overline{C}:C\subseteq A$ is countable$\}$.   If
$\sequencen{y_n}$ is a sequence in $B$ converging to $y\in X$, then for
each $n\in\Bbb N$ we can find a countable set $C_n\subseteq A$ such that
$y_n\in\overline{C}_n$, and now $C=\bigcup_{n\in\Bbb N}C_n$ is a
countable subset of $A$ such that $y\in\overline{C}\subseteq B$.   So
$B$ is sequentially closed, therefore closed, and $x\in B$.   As $A$ and
$x$ are arbitrary, $X$ is countably tight.\ \Qed}

\spheader 4A2Kd If $X$ is a sequential space, $Y$ is a topological space
and $f:X\to Y$ is sequentially continuous, then $f$ is continuous.
\prooflet{(\Engelking, 1.6.15.)}
%4@36

\spheader 4A2Ke First-countable spaces are sequential.
\prooflet{(\Engelking, 1.6.14.)}
%4@A2N

%any subspace of a first-countable space is first-countable

\spheader 4A2Kf Let $X$ be a locally compact Hausdorff space in which
every singleton set is G$_{\delta}$.   Then $X$ is first-countable.
\prooflet{\Prf\ If $\{x\}=\bigcap_{n\in\Bbb N}G_n$ where each $G_n$ is
open, then for each $n\in\Bbb N$ we can find a compact set $F_n$ such
that $x\in\interior F_n\subseteq G_n$.   By 4A2Gd,
$\{\bigcap_{i\le n}F_i:n\in\Bbb N\}$ is a base of neighbourhoods of
$x$.\ \Qed}
%4@39

\leader{4A2L}{(Pseudo-\nobreak)metrizable spaces}
\cmmnt{`Pseudometrizable'
spaces, as such, hardly appear in this volume, for the usual reasons;
they surface briefly in \S463.   It is perhaps worth noting, however,
that all the ideas, and very nearly all the results, in this paragraph
apply equally well to pseudometrics and pseudometrizable topologies.
If $X$ is a set and $\rho$ is a pseudometric on $X$,
set $U(x,\delta)=\{y:\rho(x,y)<\delta\}$ for $x\in X$ and $\delta>0$.}

\spheader 4A2La Any subspace of a (pseudo-\nobreak)metrizable space is
(pseudo-)\discretionary{}{}{}metrizable\cmmnt{ (2A3J)}.
A topological space is metrizable iff it is pseudometrizable and
Hausdorff\prooflet{ (2A3L)}.

\spheader 4A2Lb Metrizable spaces are
paracompact\prooflet{ (\Engelking, 5.1.3;  \Csaszar, 8.3.16;  \Cech,
30C.2; \Gaal, p.\ 155)},
%4@26
therefore hereditarily metacompact\cmmnt{ ((a) above and 4A2F(g-i))}.
%4@25 4@38

\spheader 4A2Lc A metrizable space is
perfectly normal\prooflet{ (\Engelking, 4.1.13;  \Csaszar, 8.4.5.)},
so every closed set is a zero set and every open set is a cozero set (in
particular, is F$_{\sigma}$).
%4@23 %4@11, %4@12
\leaveitout{A metrizable space is monotonically normal.    \Prf\ If
$(X,\rho)$ is a metric space, then for disjoint closed sets $E$,
$F\subseteq X$ set $W(E,F)=\{x:\rho(x,E)<\rho(x,F)\}$, writing
$\rho(x,E)=\inf_{y\in E}\rho(x,y)$ (with $\rho(x,\emptyset)=\infty$).
Now $W$ satisfies the conditions of the definition in 4A2A.\ \Qed
}%end of leaveitout

\spheader 4A2Ld If $X$ is a pseudometrizable space, it is
first-countable.
\prooflet{(If $\rho$ is a pseudometric defining the topology of $X$, and
$x$ is any point of $X$, then $\{\{y:\rho(y,x)<2^{-n}\}:n\in\Bbb N\}$ is
a base of neighbourhoods of $x$.)}   So $X$ is sequential and countably
tight\prooflet{ (4A2Ke, 4A2Kc)}, and if $Y$ is another topological space
and $f:X\to Y$ is sequentially continuous, then $f$ is
continuous\prooflet{ (4A2Kd)}.

\spheader 4A2Le {\bf Relative compactness} Let $X$ be a
pseudometrizable space and $A$ a subset of $X$.   Then the following are
equiveridical:  ($\alpha$) $A$ is relatively compact;  ($\beta$) $A$ is
relatively countably compact;  ($\gamma$) every sequence in $A$ has a
subsequence with a limit in $X$.
\prooflet{\Prf\ Fix a pseudometric $\rho$ defining the topology of $X$.
($\alpha$)$\Rightarrow$($\beta$) by 4A2G(f-iii).   If $\sequencen{x_n}$
is a sequence in $A$ with a cluster point $x\in X$, then we can choose
$\sequence{i}{n_i}$ inductively such that $\rho(x_{n_i},x)\le 2^{-i}$
and $n_{i+1}>n_i$ for every $i$;  now $\sequence{i}{x_{n_i}}\to x$;  it
follows that ($\beta$)$\Rightarrow$($\gamma$).   Now assume that
($\alpha$) is false.   Then there is an ultrafilter $\Cal F$ on $X$
containing $A$ which has no limit in $X$ (3A3Be, 3A3De).   If $\Cal F$
is a Cauchy
filter, choose $F_n\in\Cal F$ such that $\rho(x,y)\le 2^{-n}$ whenever
$x$, $y\in F_n$, and $x_n\in A\cap\bigcap_{i\le n}F_i$ for each $n$;
then it is easy to see that $\sequencen{x_n}$ is a sequence in $A$ with
no convergent subsequence.   If $\Cal F$ is not a Cauchy filter, let
$\epsilon>0$ be such that there is no $F\in\Cal F$ such that
$\rho(x,y)\le\epsilon$ for every $x$, $y\in F$.
Then $X\setminus U(x,\bover12\epsilon)\in\Cal F$ for every $x\in X$, so we can choose
$\sequencen{x_n}$ inductively such that
$x_n\in A\setminus\bigcup_{i<n}U(x_i,\bover12\epsilon)$ for every
$n\in\Bbb N$, and again we have a sequence $\sequencen{x_n}$ in $A$ with
no convergent subsequence in $X$.   Thus
not-($\alpha$)$\Rightarrow$ not-($\gamma$) and the proof is complete.\
\Qed}

\spheader 4A2Lf {\bf Compactness} If $X$ is a pseudometrizable space,
it is compact iff it is countably compact iff it is sequentially
compact.
\prooflet{((e) above, using 4A2G(f-i).   Compare \Engelking, 4.1.17, and
\Csaszar, 5.3.33 \& 5.3.47.)}

\spheader 4A2Lg(i) If $(X,\rho)$ is a metric space, its topology has a
base which is $\sigma$-metrically-discrete.
\prooflet{\Prf\ Enumerate $X$ as $\ofamily{\xi}{\kappa}{x_{\xi}}$ where
$\kappa$ is a cardinal.   Let $\sequencen{(q_n,q'_n)}$ be a sequence
running over $\{(q,q'):q,\,q'\in\Bbb Q,\,0<q<q'\}$ in such a way that
$q'_n-q_n\ge 2^{-n}$ for every $n\in\Bbb N$.   For $n\in\Bbb N$,
$\xi<\kappa$ set $G_{n\xi}
=\{x:\rho(x,x_{\xi})<q_n,\,\inf_{\eta<\xi}\rho(x,x_{\eta})>q'_n\}$
(interpreting $\inf\emptyset$ as $\infty$).   Then
$\Cal U=\langle G_{n\xi}\rangle_{\xi<\kappa,n\in\Bbb N}$ is a
$\sigma$-metrically-discrete family of open sets.   If $G\subseteq X$ is
open and $x\in G$,
let $\epsilon>0$ be such that $U(x,2\epsilon)\subseteq G$.   Let
$\xi<\kappa$ be minimal such that $\rho(x,x_{\xi})<\epsilon$, and let
$n\in\Bbb N$ be such that $\rho(x,x_{\xi})<q_n<q'_n<\epsilon$;  then
$x\in G_{n\xi}\subseteq G$.   As $x$ and $G$ are arbitrary, $\Cal U$ is
a base for the topology of $X$.\ \Qed}

\quad (ii) Consequently, any metrizable space has a
$\sigma$-disjoint base.
\cmmnt{(Compare \Engelking, 4.4.3;  \Csaszar, 8.4.5;
\Kuratowski, \S21.XVII.)}
%4@17 %4@25

\spheader 4A2Lh The product of a countable family of metrizable spaces
is metrizable.
\prooflet{(\Engelking, 4.2.2; \Csaszar, 7.3.27.)}

\spheader 4A2Li Let $X$ be a metrizable space and $\kappa\ge\omega$ a
cardinal.   Then $w(X)\le\kappa$ iff $X$ has a dense subset with cardinal
at most $\kappa$.
\prooflet{(\Engelking, 4.1.15.)}

\spheader 4A2Lj If $(X,\rho)$ is any metric space, then the balls
$B(x,\delta)=\{y:\rho(y,x)\le\delta\}$ are all closed
sets\cmmnt{ (cf.\ 1A2G)}.
In particular, in a normed space $(X,\|\,\|)$, the balls
$B(x,\delta)=\{y:\|y-x\|\le\delta\}$ are closed.
%4@46

\leader{4A2M}{Complete metric spaces (a)}
%\spheader 4A2Ma
{\bf Baire's theorem for complete metric spaces} Every
complete metric space is a Baire space.
\prooflet{(\Engelking, 4.3.36 \& 3.9.4;  \Kechris, 8.4;
\Csaszar, 9.2.1 \& 9.2.8;  \Gaal, p.\ 287.)}
So a non-empty complete metric space is not
meager\cmmnt{ (cf.\ 3A3Ha)}.
%4@41

\spheader 4A2Mb Let $\familyiI{(X_i,\rho_i)}$ be a countable family of
complete metric spaces.   Then there is a complete metric on
$X=\prod_{i\in I}X_i$ which defines the product topology on $X$.
\prooflet{(\Engelking, 4.3.12;  \Kuratowski, \S33.III.)}

\spheader 4A2Mc Let $(X,\rho)$ be a complete metric space, and
$E\subseteq X$ a G$_{\delta}$ set.   Then there is a complete metric on
$E$ which defines the subspace topology of $E$.
\prooflet{(\Engelking, 4.3.23;  \Kuratowski, \S33.VI;  \Kechris, 3.11.)}

\spheader 4A2Md Let $(X,\rho)$ be a complete metric space.   Then it is
\v{C}ech-complete.   \prooflet{(\Engelking, 4.3.26.)}

\spheader 4A2Me A non-empty complete metric space without isolated
points is uncountable.  %surely has cardinal \frakc
%4@39S
\prooflet{(If $x$ is not isolated, $\{x\}$ is nowhere dense.)}
%not Kechris not Engelking I think

\leader{4A2N}{Countable networks:  Proposition} (a) If $X$ is a
topological space with a
countable network, any subspace of $X$ has a countable network.

(b) Let $X$ be a space with a countable network.    Then $X$ is
hereditarily Lindel\"of.   If it is regular, it is perfectly normal.
%4@35

(c) If $X$ is a topological space, and $\sequencen{A_n}$ is a sequence
of subsets of $X$ each of which has a countable network (for its
subspace topology), then $A=\bigcup_{n\in\Bbb N}A_n$ has a countable
network.

(d) A continuous image of a space with a countable network has a
countable network.

(e) Let $\familyiI{X_i}$ be a countable family of topological spaces
with countable networks, with product $X$.   Then $X$ has a countable
network.

(f) If $X$ is a Hausdorff space with a countable network, there is a
countable family $\Cal G$ of open sets such that whenever $x$, $y$ are
distinct points in $X$ there are disjoint $G$, $H\in\Cal G$ such that
$x\in G$ and $y\in H$.
%4@23 %4@33

(g) If $X$ is a regular topological space with a countable network, it
has a countable network consisting of closed sets.

(h) A compact Hausdorff space with a countable network
is second-countable.

(i) If a topological space $X$ has a countable network, then any dense
set in $X$ includes a countable dense set;  in particular, $X$ is
separable.

(j) If a topological space $X$ has a countable network, then $C(X)$,
with the topology of pointwise convergence inherited from the product
topology of $\Bbb R^X$, has a countable network.

\proof{{\bf (a)} If $\Cal E$ is a countable network for the topology of
$X$, and $Y\subseteq X$, then $\{Y\cap E:E\in\Cal E\}$ is a countable
network for the topology of $Y$.

\medskip

{\bf (b)} By \Engelking, 3.8.12 $X$ is Lindel\"of.
Since any subspace of $X$ has a countable network ((a) above), it also
is Lindel\"of, and $X$ is hereditarily Lindel\"of.
By 4A2H(c-ii), if $X$ is regular, it is perfectly normal.

\medskip

{\bf (c)} If $\Cal E_n$ is a countable network for the topology of $A_n$
for each $n$, then $\bigcup_{n\in\Bbb N}\Cal E_n$ is a countable network
for the topology of $A$.

\medskip

{\bf (d)} Let $X$ be a topological space with a countable network
$\Cal E$, and $Y$ a continuous image of $X$.   Let $f:X\to Y$ be a
continuous
surjection.   Then $\{f[E]:E\in\Cal E\}$ is a network for the topology
of $Y$.   \Prf\ If $H\subseteq Y$ is open and $y\in H$, then $f^{-1}[H]$
is an open subset of $X$ and there is an $x\in X$ such that $f(x)=y$.
Now there must be an $E\in\Cal E$ such that $x\in E\subseteq f^{-1}[H]$,
so that $y\in f[E]\subseteq H$.\ \QeD\   But $\{f[E]:E\in\Cal E\}$ is
countable, so $Y$ has a countable network.

\medskip

{\bf (e)} For each $i\in I$ let $\Cal E_i$ be a countable network for
the topology of $X_i$.   For each finite $J\subseteq I$, let $\Cal C_J$
be the family of sets expressible as $\prod_{i\in I}E_i$ where
$E_i\in\Cal E_i$ for each $i\in J$ and $E_i=X_i$ for $i\in
I\setminus J$; then $\Cal C_J$ is countable because $\Cal
E_i$ is countable for
each $i\in J$.  Because the family $[I]^{<\omega}$ of finite subsets of
$I$ is countable (3A1Cd), $\Cal E=\bigcup\{\Cal C_J:J\in[I]^{<\omega}\}$
is countable.   But $\Cal E$ is a network for the topology of $X$.
\Prf\ If $G\subseteq X$ is open and $x\in G$, then there is a family
$\familyiI{G_i}$ such that every $G_i\subseteq X_i$ is open,
$J=\{i:G_i\ne X_i\}$ is finite, and $x\in\prod_{i\in I}G_i$.   For $i\in
J$, there is an $E_i\in\Cal E_i$ such that $x(i)\in E_i\subseteq G_i$;
set $E_i=X_i$ for $i\in I\setminus J$.   Then

\Centerline{$E=\prod_{i\in I}E_i\in\Cal C_J\subseteq\Cal E$}

\noindent and $x\in E\subseteq G$.\ \Qed

So $\Cal E$ is a countable network for the topology of $X$.

\medskip

{\bf (f)} By (b) and (e), $X\times X$ is hereditarily Lindel\"of.   In
particular, $W=\{(x,y):x\ne y\}$ is Lindel\"of.   Set

\Centerline{$\Cal V=\{G\times H:G$, $H\subseteq X$ are open, $G\cap
H=\emptyset\}$.}

\noindent Because $X$ is Hausdorff, $\Cal V$ is a cover of $W$.   So
there is a countable $\Cal V_0\subseteq\Cal V$ covering $W$.   Set

\Centerline{$\Cal G=\{G:G\times H\in\Cal V_0\}\cup\{H:G\times H\in\Cal
V_0\}$.}

\noindent Then $\Cal G$ is a countable family of open sets separating
the points of $X$.

\medskip

{\bf (g)} Let $\Cal E$ be a countable network
for the topology of $X$.   Set $\Cal E'=\{\overline{E}:E\in\Cal E\}$.
If $G\subseteq X$ is open and $x\in G$, then (because the topology is
regular) there is an open set $H$ such that
$x\in H\subseteq\overline{H}\subseteq G$.   Now there is an $E\in\Cal E$
such that $x\in E\subseteq H$, in
which case $\overline{E}\in\Cal E'$ and $x\in\overline{E}\subseteq G$.
So $\Cal E'$ is a countable network for $X$ consisting of closed
sets.

\medskip

{\bf (h)} \Engelking, 3.1.19.

\medskip

{\bf (i)} Let $D\subseteq X$ be dense, and $\Cal E$ a countable network
for the topology of $X$.   Let $D'\subseteq D$ be a countable set such
that $D'\cap E\ne\emptyset$ whenever $E\in\Cal E$ and
$D\cap E\ne\emptyset$.   If $G\subseteq X$ is open and not empty, there
is an $x\in D\cap G$;  now there is an $E\in\Cal E$ such that
$x\in E\subseteq G$, and as $x\in D\cap E$ there must be an
$x'\in D'\cap E$, so that $x'\in D'\cap G$.   As $G$ is arbitrary, $D'$
is dense in $X$.

Taking $D=X$, we see that $X$ has a countable dense subset.

\medskip

{\bf (j)} Let $\Cal E$ be a countable network for the topology of $X$
and $\Cal U$ a countable base for the topology of
$\Bbb R$\cmmnt{ (4A2Ua)}.   For
$E\in\Cal E$ and $U\in\Cal U$ set
$H(E,U)=\{f:f\in C(X),\,E\subseteq f^{-1}[U]\}$.   Then the set of
finite intersections of sets of the form $H(E,U)$ is a countable network
for the topology of pointwise convergence on $C(X)$.
(Compare 4A2Oe.)
}%end of proof of 4A2N

\leader{4A2O}{Second-countable spaces (a)} Let $(X,\frak T)$ be a
topological space and $\Cal U$ a countable subbase for $\frak T$.
Then $\frak T$ is second-countable.
\prooflet{
($\{X\}\cup\{U_0\cap U_1\cap\ldots\cap U_n:U_0,\ldots,U_n\in\Cal U\}$ is
countable and is a base for $\frak T$,
by 4A2B(a-i).)}
%4@24

\spheader 4A2Ob Any base of a second-countable space includes a
countable base.
\prooflet{(\Csaszar, 2.4.17.)}

\spheader 4A2Oc A second-countable space has a countable
network\cmmnt{ (because a base is also a network)}, so is separable
and hereditarily Lindel\"of\cmmnt{ (\Engelking, 1.3.8 \& 3.8.1, 4A2Nb,
4A2Ni)}.

\spheader 4A2Od The product of a countable family of second-countable
spaces is second-countable.
\prooflet{(\Engelking, 2.3.14.)}

\spheader 4A2Oe If $X$ is a second-countable space then $C(X)$, with the
topology of uniform convergence on compact sets, has a
countable network.   \prooflet{\Prf\ (See \Engelking, Ex.\ 3.4H.)
Let $\Cal U$ be a countable base for the topology of $X$
and $\Cal V$ a countable base for the topology of
$\Bbb R$\cmmnt{ (4A2Ua)}.   For $U\in\Cal U$, $V\in\Cal V$ set
$H(U,V)=\{f:f\in C(X)$, $U\subseteq f^{-1}[V]\}$.   Then the set of
finite intersections of sets of the form $H(U,V)$ is a countable network
for the topology of uniform convergence on compact subsets of
$X$.\ \Qed}

\leader{4A2P}{Separable metrizable spaces (a)}(i) A metrizable space is
second-countable iff it is separable.
\prooflet{(\Engelking, 4.1.16;  \Csaszar, 2.4.16;  \Gaal\ p.\ 120.)}

\quad(ii) A compact metrizable space is separable\prooflet{ (\Engelking,
4.1.18;
\Csaszar, 5.3.35; \Kuratowski, \S21.IX)}, so is second-countable and has
a countable network.

\quad(iii)
Any base of a separable metrizable space includes a countable
base\cmmnt{ (4A2Ob)},
%4@17 %4@14 %4@27
which is also a countable network,
%4@17
so the space is hereditarily Lindel\"of\cmmnt{ (4A2Nb)}.

\quad(iv)
Any subspace of a separable metrizable space is separable and
metrizable\cmmnt{ (4A2La, 4A2Na, 4A2Ni)}.

\quad(v)
A countable product of separable metrizable spaces is separable and
metrizable\cmmnt{ (4A2B(e-ii), 4A2Lh)}.
%4@15 %4@33 %4@25 %4@24 %4@18

\spheader 4A2Pb A topological space is separable and metrizable iff it
is second-countable, regular and Hausdorff.
\prooflet{(\Engelking, 4.2.9;  \Csaszar, 7.1.57;
\Kuratowski, \S22.II.)}
%4@19

\spheader 4A2Pc A Hausdorff continuous image of a compact metrizable space
is metrizable.
\prooflet{(It is a compact Hausdorff space, by 2A3N(b-ii), with a
countable network, by 4A2Nd, so is metrizable, by 4A2Nh.)}

\spheader 4A2Pd A metrizable space is separable iff it is ccc iff it is
Lindel\"of.
\prooflet{(\Engelking, 4.1.16.)}
%4@18

\spheader 4A2Pe If $X$ is a compact metrizable space, then $C(X)$ is
separable under its usual norm topology defined from the norm
$\|\,\,\|_{\infty}$.
\prooflet{(4A2Oe, or \Engelking, 3.4.16.)}

\vleader{48pt}{4A2Q}{Polish spaces:  Proposition}
(a) A countable discrete space is Polish.

(b) A compact metrizable space is Polish.
%4@35

(c) The product of a countable family of Polish spaces is Polish.
%4@24

(d) A G$_{\delta}$ subset of a Polish space is Polish\cmmnt{ in its
subspace topology};  in particular, a set which is either open or closed
is Polish.
%4@24

(e) The disjoint union of countably many Polish spaces is Polish.
%for (g)

(f) If $X$ is any set and $\sequencen{\frak T_n}$ is a sequence of
Polish topologies on $X$ such that $\frak T_m\cap\frak T_n$ is Hausdorff
for all $m$, $n\in\Bbb N$, then the
topology\cmmnt{ $\frak T_{\infty}$}
generated by $\bigcup_{n\in\Bbb N}\frak T_n$ is Polish.
%4@23 4@A3

(g) If $X$ is a Polish space, it is homeomorphic to a G$_{\delta}$ set
in a compact metrizable space.

(h) If $X$ is a locally compact Hausdorff space, it is Polish iff it has a
countable network iff it is metrizable and $\sigma$-compact.

\proof{{\bf (a)} Any set $X$ is complete under the discrete
metric $\rho$ defined by setting $\rho(x,y)=1$ whenever $x$, $y\in X$
are distinct.   This defines the discrete topology, and if $X$ is
countable it is separable, therefore Polish.

\medskip

{\bf (b)} By 4A2P(a-ii), it is separable;  by 4A2Je, any metric defining
the topology is complete.

\medskip

{\bf (c)} If $\familyiI{X_i}$ is a countable family of Polish spaces
with product $X$, then surely $X$ is separable (4A2B(e-ii));  and 4A2Mb
tells us that its topology is defined by a complete metric.

\medskip

{\bf (d)} If $X$ is Polish and $E$ is a G$_{\delta}$ set in $X$, then
$E$ is separable, by 4A2P(a-iv),
and its topology is defined by a complete metric, by 4A2Mc.   So $E$ is
Polish.   Any open set is of course a G$_{\delta}$ set, and any closed
set is a G$_{\delta}$ set by 4A2Lc.

\medskip

{\bf (e)} Let $\familyiI{X_i}$ be a countable disjoint family of Polish
spaces, and $X=\bigcup_{i\in I}X_i$.   For each $i\in I$ let $\rho_i$ be
a complete metric on $X_i$ defining the topology of $X_i$.   Define
$\rho:X\times X\to\coint{0,\infty}$ by setting
$\rho(x,y)=\min(1,\rho_i(x,y))$ if $i\in I$ and $x$, $y\in X_i$,
$\rho(x,y)=1$ otherwise.   It is easy to check that $\rho$ is a complete
metric on $X$ defining the disjoint union topology on $X$.   $X$ is
separable, by 4A2B(e-i), therefore Polish.

\medskip

{\bf (f)} This result is in \Kechris, 13.3;  but I spell out the proof
because it is an essential element of some measure-theoretic arguments.
On $X^{\Bbb N}$ take the product topology $\frak T$ of the topologies
$\frak T_n$.   This is Polish, by (c).   Consider the diagonal
$\Delta=\{x:x\in X^{\Bbb N}$, $x(m)=x(n)$ for all $m$, $n\in\Bbb N\}$.
This is closed in $X^{\Bbb N}$.   \Prf\ If
$x\in X^{\Bbb N}\setminus\Delta$, let $m$, $n\in\Bbb N$ be such that
$x(m)\ne x(n)$.   Because $\frak T_m\cap\frak T_n$ is Hausdorff, there
are disjoint $G$, $H\in\frak T_m\cap\frak T_n$ such that $x(m)\in G$ and
$x(n)\in H$.   Now $\{y:y\in X^{\Bbb N}$, $y(m)\in G$, $y(n)\in H\}$ is
an open set in $X^{\Bbb N}$ containing $x$ and disjoint from $\Delta$.
As $x$ is arbitrary, $\Delta$ is closed.\ \Qed

By (d), $\Delta$, with its subspace topology, is a Polish space.   Let
$f:X\to\Delta$ be the natural bijection, setting $f(t)=x$ if $x(n)=t$
for every $n$, and let $\frak S$ be the topology on $X$ which makes $f$
a homeomorphism.   The topology on $\Delta$ is generated by
$\{\{x:x\in\Delta$, $x(n)\in G\}:n\in\Bbb N$, $G\in\frak T_n\}$, so
$\frak S$ is generated by
$\{\{t:t\in X$, $t\in G\}:n\in\Bbb N$, $G\in\frak T_n\}
=\bigcup_{n\in\Bbb N}\frak T_n$.   Thus $\frak S=\frak T_{\infty}$ and
$\frak T_{\infty}$ is Polish.

\medskip

{\bf (g)} \Kechris, 4.14.

\medskip

{\bf (h)} If $X$ is Polish, then it is separable,
therefore Lindel\"of (4A2P(a-iii)).   Since the family
$\Cal G$ of relatively compact open subsets of $X$ covers $X$, there is a
countable $\Cal G_0\subseteq\Cal G$ covering $X$, and
$\{\overline{G}:G\in\Cal G_0\}$ witnesses that $X$ is $\sigma$-compact.
Also, of course, $X$ is metrizable.

If $X$ is metrizable and $\sigma$-compact, let $\sequencen{K_n}$ be a
sequence of compact sets covering $X$;  each $K_n$ has a countable network
(4A2P(a-ii)), so $X=\bigcup_{n\in\Bbb N}K_n$ has a countable network
(4A2Nc).

If $X$ has a countable network, let $Z=X\cup\{\infty\}$ be its one-point
compactification (3A3O).   This is compact and Hausdorff and has a
countable network, by 4A2Nc again, so is second-countable (4A2Nh)
and metrizable (4A2Pb) and Polish ((b) above).
So $X$ also, being an open set in $Z$, is Polish ((d) above).
}%end of proof of 4A2Q

\leader{4A2R}{Order topologies} Let $(X,\le)$ be a totally ordered set
and $\frak T$ its order topology.

(a) The set $\Cal U$ of open intervals in $X$\cmmnt{ (definition:
4A2A)} is a base for $\frak T$.

(b) $[x,y]$,
$\coint{x,\infty}$ and $\ocint{-\infty,x}$ are closed sets for all $x$,
$y\in X$.

(c) $\frak T$ is Hausdorff,
%and monotonically normal, therefore
normal and countably paracompact.
%4@39

(d) If $A\subseteq X$ then $\overline{A}$ is the set of
elements of $X$ expressible as either suprema or infima of
non-empty subsets of $A$.

(e) A subset of $X$ is closed iff it is order-closed.

(f) If $\sequencen{x_n}$ is a non-decreasing sequence in $X$
with supremum $x$, then $x=\lim_{n\to\infty}x_n$.
%4@16 %4@A2U

(g) A set $K\subseteq X$ is compact iff $\sup A$ and
$\inf A$ are defined in
$X$ and belong to $K$ for every non-empty $A\subseteq K$.

(h) $X$ is Dedekind complete iff $[x,y]$ is compact for
all $x$, $y\in X$.

(i) $X$ is compact iff it is either empty or Dedekind complete
with greatest and least elements.

(j) Any open set $G\subseteq X$ is
expressible as a union of disjoint open order-convex sets;  if
$X$ is Dedekind complete, these will be open intervals.
%4@16  5@37

(k) If $X$ is well-ordered it is locally compact.

(l) In $X\times X$, $\{(x,y):x<y\}$ is open and $\{(x,y):x\le y\}$ is
closed.
%4@39

(m) If $F\subseteq X$ and {\it either} $F$ is order-convex {\it or} $F$
is compact {\it or} $X$ is Dedekind complete and $F$ is closed, then
the subspace topology on $F$ is induced by the inherited order of $F$.
%4@19 5@33  also if  X  is Ded cpletion of  F

(n) If $X$ is ccc it is hereditarily Lindel\"of, therefore perfectly
normal.

(o) If $Y$ is another totally ordered set with its order topology, an
order-preserving function from $X$ to $Y$ is continuous iff it is
order-continuous.
%4@16

\proof{{\bf (a)} Put the definition of `order topology' (4A2A) together
with 4A2B(a-i).

\medskip

{\bf (b)} Their complements are either $X$, or members of $\Cal U$, or
unions of two members of $\Cal U$.

\medskip

{\bf (c)} Fix a well-ordering $\preccurlyeq$ of $X$.

\medskip

\quad{\bf (i)} If $x<y\in X$,
define $U_{xy}$, $U_{yx}$ as follows:  if $\ooint{x,y}$ is empty,
$U_{xy}=\ocint{-\infty,x}=\ooint{-\infty,y}$ and
$U_{yx}=\coint{y,\infty}=\ooint{x,\infty}$;  otherwise, let $z$ be the
$\preccurlyeq$-least member of $\ooint{x,y}$ and set
$U_{xy}=\ooint{-\infty,z}$, $U_{yx}=\ooint{z,\infty}$.

This construction ensures that if $x$, $y$ are any distinct points of
$X$, $U_{xy}$ and $U_{yx}$ are disjoint open sets containing $x$, $y$
respectively, so $\frak T$ is Hausdorff.

\medskip

\quad{\bf (ii)} Now suppose that $F\subseteq X$ is closed and that
$x\in X\setminus F$.
Then $V_{xF}=\interior(X\cap\bigcap_{y\in F}U_{xy})$ contains $x$.
\Prf\ There are $u$, $v\in X\cup\{-\infty,\infty\}$ such that
$x\in\ooint{u,v}\subseteq X\setminus F$.   If $\ooint{u,x}=\emptyset$,
set $u'=u$;  otherwise, let $u'$ be the $\preccurlyeq$-least member of
$\ooint{u,x}$.   Similarly,
if $\ooint{x,v}=\emptyset$, set $v'=v$;  otherwise, let $v'$ be the
$\preccurlyeq$-least member of $\ooint{x,v}$.   Then $u'<x<v'$.   Now
suppose that
$y\in F$ and $y>x$.   If $\ooint{x,v}=\emptyset$, then
$U_{xy}\supseteq\ocint{-\infty,x}=\ooint{-\infty,v'}$.   Otherwise,
$v'\in\ooint{x,v}\subseteq\ooint{x,y}$, so $U_{xy}=\ooint{-\infty,z}$
where $z$ is the $\preccurlyeq$-least member of $\ooint{x,y}$.   But
this means that $z\preccurlyeq v'$ and either $z=v'$ or
$z\notin\ooint{x,v}$;  in either case, $v'\le z$ and
$\ooint{-\infty,v'}\subseteq U_{xy}$.

Similarly, $\ooint{u',\infty}\subseteq U_{xy}$ whenever $y\in F$ and
$y<x$.   So $x\in\ooint{u',v'}\subseteq V_{xF}$.\ \Qed

\medskip

\leaveitout{
\quad{\bf (iii)} It follows that $X$ is regular.   \Prf\
If $F\subseteq X$ is open and $x\in X\setminus F$, then
$V_{xF}$ is an open set containing $x$.   But $\bigcup_{y\in F}U_{yx}$ is
an open set including $F$ and disjoint from $V_{xF}$.   So $x$ and $F$ can
be separated by open sets.\ \Qed

\medskip

\quad{\bf (iv)} For open $G\subseteq X$ and $x\in G$, set
$W(x,G)=V_{x,X\setminus G}$.   Then $W(x,G)$ is an open set containing $x$
and included in $G$.   If $x\in G\subseteq H$ then

\Centerline{$W(x,G)=\interior(X\cap\bigcap_{y\in X\setminus G}U_{xy})
\subseteq\interior(X\cap\bigcap_{y\in X\setminus H}U_{xy})
=W(x,H)$.}

\noindent If $x$, $y\in X$ are distinct then
$W(x,X\setminus\{y\})\subseteq U_{xy}$ is
disjoint from $W(y,X\setminus\{x\})\subseteq U_{yx}$.   So all the
conditions of 4A2F(?-ii) are satisfied and $X$ is monotonically normal.
By 4A2F?, it is normal and countably paracompact.
}%end of leaveitout

\quad{\bf (iii)} Let $E$ and $F$ be disjoint closed sets.   Set
$G=\bigcup_{x\in E}V_{xF}$, $H=\bigcup_{y\in F}V_{yE}$.   Then $G$ and
$H$ are open sets including $E$, $F$ respectively.   If $x\in E$ and
$y\in F$, then $V_{xF}\cap V_{yE}\subseteq U_{xy}\cap U_{yx}=\emptyset$,
so $G\cap H=\emptyset$.   As $E$ and $F$ are arbitrary, $\frak T$ is
normal.

\medskip

\quad{\bf (iv)} Let $\sequencen{F_n}$ be a non-increasing sequence of
closed sets with empty intersection.   Let $\Cal I$ be the family of
open intervals $I\subseteq X$ such that $I\cap F_n=\emptyset$ for some
$n$.   Because the $F_n$ are closed and have empty intersection,
$\Cal I$ covers $X$.   If $I$, $I'\in\Cal I$ are not
disjoint, $I\cup I'\in\Cal I$;  so we have an equivalence relation
$\sim$ on $X$ defined by saying that $x\sim y$ if there is some
$I\in\Cal I$ containing both $x$ and $y$.   The corresponding
equivalence classes are open and therefore closed, and are order-convex.
Let $\Cal G$ be the set of equivalence classes for $\sim$.

For each $G\in\Cal G$, fix $x_G\in G$.   Set
$G^+=G\cap\coint{x_G,\infty}$.   Then we have a
non-decreasing sequence $\sequencen{G^+_n}$ of closed sets, with union
$G^+$, such that $G^+_n\cap F_n=\emptyset$ for each $n$.
\Prf\ If there is some $m\in\Bbb N$ such that
$G^+\cap F_m=\emptyset$, set $G^+_n=\emptyset$ if
$G^+\cap F_n\ne\emptyset$, $G^+$ if $G^+\cap F_n=\emptyset$.   Otherwise,
given $x\in G^+$ and $n\in\Bbb N$, there is some $m$ such that $[x_G,x]$
does not meet $F_m$, and an $x'\in G^+\cap F_{\max(m,n)}$, so that
$x'\in F_n$ and $x'>x$.   We can therefore choose a strictly increasing
sequence $\sequence{k}{x_k}$ such that $x_0=x_G$ and
$x_{k+1}\in G^+\cap F_k$ for each $k$.
If $x$ is any upper bound of $\{x_k:k\in\Bbb N\}$ then $x\not\sim x_G$,
so $G^+=\bigcup_{k\in\Bbb N}[x_G,x_k]$.   Now, for each $n$, there is a
least $k(n)$ such that $[x_G,x_{k(n)}]\cap F_n\ne\emptyset$;  set
$G^+_n=\emptyset$ if $k(n)=0$, $[x_G,x_{k(n-1)}]$ otherwise.
As $F_{n+1}\subseteq F_n$, $k(n+1)\ge k(n)$
for each $n$.   Since each $[x_G,x_k]$ is disjoint from some
$F_n$, and therefore from all but finitely many $F_n$,
$\lim_{n\to\infty}k(n)=\infty$ and $G^+=\bigcup_{n\in\Bbb N}G^+_n$.\ \Qed

Similarly, $G^-=G\cap\ocint{-\infty,x_G}$ can be expressed as the union
of a non-decreasing sequence $\sequencen{G^-_n}$ of closed sets such that
$G^-_n\cap F_n=\emptyset$ for every $n$.   Now set
$F'_n=\bigcup_{G\in\Cal G}G^+_n\cup G^-_n$ for each $n$.   Because every
$G^+_n$ and $G^-_n$ is closed, and every $G$ is open-and-closed, $F'_n$
is closed.   So $\sequencen{F'_n}$ is a non-decreasing sequence of closed
sets with union $X$, and $F'_n$ is disjoint from $F_n$ for each $n$.
Accordingly $\sequencen{X\setminus F'_n}$ is a non-increasing sequence of
open sets with empty intersection enveloping the $F_n$.   As
$\sequencen{F_n}$ is arbitrary, $\frak T$ is countably paracompact (4A2Ff).
%should check Rudin 84 for the `monotonically normal' case, the
%proof may be more useful
%not sure about that, but if there is any other reason to talk about
%monotonically normal spaces then this would be the right thing to do

\medskip

{\bf (d)} Let $B$ be the set of such suprema and infima.   For
$x\in X$ set $A_x=A\cap\ocint{-\infty,x}$, $A'_x=A\cap\coint{x,\infty}$.
Then $x\in B$ iff either $x=\sup A_x$ or $x=\inf A'_x$, so

$$\eqalign{x\notin B
&\iff x\ne\sup A_x\text{ and }x\ne\inf A'_x\cr
&\iff\text{ there are }u<x,\,v>x\text{ such that }
  A_x\subseteq\ocint{-\infty,u}
  \text{ and }A'_x\subseteq\coint{v,\infty}\cr
&\iff\text{ there are }u,\,v\text{ such that }
  x\in\ooint{u,v}\subseteq X\setminus A\cr
&\iff x\notin\overline{A}.\cr}$$

\noindent Thus $B=\overline{A}$, as claimed.

\medskip

{\bf (e)} Because $X$ is totally ordered, all its subsets are both
upwards-directed and downwards-directed;  so we have only to join
the definition in 313Da to (d) above.

\medskip

{\bf (f)} If $x\in\ooint{u,v}$ then there is some $n\in\Bbb N$ such
that $x_n\ge u$, and now $x_i\in\ooint{u,v}$ for every $i\ge n$.

\medskip

{\bf (g)(i)} If $K$ is compact and $A\subseteq K$ is non-empty,
let $B$ be the set of upper bounds for $A$ in $X\cup\{-\infty,\infty\}$,
and set
$\Cal G=\{\ooint{-\infty,a}:a\in A\}\cup\{\ooint{b,\infty}:b\in B\}$.
Then no finite subfamily of $\Cal G$ can cover $K$;  and if
$c\in K\setminus\bigcup\Cal G$ then $c=\sup A$.   Similarly, any
non-empty subset of $K$ has an infimum in $X$ which belongs to $K$.

\medskip

\quad{\bf (ii)} Now suppose that $K$ satisfies the condition.   By (d)
above, it
is closed.   If it is empty it is certainly compact.   Otherwise,
$a_0=\inf K$ and $b_0=\sup K$ are defined in $X$ and belong to $K$.
Let $\Cal G$ be an open cover of $K$.   Set

\Centerline{$A=\{x:x\in X$,
$K\cap[a_0,x]$ is not covered by any finite
$\Cal G_0\subseteq\Cal G\}$.}

\noindent Note that $A$ is bounded below by $a_0$.   \Quer\ If
$b_0\in A$, then $c=\inf A$ is defined and belongs to $[a_0,b_0]$,
because $X$ is Dedekind complete.   If $c\notin K$ then
there are $u$, $v$ such that $c\in\ooint{u,v}\subseteq X\setminus K$;
if $c\in K$ then there are $u$, $v$ such that
$c\in\ooint{u,v}\subseteq G$ for some $G\in\Cal G$.   In either case,
$u\notin A$, so that
$K\cap\coint{a_0,v}\subseteq(K\cap[a_0,u])\cup\ooint{u,v}$
is covered by a finite subset of $\Cal G$, and $A$ does not meet
$\coint{a_0,v}$, that is, $A\subseteq\coint{v,\infty}$ and $v$ is a
lower bound of $A$.\ \BanG\   Thus $b_0\notin A$, and $K=K\cap[a_0,b_0]$
is covered by a finite subset of $\Cal G$.   As $\Cal G$ is arbitrary,
$K$ is compact.

\medskip

{\bf (h)} Use (g).

\medskip

{\bf (i)} Use (h).

\medskip

{\bf (j)} (Compare 2A2I.)
For $x$, $y\in G$ write $x\sim y$ if either $x\le y$ and
$[x,y]\subseteq G$ or $y\le x$ and $[y,x]\subseteq G$.   It is easy to
check that $\sim$ is an equivalence relation on $G$.   Let $\Cal C$ be
the set of equivalence classes under $\sim$.   Then $\Cal C$ is a
partition of $G$ into order-convex sets.
Now every $C\in\Cal C$ is open.   \Prf\ If $x\in C\in\Cal C$ then
there are $u$, $v\in X\cup\{-\infty,\infty\}$ such that
$x\in\ooint{u,v}\subseteq G$;  now $\ooint{u,v}\subseteq C$.\ \QeD\   So we
have our partition of $G$ into disjoint open order-convex sets.

If $X$ is Dedekind complete, then every member of $\Cal C$ is an open
interval.   \Prf\ Take $C\in\Cal C$.   Set

\Centerline{$A=\{u:u\in X\cup\{-\infty\},\,u<x$ for every $x\in C\}$,}

\Centerline{$B=\{v:v\in X\cup\{\infty\},\,x<v$ for every $x\in C\}$,}

\Centerline{$a=\sup A$,\quad$b=\inf B$;}

\noindent these are defined because $X$ is Dedekind complete.   If
$a<x<b$, there are $y$, $z\in C$ such that $y\le x\le z$, so that
$[y,x]\subseteq[y,z]\subseteq G$ and $y\sim x$ and $x\in C$;  thus
$\ooint{a,b}\subseteq C$.   If $x\in C$, there is an open interval
$\ooint{u,v}$ containing $x$ and included in $G$;  now
$\ooint{u,v}\subseteq C$, so $a\le u<x<v\le b$
and $x\in\ooint{a,b}$.   Thus $C=\ooint{a,b}$ is an open interval.\
\Qed

\medskip

{\bf (k)} Use (h).

\medskip

{\bf (l)} Write $W$ for $\{(x,y):x<y\}$.   If $x<y$, then either there
is a $z$ such that $x<z<y$, in which case
$\ooint{-\infty,z}\times\ooint{z,\infty}$ is an open set containing
$(x,y)$ and included in $W$, or $\ooint{x,y}=\emptyset$,
so $\ooint{-\infty,y}\times\ooint{x,\infty}$ is an open set containing
$(x,y)$ and included in $W$.   Thus $W$ is open.

Now $\{(x,y):x\le y\}=(X\times X)\setminus\{(x,y):y<x\}$ is closed.

\medskip

{\bf (m)} The subspace topology $\frak T_F$ on $F$ is generated by sets
of the form $F\cap\ooint{-\infty,x}$, $F\cap\ooint{x,\infty}$ where
$x\in X$ (4A2B(a-vi)), while the order topology $\frak S$ on $F$ is
generated by sets of the form $F\cap\ooint{-\infty,x}$,
$F\cap\ooint{x,\infty}$ where $x\in F$.   So
$\frak S\subseteq\frak T_F$.

Now suppose that one of the three conditions is satisfied, and that
$x\in X$.   If $x\in F$, or $F\cap\ooint{-\infty,x}$ is either $F$ or
$\emptyset$, then of course
$F\cap\ooint{-\infty,x}\in\frak S$.   Otherwise, $F$ meets both
$\ooint{-\infty,x}$ and $\coint{x,\infty}$ and does not contain $x$, so
is not order-convex.   In this case
$x'=\inf(F\cap\coint{x,\infty})$ is defined and belongs to $F$.   \Prf\
If $F$ is compact, this is covered by (g).   If $X$ is Dedekind complete
and $F$ is closed, then $x'$ is defined, and belongs to $F$ by (e).\
\QeD\   Now $F\cap\ooint{-\infty,x}=F\cap\ooint{-\infty,x'}\in\frak S$.

Similarly, $F\cap\ooint{x,\infty}\in\frak S$ for every $x\in X$.   But
this means that $\frak T_F\subseteq\frak S$, so the two topologies are
equal, as stated.

\medskip

{\bf (n)(i)} Let $\Cal G$ be a family of open subsets of $X$ with
union $H$.   Set

\Centerline{$\Cal A=\{A:A\subseteq\bigcup\Cal G_0$ for some countable
$\Cal G_0\subseteq\Cal G\}$.}

\noindent   (I seek to show that
$H\in\Cal A$.)   Of course $\bigcup\Cal A_0\in\Cal A$ for every
countable subset $\Cal A_0$ of $\Cal A$.

\medskip

\quad{\bf (ii)} Let $\Cal C$ be the family of order-convex members
of $\Cal A$, and $\Cal C^*$ the family
of maximal members of $\Cal C$.   If $C\in\Cal C$ is not included in any
member of $\Cal C^*$, then there is a $C'\in\Cal C$ such that
$C\subseteq C'$ and $\interior(C'\setminus C)\ne\emptyset$.   \Prf\
Since no member of $\Cal C$ including $C$ can be maximal, we can find
$C'\in\Cal C$ such that $C\subseteq C'$ and $\#(C'\setminus C)\ge 5$.
Because $C$ is order-convex, every point of $X\setminus C$ is either a
lower bound or an upper bound of $C$, and there must be three points
$x<y<z$ of $C'\setminus C$ on the same side of $C$.   In this case,

\Centerline{$y\in\ooint{x,z}
\subseteq\interior(C'\setminus C)$,}

\noindent so we have an appropriate $C'$.\ \Qed

\medskip

\quad{\bf (iii)} In fact, every member of $\Cal C$ is included in a
member of
$\Cal C^*$.   \Prf\Quer\ Suppose, if possible, otherwise.   Then we can
choose a strictly increasing family $\ofamily{\xi}{\omega_1}{C_{\xi}}$
in $\Cal C$  inductively, as follows.   Start from any non-empty
$C_0\in\Cal C$ not included in any member of $\Cal C^*$.   Given that
$C_0\subseteq C_{\xi}\in\Cal C$, then $C_{\xi}$ cannot be included in
any member of $\Cal C^*$, so by ($\beta$) above there is a
$C_{\xi+1}\in\Cal C$ such that
$\interior(C_{\xi+1}\setminus C_{\xi})$ is non-empty.   Given
$\ofamily{\eta}{\xi}{C_{\eta}}$ where $\xi<\omega_1$ is a non-zero
countable limit ordinal, set $C_{\xi}=\bigcup_{\eta<\xi}C_{\eta}$;  then
$C_{\xi}$ is order-convex, because $\{C_{\eta}:\eta<\xi\}$ is
upwards-directed, and belongs to $\Cal A$, because $\Cal A$ is closed
under countable unions, so $C_{\xi}\in\Cal C$ and the induction
proceeds.

Now, however,
$\ofamily{\xi}{\omega_1}{\interior(C_{\xi+1}\setminus C_{\xi})}$ is an
uncountable disjoint family of non-empty open sets, and $X$ is not ccc.\
\Bang\Qed

\medskip

\quad{\bf (iv)} Since $C\cup C'$ is order-convex whenever $C$,
$C'\in\Cal C$ and $C\cap C'\ne\emptyset$, $\Cal C^*$ is a disjoint
family.   Moreover, if $x\in H$, there is some open interval containing
$x$ and belonging to $\Cal C$, so $x\in\interior C$ for some
$C\in\Cal C^*$;  this shows that $\Cal C^*$ is an open cover of $H$.
Because $X$ is ccc, $\Cal C^*$ is countable, so
$H=\bigcup\Cal C^*\in\Cal A$.   Thus there is some countable
$\Cal G_0\subseteq\Cal G$ with union $H$;  as $\Cal G$ is arbitrary, $X$
is hereditarily Lindel\"of, by 4A2H(c-i).

\medskip

\quad{\bf (v)} By 4A2H(c-ii), $X$ is perfectly normal.

\medskip

{\bf (o)(i)} Suppose that $f$ is continuous.   If $A\subseteq X$
is a non-empty set with supremum $x$ in $X$, then $x\in\overline{A}$, by
(d), so $f(x)\in\overline{f[A]}$ (3A3Cc) and $f(x)$ is less than or
equal to any upper bound of $f[A]$;  but $f(x)$ is an upper bound of
$f[A]$, because $f$ is order-preserving, so $f(x)=\sup f[A]$.
Similarly, $f(\inf A)=\inf f[A]$ whenever $A$ is non-empty and has an
infimum, so $f$ is order-continuous.

\medskip

\quad{\bf (ii)} Now suppose that $f$ is order-continuous.   Take any
$y\in Y$ and consider $A=f^{-1}[\,\ooint{-\infty,y}\,]$,
$B=X\setminus A$.   If
$x\in A$ then $f(x)$ cannot be $\inf f[B]$ so $x$ cannot be $\inf B$ and
there is an $x'\in X$ such that $x<x'\le z$ for every $z\in B$;  in
which case
$x\in\ooint{-\infty,x'}\subseteq A$.   So $A$ is open.   Similarly,
$f^{-1}[\,\ooint{y,\infty}\,]$ is open.   By 4A2B(a-ii), $f$ is
continuous.
}%end of proof of 4A2R

\vleader{48pt}{4A2S}{Order topologies on ordinals (a)}
Let $\zeta$ be an ordinal with its order topology.

\quad(i) $\zeta$ is locally compact\cmmnt{ (4A2Rk)};  all the sets
$[0,\eta]=\ooint{-\infty,\eta+1}$, for $\eta<\zeta$, are open and
compact\cmmnt{ (4A2Rh)}.
If $\zeta$ is a successor ordinal, it is compact\cmmnt{, being of the
form $[0,\eta]$ where $\zeta=\eta+1$}.
%4@35

\quad(ii) For any $A\subseteq\zeta$,
$\overline{A}=\{\sup B:\emptyset\ne B\subseteq A,\,\sup B<\zeta\}$.
\prooflet{(4A2Rd, because $\inf B\in B\subseteq A$ for every non-empty
$B\subseteq A$.)}

\quad (iii) If $\xi\le\zeta$, then the subspace topology on
$\xi$ induced by the order topology of $\zeta$ is the order topology of
$\xi$.  \prooflet{(4A2Rm.)}

\spheader 4A2Sb Give $\omega_1$ its order topology.

\quad(i) $\omega_1$ is first-countable.   \prooflet{\Prf\ If
$\xi<\omega_1$ is either zero or a successor ordinal, then $\{\xi\}$ is
open so $\{\{\xi\}\}$ is a base of neighbourhoods of $\xi$.   If $\xi$
is a non-zero limit ordinal, there is a sequence $\sequencen{\xi_n}$ in
$\xi$ with supremum $\xi$, and $\{\ocint{\xi_n,\xi}:n\in\Bbb N\}$ is a
base of neighbourhoods of $\xi$.\ \Qed}
%4@25

\quad(ii) Singleton subsets of $\omega_1$ are zero sets.
\prooflet{(Assemble 4A2F(d-v), 4A2Rc and (i) above.)}
%4@19

\quad(iii) If $f:\omega_1\to\Bbb R$ is continuous, there is a
$\xi<\omega_1$ such
that $f(\eta)=f(\xi)$ for every $\eta\ge\xi$.
\prooflet{\Prf\Quer\ Otherwise, we may define a strictly increasing
family $\langle\zeta_{\xi}\rangle_{\xi<\omega_1}$ in $\omega_1$ by
saying that $\zeta_0=0$,

\Centerline{$\zeta_{\xi+1}=\min\{\eta:\eta>\zeta_{\xi},\,f(\eta)\ne
f(\zeta_{\xi})\}$}

\noindent for every $\xi<\omega_1$,

\Centerline{$\zeta_{\xi}=\sup\{\zeta_{\eta}:\eta<\xi\}$}

\noindent for non-zero countable limit ordinals $\xi$.   Now

\Centerline{$\omega_1
=\bigcup_{k\in\Bbb N}\{\xi:|f(\zeta_{\xi+1})-f(\zeta_{\xi})|
\ge 2^{-k}\}$,}

\noindent so there is a $k\in\Bbb N$ such that
$A=\{\xi:|f(\zeta_{\xi+1})-f(\zeta_{\xi})|\ge 2^{-k}\}$ is infinite.
Take a strictly increasing sequence $\langle\xi_n\rangle_{n\in\Bbb N}$
in $A$ and set
$\xi=\sup_{n\in\Bbb N}\xi_n=\sup_{n\in\Bbb N}(\xi_n+1)$.   Then
$\sequencen{\zeta_{\xi_n}}$ and
$\sequencen{\zeta_{\xi_n+1}}$ are strictly increasing sequences with
supremum $\zeta_{\xi}$, so both converge to $\zeta_{\xi}$ in the order
topology of $\omega_1$ (4A2Rf), and

\Centerline{$f(\zeta_{\xi})
=\lim_{n\to\infty}f(\zeta_{\xi_n})
=\lim_{n\to\infty}f(\zeta_{\xi_n+1})$.}

\noindent But this means that

\Centerline{$\lim_{n\to\infty}f(\zeta_{\xi_n})-f(\zeta_{\xi_n+1})=0$,}

\noindent which is impossible, because
$|f(\zeta_{\xi_n})-f(\zeta_{\xi_n+1})|\ge 2^{-k}$ for every
$n$.\ \Bang\Qed}%end of prooflet

\leader{4A2T}{Topologies on spaces of subsets}\cmmnt{ In \S\S446, 476 and
479 it will be useful to be able to discuss topologies on spaces of closed
sets.   In fact everything we really need can be expressed in terms of Fell
topologies ((a-ii) here), but it may help if I put these in the context of
two other constructions, Vietoris topologies and Hausdorff metrics (see (a)
and (g) below), which
may be more familiar to some readers.}    Let $X$ be a topological
space, and $\Cal C=\Cal C_X$ the family of closed subsets of $X$.

\spheader 4A2Ta(i) The {\bf Vietoris topology} on $\Cal C$ is the
topology generated by sets of the forms

\Centerline{$\{F:F\in\Cal C,\,F\cap G\ne\emptyset\}$,
\quad$\{F:F\in\Cal C,\,F\subseteq G\}$}

\noindent where $G\subseteq X$ is open.
%4@41Xr 4@76A 5@26I
%Engelking 2.7.20  Kuratowski I.17 ("exponential topology")

\quad(ii) The {\bf Fell topology} on
$\Cal C$ is the topology generated by sets of the forms

\Centerline{$\{F:F\in\Cal C,\,F\cap G\ne\emptyset\}$,
\quad$\{F:F\in\Cal C,\,F\cap K=\emptyset\}$}

\noindent where $G\subseteq X$ is open and $K\subseteq X$ is compact.
If $X$ is Hausdorff then the Fell topology is coarser than the
Vietoris topology.
If $X$ is compact and Hausdorff the two topologies agree.
%5@26I

\quad(iii) Suppose $X$ is metrizable, and
that $\rho$ is a metric on $X$ inducing its
topology.   For a non-empty subset $A$ of $X$, write
$\rho(x,A)=\inf_{y\in A}\rho(x,y)$ for every $x\in X$.
Note that\cmmnt{ $\rho(x,A)\le\rho(x,y)+\rho(y,A)$ for all $x$,
$y\in X$, so that} $x\mapsto\rho(x,A):X\to\Bbb R$ is $1$-Lipschitz.

For $E$, $F\in\Cal C\setminus\{\emptyset\}$, set

\Centerline{$\tilde\rho(E,F)
=\min(1,\max(\sup_{x\in E}\rho(x,F),\sup_{y\in F}\rho(y,E)))$.}

\noindent\cmmnt{If $E$, $F\in\Cal C\setminus\{\emptyset\}$ and $z\in X$, then
$\rho(z,F)\le\rho(z,E)+\sup_{x\in E}\rho(x,F)$;  from this it
is easy to
see that }$\tilde\rho$ is a metric on $\Cal C\setminus\{\emptyset\}$,
the {\bf Hausdorff metric}.   \cmmnt{Observe that}
$\tilde\rho(\{x\},\{y\})=\min(1,\rho(x,y))$ for all $x$, $y\in X$.
%4@76A

\cmmnt{\medskip

\noindent{\bf Remarks} The formula I give for $\tilde\rho$ has a somewhat
arbitrary feature `$\min(1,\ldots)$'.
Any number strictly greater than $0$ can be used in place of `$1$' here.
Many authors prefer to limit themselves to the family of non-empty closed
sets of finite diameter, rather than the whole of
$\Cal C\setminus\{\emptyset\}$;
this makes it more more natural to omit the truncation, and work with
$(E,F)\mapsto\max(\sup_{x\in E}\rho(x,F),\sup_{y\in F}\rho(y,E))$.
All such variations produce uniformly equivalent metrics.
A more radical approach redefines `metric' to allow functions which take
the value $\infty$;  but this seems a step too far.

Given that I am truncating my Hausdorff metrics by the value $1$,
there would be no extra problems if I defined $\tilde\rho(\emptyset,E)=1$
for every non-empty closed set $E$;
but I think I am following the majority in regarding Hausdorff distance
as defined only for non-empty sets.
}%end of comment

\spheader 4A2Tb(i) The Fell topology is T$_1$.
\prooflet{\Prf\ If $F\subseteq X$ is closed and $x\in X$, then
$\{E:E\in\Cal C$, $E\cap(X\setminus F)=\emptyset\}$ and
$\{E:E\in\Cal C$, $E\cap\{x\}\ne\emptyset\}$ are complements of open sets,
so are closed.   Now if $F\in\Cal C$ then

\Centerline{$\{F\}=\{E:E\subseteq F\}\cap\bigcap_{x\in F}\{E:x\in E\}$}

\noindent is closed.\ \Qed}

\quad(ii) The map $(E,F)\mapsto E\cup F:\Cal C\times\Cal C\to\Cal C$
is continuous for the Fell topology.
%5@26I
\prooflet{\Prf\ If $G\subseteq X$ is open and $K\subseteq X$ is compact,
then

\Centerline{$\{(E,F):(E\cup F)\cap G\ne\emptyset\}
=\{(E,F):E\cap G\ne\emptyset$ or $F\cap G\ne\emptyset\}$,}

\Centerline{$\{(E,F):(E\cup F)\cap K=\emptyset\}
=\{E:E\cap K=\emptyset\}\times\{F:F\cap K=\emptyset\}$}

\noindent are open in the product topology.   So the result follows by
4A2B(a-ii).\ \Qed}

%Similarly, $(E,F)\mapsto E\cup F:\Cal C\times\Cal C\to\Cal C$
%is continuous for the Vietoris topology.}

\quad(iii) $\Cal C$ is compact in the Fell topology.
\prooflet{\Prf\ Let $\frak F$ be an ultrafilter on $\Cal C$.
Set $F_0=\bigcap_{\Cal E\in\frak F}\overline{\bigcup\Cal E}$.
($\alpha$) If $G\subseteq X$ is open and $G\cap F_0\ne\emptyset$, set
$\Cal E=\{F:F\in\Cal C,\,F\cap G=\emptyset\}$.   Then
$\overline{\bigcup\Cal E}$ does not meet $G$, so does not include $F_0$,
and $\Cal E\notin\frak F$.   Accordingly
$\{F:F\cap G\ne\emptyset\}=\Cal C\setminus\Cal E$ belongs to $\frak F$.
($\beta$) If $K\subseteq X$ is compact and $F_0\cap K=\emptyset$, then
$\{K\cap\overline{\bigcup\Cal E}:\Cal E\in\frak F\}$ is a
downwards-directed family of relatively closed subsets of $K$ with empty
intersection so must contain the empty set, and there is an
$\Cal E\in\frak F$ such that $K\cap F=\emptyset$ for every $F\in\Cal E$,
that is, $\{F:F\cap K=\emptyset\}$ belongs to $\frak F$.   ($\gamma$) By
4A2B(a-iv), $\frak F\to F_0$.   As $\frak F$ is arbitrary, $\Cal C$ is
compact.\ \Qed}
%4@76F 4@76G

\spheader 4A2Tc If $X$ is Hausdorff,
$x\mapsto\{x\}$ is continuous for the Fell topology on $\Cal C$.
\prooflet{\Prf\ If $G\subseteq X$ is open, then
$\{x:\{x\}\cap G\ne\emptyset\}=G$ is open.   If $K\subseteq X$ is compact,
then $\{x:\{x\}\cap K=\emptyset\}=X\setminus K$ is open (3A3Dc).\ \Qed}

\spheader 4A2Td If $X$ and another topological space $Y$ are regular,
and $\Cal C_Y$, $\Cal C_{X\times Y}$ are the families of closed subsets
of $Y$ and $X\times Y$ respectively, then
$(E,F)\mapsto E\times F:\Cal C_X\times\Cal C_Y\to\Cal C_{X\times Y}$ is
continuous when each space is given its Fell topology.
%4@46G 4@76G 4@76K
\prooflet{\Prf\
(i) Suppose that $W\subseteq X\times Y$ is open, and consider $\Cal V_W
=\{(E,F):E\in\Cal C_X$, $F\in\Cal C_Y$, $(E\times F)\cap W\ne\emptyset\}$.
If $(E_0,F_0)\in\Cal V_W$, take $(x_0,y_0)\in (E_0\times F_0)\cap W$.   Let
$G\subseteq X$ and $H\subseteq Y$ be open sets such that
$(x_0,y_0)\in G\times H\subseteq W$.   Then $\{(E,F):E\in\Cal C_X$,
$F\in\Cal C_Y$, $E\cap G\ne\emptyset$, $F\cap H\ne\emptyset\}$ is an open
set in $\Cal C_X\times\Cal C_Y$ containing $(E_0,F_0)$ and included in
$\Cal V_W$.   As $(E_0,F_0)$ is arbitrary, $\Cal V_W$ is open in
$\Cal C_X\times\Cal C_Y$.   (ii) Suppose that $K\subseteq X\times Y$ is
compact, and consider $\Cal W_K
=\{(E,F):E\in\Cal C_X$, $F\in\Cal C_Y$, $(E\times F)\cap K=\emptyset\}$.
If $(E_0,F_0)\in\Cal W_K$, then set $K_1=K\cap(E_0\times Y)$ and
$K_2=K\cap(X\times F_0)$.   These are disjoint closed subsets of $K$, so
there are disjoint open subsets $U_1$, $U_2$ of $X\times Y$ including
$K_1$, $K_2$ respectively (4A2F(h-ii)).   Now $K'_1=K\setminus U_1$ and
$K'_2=K\setminus U_2$ are compact subsets of $K$ with union $K$.

Let $\pi_1$, $\pi_2$ be the projections from $X\times Y$
onto $X$, $Y$ respectively;  then
$\{(E,F):E\in\Cal C_X$, $E\cap\pi_1[K'_1]=F\cap\pi_2[K'_2]=\emptyset\}$
is an open set in $\Cal C_X\times\Cal C_Y$ containing $(E_0,F_0)$ and
included in $\Cal W_K$.   As $(E_0,F_0)$ is arbitrary, $\Cal W_K$ is open.
(iii) Putting these together with 4A2B(a-ii), we see that
$(E,F)\mapsto E\times F$ is continuous.\ \Qed}

\spheader 4A2Te Suppose that $X$ is locally compact and Hausdorff.

\quad(i) The
set $\{(E,F):E$, $F\in\Cal C$, $E\subseteq F\}$ is closed in
$\Cal C\times\Cal C$ for the product topology defined from the Fell
topology on $\Cal C$.
\prooflet{\Prf\ Suppose that $E_0$, $F_0\in\Cal C$ and
$E_0\not\subseteq F_0$.   Take $x\in E_0\setminus F_0$.
Because $X$ is locally compact and Hausdorff,
there is a relatively compact open set $G$ such that $x\in G$ and
$\overline{G}\cap F_0=\emptyset$.   Now
$\Cal V=\{E:E\cap G\ne\emptyset\}$ and
$\Cal W=\{F:F\cap\overline{G}=\emptyset\}$ are open sets in $\Cal C$
containing $E_0$, $F_0$ respectively, and $E\not\subseteq F$ for every
$E\in\Cal V$ and $F\in\Cal W$.   This shows that
$\{(E,F):E\not\subseteq F\}$ is open, so that its complement is closed.\
\Qed}
%4@46G
\cmmnt{

It follows that }$\{(x,F):x\in F\}\cmmnt{=\{(x,F):\{x\}\subseteq F\}}$ is
closed in $X\times\Cal C$ when $\Cal C$ is given its
Fell topology\cmmnt{, since $x\mapsto\{x\}$ is continuous,
by (c) above}.
%4@46

\quad(ii) The Fell topology on $\Cal C$ is Hausdorff.
%4@95N 4@95O
\prooflet{\Prf\ The set

\Centerline{$\{(E,E):E\in\Cal C\}
=\{(E,F):E\subseteq F$ and $F\subseteq E\}$}

\noindent is closed in $\Cal C\times\Cal C$, by (i).   So 4A2F(a-iii)
applies.\ \Qed}\cmmnt{

}  It follows that if $\familyiI{F_i}$ is a family in $\Cal C$,
and $\Cal F$
is an ultrafilter on $I$, then we have a well-defined limit
$\lim_{i\to\Cal F}F_i$ defined in $\Cal C$ for the Fell
topology\cmmnt{, because $\Cal C$ is compact ((b-iii) above)}.
%4@46G 4@46O

\quad(iii) If $\Cal L\subseteq\Cal C$ is compact, then $\bigcup\Cal L$ is a
closed subset of $X$.   \prooflet{\Prf\ Take
$x\in X\setminus\bigcup\Cal L$.   For every $C\in\Cal L$, there is a
relatively compact open set $G$ containing $x$ such that
$C\cap\overline{G}$ is empty;  now finitely many such open sets $G$ must
suffice for every $C\in\Cal L$, and the intersection of these $G$ is a
neighbourhood of $x$ not meeting $\bigcup\Cal L$.\ \QeD\   (Compare
4A2Gm.)}


\spheader 4A2Tf Suppose that $X$ is metrizable, locally compact and
separable.   Then the Fell topology on $\Cal C$ is metrizable.
%5@26I
\prooflet{\Prf\ $X$ is second-countable (4A2P(a-i));
let $\Cal U$ be a countable base for the topology of $X$ consisting of
relatively compact open sets (4A2Ob) and closed under finite unions.
Let $\Bbb V$ be the set of open sets in $\Cal C$ expressible in the form

\Centerline{$\{F:F\cap U\ne\emptyset$ for every $U\in\Cal U_0$,
$F\cap\overline{V}=\emptyset\}$}

\noindent where $\Cal U_0\subseteq\Cal U$ is finite and $V\in\Cal U$.
Then $\Bbb V$ is countable.   If $\Cal V\subseteq\Cal C$ is open and
$F_0\in\Cal V$, then there are a finite family $\Cal G$ of open sets in $X$
and a compact $K\subseteq X$ such that

\Centerline{$F_0\in\{F:F\cap G\ne\emptyset$ for every $G\in\Cal G$,
$F\cap K=\emptyset\}\subseteq\Cal V$.}

\noindent For each $G\in\Cal G$ there is a $U_G\in\Cal U$ such that
$U_G\subseteq G$ and $F\cap U_G\ne\emptyset$.   Next, each point of $K$
belongs to a member of $\Cal U$ with closure disjoint from $F_0$, so
(because $K$ is compact and
$\Cal U$ is closed under finite unions) there is a $V\in\Cal U$
such that $K\subseteq V$ and $F_0\cap\overline{V}=\emptyset$.   Now

\Centerline{$\Cal V'=\{F:F\cap U_G\ne\emptyset$ for every $G\in\Cal G$,
$F\cap\overline{V}=\emptyset\}$}

\noindent belongs to $\Bbb V$, contains $F_0$ and is included in $\Cal V$.
This shows that $\Bbb V$ is a base for the Fell topology, and the Fell
topology is second-countable.   Since we already know that it is compact
and Hausdorff, therefore regular, it is metrizable (4A2Pb).\ \Qed}

\spheader 4A2Tg Suppose that $X$ is metrizable, and
that $\rho$ is a metric
inducing the topology of $X$;  let $\tilde\rho$ be the corresponding
Hausdorff metric on $\Cal C\setminus\{\emptyset\}$.

\quad(i) The topology $\frak S_{\tilde\rho}$ defined by
$\tilde\rho$ is finer than the Fell topology $\frak S_F$
on $\Cal C\setminus\{\emptyset\}$.
\prooflet{\Prf\ Let $G\subseteq X$ be open, and consider the set
$\Cal V_G=\{F:F\in\Cal C$, $F\cap G\ne\emptyset\}$.   If $E\in\Cal V_G$,
take $x\in E\cap G$ and $\epsilon>0$ such that $U(x,\epsilon)\subseteq G$;  then
$\{F:\tilde\rho(F,E)<\epsilon\}\subseteq\Cal V_G$.   As $E$ is arbitrary,
$\Cal V_G$ is $\frak S_{\tilde\rho}$-open.   Next, suppose that
$K\subseteq X$ is compact, and consider the set
$\Cal W_K=\{F:F\in\Cal C\setminus\{\emptyset\}$, $F\cap K=\emptyset\}$.
If $E\in\Cal W_K$, the function $x\mapsto\rho(x,E):K\to\ooint{0,\infty}$
is continuous, so has a non-zero lower bound $\epsilon$ say;  now
$\{F:\tilde\rho(F,E)<\epsilon\}\subseteq\Cal W_K$.   As $E$ is arbitrary,
$\Cal W_K$ is $\frak S_{\tilde\rho}$-open.   So $\frak S_{\tilde\rho}$ is
finer than the topology $\frak S_F$ generated by the sets $\Cal V_G$ and
$\Cal W_K$.\ \Qed}

\quad(ii) If $X$ is compact, then $\frak S_{\tilde\rho}$ and
$\frak S_F$ are the same, and both are compact.
\prooflet{\Prf\ Suppose that $E\in\Cal V\in\frak S_{\tilde\rho}$.
Let $\epsilon\in\ooint{0,1}$ be such that
$F\in\Cal V$ whenever $F\in\Cal C\setminus\{\emptyset\}$ and
$\tilde\rho(E,F)<2\epsilon$.
Because $X$ is compact, $E$ is $\rho$-totally bounded (4A2Je) and there is
a finite set $I\subseteq E$ such that
$E\subseteq\bigcup_{x\in I}U(x,\epsilon)$.   Because $x\mapsto\rho(x,E)$ is
continuous, $K=\{x:\rho(x,E)\ge\epsilon\}$ is closed, therefore compact;
now

\Centerline{$\Cal W=\{F:F\in\Cal C$, $F\cap K=\emptyset$,
$F\cap U(x,\epsilon)\ne\emptyset$ for every $x\in I\}$}

\noindent is a neighbourhood of $E$ for $\frak S_F$ included in
$\Cal V$.   Thus $\Cal V$ is a neighbourhood of $E$ for
$\frak S_F$;  as $E$ and $\Cal V$ are arbitrary, $\frak S_F$ is finer than
$\frak S_{\tilde\rho}$.   So the two topologies are equal.

Observe finally that $\{\emptyset\}=\{F:F\in\Cal C$, $F\cap X=\emptyset\}$
is open for the Fell topology on $\Cal C$, so
$\Cal C\setminus\{\emptyset\}$ is closed, therefore compact, by (b-iii).
So $\frak S_{\tilde\rho}=\frak S_F$ is compact.\ \Qed}

\leader{4A2U}{Old friends (a)}
$\Bbb R$, with its usual topology, is metrizable\cmmnt{ (2A3Ff)} and
separable\cmmnt{ (the countable set $\Bbb Q$ is dense)}, so is
second-countable\cmmnt{ (4A2P(a-i))}.
Every subset of $\Bbb R$ is separable\cmmnt{ (4A2P(a-iv))};
\cmmnt{in particular,}
every dense subset of $\Bbb R$ has a countable subset which is still
dense.
%4@18 %4@33 %4@25 %4@39

\spheader 4A2Ub $\BbbN^{\Bbb N}$ is Polish in its usual
topology\cmmnt{ (4A2Qc)},
%4@23
so has a countable network\cmmnt{ (4A2P(a-iii) or 4A2Ne)},
%4@A3
and is hereditarily Lindel\"of\cmmnt{ (4A2Nb or 4A2Pd)}.
%4@22
Moreover, it is homeomorphic to
$[0,1]\setminus\Bbb Q$ 
and $\Bbb R\setminus\Bbb Q$\dvAnew{2023}\prooflet{ 
(\Kuratowski, \S36.II;  \Kechris, 7.7)}.
%4@39

\spheader 4A2Uc The map
$x\mapsto\bover23\sum_{j=0}^{\infty}3^{-j}x(j)$\cmmnt{ (cf.\ 134Gb)}
is a homeomorphism between $\{0,1\}^{\Bbb N}$ and the Cantor set
$C\subseteq[0,1]$.   \prooflet{(It is a continuous bijection.)}
%4@39

\spheader 4A2Ud If $I$ is any set, then the map
$A\mapsto\chi A:\Cal PI\to\{0,1\}^I$ is a homeomorphism\cmmnt{ (for
the usual topologies on $\Cal PI$ and $\{0,1\}^I$, as described in 4A2A
and 3A3K)}.   So $\Cal PI$ is zero-dimensional, compact\cmmnt{ (3A3K)}
and Hausdorff.   If $I$ is countable, then $\Cal PI$ is metrizable,
therefore Polish\cmmnt{ (4A2Qb)}.

\spheader 4A2Ue Give the space $C(\coint{0,\infty})$ the topology
$\frak T_c$ of uniform convergence on compact sets.

\medskip

\quad{\bf (i)} $C(\coint{0,\infty})$ is a Polish locally convex
linear topological space.   \prooflet{\Prf\ $\frak T_c$ is determined by
the seminorms $f\mapsto\sup_{t\le n}|f(t)|$ for $n\in\Bbb N$, so it is a
metrizable linear space topology.   By 4A2Oe, it has a countable network,
so is separable.   Any function which is continuous on
every set $[0,n]$ is continuous on $\coint{0,\infty}$, so
$C(\coint{0,\infty})$ is complete under the metric
$(f,g)\mapsto\sum_{n=0}^{\infty}\min(2^{-n},\sup_{t\le n}|f(t)-g(t)|)$;  as
this metric defines $\frak T_c$, $\frak T_c$ is Polish.\ \Qed}

\medskip

\quad{\bf (ii)} Suppose that $A\subseteq C(\coint{0,\infty})$ is such that
$\{f(0):f\in A\}$ is bounded and for every $a\ge 0$
and $\epsilon>0$ there is
a $\delta>0$ such that $|f(s)-f(t)|\le\epsilon$ whenever
$f\in A$, $s$, $t\in[0,a]$
and $|s-t|\le\delta$.   Then $A$ is relatively compact for $\frak T_c$.
\prooflet{
\Prf\ Note first that $\{f(a):f\in A\}$ is bounded for every $a\ge 0$,
since if $\delta>0$ is such that $|f(s)-f(t)|\le 1$ whenever $f\in A$,
$s$, $t\in[0,a]$ and $|s-t|\le\delta$, then
$|f(a)|\le|f(0)|+\lceil\Bover{a}{\delta}\rceil$ for every $f\in A$.   So if
$\Cal F$ is an ultrafilter on $C(\coint{0,\infty})$ containing $A$,
$g(a)=\lim_{f\to\Cal F}f(a)$ is defined for every $a\ge 0$.   If $a\ge 0$
and $\epsilon>0$, let $\delta\in\ocint{0,1}$ be such that
$|f(s)-f(t)|\le\epsilon$ whenever $f\in A$,
$s$, $t\in[0,a+1]$ and $|s-t|\le\delta$;
then $|g(s)-g(a)|\le\epsilon$ whenever $|s-a|\le\delta$;  as $a$ and
$\epsilon$ are arbitrary, $g\in C(\coint{0,\infty})$.   If $a\ge 0$ and
$\epsilon>0$, let $\delta>0$ be such that
$|f(s)-f(t)|\le\bover13\epsilon$ whenever $f\in A$, $s$, $t\in[0,a]$
and $|s-t|\le\delta$.   Then

\Centerline{$A'
=\{f:f\in A$, $|f(i\delta)-g(i\delta)|\le\Bover13\epsilon$ for every
$i\le\lceil\Bover{a}{\delta}\rceil\}$}

\noindent belongs to $\Cal F$.   Now $|f(t)-g(t)|\le\epsilon$ for every
$f\in A'$ and $t\in[0,a]$.   As $a$ and $\epsilon$ are arbitrary,
$\Cal F\to g$ for $\frak T_c$.   As $\Cal F$ is arbitrary, $A$ is
relatively compact (3A3De).\ \Qed}

\discrpage

\frfilename{mt4a3.tex}
\versiondate{7.1.17}
\copyrightdate{2007}

\def\chaptername{Appendix}
\def\sectionname{Topological $\sigma$-algebras}

\def\tildeClll{\hbox{$\tilde C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.7pt\vrule height6.3pt width.5pt depth0pt
  \hskip.7pt\vrule height6.3pt width.5pt depth0pt
  \hskip-2.5pt\lower5pt\hbox{\char'23}}}}

\newsection{4A3}

I devote a section to some $\sigma$-algebras which can be defined on
topological spaces.   While `measures' will not be mentioned here, the
manipulation of these $\sigma$-algebras is an essential part of the
technique of measure theory, and I will give proofs and exercises as if
this were part of the main work.   I look at Borel $\sigma$-algebras
(4A3A-4A3J), % 4A3A 4A3B 4A3C 4A3D 4A3E 4A3F 4A3G 4A3H 4A3I 4A3J
Baire $\sigma$-algebras (4A3K-4A3P), % 4A3K 4A3L 4A3M 4A3N 4A3O 4A3P
spaces of \cadlag\ functions (4A3Q),
Baire-property algebras (4A3R, 4A3S) and cylindrical $\sigma$-algebras
on linear spaces (4A3U-4A3W). % 4A3U 4A3V 4A3W

\leader{4A3A}{Borel sets} If $(X,\frak T)$ is a topological space, the
{\bf Borel $\sigma$-algebra} of $X$ is the $\sigma$-algebra $\Cal B(X)$
of subsets
of $X$ generated by $\frak T$.   Its elements are the {\bf Borel sets}
of $X$.   If $(Y,\frak S)$ is another topological space with Borel
$\sigma$-algebra $\Cal B(Y)$, a function $f:X\to Y$ is
{\bf Borel measurable} if
$f^{-1}[H]\in\Cal B(X)$ for every $H\in\frak S$, and is a
{\bf Borel isomorphism} if it is a bijection and
$\Cal B(Y)=\{F:F\subseteq Y,\,f^{-1}[F]\in\Cal B(X)\}$\cmmnt{, that
is, $f$ is an isomorphism between the structures $(X,\Cal B(X))$ and
$(Y,\Cal B(Y))$}.
%4@11 %4@33
%Engelking 1.3

\leader{4A3B}{$(\Sigma,\text{\rm{T}})$-measurable
functions }\cmmnt{It is time I put the following idea into bold type.

\medskip

}{\bf (a)} Let $X$ and $Y$ be sets, with $\sigma$-algebras
$\Sigma\subseteq\Cal PX$ and
$\Tau\subseteq\Cal PY$.   A function $f:X\to Y$ is
$(\Sigma,\Tau)${\bf-measurable} if $f^{-1}[F]\in\Sigma$ for every
$F\in\Tau$.

\spheader 4A3Bb If $\Sigma$, $\Tau$ and $\Upsilon$ are $\sigma$-algebras
of subsets of $X$, $Y$ and $Z$ respectively, and $f:X\to Y$ is
$(\Sigma,\Tau)$-measurable while $g:Y\to Z$ is
$(\Tau,\Upsilon)$-measurable, then $gf:X\to Z$ is
$(\Sigma,\Upsilon)$-measurable.  \prooflet{ (If $H\in\Upsilon$,
$g^{-1}[H]\in\Tau$ so $(gf)^{-1}[H]=f^{-1}[g^{-1}[H]]\in\Sigma$.)}

\spheader 4A3Bc Let $\familyiI{X_i}$ be a family of sets with product
$X$, $Y$ another set, and $f:X\to Y$ a function.   If
$\Tau\subseteq\Cal PY$, $\Sigma_i\subseteq\Cal PX_i$ are
$\sigma$-algebras for each $i$, then $f$ is
$(\Tau,\Tensorhat_{i\in I}\Sigma_i)$-measurable iff $\pi_if:Y\to X_i$ is
$(\Tau,\Sigma_i)$-measurable for every $i$, where $\pi_i:X\to X_i$ is
the coordinate map.   \prooflet{\Prf\ $\pi_i$ is
$(\Tensorhat_{j\in I}\Sigma_j,\Sigma_i)$-measurable, so if $f$ is
$(\Tau,\Tensorhat_{j\in I}\Sigma_j)$-measurable then $\pi_if$ must be
$(\Tau,\Sigma_i)$-measurable.   In the other direction, if every
$\pi_if$ is measurable, then $\{H:H\subseteq X$, $f^{-1}[E]\in\Tau\}$ is
a $\sigma$-algebra of subsets of $X$ containing $\pi_i^{-1}[E]$ whenever
$i\in I$ and $E\in\Sigma_i$, so includes $\Tensorhat_{i\in I}\Sigma_i$,
and $f$ is measurable.\ \Qed}

\leader{4A3C}{Elementary facts (a)} If $X$ is a topological
space and $Y$ is a subspace of $X$, then $\Cal B(Y)$ is just the
subspace $\sigma$-algebra $\{E\cap Y:E\in\Cal B(X)\}$.
\prooflet{\Prf\ $\{E:E\subseteq X,\,E\cap Y\in\Cal B(Y)\}$ and
$\{E\cap Y:E\in\Cal B(X)\}$
are $\sigma$-algebras containing all open sets, so include
$\Cal B(X)$, $\Cal B(Y)$ respectively.\ \Qed}
%4@12 %4@27 %4@33 %4@36

\spheader 4A3Cb If $X$ is a set, $\Sigma$ is a $\sigma$-algebra of
subsets of $X$, $(Y,\frak S)$ is a topological space and $f:X\to Y$ is a
function, then $f$ is $(\Sigma,\Cal B(Y))$-measurable iff
$f^{-1}[H]\in\Sigma$ for every $H\in\frak S$.   \prooflet{\Prf\ If $f$
is $(\Sigma,\Cal B(Y))$-measurable then $f^{-1}[H]\in\Sigma$ for every
$H\in\frak S$ just because $\frak S\subseteq\Cal B(Y)$.   If
$f^{-1}[H]\in\Sigma$ for every $H\in\frak S$, then
$\{F:F\subseteq Y,\,f^{-1}[F]\in\Sigma\}$ is a
$\sigma$-algebra of subsets of $Y$ (111Xc;  cf.\ 234C)
including $\frak S$, so
includes $\Cal B(Y)$, and $f$ is $(\Sigma,\Cal B(Y))$-measurable.\ \Qed}

\spheader 4A3Cc If $X$ and $Y$ are topological spaces, and $f:X\to Y$ is
a function, then $f$ is Borel measurable iff it is
$(\Cal B(X),\Cal B(Y))$-measurable.
\prooflet{(Apply (b) with $\Sigma=\Cal B(X)$.)}
So if $X$, $Y$ and $Z$ are topological spaces and $f:X\to Y$,
$g:Y\to Z$ are Borel measurable functions, then $gf:X\to Z$ is Borel
measurable.
\prooflet{(Use 4A3Bb.)}
%4@39

\spheader 4A3Cd If $X$ and $Y$ are topological spaces and $f:X\to Y$
is continuous, it is Borel measurable.   \prooflet{(Immediate from the
definitions in 4A3A.)}
%4@13, %4@16, %4@17, %4@21

\spheader 4A3Ce If $X$ is a topological space and
$f:X\to[-\infty,\infty]$ is lower semi-continuous, then it is Borel
measurable.
\prooflet{(The inverse image of a half-open interval
$\ocint{\alpha,\beta}$ is a difference of open sets, so is a Borel set,
and every open subset of $[-\infty,\infty]$ is a countable union of such
half-open intervals.)}
%4@45

\spheader 4A3Cf If $\familyiI{X_i}$ is a family of topological spaces
with product $X$, then
$\Cal B(X)\supseteq\Tensorhat_{i\in I}\Cal B(X_i)$.
\prooflet{(Put (d) and 4A3Bc together.)}
%4@25

\spheader 4A3Cg Let $X$ be a topological space.

\medskip

\quad{\bf (i)} The algebra $\frak A$ of
subsets generated by the open sets is precisely the family of sets
expressible as a disjoint union $\bigcup_{i\le n}G_i\cap F_i$ where
every $G_i$ is open and every $F_i$ is closed.
\prooflet{\Prf\ Write $\Cal A$ for the family of sets expressible in
this form.   Of course $\Cal A\subseteq\frak A$.   In the other
direction, observe that

\inset{$X\in\Cal A$,

if $E$, $E'\in\Cal A$ then $E\cap E'\in\Cal A$,

if $E\in\Cal A$ then $X\setminus E\in\Cal A$}

\noindent because if $G_i$ is open and $F_i$ is closed for $i\le n$,
then (identifying $\{0,\ldots,n\}$ with $n+1$)

$$\eqalign{X\setminus\bigcup_{i\le n}(G_i\cap F_i)
&=\bigcap_{i\le n}(X\setminus G_i)\cup(G_i\setminus F_i)\cr
&=\bigcup_{I\subseteq n+1}\bigl(\bigcap_{i\in I}(G_i\setminus F_i)
  \cap\bigcap_{i\in(n+1)\setminus I}(X\setminus G_i)\bigr)\cr}$$

\noindent belongs to $\Cal A$.   So $\Cal A$ is an algebra of sets and
must be equal to $\frak A$.\ \Qed}

\medskip

\quad{\bf (ii)} $\Cal B(X)$ is the smallest
family $\Cal E\supseteq\frak A$ such that
$\bigcup_{n\in\Bbb N}E_n\in\Cal E$ for every non-decreasing sequence
$\sequencen{E_n}$ in
$\Cal E$ and $\bigcap_{n\in\Bbb N}E_n\in\Cal E$ for every non-increasing
sequence $\sequencen{E_n}$ in $\Cal E$.
\prooflet{(136G.)}
%4@44

\leader{4A3D}{Hereditarily Lindel\"of spaces (a)} Suppose that $X$ is a
hereditarily Lindel\"of space and $\Cal U$ is a subbase for the topology
of $X$.   Then $\Cal B(X)$ is the $\sigma$-algebra of subsets of $X$
generated by $\Cal U$.   \prooflet{\Prf\ Write $\Sigma$ for the
$\sigma$-algebra generated by $\Cal U$.   Of course
$\Sigma\subseteq\Cal B(X)$ just because every member of $\Cal U$ is
open.   In the other direction, set

\Centerline{$\Cal V
=\{X\}\cup\{U_0\cap U_1\cap\ldots\cap U_n:U_0,\ldots,U_n\in\Cal U\}$;}

\noindent then $\Cal V\subseteq\Sigma$ and $\Cal V$ is a base for the
topology of $X$ (4A2B(a-i)).   If $G\subseteq X$ is open, set
$\Cal V_1=\{V:V\in\Cal V$, $V\subseteq G\}$;  then $G=\bigcup\Cal V_1$.
Because $X$ is hereditarily Lindel\"of, there is a countable set
$\Cal V_0\subseteq\Cal V_1$ such that $G=\bigcup\Cal V_0$ (4A2H(c-i)),
so that $G\in\Sigma$.   Thus every open set belongs to $\Sigma$ and
$\Cal B(X)\subseteq\Sigma$.\ \Qed}

\spheader 4A3Db Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets
of $X$, $Y$ a hereditarily Lindel\"of space, $\Cal U$ a subbase for the
topology of $Y$, and $f:X\to Y$ a function.   If $f^{-1}[U]\in\Sigma$
for every $U\in\Cal U$, then $f$ is $(\Sigma,\Cal B(Y))$-measurable.
\prooflet{\Prf\ $\{F:F\subseteq Y$, $f^{-1}[F]\in\Sigma\}$ is a
$\sigma$-algebra of subsets of $Y$ including $\Cal U$, so contains every
open set, by (a), and therefore every Borel set, as in 4A3Cb.\ \Qed}

\spheader 4A3Dc Let $\familyiI{X_i}$ be a family of topological spaces
with product $X$.   Suppose that $X$ is hereditarily Lindel\"of.

\medskip

\quad{\bf (i)} $\Cal B(X)=\Tensorhat_{i\in I}\Cal B(X_i)$.   \prooflet{\Prf\
By 4A3Cf, $\Cal B(X)\supseteq\Tensorhat_{i\in I}\Cal B(X_i)$.   On the
other hand, $\Tensorhat_{i\in I}\Cal B(X_i)$ is a $\sigma$-algebra
including

\Centerline{$\Cal U
=\{\pi_i^{-1}[G]:i\in I$, $G\subseteq X_i$ is open$\}$,}

\noindent where $\pi_i(x)=x(i)$ for $i\in I$ and $x\in X$;  since
$\Cal U$ is a subbase for the topology of $X$, (a) tells us that
$\Tensorhat_{i\in I}\Cal B(X_i)$ includes $\Cal B(X)$.\ \Qed}

\medskip

\quad{\bf (ii)} If $Y$ is another topological space, then a function
$f:Y\to X$ is Borel measurable iff $\pi_if:Y\to X_i$ is Borel measurable
for every $i\in I$, where $\pi_i:X\to X_i$ is the canonical map.
\prooflet{(Use 4A3Bc.)}
%4@A2O

\cmmnt{
\leader{4A3E}{Applications} Recall that any topological space with a
countable network (in particular, any second-countable space, any
separable metrizable space) is hereditarily
Lindel\"of (4A2Nb, 4A2Oc, 4A2P(a-iii));  and so is any ccc
totally ordered space (4A2Rn).   So 4A3Da-4A3Db will be
applicable to these.   As for product
spaces, the product of a countable family of spaces with countable
networks again has a countable network (4A2Ne), so for such
spaces we shall be able to use 4A3Dc.
For instance, $\Cal B(\{0,1\}^{\Bbb N})$ is the $\sigma$-algebra of
subsets of $\{0,1\}^{\Bbb N}$
generated by the sets $\{x:x(n)=1\}$ for $n\in\Bbb N$.
%4@33
}%end of comment

\leader{4A3F}{Spaces with countable networks (a)} Let $X$ be a
topological space with a countable network.   Then
$\#(\Cal B(X))\le\frak c$.
\prooflet{\Prf\ Let $\Cal E$ be a countable network for the topology of
$X$ and $\Sigma$ the $\sigma$-algebra of subsets of $X$ generated by
$\Cal E$.   Then $\#(\Sigma)\le\frak c$ (4A1O).   If $G\subseteq X$ is
open, there is a subset $\Cal E'$ of $\Cal E$ such that
$\bigcup\Cal E'=G$;  but $\Cal E'$ is necessarily countable, so
$G\in\Sigma$.   It follows that $\Cal B(X)\subseteq\Sigma$ and
$\#(\Cal B(X))\le\frak c$.\ \Qed}

\spheader 4A3Fb $\#(\Cal B(\BbbN^{\Bbb N}))=\frak c$.
\prooflet{\Prf\ $\BbbN^{\Bbb N}$ has a countable network (4A2Ub), so
$\#(\Cal B(\BbbN^{\Bbb N}))\le\frak c$.   On the other hand,
$\Cal B(\BbbN^{\Bbb N})$ contains all singletons, so

\Centerline{$\#(\Cal B(\BbbN^{\Bbb N}))\ge\#(\BbbN^{\Bbb N})
\ge\#(\{0,1\}^{\Bbb N})=\frak c$. \Qed}}
%4@24

\leader{4A3G}{Second-countable spaces (a)} Suppose that $X$ is a
second-countable space and $Y$ is any topological space.   Then
$\Cal B(X\times Y)=\Cal B(X)\tensorhat\Cal B(Y)$.
\prooflet{\Prf\ By 4A3Cf,
$\Cal B(X\times Y)\supseteq\Cal B(X)\tensorhat\Cal B(Y)$.   On the other
hand, let $\Cal U$ be a countable base for the topology of $X$.   If
$W\subseteq X\times Y$ is open, set

\Centerline{$V_U
=\bigcup\{H:H\subseteq Y$ is open, $U\times H\subseteq W\}$}

\noindent for $U\in\Cal U$.   Then $W=\bigcup_{U\in\Cal U}U\times V_U$
belongs to $\Cal B(X)\tensorhat\Cal B(Y)$.   As $W$ is arbitrary,
$\Cal B(X\times Y)\subseteq\Cal B(X)\tensorhat\Cal B(Y)$.\ \Qed}
%countable network for X will do if we have a network consisting of
%Borel sets

\spheader 4A3Gb If $X$ is any topological space, $Y$ is a T$_0$
second-countable space, and $f:X\to Y$ is Borel measurable, then
(the graph of) $f$ is a Borel set in $X\times Y$.   \prooflet{\Prf\
Let $\Cal U$ be a countable base for the topology of $Y$.   Because $Y$ is
T$_0$,

\Centerline{$f=\bigcap_{U\in\Cal U}
 (\{(x,y):x\in f^{-1}[U]$, $y\in U\}
 \cup\{(x,y):x\in X\setminus f^{-1}[U]$, $y\in Y\setminus U\})$}

\noindent which is a Borel subset of $X\times Y$ by 4A3Cc and 4A3Cf.\ \Qed}

\leader{4A3H}{Borel sets in Polish spaces:  Proposition} Let
$(X,\frak T)$ be a Polish space and $E\subseteq X$ a
Borel set.   Then there is a Polish topology $\frak S$ on $X$, including
$\frak T$, for which $E$ is open.

\proof{ Let $\Cal E$ be the union of all the Polish topologies on $X$
including $\frak T$.   Of course $X\in\Cal E$.   If $E\in\Cal E$ then
$X\setminus E\in\Cal E$.   \Prf\ There is a Polish topology
$\frak S\supseteq\frak T$ such that $E\in\frak S$.   As both $E$ and
$X\setminus E$ are Polish in the subspace topologies $\frak S_E$,
$\frak S_{X\setminus E}$ induced by $\frak S$ (4A2Qd), the disjoint
union topology $\frak S'$ of $\frak S_E$ and $\frak S_{X\setminus E}$ is
also Polish (4A2Qe).   Now $\frak S'\supseteq\frak S\supseteq\frak T$
and $X\setminus E\in\frak S'$, so $X\setminus E\in\Cal E$.\ \QeD\
Moreover, the union of any sequence $\sequencen{E_n}$ in $\Cal E$
belongs to $\Cal E$.   \Prf\ For each $n\in\Bbb N$ let
$\frak S_n\supseteq\frak T$ be a Polish topology containing $E_n$.   If
$m$, $n\in\Bbb N$ then $\frak S_m\cap\frak S_n$ includes $\frak T$, so
is Hausdorff.   By 4A2Qf, the topology $\frak S$ generated by
$\bigcup_{n\in\Bbb N}\frak S_n$ is Polish.   Of course
$\frak S\supseteq\frak T$, and $\bigcup_{n\in\Bbb N}E_n\in\frak S$, so
$\bigcup_{n\in\Bbb N}E_n\in\Cal E$.\ \Qed

Thus $\Cal E$ is a $\sigma$-algebra.   Since it surely includes
$\frak T$, it includes $\Cal B(X,\frak T)$, as claimed.
}%end of proof of 4A3H
%4@23

\leader{4A3I}{Corollary} If $(X,\frak T)$ is a Polish space and
$\sequencen{E_n}$ is a
sequence of Borel subsets of $X$, then there is a
zero-dimensional Polish topology
$\frak S$ on $X$, including $\frak T$, for which every $E_n$ is
open-and-closed.

\proof{{\bf (a)} By 4A3H, we can find for each $n\in\Bbb N$ a Polish topology
$\frak T_n\supseteq\frak T$ containing $E_k$ (if $n=2k$ is even) or
$X\setminus E_k$ (if $n=2k+1$ is odd);  now the topology
$\frak T'$ generated by $\bigcup_{n\in\Bbb N}\frak T_n$ is
Polish, by 4A2Qf, and every $E_n$ is open-and-closed for $\frak T'$.

\medskip

{\bf (b)} Now choose $\sequencen{\Cal V_n}$ and $\sequencen{\frak T'_n}$
inductively such that

\inset{$\frak T'_0=\frak T'$,

given that $\frak T'_n$ is a Polish topology on $X$, then $\Cal V_n$ is a
countable base for $\frak T'_n$ and
$\frak T'_{n+1}$ is a Polish topology on $X$ including
$\frak T'_n\cup\{X\setminus V:V\in\Cal V_n\}$}

\noindent (using (a) for the inductive step).   Now the topology
$\frak S$ generated by $\bigcup_{n\in\Bbb N}\frak T'_n$ is Polish,
includes $\frak T$, contains every $E_n$ and its complement and has a base
$\bigcup_{n\in\Bbb N}\Cal V_n$ consisting of open-and-closed sets.
}%end of proof of 4A3I

\leader{4A3J}{Borel sets in $\omega_1$:  Proposition} A set
$E\subseteq\omega_1$ is a Borel set iff either $E$ or its complement
includes a closed cofinal set.

\proof{{\bf (a)} Let $\Sigma$ be the family of all those sets
$E\subseteq\omega_1$ such that either $E$ or $\omega_1\setminus E$
includes a closed cofinal set.   Of course $\Sigma$ is closed under
complements.
Because the intersection of a sequence of closed cofinal sets is a
closed cofinal set (4A1Bd), the
union of any sequence in $\Sigma$ belongs to $\Sigma$;  so $\Sigma$ is a
$\sigma$-algebra.   If $E$ is closed, then either it is cofinal with
$\omega_1$, and is a closed cofinal set, or there is a $\xi<\omega_1$
such that
$E\subseteq\xi$, in which case $\omega_1\setminus E$ includes the
closed cofinal set
$\omega_1\setminus\xi$;  in either case, $E\in\Sigma$.   Thus every open
set belongs to $\Sigma$, and $\Sigma$ includes $\Cal B(\omega_1)$.

\medskip

{\bf (b)} Now suppose that $E\subseteq\omega_1$ is such that there is a
closed cofinal set $F\subseteq\omega_1\setminus E$.   For each
$\xi<\omega_1$ let
$f_{\xi}:\xi\to\Bbb N$ be an injective function.   Define $g:E\to\Bbb N$
by setting $g(\eta)=f_{\alpha(\eta)}(\eta)$, where
$\alpha(\eta)=\min(F\setminus\eta)$ for $\eta\in E$.   Set
$A_n=g^{-1}[\{n\}]$ for $n\in\Bbb N$, so that
$E=\bigcup_{n\in\Bbb N}A_n$.   If $n\in\Bbb N$,
$\xi\in\overline{A}_n\setminus A_n$
and $\xi'<\xi$, there must be $\eta$, $\eta'\in A_n$ such that
$\xi'\le\eta<\eta'<\xi$.   But now, because $f_{\alpha(\eta')}$ is
injective, while $g(\eta)=g(\eta')=n$, $\alpha(\eta)\ne\alpha(\eta')$, so
$\alpha(\eta)\in F\cap\ooint{\xi',\xi}$.   As $\xi'$ is arbitrary,
$\xi\in\overline{F}=F$.   This shows that
$\overline{A}_n\subseteq A_n\cup F$ and $A_n=\overline{A}_n\setminus F$
is a Borel set.   This is true for every $n\in\Bbb N$, so
$E=\bigcup_{n\in\Bbb N}A_n$ is a Borel set.

\medskip

{\bf (c)} If $E\subseteq\omega_1$ includes a closed cofinal set, then
(b) tells us that $\omega_1\setminus E$ and $E$ are Borel sets.
Thus $\Sigma\subseteq\Cal B(\omega_1)$ and $\Sigma=\Cal B(\omega_1)$, as
claimed.
}%end of proof of 4A3J

\leader{4A3K}{Baire sets }\cmmnt{When we come to study
measures in terms of the integrals of
continuous functions (\S436), we find that it is sometimes inconvenient
or even impossible to apply them to arbitrary Borel sets, and we need to
use a smaller $\sigma$-algebra, as follows.

\medskip

}{\bf (a) Definition} Let $X$ be a topological space.
The {\bf Baire $\sigma$-algebra} $\CalBa(X)$ of $X$ is the
$\sigma$-algebra generated
by the zero sets.   Members of $\CalBa(X)$ are called
{\bf Baire} sets.   \cmmnt{({\bf Warning!}
Do not confuse `Baire sets' in this sense
with `sets with the Baire property' in the sense of 4A3R, nor
with `sets which are Baire spaces in their subspace topologies'.)}

\spheader 4A3Kb For any topological space $X$,
$\CalBa(X)\subseteq\Cal B(X)$\cmmnt{ (because every zero set is
closed, therefore Borel)}.   If $\frak T$ is perfectly
normal\cmmnt{ -- for instance, if it is metrizable (4A2Lc), or is
regular and hereditarily Lindel\"of (4A2H(c-ii)) --} then
$\CalBa(X)=\Cal B(X)$\cmmnt{ (because
every closed set is a zero set, by 4A2Fi, so every open set belongs
to $\CalBa(X)$)}.

\spheader 4A3Kc Let $X$ and $Y$ be topological
spaces, with Baire $\sigma$-algebras $\CalBa(X)$, $\CalBa(Y)$
respectively.   If $f:X\to Y$
is continuous, it is $(\CalBa(X),\CalBa(Y))$-measurable.
\prooflet{\Prf\ Let $\Tau$ be the $\sigma$-algebra
$\{F:F\subseteq Y,\,f^{-1}[F]\in\CalBa(X)\}$.   If
$g:Y\to\Bbb R$ is continuous, then $gf:X\to\Bbb R$
is continuous, so

\Centerline{$f^{-1}[\{y:g(y)=0\}]
=\{x:gf(x)=0\}\in\CalBa(X)$,}

\noindent and $\{y:g(y)=0\}\in\Tau$.   Thus every zero set belongs to
$\Tau$, and $\Tau\supseteq\CalBa(Y)$.\ \Qed}

\spheader 4A3Kd In particular, if $X$ is a subspace of $Y$, then
$E\cap X\in\CalBa(X)$ whenever
$E\in\CalBa(Y)$.   \cmmnt{More fundamentally,} $F\cap X$ is a
zero set in $X$ for every zero set $F\subseteq Y$\cmmnt{, just because
$g\restrp X$ is continuous for any continuous $g:Y\to\Bbb R$}.

\spheader 4A3Ke If $X$ is a topological space and $Y$ is a separable
metrizable space, a function $f:X\to Y$ is {\bf Baire measurable} if
$f^{-1}[H]\in\CalBa(X)$ for every open $H\subseteq Y$.   \cmmnt{Observe that}
$f$ is Baire measurable\cmmnt{ in this sense} iff it is
$(\CalBa(X),\Cal B(Y))$-measurable iff it is
$(\CalBa(X),\CalBa(Y))$-measurable.

\leader{4A3L}{Lemma} Let $(X,\frak T)$ be a topological space.   Then
$\CalBa(X)$ is just the smallest $\sigma$-algebra of subsets of $X$ with
respect to
which every continuous real-valued function on $X$ is measurable.

\proof{{\bf (a)} Let $f:X\to\Bbb R$ be a continuous function and
$\alpha\in\Bbb R$.   Set $g(x)=\max(0,f(x)-\alpha)$ for $x\in X$;  then
$g$ is continuous, so

\Centerline{$\{x:f(x)\le\alpha\}=\{x:g(x)=0\}$}

\noindent is a zero set and belongs to $\CalBa(X)$.   As $\alpha$ is
arbitrary, $f$ is $\CalBa(X)$-measurable.

\medskip

{\bf (b)} On the other hand, if $\Sigma$ is any $\sigma$-algebra of
subsets of $X$ such that every continuous real-valued function on $X$ is
$\Sigma$-measurable, and $F\subseteq X$ is a zero set, then there is a
continuous $g$ such that $F=g^{-1}[\{0\}]$, so that $F\in\Sigma$;  as
$F$ is arbitrary, $\Sigma\supseteq\CalBa(X)$.
}%end of proof of 4A3L

\leader{4A3M}{Product spaces} Let $\familyiI{X_i}$ be a family of
topological spaces with product $X$.

\spheader 4A3Ma $\CalBa(X)\supseteq\Tensorhat_{i\in I}\CalBa(X_i)$.
\prooflet{(Apply 4A3Bc to the identity map from $X$ to itself;
compare 4A3Cf.)}

\spheader 4A3Mb Suppose that $X$ is ccc.   Then every Baire subset of
$X$ is determined by coordinates in a countable set.
\prooflet{\Prf\ By 254Mb, the family $\Cal W$ of sets determined by
coordinates in countable sets is a $\sigma$-algebra of subsets of $X$.
By 4A2E(b-ii), every continuous real-valued function is
$\Cal W$-measurable, so $\Cal W$ contains every zero set and every Baire
set.\ \Qed}

\leader{4A3N}{Products of separable metrizable spaces:  Proposition} Let
$\familyiI{X_i}$ be a family of separable metrizable spaces, with
product $X$.

(a) $\CalBa(X)=\Tensorhat_{i\in I}\CalBa(X_i)
=\Tensorhat_{i\in I}\Cal B(X_i)$.

(b) $\CalBa(X)$ is the family of those Borel subsets of $X$ which are
determined by coordinates in countable sets.

(c) A set $Z\subseteq X$ is a zero set iff it is closed and
determined by coordinates in a countable set.

(d) If $Y$ is a dense subset of $X$, then the Baire
$\sigma$-algebra $\CalBa(Y)$ of $Y$ is just the subspace
$\sigma$-algebra $\CalBa(X)_Y$ induced by $\CalBa(X)$.

(e) If $Y$ is a set, $\Tau$ is a $\sigma$-algebra of subsets of $Y$, and
$f:Y\to X$ is a function, then $f$ is $(\Tau,\CalBa(X))$-measurable iff
$\pi_if:Y\to X_i$ is $(\Tau,\Cal B(X_i))$-measurable for every $i\in I$,
where $\pi_i(x)=x(i)$ for $x\in X$ and $i\in I$.

\proof{{\bf (a)} $X$ is ccc (4A2E(a-iii)), so if $f:X\to\Bbb R$ is
continuous,
there are a countable set $J\subseteq I$ and a continuous function
$g:X_J\to\Bbb R$ such that $f=g\tilde\pi_J$, where
$X_J=\prod_{i\in J}X_i$ and
$\tilde\pi_J:X\to X_J$ is the canonical map (4A2E(b-ii)).   Now $X_J$ is
separable and metrizable (4A2P(a-v)), therefore hereditarily Lindel\"of
(4A2P(a-iii)),
so $\Cal B(X_J)=\Tensorhat_{i\in J}\Cal B(X_i)$, by
4A3D(c-i).   By 4A3Bc, $\tilde\pi_J$ is
$(\Tensorhat_{i\in I}\Cal B(X_i),
\Tensorhat_{j\in J}\Cal B(X_j))$-measurable, so $f$ is
$\Tensorhat_{i\in I}\Cal B(X_i)$-measurable.   As $f$ is arbitrary,
$\CalBa(X)\subseteq\Tensorhat_{i\in I}\Cal B(X_i)$ (4A3L).
Also $\Cal B(X_i)=\CalBa(X_i)$ for every $i$ (4A3Kb), so
$\Tensorhat_{i\in I}\Cal B(X_i)=\Tensorhat_{i\in I}\CalBa(X_i)$.
With 4A3Ma, this proves the result.

\medskip

{\bf (b)(i)} If $W\in\CalBa(X)$ it is certainly a Borel
set (4A3Kb), and by 4A3Mb it is determined by coordinates in a countable
set.

\medskip

\quad{\bf (ii)} If $W$ is a Borel subset of $X$ determined by
coordinates in a countable subset $J$ of $X$, then write
$X_J=\prod_{i\in J}X_i$ and
$X_{I\setminus J}=\prod_{i\in I\setminus J}X_i$;  let
$\tilde\pi_J:X\to X_J$ be the canonical map.   We can identify $W$
with $W'\times X_{I\setminus J}$, where $W'$ is some subset of $X_J$.
Now if
$z\in X_{I\setminus J}$, $W'=\{w:w\in X_J,\,(w,z)\in W\}$ is a Borel
subset of $X_J$, because $w\mapsto(w,z):X_J\to X$ is continuous.   (I am
passing over the trivial case $X=\emptyset$.)   Since $X_J$ is
metrizable (4A2P(a-v) again), $W'\in\CalBa(X_J)$ (4A3Kb) and
$W=\tilde\pi_J^{-1}[W']$ is a Baire set (4A3Kc).

\medskip

{\bf (c)(i)} If $Z$ is a zero set, it is surely closed;
and it is determined by coordinates in a
countable set by (b) above, or directly from 4A2E(b-ii) again.

\medskip

\quad{\bf (ii)} If $Z$ is closed and determined by
coordinates in a countable set $J$, then (in the language of (b))
it can be identified with
$Z'\times X_{I\setminus J}$ for some $Z'\subseteq X_J$.   As in the
proof of (b), $Z'$ is closed (at least, if
$X_{I\setminus J}\ne\emptyset$), so is a zero set (4A2Lc), and
$Z=\tilde\pi_J^{-1}[Z']$ is a zero set (4A2C(b-iv)).

\medskip

{\bf (d)(i)} $\CalBa(Y)\supseteq\CalBa(X)_Y$ by 4A3Kd.

\medskip

\quad{\bf (ii)} Let $f:Y\to\Bbb R$ be any continuous function.   For each
$n\in\Bbb N$, there is an open set $G_n\subseteq X$ such that
$G_n\cap Y=\{y:f(y)>2^{-n}\}$.   Now $\overline{G}_n$ is determined by
coordinates in a countable set (4A2E(b-i)), so is a zero set,
by (c) here.   Because $Y$ is dense in $X$,
$\overline{G}_n=\overline{G_n\cap Y}$ does not meet $\{y:f(y)=0\}$, and
$\{y:f(y)>0\}=Y\cap\bigcup_{n\in\Bbb N}\overline{G}_n$ belongs to
$\CalBa(X)_Y$.   Thus $\CalBa(X)_Y$ contains every cozero subset of
$Y$ and includes $\CalBa(Y)$.

\medskip

{\bf (e)} Put (a) and 4A3Bc together.
}%end of proof of 4A3N

\leader{4A3O}{Compact spaces (a)} Let $(X,\frak T)$ be a topological
space, $\Cal U$ a subbase for
$\frak T$, and $\frak A$ the algebra of subsets of $X$ generated by
$\Cal U$.   If $H\subseteq X$ is open and $K\subseteq H$ is compact,
there is an open $E\in\frak A$ such that $K\subseteq E\subseteq H$.
\prooflet{\Prf\ Set $\Cal V
=\{X\}\cup\{U_0\cap U_1\cap\ldots\cap U_n:U_0,\ldots,U_n\in\Cal U\}$, so
that $\Cal V$ is a base for $\frak T$ and
$\Cal V\subseteq\frak A$.   $\{U:U\in\Cal V,\,U\subseteq H\}$ is an
open cover of the compact set $K$, so there is a finite set
$\Cal U_0\subseteq\Cal V$ such that $E=\bigcup\Cal U_0$ includes $K$
and is included in $H$;  now $E\in\frak A$.\ \Qed}

\spheader 4A3Ob Let $(X,\frak T)$ be a compact space and $\Cal U$ a
subbase for $\frak T$.   Then every open-and-closed subset of $X$
belongs to the algebra of subsets of $X$ generated by $\Cal U$.
\prooflet{(If $F\subseteq X$ is open-and-closed, it is also compact;
apply (a) here with $K=H=F$.)}
%4@54

\spheader 4A3Oc Let $(X,\frak T)$ be a compact space and $\Cal U$ a
subbase for $\frak T$.   Then $\CalBa(X)$ is included in the
$\sigma$-algebra of subsets of $X$ generated by $\Cal U$.
\prooflet{\Prf\ Let $\Sigma$ be the $\sigma$-algebra generated by
$\Cal U$.   If $Z\subseteq X$ is a zero set, there is a sequence
$\sequencen{H_n}$ of open sets with intersection $Z$ (4A2C(b-vi)); now
we can find a sequence $\sequencen{E_n}$ in $\Sigma$ such that
$Z\subseteq E_n\subseteq H_n$ for every $n$, by (a), so that
$Z=\bigcap_{n\in\Bbb N}E_n\in\Sigma$.   This shows that every zero set
belongs to $\Sigma$, so $\Sigma$ must include $\CalBa(X)$.\ \Qed}

\spheader 4A3Od In a compact Hausdorff zero-dimensional space the Baire
$\sigma$-algebra is the $\sigma$-algebra generated by the
open-and-closed sets.
\prooflet{(Apply (c) with $\Cal U$ the family of open-and-closed
sets.)}

\spheader 4A3Oe Let $\familyiI{X_i}$ be a family of compact Hausdorff
spaces with product $X$.
Then $\CalBa(X)=\Tensorhat_{i\in I}\CalBa(X_i)$.
\prooflet{\Prf\ By 4A3Ma,
$\CalBa(X)\supseteq\Tensorhat_{i\in I}\CalBa(X_i)$.   On the
other hand, let $\Cal U_i$ be the family of cozero sets in $X_i$ for
each $i$.
Because $X_i$ is completely regular (3A3Bb), $\Cal U_i$ is a base for
its topology (4A2Fc).   Set

\Centerline{$\Cal W
=\{\prod_{i\in I}U_i:U_i\in\Cal U_i$ for every $i\in I$,
  $\{i:U_i\ne X_i\}$ is finite$\}$,}

\noindent so that $\Cal W\subseteq\Tensorhat_{i\in I}\CalBa(X_i)$ is a
base for the topology of $X$.   By (c) above, $\CalBa(X)$ is the
$\sigma$-algebra generated by $\Cal W$, and
$\CalBa(X)\subseteq\Tensorhat_{i\in I}\CalBa(X_i)$.\ \Qed}
%4@63
%4@52 %4@39?

\spheader 4A3Of In particular, for any set $I$, $\CalBa(\{0,1\}^I)$ is
the $\sigma$-algebra generated by sets of the form $\{x:x(i)=1\}$ as $i$
runs over $I$.
%for 4@53

\leader{4A3P}{Proposition} The Baire $\sigma$-algebra $\CalBa(\omega_1)$
of $\omega_1$ is\cmmnt{ just} the
countable-cocountable algebra\cmmnt{ (211R)}.

\proof{ We
see from 4A2S(b-iii) that every continuous function is measurable with
respect to
the countable-cocountable algebra, so $\CalBa(\omega_1)$ is included in
the countable-cocountable algebra.   On the other hand,

\Centerline{$[0,\xi]=\{\eta:\eta\le\xi\}=\coint{0,\xi+1}
=\omega_1\setminus\ooint{\xi,\omega_1}$}

\noindent is an open-and-closed set
(4A2S(a-i)), therefore a zero set, therefore belongs to
$\CalBa(\omega_1)$, for
every $\xi<\omega_1$.   Now if $\xi<\omega_1$, it is itself a countable
set, so

\Centerline{$\coint{0,\xi}=\bigcup_{\eta<\xi}[0,\eta]
\in\CalBa(\omega_1)$,
\quad$\{\xi\}=[0,\xi]\setminus\coint{0,\xi}\in\CalBa(\omega_1)$.}

\noindent It follows that every countable set belongs to
$\CalBa(\omega_1)$ and
the countable-cocountable algebra is included in $\CalBa(\omega_1)$.
}%end of proof of 4A3P

\leader{4A3Q}{C\`adl\`ag functions}\dvAformerly{4A3W}
Let $X$ be a Polish space, and $\Cdlg$ the set of \cadlag\
functions\cmmnt{ (definition:  4A2A)} from $\coint{0,\infty}$ to $X$,
with its topology of
pointwise convergence inherited from $X^{\coint{0,\infty}}$.

(a) $\CalBa(\Cdlg)$ is the subspace $\sigma$-algebra induced by
$\CalBa(X^{\coint{0,\infty}})$.

(b) $(\Cdlg,\CalBa(\Cdlg))$ is a standard Borel space.

(c)(i) For any $t\ge 0$, let $\CalBa_t(\Cdlg)$ be the $\sigma$-algebra
of subsets of $\Cdlg$ generated by the functions $\omega\mapsto\omega(s)$
for $s\le t$.   Then
$(\omega,s)\mapsto\omega(s):\Cdlg\times[0,t]\to X$ is
$\CalBa_t(\Cdlg)\tensorhat\Cal B([0,t])$-measurable.

\quad(ii) $(\omega,t)\mapsto\omega(t):\Cdlg\times\coint{0,\infty}\to X$ is
$\CalBa(\Cdlg)\tensorhat\Cal B(\coint{0,\infty})$-measurable.

(d) The set $C(\coint{0,\infty};X)$ of continuous functions from
$\coint{0,\infty}$ to $X$ belongs to $\CalBa(\Cdlg)$.

\proof{{\bf (a)} Use 4A3Nd.

\medskip

{\bf (b)(i)}
Fix a complete metric $\rho$ on $X$ defining its topology.
For $A\subseteq B\subseteq\Bbb R$,
$f\in X^B$ and $\epsilon>0$, set

$$\eqalign{\jump_A(f,\epsilon)
&=\sup\{n:\text{ there is an }I\in[A]^n
\text{ such that }\rho(f(s),f(t))>\epsilon\cr
&\mskip180mu
\text{ whenever }s<t\text{ are successive elements of }I\}\cr}$$

\noindent (see 438P).   Set $D=\Bbb Q\cap\coint{0,\infty}$.
Then any set of the form
$\{f:\jump_A(f,\epsilon)>m\}$ is open, so

$$\eqalign{E
&=\bigcap_{n\in\Bbb N}\bigcup_{m\in\Bbb N}
  \{f:f\in X^D,\,\jump_{D\cap[0,n]}(f,2^{-n})\le m\}\cr
&\mskip150mu  \cap\bigcap_{\Atop{q\in D}{n\in\Bbb N}}
   \bigcup_{m\ge n}
   \{f:f\in X^D,\,\rho(f(q+2^{-m}),f(q))\le 2^{-n}\}\cr}$$

\noindent is Borel in the Polish space $X^D$.

\medskip

\quad{\bf (ii)} If $\omega\in\Cdlg$, then $\jump_{[0,n]}(\omega,2^{-n})$
is finite for every $n\in\Bbb N$.   (Apply 438Pa to an extension of
$\omega$ to a member of $X^{\Bbb R}$ which is constant on
$\ooint{-\infty,0}$;  or make the trifling required changes to the
argument of 438Pa.)   Since $\lim_{n\to\infty}\omega(q+2^{-n})=\omega(q)$
for every $q\in D$, $\omega\restr D\in E$.

\medskip

\quad{\bf (iii)} Conversely, given $f\in E$, $\jump_{[0,n]}(f,\epsilon)$
is finite for every $n\in\Bbb N$ and $\epsilon>0$, so
$\lim_{q\in D,q\downarrow t}f(q)$ is defined in $X$ for every $t\ge 0$,
and $\lim_{q\in D,q\uparrow t}f(q)$ is defined in $X$ for every $t>0$.
(Apply the argument of (a-ii) of the proof of 438P.)   Set
$\omega_f(t)=\lim_{q\in D,q\downarrow t}f(q)$ for $t\ge 0$.
Because $f(q)$ is a cluster point of $\sequencen{f(q+2^{-n})}$ for every
$q\in D$, $\omega_f$ extends $f$.   It is easy to see that
$\jump_{\coint{0,n}}(\omega_f,\epsilon)=\jump_{\coint{0,n}}(f,\epsilon)$
is finite for every $n\in\Bbb N$ and $\epsilon>0$, so
$\lim_{s\downarrow t}\omega_f(s)$ is defined for every $t\ge 0$ and
$\lim_{s\uparrow t}\omega_f(s)$ is defined for every $t>0$.
Also, for $t\ge 0$,

\Centerline{$\lim_{s\downarrow t}\omega_f(s)
=\lim_{q\in D,q\downarrow t}\omega_f(q)
=\lim_{q\in D,q\downarrow t}f(q)=\omega_f(t)$.}

\noindent Thus $\omega_f\in\Cdlg$.   Clearly a member of $\Cdlg$ is
uniquely determined by its values on $D$, so $f\mapsto\omega_f$ and
$\omega\mapsto\omega\restr D$ are the two halves of a
bijection between $E$ and $\Cdlg$.

\medskip

\quad{\bf (iv)} By 424G, $E$, with its Borel (or Baire)
$\sigma$-algebra $\Cal B(E)$, is a standard Borel space.
Of course the map $\omega\mapsto\omega\restr D$ is
$(\CalBa(\Cdlg),\Cal B(E))$-measurable.   But also the map
$f\mapsto\omega_f(t):E\to X$ is $\Cal B(E)$-measurable for every
$t\ge 0$.   \Prf\ If $\sequencen{q_n}$ is a sequence in $D$
decreasing to $t$, $\omega_f(t)=\lim_{n\to\infty}f(q_n)$ for every
$f\in E$, and we can use 418Ba.\ \QeD\  Since $\CalBa(\Cdlg)$ is the
$\sigma$-algebra induced by $\CalBa(X^{\coint{0,\infty}})$ (4A3Nd), and
$\CalBa(X^{\coint{0,\infty}})
=\Tensorhat_{\coint{0,\infty}}\Cal B(X)$ (4A3Na),
this is enough to show that
$f\mapsto\omega_f$ is $(\Cal B(E),\CalBa(\Cdlg))$-measurable.

Thus $(\Cdlg,\CalBa(\Cdlg))\cong(E,\Cal B(E))$ is a standard Borel
space.

\medskip

{\bf (c)(i)} For $n\in\Bbb N$, $\omega\in\Cdlg$ and $s\in[0,t]$, set
$h_n(\omega,s)=\omega(\min(t,2^{-n}i))$ if $i\in\Bbb N$ and
$2^{-n}(i-1)<s\le 2^{-n}i$.   If $G\subseteq X$ is open then

$$\eqalign{\{(\omega,s):s\le t,\,h_n(\omega,s)\in G\}
&=\bigcup_{i\in\Bbb N}\{\omega:\omega(\min(t,2^{-n}i))\in G\}\cr
&\mskip100mu
\times([0,t]\cap\ocint{2^{-n}(i-1),2^{-n}i})\cr
&\in\CalBa_t(\Cdlg)\tensorhat\Cal B([0,t]),\cr}$$

\noindent so
$h_n$ is $\CalBa_t(\Cdlg)\tensorhat\Cal B([0,t])$-measurable.
Now $\omega(s)=\lim_{n\to\infty}h_n(\omega,s)$ for every
$\omega\in\Cdlg$ and $s\in[0,t]$, so $(\omega,s)\mapsto\omega(s)$ is
measurable in the same sense, by 418Ba again.

\medskip

\quad{\bf (ii)} If $G\subseteq X$ is open then

$$\eqalign{\{(\omega,t):t\ge 0,\,\omega(t)\in G\}&
=\bigcup_{n\in\Bbb N}\{(\omega,t):t\in[0,n],\,\omega(t)\in G\}\cr&
\in\CalBa(\Cdlg)\tensorhat\Cal B(\coint{0,\infty}).\cr}$$

\medskip

{\bf (d)}

$$\eqalign{C(\coint{0,\infty};X)
&=\bigcap_{n\in\Bbb N}\bigcup_{m\in\Bbb N}
  \{\omega:\omega\in\Cdlg,\,\rho(\omega(q),\omega(q'))\le 2^{-n}\cr
&\mskip100mu
  \text{ whenever }q,\,q'\in\Bbb Q\cap[0,n]\text{ and }
    |q-q'|\le 2^{-m}\}.\cr}$$
}%end of proof of 4A3Q

\leader{4A3R}{Baire property}\dvAformerly{4A3Q}
Let $X$ be a topological space, and $\Cal M$
the ideal of meager subsets of $X$.   A subset
$X$ has the {\bf Baire property} if it is expressible in the form
$G\symmdiff M$
where $G\subseteq X$ is open and $M\in\Cal M$;
\cmmnt{that is,}
$A\subseteq X$ has the Baire property if there is an open set
$G\subseteq X$ such that $G\symmdiff A$ is meager.   \prooflet{(For
$A=G\symmdiff M$ iff $M=G\symmdiff A$.)}   The family
$\widehat{\Cal B}(X)$ of all such sets is the {\bf Baire-property
algebra} of $X$.
\cmmnt{(See 4A3S.)}   \cmmnt{({\bf Warning!} do not confuse the
`Baire-property algebra' $\widehat{\Cal B}$ with the
`Baire $\sigma$-algebra' $\CalBa$ as defined in 4A3K.)}
The quotient algebra $\widehat{\Cal B}(X)/\Cal M$ is the {\bf category
algebra} of $X$.

\leader{4A3S}{Proposition}\dvAformerly{4A3R}
Let $X$ be a topological space.

(a) Let $A\subseteq X$ be any set.

\quad(i) There is a largest open set $G\subseteq X$ such that $A\cap G$ is
meager.

\quad(ii) $H=X\setminus\overline{G}$ is the smallest regular open subset
of $X$ such that $A\setminus H$ is meager;  $H\subseteq\overline A$.

\quad(iii) $H$ is in itself a Baire space.

\quad(iv) If $A\in\widehat{\Cal B}(X)$, $H\symmdiff A$ is meager.

\quad(v) If $X$ is a Baire space and $A\in\widehat{\Cal B}(X)$, then
$H$ is the largest open subset of $X$ such that $H\setminus A$ is meager.

(b)(i) $\widehat{\Cal B}(X)$ is a $\sigma$-algebra of subsets of $X$
including $\Cal B(X)$.

\quad(ii) $\widehat{\Cal B}(X)=\{G\symmdiff M:G\subseteq X$ is
a regular open set, $M\in\Cal M\}$.

(c) If $X$ has a
countable network, its category algebra has a countable
order-dense set\cmmnt{ (definition:  313J)}.

\proof{{\bf (a)} (See {\smc Kechris 95}, 8.29.)

\medskip

\quad{\bf (i)} Set
$\Cal U=\{U:U\subseteq X$ is open, $A\cap U$ is meager$\}$.   Let
$\Cal U_0\subseteq\bigcup\Cal U$ be a maximal disjoint set, and
$G_0=\bigcup\Cal U_0$.   Then $A\cap G_0$ is meager.   \Prf\ For each
$U\in\Cal U_0$, let $\sequencen{F_{Un}}$ be a sequence of nowhere dense
closed sets covering $A\cap U$.   Set
$A_n=\bigcup_{U\in\Cal U_0}F_{Un}$.   If $V\subseteq X$ is any
non-empty open set, either $V\cap A_n$ is empty or there is a
$U\in\Cal U_0$ such that $V\cap U\ne\emptyset$, in which case
$V\cap U\setminus F_{Un}$ is a non-empty open subset of $U$ not meeting
$A_n$.   Thus $A_n$ is nowhere dense.   This is true for every $n$, so
$A\cap G_0\subseteq\bigcup_{n\in\Bbb N}A_n$ is meager.\ \Qed

If $U\in\Cal U$, then $V=U\setminus\overline{G}_0$ belongs to $\Cal U$ and
cannot meet any member of $\Cal U_0$, so must be empty.   Thus
$U\subseteq\overline{G}_0$;  as $U$ is arbitrary, $G=\bigcup\Cal U$ is included
in $\overline{G}_0$ and $G\setminus G_0$ is nowhere dense.   It follows that
$A\cap G$ is meager, and $G$ is the largest open set for which this is true.

\medskip

\quad{\bf (ii)} $H=X\setminus\overline{G}=\interior(X\setminus G)$ is now a
regular open set (314O), and
$A\setminus H=(A\cap G)\cup(\overline G\setminus G)$ is meager.
If $H'$ is another regular open set such that $A\setminus H'$ is
meager, then $H\setminus\overline{H'}$ is open and meets $A$ in a meager set, so
is included in $G$ and must be empty.   So
$H\subseteq\interior\overline{H'}=H'$.
Thus $H$ is the smallest regular open set such that $A\setminus H$ is
meager.   Of course $X\setminus\overline A$ belongs to $\Cal U$ so is included
in $G$ and does not meet $H$, that is, $H\subseteq\overline{A}$.

\medskip

\quad{\bf (iii)} If now $\sequencen{H_n}$ is a sequence of open subsets of
$H$ which are dense in $H$, and $H'\subseteq H$ is any non-empty open set,
$H'\setminus H_n$ is nowhere dense for every $n$, so
$H'\setminus\bigcap_{n\in\Bbb N}H_n$ is meager.   On the other hand,
$A\cap H'$ is non-meager so $H'$ also is, and $H'$ must meet
$\bigcap_{n\in\Bbb N}H_n$.   As $H'$ is arbitrary,
$\bigcap_{n\in\Bbb N}H_n$ is dense in
$H$;  as $\sequencen{H_n}$ is arbitrary, $H$ is a Baire space in its
subspace topology.

\medskip

\quad{\bf (iv)} If $A$ has the Baire property, there is an open set $U$
such that $A\symmdiff U$ is meager.   In this case,
$A\cap H\setminus\overline U$ must be meager, so
$H\subseteq\overline U$ and
$H\setminus A\subseteq(U\setminus A)\cup(\overline U\setminus G)$ is
meager and $H\symmdiff A=(A\setminus H)\cup(H\setminus A)$ is meager.

\medskip

\quad{\bf (v)} By (iv), $H\setminus A$ is meager.   If $U\subseteq X$ is
open and $U\setminus A$ is meager, set $V=U\setminus\overline{H}$.
Then $V\setminus A$ and $V\cap A$ are both meager, so $V$ is
meager, and must be empty, since $X$ is a Baire space.   Thus
$U\subseteq\overline{H}$ and $G\subseteq H$, because $H$ is a regular open
set.

\medskip

{\bf (b)(i)} (See {\smc \v{C}ech 66}, \S22C;  {\smc Kuratowski 66},
\S11.III;  {\smc Kechris 95}, 8.22.)
Of course $X=X\symmdiff\emptyset$ belongs to $\widehat{\Cal B}(X)$.   If
$E\in\widehat{\Cal B}(X)$, let $G\subseteq X$ be an open set such that
$E\symmdiff G$ is meager.   Then

\Centerline{$(X\setminus\overline{G})\symmdiff(X\setminus E)
\subseteq(\overline{G}\setminus G)\cup(G\symmdiff E)$}

\noindent is meager, so $X\setminus E\in\widehat{\Cal B}(X)$.   If
$\sequencen{E_n}$ is a sequence in $\widehat{\Cal B}(X)$, then for each
$n\in\Bbb N$ we can find an open set $G_n$ such that $G_n\symmdiff E_n$
is meager, and now

\Centerline{$(\bigcup_{n\in\Bbb N}G_n)\symmdiff(\bigcup_{n\in\Bbb N}E_n)
\subseteq\bigcup_{n\in\Bbb N}G_n\symmdiff E_n$}

\noindent is meager, so $\bigcup_{n\in\Bbb N}E_n\in\widehat{\Cal B}(X)$.

This shows that $\widehat{\Cal B}(X)$ is a $\sigma$-algebra of subsets
of $X$.   Since it contains every open set, it must include the Borel
$\sigma$-algebra.

\medskip

\quad{\bf (ii)} Of course $G\symmdiff M\in\widehat{B}(X)$ whenever $G$ is a
regular open set and $M$ is meager.   In the other direction, given a set
$A$ with the Baire property, let $G_0$ be an open set such that
$G_0\symmdiff A$ is meager.
Then $G=\interior\overline{G}_0$ is a regular open set and $G\symmdiff G_0$ is
nowhere dense, so $M=G\symmdiff A$ is meager, while $A=G\symmdiff M$.

\medskip

{\bf (c)} Suppose that $X$ has a countable network $\Cal A$.   For
$A\in\Cal A$, set $d_A=\overline{A}^{\ssbullet}$, the equivalence class
of $\overline{A}\in\widehat{\Cal B}$
in $\widehat{\Cal B}/\Cal M$.   If $b\in\widehat{\Cal B}/\Cal M$ is
non-zero, take $E\in\widehat{\Cal B}$ such that $E^{\ssbullet}=b$, and an open set $G\subseteq X$ such that
$E\symmdiff G\in\Cal M$.   Then $G$ is not meager.   Set $\Cal
A_1=\{A:A\in\Cal A$, $A\subseteq G\}$;  then $\Cal A_1$ is countable and
$G=\bigcup\Cal A_1$, so there is a non-meager $A\in\Cal A_1$.   In this
case $\overline{A}$ is not meager, so $d_A\ne 0$, while
$\overline{A}\setminus G\subseteq\overline{G}\setminus G$ is nowhere
dense, so $d_A\Bsubseteq b$.   As $b$ is arbitrary,
$\{d_A:A\in\Cal A\}$ is order-dense in $\widehat{\Cal B}/\Cal M$, and is
countable because $\Cal A$ is.
}%end of proof of 4A3S

\leader{*4A3T}{}\dvAformerly{4A3S}\cmmnt{ The following result does not 
mention any topology, but
its principal applications are with $\Cal J$ an ideal of meager sets, so I
slip it in here.   It will be useful in \S424 and in Volume 5.

\medskip

\noindent}{\bf Lemma} Let $X$ and $Y$ be sets, $\Sigma$ a
$\sigma$-algebra
of subsets of $X$, $\Tau$ a $\sigma$-algebra of subsets of $Y$ and
$\Cal J$ a
$\sigma$-ideal of $\Tau$.   Suppose that the quotient Boolean algebra
$\Tau/\Cal J$ has a countable order-dense set.

(a) $\{x:x\in X,\,W[\{x\}]\cap A\in\Cal J\}$ belongs to $\Sigma$ for any
$W\in\Sigma\tensorhat\Tau$ and $A\subseteq Y$.

(b) For every $W\in\Sigma\tensorhat\Tau$ there are sequences
$\sequencen{E_n}$ in $\Sigma$, $\sequencen{V_n}$ in $\Tau$ such that
$(W\symmdiff W_1)[\{x\}]\in\Cal J$ for every
$x\in X$, where $W_1=\bigcup_{n\in\Bbb N}E_n\times V_n$.
%not sure if (b) needed, but see 527J

\woddheader{*4A3T}{0}{0}{0}{60pt}

\proof{ (Compare {\smc Kechris 95}, 16.1.)

\medskip

{\bf (a)} Let $D\subseteq\Tau/\Cal J$ be a countable order-dense set.
Let $\Cal V\subseteq\Tau$ be a countable set such that
$D=\{V^{\ssbullet}:V\in\Cal V\}$.
Let $\Lambda$ be the family of subsets $W$ of $X\times Y$  such
that

\Centerline{$W[\{x\}]\in\Tau$ for every $x\in X$,}

\Centerline{$\{x:x\in X,\,W[\{x\}]\cap A\in\Cal J\}\in\Sigma$
for every $A\subseteq Y$.}

\medskip

\quad{\bf (i)} Of course $\emptyset\in\Lambda$, because if $W=\emptyset$
then $W[\{x\}]=\emptyset$ for every $x\in X$, and
$\{x:W[\{x\}]\cap A\in\Cal J\}=X$ for every $A\subseteq Y$.

\medskip

\quad{\bf (ii)} Suppose that $W\in\Lambda$, and set
$W'=(X\times Y)\setminus W$.   Then

\Centerline{$W'[\{x\}]=Y\setminus W[\{x\}]\in\Tau$}

\noindent for every $x\in X$.   Now suppose that $A\subseteq Y$.   Set
$\Cal V^*=\{V:V\in\Cal V,\,A\cap V\notin\Cal J\}$,

\Centerline{$E=\{x:W'[\{x\}]\cap A\in\Cal J\}$,}

\Centerline{$E'
=\{x:V\cap W[\{x\}]\notin\Cal J\text{ for every }V\in\Cal V^*\}$.}

\noindent Then $E=E'$.   \Prf\ ($\alpha$) If $x\in E$ and
$V\in\Cal V^*$, then $A\cap W'[\{x\}]=A\setminus W[\{x\}]$ belongs to
$\Cal J$, so
$V\cap A\setminus W[\{x\}]\in\Cal J$ and
$V\cap W[\{x\}]\supseteq V\cap A\cap W[\{x\}]$ is not in $\Cal J$.   As
$V$
is arbitrary, $x\in E'$;  as $x$ is
arbitrary, $E\subseteq E'$.   ($\beta$) If $x\notin E$, then
$W'[\{x\}]\cap A\notin\Cal J$.   Set $\Cal V_1=\{V:V\in\Cal V$,
$V\setminus W'[\{x\}]\in\Cal J\}$, $D_1=\{V^{\ssbullet}:V\in\Cal V_1\}$.
Then $D_1=\{d:d\in D$, $d\Bsubseteq W'[\{x\}]^{\ssbullet}\}$, so
$W'[\{x\}]^{\ssbullet}=\sup D_1$ in $\Tau/\Cal J$ (313K).   Because
$\Cal J$ is
a $\sigma$-ideal, the map $F\mapsto F^{\ssbullet}:\Tau\to\Tau/\Cal J$ is
sequentially order-continuous (313Qb), and
$(\bigcup\Cal V_1)^{\ssbullet}=\sup D_1$
(313Lc), that is, $W'[\{x\}]\symmdiff\bigcup\Cal V_1\in\Cal J$.   There
must therefore be a $V\in\Cal V_1$ such that $V\cap A\notin\Cal J$, that
is, $V\in\Cal V^*$.   At the same time,
$V\cap W[\{x\}]=V\setminus W'[\{x\}]$ belongs to $\Cal J$, so $V$
witnesses that $x\notin E'$.   As $x$ is arbitrary, $E'\subseteq E$.\
\Qed

Since $W\in\Lambda$,

\Centerline{$E=E'
=X\setminus\bigcup_{V\in\Cal V^*}\{x:V\cap W[\{x\}]\in\Cal J\}$}

\noindent belongs to $\Tau$.   As $A$ is arbitrary,
$W'\in\Lambda$.   Thus the complement of any member of $\Lambda$ belongs
to $\Lambda$.

\medskip

\quad{\bf (iii)} If $\sequencen{W_n}$ is any sequence in $\Lambda$ with
union $W$, then

\Centerline{$W[\{x\}]
=\bigcup_{n\in\Bbb N}W_n[\{x\}]\in\Tau$}

\noindent for every $x$, while

\Centerline{$\{x:W[\{x\}]\cap A\in\Cal J\}
=\bigcap_{n\in\Bbb N}\{x:W_n[\{x\}]\cap A\in\Cal J\}\in\Sigma$}

\noindent for every $A\subseteq Y$.   So $W\in\Lambda$.

\medskip

\quad{\bf (iv)} What this shows is that $\Lambda$ is a $\sigma$-algebra
of
subsets of $X\times Y$.   But if $W=E\times F$, where $E\in\Sigma$ and
$F\in\Tau$, then

\Centerline{$W[\{x\}]\in\{\emptyset,F\}\subseteq\Tau$ for
every $x\in X$,}

\Centerline{$\{x:W[\{x\}]\cap A\in\Cal J\}\in\{X\setminus
E,X\}\subseteq\Sigma$}

\noindent for every $A\subseteq Y$;  so $W\in\Lambda$.   Accordingly
$\Lambda$ must include $\Sigma\tensorhat\Tau$, as
claimed.

\medskip

{\bf (b)} Let $\sequencen{V_n}$ be a sequence running over
$\Cal V\cup\{\emptyset\}$.   For each $n\in\Bbb N$, set

\Centerline{$E_n=\{x:V_n\setminus W[\{x\}]\in\Cal J\}$;}

\noindent by (a), $E_n\in\Sigma$.   Set
$W_1=\bigcup_{n\in\Bbb N}E_n\times V_n$.   Take any $x\in X$.
Then $W_1[\{x\}]=\bigcup_{n\in I}V_n$ where $I=\{n:x\in E_n\}$.
Since $V_n\setminus W[\{x\}]\in\Cal J$ for every $n\in I$,
$W_1[\{x\}]\setminus W[\{x\}]\in\Cal J$.   \Quer\ If
$W[\{x\}]\setminus W_1[\{x\}]\notin\Cal J$, there is a
$d\in D$ such that
$0\ne d\Bsubseteq W[\{x\}]^{\ssbullet}\Bsetminus W_1[\{x\}]^{\ssbullet}$.   Let $n\in\Bbb N$ be such that
$V_n^{\ssbullet}=d$;  then $V_n\setminus W[\{x\}]\in\Cal J$, so $n\in I$
and $V_n\subseteq W_1[\{x\}]$ and $d\Bsubseteq W_1[\{x\}]^{\ssbullet}$,
which is impossible.\ \Bang

As $x$ is arbitrary, $W_1$ has the required properties.
}%end of proof of 4A3T
%4@24

\leaveitout{\spheader 4{}A2Oc Let $X$ be any topological space and
$\widehat{\Cal B}(X)$ its Baire-property algebra.   If $Y$ is a
second-countable space and $f:X\to Y$ is
$\widehat{\Cal B}(X)$-measurable,
then there is a Borel measurable function $g:X\to Y$ such that
$\{x:x\in X$, $g(x)\ne f(x)\}$ is meager.
\prooflet{\Prf\ Let $\Cal U$ be a countable base for the topology of
$Y$.   For each $U\in\Cal U$, let $G_U\subseteq X$ be an open set such
that $f^{-1}[U]\symmdiff G_U$ is meager;  let
$F_U\supseteq f^{-1}[U]\symmdiff G_U$ be a meager F$_{\sigma}$ set.
Then $Q=X\setminus\bigcup_{U\in\Cal U}F_U$ is a comeager G$_{\delta}$
set.   Take any $y_0\in Y$ (I am passing over the trivial case
$Y=\emptyset$), and set $g(x)=f(x)$ for $x\in Q$, $g(x)=y_0$ for
$x\in X\setminus Q$.   Then $\{x:g(x)=f(x)\}\supseteq Q$ is comeager.
If $U\in\Cal U$,   $g^{-1}[U]$ is either $Q\cap f^{-1}[U]=Q\cap G_U$ or
$(Q\cap G_U)\cup(X\setminus Q)$, and in either case is Borel.   So $g$
is Borel measurable, by 4A3?\ \Qed}
}%not needed I think, until vol 5 anyway

\leader{4A3U}{Cylindrical \dvrocolon{$\sigma$-algebras}}
\dvAformerly{4A3T}\cmmnt{ I
offer a note on a particular type of Baire $\sigma$-algebra.

\medskip

\noindent}{\bf Definition} Let $X$ be a linear topological
space.   Then the {\bf cylindrical $\sigma$-algebra} of $X$ is the
smallest $\sigma$-algebra $\Sigma$ of subsets of $X$ such that every
continuous linear functional on $X$ is $\Sigma$-measurable.

\leader{4A3V}{Proposition}\dvAformerly{4A3U} Let $X$ be a linear
topological space and $\frak T_s=\frak T_s(X,X^*)$ its weak topology.
Then the cylindrical $\sigma$-algebra of $X$ is just the Baire
$\sigma$-algebra of $(X,\frak T_s)$.

\proof{{\bf (a)} Let $\familyiI{f_i}$ be a Hamel basis for $X^*$
(4A4Ab).   For $x\in X$, set $Tx=\familyiI{f_i(x)}$;  then
$T:X\to\BbbR^I$ is a linear operator.   Now
$Y=T[X]$ is dense in $\BbbR^I$.   \Prf\ $Y$ is a linear
subspace of $\BbbR^I$, so its closure $\overline{Y}$ also is (2A5Ec).
If $\phi\in(\BbbR^I)^*$ is such that $\phi(Tx)=0$ for every
$x\in X$, there are a finite set $J\subseteq I$ and a family
$\family{i}{J}{\alpha_i}\in\BbbR^J$ such that
$\phi(y)=\sum_{i\in J}\alpha_iy(i)$ for every $y\in\BbbR^I$ (4A4Be).
In this case, $\sum_{i\in J}\alpha_if_i(x)=\phi(Tx)=0$ for every
$x\in X$;  but $\familyiI{f_i}$ is linearly independent, so $\alpha_i=0$
for every $i\in J$ and $\phi=0$.   By 4A4Eb, $\overline{Y}$ must be the
whole of $\BbbR^I$.\ \Qed

\medskip

{\bf (b)(i)} Set $\frak T'_s=\{T^{-1}[H]:H\subseteq Y$ is open$\}$.
Then $\frak T'_s=\frak T_s$.   \Prf\ Because every $f_i$ is continuous,
$T$ is continuous, so $\frak T'_s\subseteq\frak T_s$.   On the other
hand, any $f\in X^*$ is a linear combination of the $f_i$, so is
$\frak T'_s$-continuous, and $\frak T_s\subseteq\frak T'_s$.\ \Qed

\medskip

\quad{\bf (ii)} If $g:X\to\Bbb R$ is $\frak T_s$-continuous, there is a
continuous $g_1:Y\to\Bbb R$ such that $g=g_1T$.   \Prf\ If $x$,
$x'\in X$ are such that $Tx=Tx'$, then every $\frak T_s$-open set
containing one must contain the other, so $g(x)=g(x')$.   This means
that there is a function $g_1:Y\to\Bbb R$ such that $g=g_1T$.
Next, if $U\subseteq\Bbb R$ is open, $T^{-1}[g_1^{-1}[U]]=g^{-1}[U]$
belongs to $\frak T_s=\frak T'_s$, so $g_1^{-1}[U]$ must be open in $Y$;
as $U$ is arbitrary, $g_1$ is continuous.\ \Qed

\medskip

{\bf (c)} Now let $\Sigma$ be the cylindrical $\sigma$-algebra of $X$.
Then every $f_i:X\to\Bbb R$ is $\Sigma$-measurable, so $T:X\to\Bbb R^I$
is $(\Sigma,\CalBa(\BbbR^I))$-measurable, by 4A3Ne.
Because $Y$ is dense in $\BbbR^I$, $\CalBa(Y)$ is the subspace
$\sigma$-algebra induced by
$\CalBa(\BbbR^I)$ (4A3Nd).   So $T:X\to Y$ is
$(\Sigma,\CalBa(Y))$-measurable.   Now if $g:X\to\Bbb R$ is a continuous
function, there is a continuous function $g_1:Y\to\Bbb R$ such that
$g=g_1T$;  $g_1$ is $\CalBa(Y)$-measurable, so $g$ is
$\Sigma$-measurable.   As this is true for every $\frak T_s$-continuous
$g:X\to\Bbb R$, $\CalBa(X)\subseteq\Sigma$.

\medskip

{\bf (d)} On the other hand, $\Sigma\subseteq\CalBa(X)$ just because
every member of $X^*$ is $\frak T_s$-continuous.   So
$\Sigma=\CalBa(X)$, as claimed.
}%end of proof of 4A3V

\leader{4A3W}{Proposition}\dvAformerly{4A3V} Let $(X,\frak T)$ 
be a separable metrizable
locally convex linear topological space, and
$\frak T_s=\frak T_s(X,X^*)$ its weak topology.   Then the cylindrical
$\sigma$-algebra of $X$ is also both the Baire $\sigma$-algebra and the
Borel $\sigma$-algebra for both $\frak T$ and $\frak T_s$.

\proof{{\bf (a)} For any linear topological space, we have

\Centerline{$\Sigma\subseteq\CalBa(X,\frak T_s)
\subseteq\CalBa(X,\frak T)\subseteq\Cal B(X,\frak T)$,
\quad$\CalBa(X,\frak T_s)\subseteq\Cal B(X,\frak T_s)
\subseteq\Cal B(X,\frak T)$,}

\noindent writing $\Sigma$ for the cylindrical $\sigma$-algebra and
$\CalBa(X,\frak T_s)$, $\CalBa(X,\frak T)$, $\Cal B(X,\frak T_s)$
and $\Cal B(X,\frak T)$ for the Baire and Borel $\sigma$-algebras of the
two topologies.   So all I have to do is to show that
$\Cal B(X,\frak T)\subseteq\Sigma$.

\medskip

{\bf (b)} Let $\sequencen{\tau_n}$ be a sequence of seminorms defining the
topology of $X$ (4A4Cf), and $D\subseteq X$ a countable dense set;  for
$n\in\Bbb N$, $\delta>0$ and $x\in X$, set
$U_n(x,\delta)=\{y:\tau_i(y-x)<\delta$ for every $i\le n\}$.   Then
every $U_n(x,\delta)$ is a convex open set.
Set

\Centerline{$\Cal U=\{U_n(z,2^{-m}):z\in D,\,m,\,n\in\Bbb N\}$.}

\medskip

{\bf (c)} If $G\subseteq X$ is any convex open set,
$\overline{G}\in\Sigma$.
\Prf\ Set $\Cal U_G=\{U:U\in\Cal U,\,U\cap G=\emptyset\}$.
For each $U\in\Cal U_G$, there are $f_U\in X^*$, $\alpha_U\in\Bbb R$
such that $f_U(x)<\alpha_U$ for every $x\in G$ and $f_U(x)>\alpha_U$ for
every $x\in U$ (4A4Db).   So $F=\{x:f_U(x)\le\alpha_U$ for every
$U\in\Cal U_G\}$ belongs to $\Sigma$.   Of course
$F\supseteq\overline{G}$.   On the other hand, if $x\notin\overline{G}$,
there are $m$, $n\in\Bbb N$ such that
$G\cap U_n(x,2^{-m})=\emptyset$;  if we take
$z\in D\cap U_n(x,2^{-m-1})$, $U=U_n(z,2^{-m-1})$ then
$x\in U\in\Cal U_G$, so $f_U(x)>\alpha_U$ and $x\notin F$.   Thus
$\overline{G}=F$ belongs to $\Sigma$.\ \Qed

\medskip

{\bf (d)} In particular, $\Cal V=\{\overline{U}:U\in\Cal U\}$ is
included in $\Sigma$.   But $\Cal V$ is a countable network for
$\frak T$.   So every member of $\frak T$ is a union of countably many
members of $\Sigma$ and belongs to $\Sigma$.   It follows at once that
$\Cal B(X,\frak T)\subseteq\Sigma$, as required.
}%end of proof of 4A3W

\exercises{\leader{4A3X}{Basic exercises (a)}
%\spheader 4A3Xa
Let $X$ be a regular space with a countable network and
$Y$ any topological space.   Show that
$\Cal B(X\times Y)=\Cal B(X)\tensorhat\Cal B(Y)$.  \Hint{4A2Ng.}
%4A3G

\spheader 4A3Xb Let $X$ be a topological space, and $E\subseteq X$.
Show that the following are equiveridical:   ($\alpha$) $E\in\CalBa(X)$;
($\beta$) there are a continuous function
$f:X\to[0,1]^{\Bbb N}$ and $F\in\Cal B([0,1]^{\Bbb N})$ such
that $E=f^{-1}[F]$.
%4A3L

\sqheader 4A3Xc Let $X$ be a topological space, and $K\subseteq X$ a
compact set such that $K\in\CalBa(X)$.   Show that $K$ is a zero set.
\Hint{4A3Xb.}
%4A3Xb 4A3L

\spheader 4A3Xd Let $X$ be a compact Hausdorff space such that
$\CalBa(X)=\Cal B(X)$.   Show that $X$ is perfectly normal.
%4A3L 4A3Xc

\spheader 4A3Xe Let $\frak S$ be the topology on $\Bbb R$ generated by
the usual topology and $\{\{x\}:x\in\Bbb R\setminus\Bbb Q\}$.   Show
that $\frak S$ is completely regular and Hausdorff and that $\Bbb Q$ is
a closed Baire set which is not a zero set.
%4A3Xc 4A3L
%what if we have local compactness?

\sqheader 4A3Xf Let $X$ be a ccc completely regular topological space.
Show that any nowhere dense set is included in a nowhere dense zero set.
%4A3M repeated in 5A4E(d-i)

\spheader 4A3Xg Let $\familyiI{X_i}$ be a family of spaces with
countable networks, and $X$ their product.   Show that
$\CalBa(X)=\Tensorhat_{i\in I}\CalBa(X_i)$.
%4A3N

\sqheader 4A3Xh(i) Let $I$ be an uncountable set with its discrete
topology, and $X$ the one-point compactification of $I$.   Show that
$\CalBa(I)$ is not the subspace $\sigma$-algebra generated by
$\CalBa(X)$.  (ii) Show that $\omega_1$, with its order topology,
has a subset $I$ such that
$\CalBa(I)$ is not the subspace $\sigma$-algebra induced by
$\CalBa(\omega_1)$.
%4A3Kb 4A3P

\leader{4A3Y}{Further exercises (a)}\dvAnew{2017}
%\spheader 4A3Ya
Give an example of a Hausdorff space $X$ with a countable network and a
metrizable space $Y$ such that
$\Cal B(X\times Y)\ne\Cal B(X)\tensorhat\Cal B(Y)$.
%4A3Xa 4A3G

\spheader 4A3Yb
Let $X$ be a \v{C}ech-complete space and $\sequencen{E_n}$ a sequence of
Borel sets in $X$.   Show that there is a \v{C}ech-complete topology on
$X$, finer than the given topology, with the same weight, and containing
every $E_n$.
%4A3I use argument for Polish spaces

\spheader 4A3Yc\dvAnew{2017}
Give an example of compact Hausdorff spaces $X$, $Y$ and a
function $f:X\to Y$ which is $(\CalBa(X),\CalBa(Y))$-measurable but not
Borel measurable.
%4A3K 

\spheader 4A3Yd Let $\familyiI{X_i}$ be a family of regular spaces with
countable networks, with product $X$.   Show that (i) $\CalBa(X)$ is the
family of Borel subsets of $X$ which are
determined by coordinates in countable sets;  (ii) a set $Z\subseteq X$
is a zero set iff it is closed and
determined by coordinates in a countable set;  (iii) if $Y\subseteq X$
is dense, then $\CalBa(Y)$ is just the subspace
$\sigma$-algebra $\CalBa(X)_Y$ induced by $\CalBa(X)$.
%4A3N mt4a3bits

\spheader 4A3Ye Let $X$ be a normal topological space and $Y$ a closed
subset of $X$.
Show that every Baire subset of $Y$ is the intersection of $Y$ with a
Baire subset of $X$.   %\Hint{use Tietze's theorem.}
%4A3L 4{}12P 4A3N

%Skorokhod metric?
}%end of exercises

\endnotes{
\Notesheader{4A3} Much of this section consists of easy technicalities.
It is however not always easy to guess at the exact results obtainable
by these methods.
It is important to notice that Baire $\sigma$-algebras on subspaces can
give difficulties which do not arise with Borel $\sigma$-algebras
(4A3Xh, 4A3Ca).   I have expressed 4A3D in terms of `hereditarily
Lindel\"of' spaces.   Of course the separable metrizable spaces form by
far the most important class of these, but there are others (the split
interval, for instance) which are of great interest in measure theory.
Similarly, there are important products of non-metrizable spaces which
are ccc (e.g., 417Xt(vii)), so that 4A3Mb has something to say.

4A3H-4A3I are a most useful tool in studying Borel subsets of Polish
spaces, especially in conjunction with the First Separation Theorem
(422I);  see 424G and 424H.
I include 4A3Yb and 4A3Yd to show that some more of the arguments here
can be adapted to non-separable or non-metrizable spaces.

You will note my caution in the definition of `Baire measurable'
function (4A3Ke).   This is supposed to cover the case of functions
taking values in $[-\infty,\infty]$ without taking a position on
functions between general topological spaces (4A3Yc).

It is relatively easy to show that spaces of \cadlag\ functions have
standard Borel structures (4A3Qb).   To exhibit usable complete metrics
generating these is another matter;  see chap.\ 3 of {\smc Billingsley 99}.
}%end of notes

\discrpage

\frfilename{mt4a4.tex}
\versiondate{19.6.13}
\copyrightdate{2002}

\def\Bourbaki{{\smc Bourbaki 87}}
\def\Schaefer{{\smc Schaefer 71}}
\def\Kothe{{\smc K\"othe 69}}
%\def\NB{{> Narici \& Beckenstein 85}}
\def\DS{{\smc Dunford \& Schwartz 57}}
\def\Rudin{{\smc Rudin 91}}
\def\Taylor{{\smc Taylor 64}}
\def\Jameson{{\smc Jameson 74}}

\def\chaptername{Appendix}
\def\sectionname{Locally convex spaces}

\newsection{4A4}

As in \S3A5, all the ideas, and nearly all the results as stated below,
are applicable to complex linear spaces;  but for the purposes of this
volume the real case will almost always
be sufficient, and for definiteness you may
take it that the scalar field is $\Bbb R$, except in 4A4J-4A4K.
(Complex Hilbert spaces arise naturally in \S445.)

\leader{4A4A}{Linear spaces (a)} If $U$ is a linear space, a
{\bf Hamel basis} for $U$ is a maximal linearly independent family
$\familyiI{u_i}$ in $U$, so that every member of $U$ is uniquely
expressible as $\sum_{i\in J}\alpha_iu_i$ for some finite
$J\subseteq I$ and $\family{i}{J}{\alpha_i}\in(\Bbb R\setminus\{0\})^J$.
%4A4

\spheader 4A4Ab Every linear space has a Hamel basis.
\prooflet{(\Schaefer, p.\ 10;  \Kothe, \S7.3.)}
%Bourbaki, Algebra, p 25

\spheader 4A4Ac If $U$ is a linear space, I write $U'$ for the
{\bf algebraic dual} of $U$, the linear
space of all linear functionals from $U$ to $\Bbb R$.
%4A4

\leader{4A4B}{Linear topological spaces}\cmmnt{ (see \S2A5)}
{\bf (a)} If $U$ is a linear topological space, and $V$ is a linear
subspace of $U$, then $V$, with the linear structure and topology
induced by those of $U$, is again a linear topological space.
\prooflet{(\Bourbaki, I.1.3; \Schaefer, \S I.2; \Kothe, \S15.2.)}

\spheader 4A4Bb If $\familyiI{U_i}$ is any family of linear topological
spaces, then $U=\prod_{i\in I}U_i$, with the product linear space
structure and topology, is again a linear topological space.
\prooflet{(\Bourbaki, I.1.3;  \Schaefer, \S I.2;  \Kothe, \S15.4.)}
% \NB, \S5.6; \KN, II.5.9
\cmmnt{In particular, }$\Bbb R^X$, with its usual linear and topological
structures, is a linear topological space, for any set $X$.
%4A4

\spheader 4A4Bc If $U$ and $V$ are linear topological spaces, the set of
continuous linear operators from $U$ to $V$ is a linear subspace of the
space $\eurm L(U;V)$ of all linear operators from $U$ to $V$.
\prooflet{}
If $U$, $V$ and $W$ are linear topological spaces, and $T:U\to V$ and
$S:V\to W$ are continuous linear operators, then $ST:U\to W$ is a
continuous linear operator.
%4A4Bd etc

\spheader 4A4Bd If $U$ is a linear topological space, I will write $U^*$
for the {\bf dual} of $U$, the space of all continuous linear
functionals from $U$ to
$\Bbb R$\cmmnt{ (compare 2A4H)}.   $U^*$ is a linear subspace of
$U'$\cmmnt{ as defined in 4A4Ac}.   The {\bf weak topology} on $U$,
$\frak T_s(U,U^*)$, is that defined\cmmnt{ by the method of 2A5B} from
the seminorms $u\mapsto|f(u)|$ as $f$ runs over
$U^*$\cmmnt{ (compare
2A5Ia)}. The {\bf weak* topology} on $U^*$, $\frak T_s(U^*,U)$, is
that defined from the seminorms
$f\mapsto|f(u)|$ as $u$ runs over $U$\cmmnt{ (compare 2A5Ig)}.
\cmmnt{By 2A5B, both are linear space topologies.}
If $U$ and $V$ are linear topological spaces, $T:U\to V$ is a continuous
linear operator, and $g\in V^*$, then $gT\in U^*$\prooflet{ ((c)
above)};  \cmmnt{consequently} $T$ is
$(\frak T_s(U,U^*),\frak T_s(V,V^*))$-continuous.

\spheader 4A4Be If $U=\prod_{i\in I}U_i$ is a product of linear
topological spaces, then every element of $U^*$ is of the form
$u\mapsto\sum_{i\in J}f_i(u(i))$ where $J\subseteq I$ is finite and
$f_i\in U_i^*$ for every $i\in J$.
\prooflet{(\Bourbaki, II.6.6; \Schaefer, IV.4.3;  \Kothe, \S22.5.)}
Consequently the weak topology on $U$ is the
product of the weak topologies on the $U_i$.

\spheader 4A4Bf Let $U$ be a linear topological space.   For $A\subseteq U$
write $A^{\smallcirc}$ for its {\bf polar} set
$\{f:f\in U^*$, $f(x)\le 1$ for every $x\in A\}$ in $U^*$.
If $G$ is a neighbourhood of $0$ in $U$, then $G^{\smallcirc}$ is a
$\frak T_s(U^*,U)$-compact subset of $U^*$\cmmnt{ (compare 3A5F)}.
\prooflet{(\Schaefer, III.4.3;  \Kothe, \S20.9;  \Rudin, 3.15.)}

\spheader 4A4Bg Let $U$ be a linear topological
space.   If $D\subseteq U$ is non-empty and closed under addition and
multiplication by rationals, $\overline{D}$ is a linear subspace of $U$.
\prooflet{\Prf\ The linear span

\Centerline{$V=\{\sum_{i=0}^n\alpha_iu_i:u_0,\ldots,u_n\in D,\,
\alpha_0,\ldots,\alpha_n\in\Bbb R\}$}

\noindent of

\Centerline{$D=\{\sum_{i=0}^n\alpha_iu_i:u_0,\ldots,u_n\in D,\,
\alpha_0,\ldots,\alpha_n\in\Bbb Q\}$}

\noindent is included in $\overline{D}$, because addition and scalar
multiplication are continuous;  so $\overline{D}=\overline{V}$ is a
linear subspace.\ \QeD}
If $A\subseteq U$ is separable, then the closed linear subspace
generated by $A$ is separable.   \prooflet{\Prf\ Let $D_0\subseteq A$ be
a countable dense subset;  then

\Centerline{$D=\{\sum_{i=0}^n\alpha_iu_i:u_0,\ldots,u_n\in D_0,\,
\alpha_0,\ldots,\alpha_n\in\Bbb Q\}$}

\noindent is countable, and $\overline{D}$ is separable;  but
$\overline{D}$ is the closed linear subspace generated by $A$.\ \Qed}
%5{}27

\spheader 4A4Bh If $\familyiI{u_i}$ is an indexed family in a Hausdorff
linear topological space $U$ and $u\in U$, we say that
$u=\sum_{i\in I}u_i$ if for every neighbourhood $G$ of $u$ there is a
finite set $J\subseteq I$ such that $\sum_{i\in K}u_i\in G$ whenever
$K\subseteq I$ is finite and $J\subseteq K$\cmmnt{ (compare 226Ad)}.

If $\familyiI{v_i}$ is another family with the same index set, and
$v=\sum_{i\in I}v_i$ is defined, then $\sum_{i\in I}(u_i+v_i)$ is defined
and equal to $u+v$.   \prooflet{\Prf\
If $G$ is a neighbourhood of $u+v$, there
are neighbourhoods $H$, $H'$ of $u$, $v$ respectively such that
$H+H'\subseteq G$;  there are finite sets $J$, $J'\subseteq I$ such that
$\sum_{i\in K}u_i\in H$ whenever $J\subseteq K\in[I]^{<\omega}$ and
$\sum_{i\in K}v_i\in H'$ whenever $J'\subseteq K\in[I]^{\omega}$;  now
$\sum_{i\in K}u_i+v_i\in G$ whenever
$J\cup J'\subseteq K\in[I]^{<\omega}$.\ \Qed}

If now $V$ is another Hausdorff linear topological space and $T:U\to V$
is a continuous linear operator,
$\sum_{i\in I}Tu_i=T(\sum_{i\in I}u_i)$ if the right-hand-side is
defined.  \prooflet{\Prf\ Set $u=\sum_{i\in I}u_i$.   If $H$ is an open
set containing $Tu$, then $T^{-1}[H]$ is an open set containing $u$, so
there is a $J\in[I]^{<\omega}$ (notation: 3A1J) such that
$\sum_{i\in K}u_i\in T^{-1}[H]$ and $\sum_{i\in K}Tu_i\in H$ whenever
$J\subseteq K\in[I]^{<\omega}$.\ \Qed}
%4A4Ie

\spheader 4A4Bi If $U$ is a Hausdorff linear topological space, then any
finite-dimensional linear subspace of $U$ is closed.
\prooflet{(\Schaefer I.3.3;  \Taylor, 3.12-C;  \Rudin, 1.2.1.)}
%\Bourbaki\query

\spheader 4A4Bj\dvAnew{2011}
If $U$ is a first-countable Hausdorff linear topological space which
(regarded as a linear topological space) is complete, then there is a
metric $\rho$ on $U$, defining its topology, under which $U$ is complete.
\prooflet{\Prf\ Let $\Cal W$ be the uniformity of $U$ (3A4Ad).   We know
there is a sequence $\sequencen{G_n}$ running over a base of neighbourhoods
of $0$ in $U$;  setting $W_n=\{(u,v):u-v\in G_n\}$ for each $n$,
$\{W_n:n\in\Bbb N\}$ generates $\Cal W$.   So there is a metric $\rho$ on
$U$ defining $\Cal W$ (4A2Jb).   Because $X$ is $\Cal W$-complete, it is
$\rho$-complete ({\smc Engelking 89}, 8.3.5).\ \Qed}

\leaveitout{
\spheader  If $U$ is a Hausdorff linear topological space, then its
completion $\hat U$\cmmnt{ (3A4H)} has a unique linear structure
extending that of $U$ and rendering it a linear topological space.
\prooflet{(\Bourbaki, I.1.4;  \Schaefer, I.1.5;  \Kothe, \S15.3.)}
%\NB, 5.7.3;  \KN, II.7.11
Every member of $U^*$ has a unique extension to a member of $\hat U^*$,
so $U^*$ and $\hat U^*$ can be identified as linear spaces.
\prooflet{(\Bourbaki, III.3.3;  \Kothe, \S15.9.)}
}%end of leaveitout

\leader{4A4C}{Locally convex spaces (a)} A linear topological space
is {\bf locally convex} if the convex open sets
form a base for the topology.

\spheader 4A4Cb A linear topological space is locally convex iff its
topology can be defined by a family of seminorms\cmmnt{ (2A5B, 2A5D)}.
\prooflet{(\Bourbaki, II.4.1;  \Schaefer, \S II.4;  \Kothe, \S18.1.)}
%\KN, II.6.6;  \NB, 6.5.1
%??

\spheader 4A4Cc Let $U$ be a linear space and $\tau$ a seminorm on $U$.
\cmmnt{Then} $N_{\tau}=\{u:\tau(u)=0\}$ is a linear subspace of $X$.
On the quotient space $U/N_{\tau}$ we have a norm defined by setting
$\|u^{\ssbullet}\|=\tau(u)$ for every $u\in U$.
\prooflet{(\Bourbaki, II.1.3;  \Schaefer, II.5.4;  \Rudin, 1.43.)}

\spheader 4A4Cd Let $U$ be a locally convex linear topological space,
and $\Tau$ the family of continuous seminorms on $U$.   For each
$\tau\in\Tau$, write $N_{\tau}=\{u:\tau(u)=0\}$\cmmnt{, as in (c)
above,} and $\pi_{\tau}$ for the canonical map from
$U$ to $U_{\tau}=U/N_{\tau}$.   Give each $U_{\tau}$ its norm, and set
$\Cal G_{\tau}=\{\pi_{\tau}^{-1}[H]:H\subseteq U_{\tau}$ is open$\}$.
Then $\bigcup_{\tau\in\Tau}\Cal G_{\tau}$ is a base for the topology of
$X$ closed under finite
%actually, countable
unions.
%and finite intersections.
\prooflet{(\Schaefer, II.5.4.)}

\spheader 4A4Ce A linear subspace of a locally convex linear topological
space is locally convex.  \prooflet{(\Bourbaki, II.4.3;  \Kothe,
\S18.3.)}   The product of any family of
locally convex linear topological
spaces\cmmnt{ (4A4Bb)} is locally convex.
\prooflet{(\Bourbaki, II.4.3;  \Kothe, \S18.3.)}
%\Grothendieck, \S1.6

\spheader 4A4Cf If $U$ is a metrizable locally convex linear topological
space, its topology can be defined by a sequence of seminorms.
\prooflet{(\Bourbaki, II.4.1;  \Kothe, \S18.2.)}

\spheader 4A4Cg Let $U$ be a linear space and $V$ a linear subspace
of\cmmnt{ the space} $U'$\cmmnt{ of all linear functionals on $U$}.
Let $\frak T_s(V,U)$ be the topology on $V$ generated by the
seminorms $f\mapsto|f(u)|$ as $u$ runs over $U$\cmmnt{ (compare
4A4Bd)}, and let $\phi:V\to\Bbb R$ be a $\frak T_s(V,U)$-continuous
linear functional.   Then there is a $u\in U$ such that $\phi(f)=f(u)$
for every $f\in V$.
\prooflet{(\Bourbaki, IV.1.1;  \Schaefer, IV.1.2;  \Kothe, \S20.2;
\Rudin, 3.10;  \DS, II.3.9;  \Taylor, 3.81-A.)}

\spheader 4A4Ch {\bf Grothendieck's theorem} If $U$ is a complete
locally convex Hausdorff linear topological space, and $\phi$ is a
linear functional on the dual $U^*$ such that
$\phi\restr G^{\smallcirc}$ is
$\frak T_s(U^*,U)$-continuous for every neighbourhood $G$ of $0$ in
$U$, then $\phi$ is of
the form $f\mapsto f(u)$ for some $u\in U$.
\prooflet{(\Bourbaki, III.3.6;  \Schaefer, IV.6.2;  \Kothe, \S21.9.)}

\leaveitout{
\spheader  The completion of a Hausdorff locally convex linear
topological space\cmmnt{ ()} is locally convex.
\prooflet{(\Bourbaki, II.4.1;  \Schaefer, II.4.1;  \Kothe, \S18.4.)}
%\Grothendieck, \S1.6
If $U$ is a complete Hausdorff locally convex linear topological space
it is isomorphic to a closed subspace of a product of Banach spaces.
\prooflet{(\Bourbaki, II.4.3;  \Schaefer, II.5.4;  \Kothe, \S18.3.)}
}%end of leaveitout

\leader{4A4D}{Hahn-Banach theorem (a)} Let $U$ be a linear space and
$\theta:U\to\coint{0,\infty}$ a seminorm.

\quad(i) If $V\subseteq U$ is a
linear subspace and $g:V\to\Bbb R$ is a linear functional such that
$|g(v)|\le\theta(v)$ for every $v\in V$, then there is a linear
functional $f:U\to\Bbb R$, extending $g$, such that $|f(u)|\le\theta(u)$
for every $u\in U$.

\quad(ii) If $u_0\in U$ then there is
a linear functional $f:U\to\Bbb R$ such that $f(u_0)=\theta(u_0)$ and
$|f(u)|\le\theta(u)$ for every $u\in U$.
\prooflet{(\Bourbaki, II.3.2;  \Rudin, 3.3;
\DS, II.3.11;  \Taylor, 3.7-C;  or use 3A5Aa.)}  
%\NB, 8.4.5;  \KN, I.3.4
%\query check exact refs

\spheader 4A4Db Let $U$ be a linear topological space and $G$, $H$ two
disjoint convex sets in $U$, of which one has non-empty interior.   Then
there are a non-zero $f\in U^*$ and an
$\alpha\in\Bbb R$ such that $f(u)\le\alpha\le f(v)$ for every $u\in G$,
$v\in H$, so that $f(u)<\alpha$ for every $u\in\interior G$ and
$\alpha<f(v)$ for every $u\in\interior H$.
\prooflet{(\Bourbaki, II.5.2;  \Schaefer, II.9.1;  \Kothe,
\S17.1.)}

\leader{4A4E}{The Hahn-Banach theorem in locally convex spaces} Let $U$
be a locally convex linear topological space.

\spheader 4A4Ea If $V\subseteq U$ is a linear subspace, then every
member of $V^*$ extends to a member of $U^*$\cmmnt{ (compare 3A5Ab)}.
\prooflet{(\Bourbaki, II.4.1;  \Schaefer, II.4.2;  \Kothe, \S20.1;
\Rudin, 3.6;  \Taylor, 3.8-D.)}
%\NB, 8.4.6
\cmmnt{

Consequently }$\frak T_s(V,V^*)$ is\cmmnt{ just} the subspace
topology on $V$ induced by $\frak T_s(U,U^*)$.  \prooflet{}

\spheader 4A4Eb Let $C\subseteq U$ be a non-empty closed convex set.
If $u\in U$ then $u\in C$ iff $f(u)\le\sup_{v\in C}f(v)$ for every
$f\in U^*$ iff $f(u)\ge\inf_{v\in C}f(v)$ for every $f\in U^*$.
\prooflet{(\Bourbaki, II.5.3;  \Schaefer, II.9.2;  \Kothe, \S20.7;  \DS,
V.2.12.)}

If $V\subseteq U$ is a closed linear subspace and $u\in U\setminus V$
there
is an $f\in U^*$ such that $f(u)\ne 0$ and $f(v)=0$ for every $v\in V$.
\prooflet{(\Bourbaki, II.5.3;  \Kothe, \S20.1;  \Rudin, 3.5;  \Taylor,
3.8-E.)}
%\NB, 8.5.5

\spheader 4A4Ec If $U$ is Hausdorff, $U^*$ separates its
points\cmmnt{ (compare 3A5Ae)}.
\prooflet{(\Bourbaki, II.4.1;  \Rudin, 3.4.)}
%\NB, 8.5.5

\spheader 4A4Ed If $u\in U$ belongs to the $\frak T_s(U,U^*)$-closure of
a convex set $C\subseteq U$, it belongs to the closure of
$C$\cmmnt{ (compare 3A5Ee)}.
\prooflet{(\Schaefer, II.9.2;  \Kothe, \S20.7;  \Rudin, 3.12.)}
%\NB, 9.7.1
In particular, if $C$ is closed, it is $\frak T_s(U,U^*)$-closed.
\prooflet{(\DS, V.2.13.)}

\spheader 4A4Ee If $C$, $C'\subseteq U$ are disjoint non-empty closed
convex sets, of
which one is compact, there is an $f\in U^*$ such that
$\sup_{u\in C}f(u)<\inf_{u\in C'}f(u)$.
\prooflet{(Apply (b) to $C-C'$.  See \Bourbaki, II.5.3;  \Schaefer,
II.9.2;  \Kothe, \S20.7; \Rudin, 3.4;  \DS, V.3.13.)}
%\NB, 8.6.5

\spheader 4A4Ef Let $V$ be a linear subspace of
$U'$.   Let $K\subseteq U$ be a non-empty $\frak T_s(U,V)$-compact
convex set, and $\phi_0:V\to\Bbb R$ a linear functional such that
$\phi_0(f)\le\sup_{u\in K}f(u)$ for every $f\in V$.   Then there is a
$u_0\in K$ such that $\phi_0(f)=f(u_0)$ for every $f\in V$.
\prooflet{\Prf\ Give $U$ the topology $\frak T_s(U,V)$ and $V'$ the
topology $\frak T_s(V',V)$
(4A4Cg).   For $u\in U$, $f\in V$ set $\hat u(f)=f(u)$;  then
$u\mapsto\hat u$ is a continuous linear operator from $U$ to $V'$
(use 2A3H), so
$\hat K=\{\hat u:u\in K\}$ is a compact convex subset of $V'$.

\Quer\ Suppose, if possible, that $\phi_0\notin\hat K$.   By (b),
there is a continuous linear functional $\pmb{\theta}:V'\to\Bbb R$
such that $\pmb{\theta}(\phi_0)>\sup_{u\in K}\pmb{\theta}(\hat u)$.   But
there is an $f\in V$ such that $\pmb{\theta}(\phi)=\phi(f)$ for every
$\phi\in V'$ (4A4Cg), so that
$\phi_0(f)>\sup_{u\in K}f(u)$, contrary to hypothesis.\ \BanG\  So there
is a $u_0\in K$ such that $\phi_0=\hat u_0$, as claimed.\ \Qed}

\spheader 4A4Eg {\bf The Bipolar Theorem} Let $A\subseteq U'$ be a
non-empty set.   Set
$A^{\smallcirc}=\{u:u\in U$, $f(u)\le 1$ for every
$f\in A\}$\cmmnt{ (compare 4A4Bf)}.   If $g\in U'$ is such that
$g(u)\le 1$ for every $u\in A^{\smallcirc}$, then $g$ belongs to the
$\frak T_s(U',U)$-closed convex hull of $A\cup\{0\}$.
\prooflet{\Prf\ Put 4A4Cg and (b) above together, as in (f).   See
\Bourbaki,
II.6.3;  \Schaefer, IV.1.5;  \Kothe, 20.8.\ \Qed}

\spheader 4A4Eh Let $W$ be a linear subspace of $U'$
separating the points of $U$.   Then $W$ is $\frak T_s(U',U)$-dense in
$U'$.   \prooflet{(For $W^0=\{0\}$.)}

\leader{4A4F}{The Mackey topology} Let $U$ be a linear space and $V$ a
linear subspace of $U'$.   The {\bf Mackey topology} $\frak T_k(V,U)$ on
$V$ is the topology of uniform convergence on convex
$\frak T_s(U,V)$-compact subsets of $U$.   Every
$\frak T_k(V,U)$-continuous linear functional on $V$ is of the form
$f\mapsto f(u)$ for some $u\in U$\prooflet{ (use 4A4Ef)}.   So every
$\frak T_k(V,U)$-closed convex set is
$\frak T_s(V,U)$-closed\prooflet{, by 4A4Ed}.   \cmmnt{(See \Bourbaki,
IV.1.1;  \Schaefer, IV.3.2;  \Kothe, 21.4.)}

\leader{4A4G}{Extreme points (a)} Let $X$ be a real linear space, and
$C\subseteq X$ a convex set.   An element of $C$ is an {\bf extreme}
point of $C$ if it is not expressible as a convex combination of two
other members of $C$\cmmnt{;  equivalently, if it is not expressible
as $\bover12(x+y)$ where $x$, $y\in C$ are distinct}.

\spheader 4A4Gb {\bf The Kre\v\i n-Mil'man theorem} Let $U$ be a
Hausdorff locally convex linear topological space and $K\subseteq U$ a
compact convex set.   Then $K$ is the closed convex hull of the set of
its extreme points.
\prooflet{(\Bourbaki, II.7.1;  \Schaefer, II.10.4;  \Kothe, \S25.1;
\Rudin, 3.22.)}
%\NB, 10.3.4

\spheader 4A4Gc Let $U$ and $V$ be Hausdorff locally convex linear
topological spaces, $T:U\to V$ a continuous linear operator,
$K\subseteq X$ a compact convex set and $v$ any extreme point of
$T[K]\subseteq V$.   Then there is an extreme point $u$ of $K$ such
that $Tu=v$.  \prooflet{\Prf\ Set $K_1=\{u':u'\in K,\,Tu'=v\}$.   Then
$K_1$ is a compact convex set so has an extreme point $u$.   \Quer\
If $u$ is not an extreme point of $K$, it is expressible as
$\alpha u_1+(1-\alpha)u_2$ where $u_1$, $u_2$ are distinct points of $K$
and $\alpha\in\ooint{0,1}$.   So $v=\alpha Tu_1+(1-\alpha)Tu_2$, and we
must
have $Tu_1=Tv=Tu_2$, because $v$ is an extreme point of $T[K]$;  but
this means that $u_1$, $u_2\in K_1$ and $u$ is not an extreme point of
$K_1$.\ \BanG\  So $u$ has the required properties.\ \Qed}

\leader{4A4H}{Proposition} Let $I$ be a set, $W$ a closed linear
subspace of $\BbbR^I$, $U$ a linear topological
space and $V$ a Hausdorff linear topological space.   Let $K\subseteq U$
be a compact set and $T:U\times\BbbR^I\to V$ a continuous linear
operator.   Then $T[K\times W]$ is closed.

\proof{ Take $v_0\in\overline{T[K\times W]}$.   Let $J\subseteq I$ be a
maximal set such that

\inset{\noindent whenever $L\subseteq J$ is finite and $H\subseteq V$ is
an open set containing $v_0$, there are a $u\in K$ and an $x\in W$
such that $x(i)=0$ for every $i\in L$ and $T(u,x)\in H$.}

\noindent Let $\Cal F$ be the filter on $U\times\BbbR^I$ generated by
the closed set $K\times W$, the sets $\{(u,x):x(i)=0\}$ for $i\in J$,
and the sets
$T^{-1}[H]$ for open sets $H$ containing $v_0$.   Then for any $j\in I$
there is an $F\in\Cal F$ such that $\{x(j):(u,x)\in F\}$ is bounded.
\Prf\Quer\ Suppose, if possible, otherwise.   Then, in particular,
$j\notin J$.   So there must be a finite set $L\subseteq J$ and an open
set $H$ containing $v_0$ such that $x(j)\ne 0$ whenever $u\in K$,
$x\in W$, $x(i)=0$ for every $i\in L$ and $T(u,x)\in H$.   By 2A5C, or
otherwise, there is a
neighbourhood $G_0$ of $0$ in $V$ such that $v_0+G_0-G_0\subseteq H$ and
$\alpha v\in G_0$ whenever $v\in G_0$ and $|\alpha|\le 1$.
Fix $u^*\in K$, $x^*\in W$ such that $x^*(i)=0$ for every $i\in L$ and
$T(u^*,x^*)\in v_0+G_0$.   If $x\in W$ and $x(i)=0$ for every
$i\in L$ and $x(j)=x^*(j)$, then $T(u^*,x^*-x)\notin v_0+G_0-G_0$ so
$T(0,x)\notin G_0$.   It follows that $T(0,x)\notin G_0$ whenever
$x\in W$ and $x(i)=0$ for every $i\in L$ and $|x(j)|\ge|x^*(j)|$.

Let $G$ be a neighbourhood of $0$ in $V$ such that $G+G-G-G\subseteq
G_0$.   We are supposing that $\{x(j):(u,x)\in F\}$ is unbounded for
every
$F\in\Cal F$.   So for every $n\in\Bbb N$ there are $u_n\in K$ and
$x_n\in W$ such that $x_n(i)=0$ for $i\in L$,
$T(u_n,x_n)\in v_0+G$ and $|x_n(j)|\ge n$.   Let $u\in K$ be a cluster
point of $\sequencen{u_n}$.  Then $T(u,0)$ is a cluster point of
$\sequencen{T(u_n,0)}$, so $M=\{n:T(u_n,0)\in T(u,0)+G\}$ is infinite.
For $n\in M$, $T(0,x_n)=T(u_n,x_n)-T(u_n,0)\in v_0-T(u,0)+G-G$;  so if
$m$,
$n\in M$, $T(0,x_m-x_n)\in G-G-(G-G)\subseteq G_0$.   But note now that
$(x_m-x_n)(i)=0$ for all $m$, $n\in\Bbb N$ and $i\in L$, and that
because $M$ is infinite there are certainly $m$, $n\in M$ such that
$|x_m(j)-x_n(j)|\ge|x^*(j)|$;  which contradicts the last paragraph.\
\Bang\Qed

Now let $\Cal G$ be any ultrafilter on $U\times\BbbR^I$ including
$\Cal F$.   Then for every $i\in I$ there is a $\gamma_i<\infty$ such
that $\Cal G$ contains $\{(u,x):|x(i)|\le\gamma_i\}$.   It follows that
$\Cal G$ has a limit $(\hat u,\hat x)$ in $K\times W$.   Now the
image filter $T[[\Cal G]]$ (2A1Ib) converges to $T(\hat u,\hat x)$;
since $T[[\Cal F]]\to v_0$, and the topology of $V$ is Hausdorff,
$v_0=T(\hat u,\hat x)\in T[K\times W]$.   As $v_0$ is arbitrary,
$T[K\times W]$ is closed.
}%end of proof of 4A4H

\leader{4A4I}{Normed spaces (a)}
Two norms $\|\,\,\|$, $\|\,\,\|'$ on a linear space $U$
give rise to the same topology iff they are {\bf equivalent} in the
sense that, for some $M\ge 0$,

\Centerline{$\|x\|\le M\|x\|'$,\quad$\|x\|'\le M\|x\|$}

\noindent for every $x\in U$.  \prooflet{(\Kothe, \S14.2;
\Taylor, 3.1-D;  \Jameson, 2.8.)}
%\Grothendieck, \S1.5

\spheader 4A4Ib If $U$ and $V$ are normed spaces, $T:U\to V$ is a linear
operator and $gT:U\to\Bbb R$ is continuous for every $g\in V^*$, then
$T$ is a bounded operator.  \prooflet{(\Jameson, 27.6.)}

\spheader 4A4Ic If $U$ is any normed space, its dual $U^*$, under its
usual norm\cmmnt{ (2A4H)}, is a Banach space.  \prooflet{(\Rudin, 4.1;
\DS, II.3.9;  \Kothe, \S14.5.)}
%4A4Id

\spheader 4A4Id If $U$ is a separable normed space, its dual $U^*$
\cmmnt{(regarded as a normed space)} is isometrically
isomorphic to a closed linear subspace of $\ell^{\infty}$.
\prooflet{\Prf\ Let $\sequencen{x_n}$ run over a dense subset of
the unit ball of $U$ (4A2P(a-iv));  define $T:U^*\to\ell^{\infty}$ by
setting $(Tf)(n)=f(x_n)$ for every $n$.   $T$ is a linear isometry
between $U^*$ and $T[U^*]$, which is closed because $U^*$ is complete
(4A4Ic, 3A4Fd).\ \Qed}

\spheader 4A4Ie Let $U$ be a Banach space.   Suppose that
$\familyiI{u_i}$ is a
family in $U$ such that $\gamma=\sum_{i\in I}\|u_i\|<\infty$.

\quad(i) $\sum_{i\in I}u_i$ is defined\cmmnt{ in the sense of 4A4Bh}.
\prooflet{\Prf\ For $J\in[I]^{<\omega}$, set
$v_J=\sum_{i\in J}u_i$.   For each $n\in\Bbb N$, there is a
$J_n\in[I]^{<\omega}$ such that $\gamma-\sum_{i\in K}\|u_i\|\le 2^{-n}$
whenever $J_n\subseteq K\in[I]^{<\omega}$.   Now

\Centerline{$\|v_{J_m}-v_{J_n}\|\le\sum_{i\in J_m\symmdiff J_n}\|u_i\|
\le 2^{-m}+2^{-n}$}

\noindent for all $m$, $n\in\Bbb N$, so $\sequencen{v_{J_n}}$ is a
Cauchy sequence and has a limit $v$ say.   If now $n\in\Bbb N$ and
$J_n\subseteq K\in[I]^{<\omega}$,

$$\eqalign{\|v-\sum_{i\in K}u_i\|
&=\lim_{m\to\infty}\|v_{J_m}-\sum_{i\in K}u_i\|
\le\limsup_{m\to\infty}\sum_{i\in J_m\symmdiff K}\|u_i\|\cr
&\le\lim_{m\to\infty}2^{-m}+2^{-n}
=2^{-n},\cr}$$

\noindent so $v=\sum_{i\in I}u_i$.\ \Qed}

\quad(ii) Now if $\family{j}{J}{I_j}$ is any partition of $I$,
$w_j=\sum_{i\in I_j}u_i$ is defined for every $j$, and
$\sum_{j\in J}w_j$ is defined and equal to $\sum_{i\in I}u_i$.

\spheader 4A4If Let $U$ be a normed space.   For $u\in U$, define
$\hat u\in U^{**}=(U^*)^*$ by setting $\hat u(f)=f(u)$ for every
$f\in U^*$.   Then $\{\hat u:u\in U,\,\|u\|\le 1\}$ is weak*-dense in
$\{\phi:\phi\in U^{**},\,\|\phi\|=1\}$.   \prooflet{(Apply 4A4Eg with
$A=\{\hat u:\|u\|\le 1\}$.)}

\leader{4A4J}{Inner product spaces (a)}
Let $U$ be an inner product space over
$\RoverC$\cmmnt{ (3A5M)}.
An {\bf orthonormal family} in $U$ is a family $\familyiI{e_i}$ in $U$
such that $\innerprod{e_i}{e_j}=0$ if $i\ne j$, $1$ if $i=j$.
An {\bf orthonormal basis} in $U$ is an orthonormal family
$\familyiI{e_i}$ in $U$ such that the closed
linear subspace of $U$ generated by $\{e_i:i\in I\}$ is $U$ itself.

\spheader 4A4Jb If $U$, $V$ are inner product spaces over $\Bbb R$
and $T:U\to V$ is an isometry such that $T(0)=0$, then
$\innerprod{Tu}{Tv}=\innerprod{u}{v}$ for all $u$, $v\in U$ and $T$ is
linear.   \prooflet{\Prf\ ($\alpha$)

\Centerline{$\innerprod{Tu}{Tv}
=\Bover12(\|Tu\|^2+\|Tv\|^2-\|Tu-Tv\|^2)
=\Bover12(\|u\|^2+\|v\|^2-\|u-v\|^2)
=\innerprod{u}{v}$.}

\noindent($\beta$) For any $u$, $v\in U$,

$$\eqalign{\|T(u+v)-Tu-Tv\|^2
&=\|T(u+v)\|^2+\|Tu\|^2+\|Tv\|^2\cr
&\hskip5em   -2\innerprod{T(u+v)}{Tu}-2\innerprod{T(u+v)}{Tv}
   +2\innerprod{Tu}{Tv}\cr
&=\|u+v\|^2+\|u\|^2+\|v\|^2\cr
&\hskip5em -2\innerprod{u+v}{u}-2\innerprod{u+v}{v}
   +2\innerprod{u}{v}\cr
&=0.\cr}$$

\noindent So $T$ is additive.   ($\gamma$) Consequently $T(qu)=qTu$ for
every $u\in U$ and $q\in\Bbb Q$;  as $T$ is continuous, it is linear.\
\Qed}

\spheader 4A4Jc If $U$, $V$ are inner product spaces over $\Bbb C$ and
$T:U\to V$ is a linear operator such that $\|Tu\|=\|u\|$ for every
$u\in U$, then $\innerprod{Tu}{Tv}=\innerprod{u}{v}$ for all $u$, $v\in U$.
\prooflet{\Prf\

\Centerline{$\Real\innerprod{Tu}{Tv}
=\Bover12(\|Tu\|^2+\|Tv\|^2-\|Tu-Tv\|^2)
=\Real\innerprod{u}{v}$,}

\Centerline{$\Imag\innerprod{Tu}{Tv}
=-\Real(i\innerprod{Tu}{Tv})
=-\Real\innerprod{T(iu)}{Tv}
=-\Real\innerprod{iu}{v}
=\Imag\innerprod{u}{v}$.  \Qed}}

\spheader 4A4Jd If $U$ is an inner product space over $\RoverC$, a
linear operator $T:U\to U$ is {\bf self-adjoint} if
$\innerprod{Tu}{v}=\innerprod{u}{Tv}$ for all $u$, $v\in U$.
%4A4M

\spheader 4A4Je If $U$ is a finite-dimensional inner product space over
$\Bbb R$, it
is isomorphic to Euclidean space $\BbbR^r$, where $r=\dim U$.
\prooflet{(\Taylor, 3.21-A.)}   In particular, any
finite-dimensional inner product space is a Hilbert space.

\spheader 4A4Jf If $U$ is an inner product space over $\RoverC$ and
$V\subseteq U$ is a linear subspace of $U$, then
$V^{\perp}=\{x:x\in U,\,\innerprod{x}{y}=0$ for every $y\in V\}$ is a
linear subspace of $U$, and $\|x+y\|^2=\|x\|^2+\|y\|^2$ for $x\in V$,
$y\in V^{\perp}$.   If $V$ is complete (in particular, if $V$ is
finite-dimensional), then $U=V\oplus V^{\perp}$.   \prooflet{(\Rudin,
12.4;  \Bourbaki, V.1.6;  \Taylor, 4.82-A.)}

\spheader 4A4Jg If $U$ is an inner product space over $\Bbb R$ and
$v_1$, $v_2\in U$ are such
that $\|v_1\|=\|v_2\|=1$, there is a linear operator $T:U\to U$ such
that $Tv_1=v_2$ and $\|Tu\|=\|u\|$ and $\|Tu-u\|\le\|v_1-v_2\|\|u\|$ for
every $u\in U$.   \prooflet{\Prf\ If $v_2$ is a multiple of $v_1$, say
$v_2=\alpha v_1$, take $Tu=\alpha u$ for every $u$.   Otherwise, set
$w=v_2-\innerprod{v_2}{v_1}v_1$ and $w_1=\Bover1{\|w\|}w$, so that
$v_2=\cos\theta v_1+\sin\theta w_1$, where
$\theta=\arccos\innerprod{v_2}{v_1}$.   Let $V$ be the two-dimensional
linear subspace of $U$ generated by $v_1$ and $w_1$, so that
$U=V\oplus V^{\perp}$.   Define a linear operator $T:U\to U$ by saying
that $Tv_1=v_2$, $Tw_1=\cos\theta w_1-\sin\theta v_1$ and $Tu=u$ for
$u\in V^{\perp}$.   Then $T$ acts on $V$ as a simple rotation through an
angle $\theta$, so $\|Tv\|=1$ and $\|Tv-v\|=\|v_2-v_1\|$ whenever
$v\in V$ and $\|v\|=1$;  generally, if $u\in U$, then $\|Tu\|=\|u\|$
and

\Centerline{$\|Tu-u\|=\|T(Pu)-Pu\|=\|v_2-v_1\|\|Pu\|
\le\|v_2-v_1\|\|u\|$,}

\noindent where $P$ is the orthogonal projection of $U$ onto $V$.\
\Qed}

\spheader 4A4Jh Let $U$ be an inner product space over $\RoverC$, and
$\familyiI{u_i}$
a countable family in $U$.   Then there is a countable orthonormal
family $\family{j}{J}{v_j}$ in $U$ such that $\{v_j:j\in J\}$ and
$\{u_i:i\in I\}$ span the same linear subspace of $U$.
\prooflet{\Prf\ We can suppose that $I\subseteq\Bbb N$;  set $u_i=0$ for
$i\in\Bbb N\setminus I$.   Define $\sequencen{v_n}$ inductively by
setting $v_n'=u_n-\sum_{i<n}\innerprod{u_n}{v_i}v_i$, $v_n=0$ if
$v'_n=0$, $\Bover1{\|v'_n\|}v'_n$ otherwise.   Set $J=\{n:v_n\ne 0\}$.\
\Qed}

\spheader 4A4Ji\dvAnew{2008}
Let $U$ be an inner product space over $\RoverC$, and
$\family{i}{I}{e_i}$ an orthonormal family in $U$.   Then
$\sum_{i\in I}|\innerprod{u}{e_i}|^2\le\|u\|^2$ for every $u\in U$.
\prooflet{(\DS, p.\ 252;  \Taylor, 3.2-D.)}
\leaveitout{Take any finite $J\subseteq I$ and set
$v=\sum_{i\in J}\innerprod{u}{e_i}e_i$.   Then
$\innerprod{v}{e_i}=\innerprod{u}{e_i}$ for every $i\in J$, so
$\innerprod{u}{v}=\|v\|^2$ and 

\Centerline{$\sum_{i\in J}|\innerprod{u}{e_i}|^2
=\|v\|^2
=2\Real\innerprod{u}{v}-\|v\|^2
=\|u\|^2-\|u-v\|^2
\le\|u\|^2$.}

\noindent As $J$ is arbitrary, 
$\sum_{i\in I}|\innerprod{u}{e_i}|^2\le\|u\|^2$.}

\spheader 4A4Jj Let $U$ be an inner product space over $\RoverC$, and
$C\subseteq U$ a convex set.   Then there is at most one point $u\in C$
such that $\|u\|\le\|v\|$ for every $v\in C$.   \prooflet{\Prf\ If $u$,
$u'$ both have this property, then $v=\bover12(u+u')\in C$, and
$\|u\|=\|u'\|\le\|v\|$;  but $4\|v\|^2+\|u-u'\|^2=2(\|u\|^2+\|u'\|^2)$,
so $\|u-u'\|=0$ and $u=u'$.\ \Qed}

For such a $u$, $\|u\|^2\le\Real\innerprod{u}{v}$ for every $v\in C$.
\prooflet{\Prf\ For $\alpha\in\ocint{0,1}$,

\Centerline{$\|u\|^2\le\|\alpha v+(1-\alpha)u\|^2
=\|u\|^2+2\alpha(\Real\innerprod{u}{v}-\|u\|^2)
  +\alpha^2\|v-u\|^2$,}

\noindent so $\Real\innerprod{u}{v}-\|u\|^2
\ge-\lim_{\alpha\downarrow 0}\bover12\alpha\|v-u\|^2$.\ \Qed}

\leader{4A4K}{Hilbert spaces (a)}
If $U$ is a real or complex Hilbert space, its unit ball is compact in
the weak topology
$\frak T_s(U,U^*)$;  any bounded set is relatively compact for
$\frak T_s(U,U^*)$.  \prooflet{(\Bourbaki, V.1.7;  \DS, IV.4.6.)}

\spheader 4A4Kb If $U$ is a real or complex Hilbert space, any
norm-bounded sequence in $U$ has a weakly convergent subsequence.
\prooflet{(462D;  \DS, IV.4.7.)}

\spheader 4A4Kc If $U$ is a real or complex Hilbert space and
$\familyiI{u_i}$ is any
orthonormal family in $U$, then it can be extended to an orthonormal
basis.   \prooflet{(\DS, IV.4.10;  \Taylor, 3.2-I.)}   In particular,
$U$ has an orthonormal basis.
% \Bourbaki\query

\leader{4A4L}{Compact operators}\cmmnt{ (see 3A5La)}
{\bf (a)} Let $U$, $V$ and $W$ be Banach spaces.   If
$T\in\eurm B(U;V)$ and $S\in\eurm B(V;W)$ and either $S$ or $T$ is a
compact operator, then $ST$ is compact.
\prooflet{(\DS, VI.5.4;  \Jameson, 34.2.)}

\spheader 4A4Lb If $U$ is a Banach space, $T\in\eurm B(U;U)$ is a
compact linear operator and $\gamma\ne 0$ then
$\{u:Tu=\gamma u\}$ is finite-dimensional.
\prooflet{(\Rudin, 4.18;  \Taylor, 5.5-C;  \DS, VII.4.5;  \Jameson,
34.8.)}

\leader{4A4M}{Self-adjoint compact operators} If $U$ is a
Hilbert space and $T:U\to U$ is a
self-adjoint compact linear operator, then $T[U]$ is included in
the closed linear span of $\{Tv:v$ is an eigenvector of $T\}$.
\prooflet{(\Taylor, 6.4-B.)}

\leader{4A4N}{Max-flow Min-cut
Theorem}\cmmnt{ ({\smc Ford \& Fulkerson 56})} Let $(V,E,\gamma)$ be a (finite) transportation network, that is,

\inset{$V$ is a finite set of `vertices',

$E\subseteq\{(v,v'):v,\,v'\in V,\,v\ne v'\}$ is a set of (directed)
`edges',

$\gamma:E\to\coint{0,\infty}$ is a function;}

\noindent we regard a member $e=(v,v')$ of $E$ as `starting' at $v$ and
`ending' at $v'$, and $\gamma(e)$ is the `capacity' of the edge $e$.
Suppose that $v_0$, $v_1\in V$ are distinct vertices
such that no edge ends at $v_0$ and no edge starts at $v_1$.
%is this needed?  yes, for calculation in (iv)
Then we have a `flow' $\phi:E\to\coint{0,\infty}$ and a `cut'
$X\subseteq E$ such that

(i) for every $v\in V\setminus\{v_0,v_1\}$,

\Centerline{$\sum_{e\in E,e\text{ starts at }v}\phi(e)
=\sum_{e\in E,e\text{ ends at }v}\phi(e)$,}

(ii) $\phi(e)\le\gamma(e)$ for every $e\in E$,

(iii) there is no path from $v_0$ to $v_1$ using only edges in
$E\setminus X$,

(iv) $\sum_{e\in E,e\text{ starts at }v_0}\phi(e)
=\sum_{e\in E,e\text{ ends at }v_1}\phi(e)
=\sum_{e\in X}\gamma(e)$.

\proof{ {\smc Bollob\'as 79}, \S III.1;  {\smc Anderson 87},
12.3.1.
}%end of proof of 4A4N

\discrpage

\frfilename{mt4a5.tex}
\versiondate{4.8.13}
\copyrightdate{2000}

\def\chaptername{Appendix}
\def\sectionname{Topological groups}

\def\Csaszar{{\smc Cs\'asz\'ar 78}}
\def\Folland{{\smc Folland 95}}
\def\HR{{\smc Hewitt \& Ross 63}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}

\newsection{4A5}

For Chapter 44 we need a variety of facts about topological groups.
Most are essentially elementary, and all the non-trivial ideas are covered
by at least one of \Csaszar\ and \HR.   In
4A5A-4A5C %4A5A 4A5B 4A5C
and 4A5I I give some simple definitions concerning groups and group
actions.   Topological groups, properly speaking, appear in 4A5D.
Their simplest properties are in 4A5E-4A5G. %\overline{Y}_0 4A5F 4A5G
I introduce `right' and `bilateral' uniformities in 4A5H;  the latter
are the more interesting (4A5M-4A5O),  %4A5M 4A5N 4A5O
but the former are also important\cmmnt{ (see the proof of 4A5P)}.
4A5J-4A5L %4A5J 4A5K 4A5L
deal with quotient spaces, including spaces of cosets of
non-normal subgroups.   I conclude with notes on metrizable groups
(4A5Q-4A5S).  %4A5Q 4A5R 4A5S

%4A5A notation
%4A5B, %4A5C actions
%4A5D top gps
%4A5E elementary props
%4A5F homos
%4A5G products
%4A5H uniformities
%4A5I continuous actions
%4A5J X/Y for subgps Y (uses 4A5C, 4A5I)
%4A5K X/\overline{\{e\}} (uses 4A5H, 4A5J)
%4A5L first iso thm (uses 4A5J)
%4A5M more on uniformities
%4A5N completion
%4A5O totally bounded sets
%4A5P notes on functions in C_k (uses 4A5H, 4A5O)
%4A5Q, 4A5R metrizable top gps
%4A5S X/Y metrizable (uses 4A5O, 4A5H, 4A5J, 4A5Q)

\leader{4A5A}{Notation} If $X$ is a group, $x_0\in X$, and $A$,
$B\subseteq X$ I write

\Centerline{$x_0A=\{x_0x:x\in A\}$,
\quad$Ax_0=\{xx_0:x\in A\}$,}

\Centerline{$AB=\{xy:x\in A,\,y\in B\}$,
\quad$A^{-1}=\{x^{-1}:x\in A\}$.}

\noindent $A$ is {\bf symmetric} if $A=A^{-1}$.
\cmmnt{Observe that $(AB)C=A(BC)$, $(AB)^{-1}=B^{-1}A^{-1}$
for any $A$, $B$, $C\subseteq X$.}

\leader{4A5B}{Group actions (a)} If $X$ is a group and $Z$ is a set, an
{\bf action} of $X$ on $Z$ is a function
$(x,z)\mapsto x\action z:X\times Z\to Z$ such that

\Centerline{$(xy)\action z=x\action(y\action z)$ for all $x$, $y\in X$ and
$z\in Z$,}

\Centerline{$e\action z=z$ for every $z\in Z$}

\noindent where $e$ is the identity of $X$.

In this context I may say that `$X$ acts on $Z$', taking the operation
$\action$ for granted.

\spheader 4A5Bb An action $\action$ of a group $X$ on a set $Z$ is
{\bf transitive} if for every $w$, $z\in Z$ there is an $x\in X$ such
that $x\action w=z$.

\spheader 4A5Bc If $\action$ is an action of a group $X$ on a set $Z$, I
write $x\action A=\{x\action z:z\in A\}$ whenever $x\in X$ and
$A\subseteq Z$.

\spheader 4A5Bd If $\action$ is an action of a group $X$ on a set $Z$,
then $z\mapsto x\action z:Z\to Z$ is a permutation for every $x\in X$.
\prooflet{(For it has an inverse $z\mapsto x^{-1}\action z$.)}   So if
$Z$ is a topological space and $z\mapsto x\action z$ is continuous for
every $x$, it is a homeomorphism for every $x$.

\spheader 4A5Be An action $\action$ of a group $X$ on a set $Z$ is {\bf
faithful} if whenever $x$, $y\in X$ are distinct there is a $z\in Z$
such that $x\action z\ne y\action z$\cmmnt{;  that is, the natural
homomorphism from $X$ to the group of permutations of $Z$ is
injective}.
%Pestov 99:  `effective action'
\cmmnt{ An action of $X$ on $Z$ is faithful iff for any
$x\in X$ which is not the identity there is a $z\in Z$ such that
$x\action z\ne z$.}

\spheader 4A5Bf If $\action$ is an action of a group $X$ on a set $Z$,
then $Y_z=\{x:x\in X,\,x\action z=z\}$ is a subgroup of $X$ (the {\bf
stabilizer} of $z$) for every $z\in Z$.   If $\action$ is transitive,
then $Y_w$ and $Y_z$ are conjugate subgroups for all $w$, $z\in Z$.
\prooflet{(If $x\action w=z$, then $Y_z=xY_wx^{-1}$.)}

\spheader 4A5Ag\dvAnew{2012}
If $\action$ is an action of a group $X$ on a set $Z$,
then sets of the form $\{a\action z:a\in X\}$ are called {\bf orbits} of
the action;  they are the equivalence classes under the equivalence
relation $\sim$, where $z\sim z'$ if there is an $a\in X$ such that
$z'=a\action z$.

\vleader{72pt}{4A5C}{Examples} Let $X$ be any group.

\spheader 4A5Ca Write

\Centerline{$x\action_ly=xy$,
\quad$x\action_ry=yx^{-1}$,
\quad$x\action_cy=xyx^{-1}$}

\noindent for $x$, $y\in X$.   These are all actions of $X$ on itself,
the {\bf left}, {\bf right} and {\bf conjugacy} actions.

\spheader 4A5Cb If $A\subseteq X$, we have an action of $X$ on the set
$\{yA:y\in X\}$ of left cosets of $A$
defined by setting $x\action(yA)=xyA$ for $x$, $y\in X$.

\spheader 4A5Cc{\bf (i)}\dvAformerly{4{}41Ac}
Let $\action$ be an action of a
group $X$ on a set $Z$.   If $f$ is any function defined on a subset of
$Z$, and $x\in X$, write $x\action f$ for the function defined by saying
that $(x\action f)(z)=f(x^{-1}\action z)$ whenever $z\in Z$ and
$x^{-1}\action z\in\dom f$.   It is easy to check that this defines an
action of $X$ on the class of all functions with domains included in
$Z$.   \cmmnt{Observe that}

\Centerline{$x\action(f+g)=(x\action f)+(x\action g)$,
\quad$x\action(f\times g)=(x\action f)\times(x\action g)$,
\quad$x\action(f/g)=(x\action f)/(x\action g)$}

\noindent whenever $x\in X$ and
$f$, $g$ are real-valued functions with domains included
in $Z$.

\medskip

\quad{\bf (ii)} In (i), if $X=Z$, we have
corresponding actions $\action_l$, $\action_r$ and $\action_c$ of
$X$ on the class of functions with domains included in $X$\cmmnt{:

\Centerline{$(x\action_lf)(y)=f(x^{-1}y)$,
\quad$(x\action_rf)(y)=f(yx)$,
\quad$(x\action_cf)(y)=f(x^{-1}yx)$}

\noindent whenever these are defined}.   These are the left, right and
conjugacy {\bf shift actions}.

Note that

\Centerline{$x\action_l\chi A=\chi(xA)$,
\quad$x\action_r\chi A=\chi(Ax^{-1})$,
\quad$x\action_c\chi A=\chi(xAx^{-1})$}

\noindent whenever $A\subseteq X$ and $x\in X$.
In this context, the following idea is sometimes useful.   If $f$ is a
function with domain included in $X$, set $\Reverse{f}(y)=f(y^{-1})$
when $y\in X$ and $y^{-1}\in\dom f$.   Then

\Centerline{$(\Reverse{f})\ssplrarrow=f$,
\quad$x\action_l\Reverse{f}=(x\action_rf)\ssplrarrow$,
\quad$x\action_r\Reverse{f}=(x\action_lf)\ssplrarrow$,
\quad$x\action_c\Reverse{f}=(x\action_cf)\ssplrarrow$}

\noindent for any such $f$ and any $x\in X$.

\spheader 4A5Cd\dvAformerly{4{}41Ad}
If $\action$ is an action of a group $X$ on a set $Z$,
$Y\subseteq X$ is a subgroup of $X$,
and $W\subseteq Z$ is $Y$-invariant in the sense that
$y\action w\in W$ whenever $y\in Y$ and $w\in W$, then
$\action\restr Y\times W$ is an action of $Y$ on $W$.   In the context
of (c-i) above, this means that if $V$ is any set of functions with
domains included in $W$ such that $y\action f\in V$ whenever $y\in Y$
and $f\in V$, then we have an action of $Y$ on $V$.

\leader{4A5D}{Definitions (a)} A {\bf topological group} is a group $X$
endowed with a topology such that the operations
$(x,y)\mapsto xy:X\times X\to X$ and $x\mapsto x^{-1}:X\to X$ are
continuous.

\spheader 4A5Db A {\bf Polish group} is a topological group in which the
topology is Polish.

\leader{4A5E}{Elementary facts} Let $X$ be any topological group.

\spheader 4A5Ea For any $x\in X$, the functions $y\mapsto xy$,
$y\mapsto yx$ and $y\mapsto y^{-1}$ are all homeomorphisms from $X$ to
itself.
\prooflet{(\HR, 4.2;  \Folland, 2.1.)}

\spheader 4A5Eb The maps $(x,y)\mapsto x^{-1}y$, $(x,y)\mapsto xy^{-1}$
and $(x,y)\mapsto xyx^{-1}$ from $X\times X$ to $X$ are continuous.

\spheader 4A5Ec $\{G:G$ is open, $e\in G$, $G^{-1}=G\}$ is a base of
neighbourhoods of the identity $e$ of $X$.
\prooflet{(\HR, 4.6;  \Folland, 2.1.)}

\spheader 4A5Ed If $G\subseteq X$ is an open set, then $AG$ and $GA$ are
open for any set $A\subseteq X$.   \prooflet{(\HR, 4.4.)}

\spheader 4A5Ee If $F\subseteq X$ is closed and $x\in X\setminus F$,
there is a neighbourhood $U$ of $e$ such that
$UxUU\cap FUU=\emptyset$.   \prooflet{\Prf\ Set
$U_1=X\setminus x^{-1}F$.   Let $U_2$ be a neighbourhood of $e$ such
that $U_2U_2U_2U_2^{-1}U_2^{-1}\subseteq U_1$.   Let $U$ be a
neighbourhood of
$e$ such that $U\subseteq U_2\cap xU_2x^{-1}$;  this works.\ \Qed}

\spheader 4A5Ef If $K\subseteq X$ is compact and $F\subseteq X$ is
closed then $KF$ and $FK$ are closed.
If $K$, $L\subseteq X$ are compact so is $KL$.
\prooflet{(\HR, 4.4.)}

\spheader 4A5Eg If there is any compact set $K\subseteq X$ such that
$\interior K$ is non-empty, then $X$ is locally compact.

\spheader 4A5Eh If $K\subseteq X$ is compact and $\Cal F$ is a
downwards-directed family of closed subsets of $X$ with intersection
$F_0$, then
$KF_0=\bigcap_{F\in\Cal F}KF$ and $F_0K=\bigcap_{F\in\Cal F}FK$.
\prooflet{\Prf\ Of course $KF_0\subseteq\bigcap_{F\in\Cal F}KF$.   If
$x\in X\setminus KF_0$, then $K^{-1}x\cap F_0$ is empty;  because
$K^{-1}x$ is compact, there is some $F\in\Cal F$ such that
$K^{-1}x\cap F=\emptyset$ (3A3Db), so that $x\notin KF$.   Accordingly
$KF_0=\bigcap_{F\in\Cal F}KF$.   Similarly,
$F_0K=\bigcap_{F\in\Cal F}FK$.\ \Qed}

\spheader 4A5Ei If $K\subseteq X$ is compact and $G\subseteq X$ is open,
then $W=\{(x,y):xKy\subseteq G\}$ is open in $X\times X$.
\prooflet{\Prf\ It is enough to deal with the case $K\ne\emptyset$.
Take $(x_0,y_0)\in W$.   For each $z\in K$, there is an
open neighbourhood $U_z$ of $e$ such that
$U_zzU_zU_z\subseteq x_0^{-1}Gy_0^{-1}$ (apply (e)
with $F=X\setminus x_0^{-1}Gy_0^{-1}$).   Now $\{zU_z:z\in K\}$ is an open
cover of $K$ so there are $z_0,\ldots,z_n\in K$ such that
$K\subseteq\bigcup_{i\le n}z_iU_{z_i}$.   Set $U=\bigcap_{i\le n}U_{z_i}$;
then $UKU\subseteq x_0^{-1}Gy_0^{-1}$ and $(x,y)\in W$ whenever $x\in x_0U$
and $y\in Uy_0$.   Accordingly $(x_0,y_0)\in\interior W$;  as $(x_0,y_0)$
is arbitrary, $W$ is open.\ \Qed}

It follows that $\{x:xK\subseteq G\}$, $\{x:Kx\subseteq G\}$ and
$\{x:xKx^{-1}\subseteq G\}$ are open in $X$.

\spheader 4A5Ej If $X$ is Hausdorff, $K\subseteq X$ is compact and $U$
is a neighbourhood of $e$, there is a neighbourhood $V$ of $e$ such that
$xy\in U$ whenever $x$, $y\in K$ and $yx\in V$;  that is,
$y^{-1}zy\in U$ whenever $z\in V$ and $y\in K$.
\prooflet{\Prf\ If $U$ is open, then $\{yx:x,\,y\in K,\,xy\notin U\}$
is a closed set not containing $e$.   Compare 4A5Oc below.\
\Qed}

\spheader 4A5Ek Any open subgroup of $X$ is also closed.
\prooflet{(\Csaszar, 11.2.12;  \HR, 5.5;  \Folland, 2.1.)}

\spheader 4A5El If $X$ is locally compact, it has an open subgroup which
is $\sigma$-compact.   \prooflet{(\HR, 5.14;  \Folland, 2.3.)}

\spheader 4A5Em\dvAnew{2013}
If $Y$ is a subgroup of $X$, its closure $\overline{Y}$ is a
subgroup of $X$.   \prooflet{(\HR, 5.3;  \Folland, 2.1.)}

\spheader 4A5En\dvAnew{2017}
Let $X$ be a group and $\Cal V$ a family of subsets of
$X$.   Then there is a topology of $X$ under which $X$ is a topological
group and $\Cal V$ is a base of neighbourhoods of the identity iff

\inset{($\alpha$) $\Cal V$ is a filter base;

($\beta$) for every $V\in\Cal V$ there is a $W\in\Cal V$ such that
$W^2\subseteq V$;

($\gamma$) for every $V\in\Cal V$ there is a $W\in\Cal V$ such that
$W^{-1}\subseteq V$;

($\delta$) for every $V\in\Cal V$ and $z\in X$ there is a $W\in\Cal V$ such
that $zWz^{-1}\subseteq V$.}

\noindent In this case, there is exactly one such topology, and it is 
Hausdorff iff $\bigcap\Cal V=\{e\}$.

\prooflet{\Csaszar, 11.2.4;  \HR, II.4.5.}

\leader{4A5F}{Proposition} (a) Let $(X,\frak T)$ and $(Y,\frak S)$ be
topological groups.   If $\phi:X\to Y$ is a group homomorphism which is
continuous at the identity of $X$, it is continuous.
\prooflet{(\Csaszar, 11.2.17;  \HR, 5.40.)}

(b) Let $X$ be a group and $\frak S$, $\frak T$ two topologies on $X$
both making $X$ a topological group.   If every $\frak S$-neighbourhood
of the identity is a $\frak T$-neighbourhood of the identity, then
$\frak S\subseteq\frak T$.  \prooflet{(Apply (a) to the identity map
from $(X,\frak T)$ to $(X,\frak S)$.)}

\leader{4A5G}{Proposition} If $\familyiI{X_i}$ is any family of
topological groups, then $\prod_{i\in I}X_i$, with the product topology and
the product group structure, is again a topological group.
\prooflet{(\HR, 6.2.)}

\leader{4A5H}{The uniformities of a topological group} Let $(X,\frak T)$
be a topological group.   Write $\Cal U$ for the set of open
neighbourhoods of the identity $e$ of $X$.

\spheader 4A5Ha For $U\in\Cal U$, set
$W_U=\{(x,y):xy^{-1}\in U\}\subseteq X\times X$.   The family
$\{W_U:U\in\Cal U\}$ is a filter base, and the filter on $X\times X$
which it generates is a uniformity on $X$, the {\bf right uniformity} of
$X$.
%Pestov 99:  `right equicontinuous uniformity'
\cmmnt{({\bf Warning!!}  %double !! deliberate
Some authors call this the `left  uniformity'.)}
This uniformity induces the topology
$\frak T$\prooflet{ (\Csaszar, 11.2.7)}.
It follows that $\frak T$ is completely regular, therefore
regular\prooflet{ (4A2Ja, or \HR, 8.4)}.

\spheader 4A5Hb For $U\in\Cal U$, set
$\tilde W_U=\{(x,y):xy^{-1}\in U,\,x^{-1}y\in U\}\subseteq X\times X$.
The family $\{\tilde W_U:U\in\Cal U\}$ is a filter base, and the filter
on $X\times X$ which it generates is a uniformity on $X$, the
{\bf bilateral uniformity}
% or {\bf lower uniformity}
of $X$.   This uniformity
induces the topology $\frak T$.   \prooflet{(\Csaszar, 11.3.c.)}

\spheader 4A5Hc\dvAnew{2010} $x\mapsto x^{-1}$ is uniformly continuous for
the bilateral uniformity.   \prooflet{(The check is elementary.)}
% for \S4{}94

\spheader 4A5Hd\dvAnew{2010} If $X$ and $Y$ are topological groups and
$\phi:X\to Y$ is a continuous homomorphism, then $\phi$ is uniformly
continuous for the bilateral uniformities.   \prooflet{\Prf\ If $V$ is a
neighbourhood of the identity in $Y$ and
$W_V=\{(y,z):yz^{-1}$, $y^{-1}z$ both belong to $V\}$ is the corresponding
member of the bilateral uniformity on $Y$, then $U=\phi^{-1}[V]$ is a
neighbourhood of the identity in $X$ and
$(\phi(x),\phi(w))\in W_V$ whenever $(x,w)\in W_U$.\ \Qed}
%\refquery\Csaszar ??}
%for 4{}94

\cmmnt{\spheader 4A5He\dvAnew{2010} If $X$ is an abelian topological group,
then the right and bilateral uniformities on $X$ coincide, and may be
called `the' topological group uniformity of $X$;  cf.\ 3A4Ad.}

\leader{4A5I}{Definitions} If $X$ is a topological group and $Z$ a
topological space, an action of $X$ on $Z$ is `continuous' or `Borel
measurable' if it is continuous, or Borel measurable, when regarded as a
function from $X\times Z$ to $Z$.

\cmmnt{Of course the left, right and conjugacy actions of a
topological group on itself are all continuous.}

\leader{4A5J}{Quotients under group actions, and quotient groups:
Theorem} (a)\dvAnew{2012}
Let $X$ be a topological space, $Y$ a topological group, and
$\action$ a continuous action of $Y$ on $X$.   Let $Z$ be the set of
orbits of the action, and for $x\in X$ write $\pi(x)\in Z$ for the orbit
containing $x$.

\quad(i)
We have a topology on $Z$ defined by saying that $V\subseteq Z$ is
open iff $\pi^{-1}[V]$ is open in $X$.   The canonical map $\pi:X\to Z$
is continuous and open.

\quad(ii)($\alpha$) If $Y$ is compact and $X$ is Hausdorff, then
$Z$ is Hausdorff.

\qquad($\beta$) If $X$ is locally compact then $Z$ is locally compact.

(b) Let $X$ be a topological
group, $Y$ a
subgroup of $X$, and $Z$ the set of left cosets of $Y$ in $X$.
Set $\pi(x)=xY$ for $x\in X$.

\quad (i) We have a topology on $Z$ defined by saying that $V\subseteq Z$
is open iff $\pi^{-1}[V]$ is open in $X$.
The canonical map $\pi:X\to Z$ is continuous and open.

\quad(ii)($\alpha$) $Z$ is Hausdorff iff $Y$ is closed.

\qquad($\beta$) If $X$ is locally compact, so is $Z$.

\qquad($\gamma$)\dvAnew{2010}
If $X$ is locally compact and Polish and $Y$ is closed, then
$Z$ is Polish.

\qquad($\delta$)\dvAnew{2010}
If $X$ is locally compact and $\sigma$-compact and $Y$ is closed
and $Z$ is metrizable, then $Z$ is Polish.

\quad(iii) We have a continuous action of $X$ on $Z$ defined by saying that
$x\action\pi(x')=\pi(xx')$ for any $x$, $x'\in X$.

\quad(iv) If $Y$ is a normal subgroup of $X$, then the group operation on
$Z$ renders it a topological group.

\proof{{\bf (a)(i)} It is elementary
to check that $\{V:\pi^{-1}[V]$ is open$\}$
is a topology such that $\pi:X\to Z$ is continuous.   To see that $\pi$ is
open, take an open set $U\subseteq X$ and consider

\Centerline{$\pi^{-1}[\pi[U]]
=\bigcup_{x'\in U}\{x:\pi(x)=\pi(x')\}
=\bigcup_{x'\in U,y\in Y}\{x:x=y\action x'\}
=\bigcup_{y\in Y}y\action U$.}

\noindent But as $x\mapsto y\action x$ is a homeomorphism
for every $y\in Y$ (4A5Ea),
every $y\action U$ is open, and the union $\pi^{-1}[\pi[U]]$ is open.
So $\pi[U]$ is open in $Z$;  as $U$ is arbitrary, $\pi$ is an open map.

\medskip

\quad{\bf (ii)}\grheada\
Set $F=\{((x,x'),y):x\in X$, $y\in Y$, $y\action x=x'\}$.
Because the function
$((x,x'),y)\mapsto(y\action x,x'):(X\times X)\times Y\to X\times X$
is continuous and
$\{(x,x):x\in X\}$ is closed in $X\times X$ (4A2F(a-iii)), $F$ is closed.
By 4A2Gm, the projection
$\{(x,x'):\,\Exists y\in Y$, $y\action x=x'\}=\{(x,x'):\pi(x)=\pi(x')\}$ is
closed in $X\times X'$ and $\{(x,x'):\pi(x)\ne\pi(x')\}$ is open.
Since $(x,x')\mapsto(\pi(x),\pi(x')):X\times X\to Z\times Z$ is an open
mapping
(4A2B(f-iv)), $\{(z,z'):z\ne z'\}$ is open in $Z\times Z$, and
$Z$ is Hausdorff by 4A2F(a-iii) in the other direction.

\medskip

\qquad\grheadb\ Use 4A2Gn.

\medskip

{\bf (b)(i)} Apply (a-i)
to the right action $(y,x)\mapsto xy^{-1}$ of $Y$ on
$X$, or see \HR, 5.15-5.16.

\medskip

\quad{\bf (ii)}\grheada\ By \HR, 5.21, $Z$ is Hausdorff iff $Y$ is closed.

\medskip

\qquad\grheadb\ Use (a-ii-$\beta$), or see
\HR, 5.22 or \Folland, 2.2.

\medskip

\qquad\grheadc\ $X$ has a countable network (4A2P(a-ii)), so $Z$ also has
(4A2Nd);  since we have just seen that $Z$ is locally compact and
Hausdorff, it must be Polish (4A2Qh).

\medskip

\qquad\grheadd\ Because $X$ is $\sigma$-compact, so is its continuous image
$Z$;  we know from ($\alpha$)-($\beta$)
that $Z$ is locally compact and Hausdorff;  we
are supposing that it is metrizable;  so it is Polish, by the other
half of 4A2Qh.

\medskip

\quad{\bf (iii)} I have noted in 4A5Cb that the formula given defines an
action.   If $V\subseteq Z$ is open and $x_0\in X$,
$z_0\in Z$ are such that $x_0\action z_0\in V$, take $x_0'\in X$ such
that $\pi(x'_0)=z_0$, and observe that $x_0x'_0\in\pi^{-1}[V]$, which is
open.   So there are open neighbourhoods $V_0$, $V'_0$ of $x_0$, $x'_0$
respectively such that $V_0V'_0\subseteq\pi^{-1}[V]$, and
$x\action z\in V$ whenever
$x\in V_0$ and $z\in\pi[V'_0]$.   Since $\pi[V'_0]$ is an open
neighbourhood of $z_0$, this is enough to show that $\action$ is
continuous at $(x_0,z_0)$.

\medskip

\quad{\bf (iv)} \Csaszar, 11.2.15;  \HR, 5.26;  \Folland, 2.2.
}%end of proof of 4A5J

\leader{4A5K}{Proposition} Let $X$ be a topological group with identity
$e$.

(a) $Y=\overline{\{e\}}$ is a closed normal subgroup of $X$.

(b) Writing $\pi:X\to X/Y$ for the canonical map,

\quad(i) a subset of $X$ is open iff it is the inverse image of an open
subset of $X/Y$,

\quad(ii) a subset of $X$ is closed iff it is the inverse image of a
closed subset of $X/Y$,

\quad(iii) $\pi[G]$ is a regular open set in $X/Y$ for every
regular open set $G\subseteq X$,

\quad(iv) $\pi[F]$ is nowhere dense in $X/Y$ for every nowhere dense set
$F\subseteq X$,

\quad(v) $\pi^{-1}[V]$ is nowhere dense in $X$ for every nowhere dense
$V\subseteq X/Y$.

\proof{{\bf (a)} \Csaszar, 11.2.13;  \HR, 5.4;  \Folland, 2.3.

\medskip

{\bf (b)(i)-(ii)} Because $\pi$ is continuous,
the inverse image of an open or
closed set is open or closed.   In the other direction,
if $G\subseteq X$ is open and $x\in G$, then
$x\overline{\{e\}}=\overline{\{x\}}\subseteq G$, because $X$ is regular
(4A5Ha).   So
$G=GY=\pi^{-1}[\pi[G]]$.   Since $\pi$ is an open map (4A5J(a-i)),
$\pi[G]$ is open and $G$ is the inverse image of an open set.   If
$F\subseteq X$ is closed,
$\pi[F]=(X/Y)\setminus\pi[X\setminus F]$ is closed and

\Centerline{$F=X\setminus\pi^{-1}[\pi[X\setminus F]]
=\pi^{-1}[(X/Y)\setminus\pi[X\setminus F]]$}

\noindent is the inverse image of a closed set.

\medskip

{\bf (iii)} If $A\subseteq X$, then $\pi[\overline{A}]$ is a closed set
included in
$\overline{\pi[A]}$ (because $\pi$ is continuous), so
$\overline{\pi[A]}=\pi[\overline{A}]$.   If $G\subseteq X$ is a regular
open set, then
$\pi^{-1}[\interior\pi[\overline{G}]]$ is an open subset of
$\pi^{-1}[\pi[\overline{G}]]=\overline{G}$, so is included in
$\interior\overline{G}=G$.   But this means that the open set $\pi[G]$
includes $\interior\pi[\overline{G}]=\interior\overline{\pi[G]}$, and
$\pi[G]=\interior\overline{\pi[G]}$ is a regular open set.

\medskip

{\bf (iv)} If $F\subseteq X$ is nowhere dense, then its closure is of
the form $\pi^{-1}[V]$ for some closed set $V\subseteq X/Y$.   Now if
$H\subseteq X/Y$ is a non-empty open set, $\pi^{-1}[H]$ is a non-empty
open subset of $X$, so is not included in $F$, and $H$ cannot be
included in $V$.   Thus $V$ is nowhere dense;  but $V\supseteq\pi[F]$,
so $\pi[F]$ is nowhere dense.

\medskip

{\bf (v)} If $V\subseteq X/Y$ is nowhere dense, and $G\subseteq X$ is
open and not empty, then $G=\pi^{-1}[H]$ for some non-empty open
$H\subseteq X/Y$.   In this case, $H\setminus\overline{V}$ is non-empty,
so $\pi^{-1}[H\setminus\overline{V}]$ is a non-empty open subset of $G$
disjoint from $\pi^{-1}[V]$.   As $G$ is arbitrary, $\pi^{-1}[V]$ is
nowhere dense.
}%end of proof of 4A5K

\leader{4A5L}{Theorem} Let $X$ be a topological group and $Y$ a normal
subgroup of $X$.   Let $\pi:X\to X/Y$ be the canonical homomorphism.

(a) If $X'$ is another topological group and $\phi:X\to X'$ a continuous
homomorphism with kernel including $Y$, then we have a continuous
homomorphism $\psi:X/Y\to X'$ defined by the formula $\psi\pi=\phi$;
$\psi$ is injective iff $Y$ is the kernel of $\phi$.

(b) Suppose that $K_1$, $K_2$ are two subgroups of $X/Y$ such that
$K_2\normalsubgroup K_1$.   Set
$Y_1=\pi^{-1}[K_1]$ and $Y_2=\pi^{-1}[K_2]$.   Then
$Y_2\normalsubgroup Y_1$ and
$Y_1/Y_2$ and $K_1/K_2$ are isomorphic as topological groups.

\proof{{\bf (a)} This is elementary group theory, except for the claim
that $\psi$ is continuous.   But if $H\subseteq X'$ is open, then
$\psi^{-1}[H]=\pi[\phi^{-1}[H]]$ is open because $\phi$ is continuous
and $\pi$ is open (4A5J(a-i));  so $\psi$ is continuous.

\medskip

{\bf (b)} See \HR, 5.35.
}%end of proof of 4A5L

\leader{4A5M}{Proposition} Let $X$ be a topological group.

(a) Let $Y$ be any subgroup of $X$.   If $X$ is given its bilateral
uniformity, then the subspace uniformity on $Y$ is the bilateral
uniformity of $Y$.  \prooflet{(\Csaszar, 11.3.13.)}

(b) If $X$ is locally compact it is complete under its right uniformity.
\prooflet{(\Csaszar, 11.3.21.)}   If $X$ is complete under its right
uniformity it is complete under its bilateral uniformity.
\prooflet{(\Csaszar, 11.3.10.)}

(c) Suppose that $X$ is Hausdorff and that $Y$ is a subgroup of $X$
which is locally compact in its subspace topology.   Then $Y$ is closed
in $X$.  \prooflet{\Prf\ Putting (a) and (b) together, we see that $Y$
is complete in its subspace uniformity, therefore closed (3A4Fd).\ \Qed}

\leader{4A5N}{Theorem} Let $X$ be a Hausdorff topological group.   Then
its completion $\widehat{X}$ under its bilateral uniformity can be
endowed (in exactly one way) with a group structure rendering it a
Hausdorff topological group in which the natural embedding of $X$ in
$\widehat{X}$ represents $X$ as a dense subgroup of $\widehat{X}$.
\prooflet{(\Csaszar, 11.3.15.)}
%`Raikov completion'
If $X$ has a neighbourhood of the identity which is totally bounded for
the bilateral uniformity, then $\widehat{X}$ is locally compact.
\prooflet{(\Csaszar, 11.3.24.)}

\leader{4A5O}{Proposition} Let $X$ be a topological group.

(a) If $A\subseteq X$, then the following are equiveridical:  (i) $A$ is
totally bounded for the bilateral uniformity of $X$;  (ii) for every
neighbourhood $U$ of the identity there is a finite set $I\subseteq X$
such that $A\subseteq IU\cap UI$.

(b) If $A$, $B\subseteq X$ are totally bounded for the bilateral
uniformity of $X$, so are $A\cup B$, $A^{-1}$ and $AB$.   In particular,
$\bigcup_{i\le n}x_iB$ is totally bounded for any $x_0,\ldots,x_n\in X$.

(c) If $A\subseteq X$ is totally bounded for the bilateral uniformity,
and $U$ is any neighbourhood of the identity, then
$\{y:xyx^{-1}\in U$ for every $x\in A\}$ is a neighbourhood of the
identity.

(d) If $X$ is the product of a family $\familyiI{X_i}$ of
topological groups, a subset $A$ of $X$ is totally bounded for the
bilateral uniformity of $X$ iff it is included in a product
$\prod_{i\in I}A_i$ where $A_i\subseteq X_i$ is totally bounded for the
bilateral uniformity of $X_i$ for every $i\in I$.

(e) If $X$ is locally compact, a subset of $X$ is totally bounded for
the bilateral uniformity iff it is relatively compact.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} Suppose that $A$ is totally
bounded, and that $U$ is a neighbourhood of the identity $e$ of $X$.
Set

\Centerline{$W=\{(x,y):xy^{-1}\in U^{-1},\,x^{-1}y\in U\}=\{(x,y):y\in Ux\cap xU\}$;}

\noindent then $W$ belongs to the uniformity, so there is a finite set $I\subseteq X$ such that
$A\subseteq W[I]$.   But $W[I]\subseteq UI\cap IU$, so
$A\subseteq UI\cap IU$.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} Now suppose that $A$ satisfies the
condition, and that $W$ belongs to the uniformity.   Then there is a
neighbourhood $U$ of $e$ such that
$\{(x,y):xy^{-1}\in U,\,x^{-1}y\in U\}\subseteq W$.   Let $V$ be a
neighbourhood of $e$ such that $VV^{-1}\subseteq U$ and
$V^{-1}V\subseteq U$.   Let $I\subseteq X$ be a finite set such that
$A\subseteq VI\cap IV$.   For $w$, $z\in I$ set
$A_{wz}=A\cap Vw\cap zV$.   If $x$, $y\in A_{wz}$,
$xy^{-1}\in Vww^{-1}V^{-1}\subseteq U$ and
$x^{-1}y\in V^{-1}z^{-1}zV\subseteq U$.   But this means that
$A_{wz}\times A_{wz}\subseteq W$.   So if we take a finite set $J$ which
meets every non-empty $A_{wz}$, $A\subseteq W[J]$.   As $W$ is
arbitrary, $A$ is totally bounded.

\medskip

{\bf (b)} Of course $A\cup B$ is totally bounded;  this is immediate
from the definition of `totally bounded'.   If $U$ is a neighbourhood of
$e$, so is $U^{-1}$, so there is a finite set $I\subseteq X$ such that
$A\subseteq IU^{-1}\cap U^{-1}I$ and
$A^{-1}\subseteq UI^{-1}\cap I^{-1}U$;  as $U$ is arbitrary, $A^{-1}$ is
totally bounded.

To see that $AB$ also is totally bounded, let $U$ be a neighbourhood of
$e$, and take a neighbourhood $V$ of $e$ such that $VV\subseteq U$.
Then there is a finite set $I\subseteq X$ such that
$A\subseteq VI$ and $B\subseteq IV$.   Let $W$ be a neighbourhood of $e$
such that
$zWz^{-1}\cup z^{-1}Wz\subseteq V$ for every $z\in I$, and $J$ a finite
set such that $B\subseteq WJ$ and $A\subseteq JW$.   Then

\Centerline{$zW\subseteq Vz$,
\quad$Wz\subseteq zV$}

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

\Centerline{$IW\subseteq VI$,
\quad$WI\subseteq IV$}

\noindent and

\Centerline{$AB\subseteq VIWJ\subseteq VVIJ\subseteq UK$,
\quad$AB\subseteq JWIV\subseteq JIVV\subseteq KU$}

\noindent where $K=IJ\cup JI$ is finite.   As $U$ is arbitrary, $AB$ is
totally bounded.

\medskip

{\bf (c)} Let $V$ be a neighbourhood of $e$ such that $VVV^{-1}\subseteq
U$.   Let $I$ be a finite set such that $A\subseteq VI$.   Let $W$ be a
neighbourhood of $e$ such that $zWz^{-1}\subseteq V$ for every $z\in I$.
If now $y\in W$ and $x\in A$, there is a $z\in I$ such that $x\in Vz$,
so that

\Centerline{$xyx^{-1}\in VzWz^{-1}V^{-1}\subseteq VVV^{-1}\subseteq U$.}

\noindent Turning this round, $\{y:xyx^{-1}\in U$ for every $x\in A\}$
includes $W$ and is a neighbourhood of $e$.

\medskip

{\bf (d)(i)} Suppose that $A$ is totally bounded.   Set
$A_i=\pi_i[A]$ for each $i\in I$, where $\pi_i(x)=x(i)$ for $x\in X$.
If $U$ is a neighbourhood of the identity in $X_i$, then
$V=\pi_i^{-1}[U]$ is a neighbourhood of the identity in $X$, so there is
a finite set $J\subseteq X$ such that $A\subseteq JV\cap VJ$;  now
$A_i\subseteq KU\cap UK$, where $K=\pi_i[J]$ is finite.   As $U$ is
arbitrary, $A_i$ is totally bounded.   This is true for every $i$, while
$A\subseteq\prod_{i\in I}A_i$.

\medskip

\quad{\bf (ii)} Suppose that $A\subseteq\prod_{i\in I}A_i$ where
$A_i\subseteq X_i$ is totally bounded for each $i\in I$.   If $A$ is
empty, of course it is totally bounded;  assume that $A\ne\emptyset$.
If $I=\emptyset$, then $X=\{\emptyset\}$ is the trivial group, and again
$A$ is totally bounded;  so assume that $I$ is non-empty.    Let $V$ be
a neighbourhood of the identity in $X$.   Then there are a non-empty
finite set $L\subseteq I$ and a family $\family{i}{L}{U_i}$ such that
$U_i$ is a neighbourhood of the identity in $X_i$ for each $i\in L$, and
$V\supseteq\bigcap_{i\in L}\pi_i^{-1}U_i$.   For each $i\in L$, let
$J_i$ be a finite subset of $X_i$ such that $A_i\subseteq J_iU_i\cap
U_iJ_i$.   Set

\Centerline{$J=\{x:x\in X,\,
x(i)$ is the identity for $i\in I\setminus L$,
$x(i)\in J_i$ for $i\in L\}$.}

\noindent Then $J$ is finite and $A\subseteq JV\cap VJ$.  As $V$ is
arbitrary, $A$ is totally bounded.

\medskip

{\bf (e)} Use (a).
}%end of proof of 4A5O

\vleader{72pt}{4A5P}{Lemma}
Let $X$ be a locally compact Hausdorff topological
group.   Take $f\in C_k(X)$\cmmnt{, the space of continuous
real-valued functions on $X$ with compact supports}.

(a) Let $K\subseteq X$ be a compact set.   Then for any $\epsilon>0$
there is a neighbourhood $W$ of the identity $e$ of $X$ such that
$|f(xay)-f(xby)|\le\epsilon$ whenever $x\in K$, $y\in X$ and
$ab^{-1}\in W$.

(b) For any $x_0\in X$, there is a non-negative $f^*\in C_k(X)$ such that
for every $\epsilon>0$ there is an open set $G$ containing $x_0$ such
that $|f(xy)-f(x_0y)|\le\epsilon f^*(y)$ for every $x\in G$ and $y\in X$.

\proof{{\bf (a)} By 4A2Jf and 4A5Ha, $f$ is uniformly continuous for the
right uniformity of $X$.   There is therefore a symmetric neighbourhood
$U$ of $e$
such that $|f(y_1)-f(y_2)|\le\epsilon$ whenever $y_1$, $y_2\in X$
and $y_1y_2^{-1}\in U$.   By 4A5Oc, there is a symmetric neighbourhood
$W$ of $e$ such that $xzx^{-1}\in U$ whenever $x\in K$ and $z\in W$.

Now suppose that $x\in K$,
$y\in X$ and $ab^{-1}\in W$.   Then
$(xay)(xby)^{-1}=xab^{-1}x^{-1}\in U$, so
$|f(xay)-f(xby)|\le\epsilon$, as required.

\medskip

{\bf (b)} We need a trifling refinement of the ideas above.

\medskip

\quad{\bf (i)} Suppose for the moment that $x_0=e$.   Set
$L=\overline{\{x:f(x)\ne 0\}}$ and let $V$ be a compact symmetric neighbourhood of
the identity $e$, so that $L$ and $VL$ are compact.
Let $f^*\in C_k(X)$ be such that $f^*\ge\chi(VL)$ (4A2G(e-i)).   Given
$\epsilon>0$, take $U$ as in (a), so that $U$ is a symmetric
neighbourhood of $e$
and $|f(y_1)-f(y_2)|\le\epsilon$ whenever $y_1y_2^{-1}\in U$;  this time
arrange further that $U\subseteq V$.   Then if $x\in U$ and $y\in X$,

\inset{{\it either} $y$ and $xy$ belong to $VL$, while
$(xy)y^{-1}\in U$, so
$|f(xy)-f(y)|\le\epsilon\le\epsilon f^*(y)$

{\it or} neither $y$ nor $xy$ belongs to $L$, so
$|f(xy)-f(y)|=0\le\epsilon f^*(y)$.}

\medskip

\quad{\bf (ii)} For the general case, set $f_0(x)=f(x_0x)$ for
$x\in X$.   Because $x\mapsto x_0x$ is a homeomorphism, $f_0\in C_k(X)$.
By (i), we have a non-negative $f^*\in C_k(X)$ such that for every
$\epsilon>0$ there is a neighbourhood $G_{\epsilon}$ of $e$ such that
$|f_0(xy)-f_0(y)|\le\epsilon f^*(y)$ whenever $x\in G_{\epsilon}$ and $y\in X$.
Now, given $\epsilon>0$, $G'=x_0G_{\epsilon}$ is a neighbourhood of $x_0$ and
$|f(xy)-f(x_0y)|\le\epsilon f^*(y)$ whenever $x\in G'$ and $y\in X$.
So $f^*$ witnesses that the result is true.
}%end of proof of 4A5P

\vleader{72pt}{4A5Q}{Metrizable groups:  Proposition} Let $(X,\frak T)$ be a
topological group.  Then the following are equiveridical:

(i) $X$ is metrizable;

(ii) the identity $e$ of $X$ has a countable neighbourhood base;

(iii) there is a metric $\rho$ on $X$, inducing the topology
$\frak T$, which is {\bf \rti}, that is, $\rho(x_1,x_2)=\rho(x_1y,x_2y)$
for all $x_1$, $x_2$, $y\in X$;

(iv) there is a \rti\ metric on $X$ which induces the right uniformity
of $X$;

(v) the bilateral uniformity of $X$ is metrizable.

\proof{ \Csaszar, 11.2.10 and 11.3.2.}

\cmmnt{\medskip

\noindent{\bf Warning!}  A Polish group (4A5Db) is of course metrizable,
so has a \rti\ metric inducing its topology.
At the same time, it has a complete metric inducing its topology.
But there is no suggestion that these two metrics should be the same,
or even induce the same uniformity (441Xr).
}%end of comment

%what we do have:  if $X$ is a complete metrizable topological
%group it is complete in its bilateral uniformity.  \Prf\ The bilateral
%uniformity is metrizable, being induced by the sum of \rti\ and \lti\
%metrics on  X .   If  \tilde X
%is the completion in the bilateral uniformity, then  X  is  dense
% G_{\delta}  therefore comeager in  \tilde X  and  X = XX^{-1}  is
%open therefore equal to  \tilde X .   (Christensen?)

\leader{4A5R}{Corollary} If $X$ is a locally compact topological group
and $\{e\}$ is a G$_{\delta}$ set in $X$, then $X$ is metrizable.
\prooflet{(Put 4A5Q and 4A2Kf together, or see \HR, 8.5.)}

\leader{4A5S}{Lemma}\dvArevised{2010}
Let $X$ be a $\sigma$-compact locally compact
Hausdorff topological group and $\sequencen{U_n}$ any sequence of
neighbourhoods of the identity in $X$.   Then $X$ has a compact normal
subgroup $Y\subseteq\bigcap_{n\in\Bbb N}U_n$ such that $Z=X/Y$ is
Polish.

\proof{ Let $\sequencen{K_n}$ be a sequence of compact sets covering
$X$.   Choose inductively a sequence $\sequencen{V_n}$ of compact
neighbourhoods of $e$ such that, for each $n\in\Bbb N$,

\inset{($\alpha$) $V_{n+1}\subseteq V_n^{-1}$,
\quad$V_{n+1}V_{n+1}\subseteq V_n$,
\quad$V_n\subseteq U_n$,

($\beta$) $xyx^{-1}\in V_n$ whenever $y\in V_{n+1}$ and
$x\in\bigcup_{i\le n}K_i$.}

\noindent (When we come to choose $V_{n+1}$, we can achieve ($\alpha$)
because inversion and multiplication are continuous, and ($\beta$) by
4A5Oc;  and we can then shrink $V_{n+1}$ to a compact neighbourhood of
$e$ because $X$ is locally compact.)   Set
$Y=\bigcap_{n\in\Bbb N}V_n$.   Then ($\alpha$) is enough to ensure that
$Y$ is a compact subgroup of $X$ included in $\bigcap_{n\in\Bbb N}U_n$,
while ($\beta$) ensures that $Y$ is normal, because for any $x\in X$
there is an $n\in\Bbb N$ such that $xV_{m+1}x^{-1}\subseteq V_m$ for
every $m\ge n$.

Let $\pi:X\to Z$ be the canonical map.   Then
$C=\bigcap_{n\in\Bbb N}\pi[\interior V_n]$ is a G$_{\delta}$ subset of
$Z$, because $\pi$ is open (4A5J(a-i) again).   But

$$\eqalign{\pi^{-1}[C]
&\subseteq\bigcap_{n\in\Bbb N}\pi^{-1}[\pi[V_{n+1}]]
=\bigcap_{n\in\Bbb N}V_{n+1}Y
\subseteq\bigcap_{n\in\Bbb N}V_n
=Y,\cr}$$

\noindent so $C=\{e_Z\}$, writing $e_Z$ for the identity of
$Z$.   Thus $\{e_Z\}$ is a G$_{\delta}$ set;  as $Z$ is locally
compact and Hausdorff (4A5J(b-ii)), it is metrizable (4A5R).
By 4A5J(b-ii-$\delta$), $Z$ is Polish.
}%end of proof of 4A5S

\leader{*4A5T}{}\cmmnt{ I shall not rely on the following fact, but it
will help you to make sense of some of the results of this volume.

\medskip

\noindent}{\bf Theorem} A compact Hausdorff topological group is dyadic.

\proof{{\smc Uspenski\v\i\ 88}.
}%end of proof of 4A5T

%Solecki's theorem:  if  X  is a complete metrizable group and for every
%neighbourhood  U  of  e  there is a finite set  F  such that  X = FUF ,
%then  X  is compact.

\discrpage

\frfilename{mt4a6.tex}
\versiondate{8.12.10}
\copyrightdate{2000}

\def\chaptername{Appendix}
\def\sectionname{Banach algebras}
\def\Folland{{\smc Folland 95}}
\def\HR{{\smc Hewitt \& Ross 63}}
\def\Rudin{{\smc Rudin 91}}

\newsection{4A6}

I give results which are needed for Chapter 44.   Those down to 4A6K
should be in any introductory text on normed algebras;  
4A6L-4A6O, %4A6L 4A6M 4A6N 4A6O
as expressed here, are a little more specialized.
As with normed spaces or linear topological spaces, Banach
algebras may be defined over either $\Bbb R$ or $\Bbb C$.   In \S445 we
need complex Banach algebras, but in \S446 I think the ideas are clearer
in the context of real Banach algebras.   Accordingly, as in \S2A4, I
express as much as possible of the theory in terms applicable equally to
either, speaking of `normed algebras' or `Banach algebras' without
qualification, and using the symbol $\RoverC$ to represent the field of
scalars.   Since (at least, if you keep to the path indicated here) the
ideas are independent of which field we work with, you will have no
difficulty in applying the arguments given in \Folland\ or \HR\ for the
complex case to the real case.   In 4A6B and 4A6I-4A6K, %4A6I 4A6J 4A6K
however, we have results which apply only to `complex' Banach algebras,
in which the underlying field is taken to be $\Bbb C$.

\leader{4A6A}{Definition (a)}\cmmnt{ I repeat a definition from
\S2A4.}   A {\bf normed algebra} is a normed
space $U$ together with a multiplication, a binary operator $\times$ on
$U$, such that

\Centerline{$u\times(v\times w)=(u\times v)\times w$,}

\Centerline{$u\times(v+w)=(u\times v)+(u\times w)$,
\quad$(u+v)\times w=(u\times w)+(v\times w)$,}

\Centerline{$(\alpha u)\times v=u\times(\alpha v)=\alpha(u\times v)$,}

\Centerline{$\|u\times v\|\le\|u\|\|v\|$}

\noindent for all $u$, $v$, $w\in U$ and $\alpha\in\RoverC$.
A normed algebra is {\bf commutative} if its
multiplication is commutative.

\spheader 4A6Ab A {\bf Banach algebra} is a normed algebra which is a
Banach space.   A {\bf unital Banach algebra} is a Banach
algebra with a multiplicative identity $e$ such that $\|e\|=1$.
\cmmnt{({\bf Warning:}  some authors %Rudin 91
reserve the term `Banach algebra' for what I call a `unital Banach
algebra'.)}

In a unital Banach algebra I will always use the letter $e$ for the
identity.
%Folland 1.1

\leader{4A6B}{Stone-Weierstrass Theorem:  fourth form} Let $X$ be a
locally compact Hausdorff space, and $C_0=C_0(X;\Bbb C)$ the complex
Banach algebra of continuous functions $f:X\to\Bbb C$ such that
$\{x:|f(x)|\ge\epsilon\}$ is compact for every $\epsilon>0$.   Let
$A\subseteq C_0$ be such that

\inset{$A$ is a linear subspace of $C_0$,}

\inset{$f\times g\in A$ for every $f$, $g\in A$,}

\inset{the complex conjugate $\bar f$ of $f$ belongs to $A$ for every
$f\in A$,}

\inset{for every $x\in X$ there is an $f\in A$ such that $f(x)\ne 0$,}

\inset{whenever $x$, $y$ are distinct points of $X$ there is an $f\in A$
such that $f(x)\ne f(y)$.}

\noindent Then $A$ is $\|\,\,\|_{\infty}$-dense in $C_0$.
%4{}45K

\proof{ Let $X_{\infty}=X\cup\{\infty\}$ be the one-point
compactification of $X$
(3A3O).   For $f\in C_0$ write $f^{\#}$ for the extension of $f$ to
$X\cup\{\infty\}$ with $f^{\#}(\infty)=0$, so that
$f^{\#}\in C_b(X_{\infty};\Bbb C)$.   Let
$B\subseteq C_b(X_{\infty};\Bbb C)$ be the set of all
functions of the form $f^{\#}+\alpha\chi X_{\infty}$ where
$f\in A$ and $\alpha\in\Bbb C$.
Then $B$ is a subalgebra of $C_b(X\cup\{\infty\})$ which contains
complex conjugates of its members and constant functions and separates
the points of $X_{\infty}$.

Take any $h\in C_0$ and $\epsilon>0$.   By the `third form' of the
Stone-Weierstrass theorem (281G), there is a $g\in B$ such that
$\|g-h^{\#}\|_{\infty}\le\bover12\epsilon$.   Express $g$ as
$f^{\#}+\alpha\chi X_{\infty}$ where $f\in A$ and $\alpha\in\Bbb C$.
Then

\Centerline{$|\alpha|=|g(\infty)|
=|g(\infty)-h^{\#}(\infty)|\le\bover12\epsilon$,}

\noindent so

\Centerline{$\|h-f\|_{\infty}=\|h^{\#}-f^{\#}\|_{\infty}
\le\|h^{\#}-g\|_{\infty}+\|g-f^{\#}\|_{\infty}
\le\bover12\epsilon+|\alpha|
\le\epsilon$.}

\noindent As $h$ and $\epsilon$ are arbitrary, $A$ is dense in $C_0$.
}%end of proof of 4A6B

\leader{4A6C}{Proposition} If $U$ is any Banach space other than
$\{0\}$, then the
space $\eurm B(U;U)$ of bounded linear operators from $U$ to itself is a
unital Banach algebra.   \prooflet{({\smc K\"othe 69}, 14.6.)}

\leader{4A6D}{Proposition} Any normed algebra $U$ can be embedded as
a subalgebra of a unital Banach algebra $V$, in such a way that if $U$
is commutative so is $V$.   \prooflet{(\Folland, \S1.3;  \HR, C.3.)}
%4A6K  4A6F

\leader{4A6E}{Proposition} Let $U$ be a unital Banach algebra and
$W\subseteq U$ a closed proper ideal.   Then $U/W$, with the quotient
linear structure, ring structure and norm, is a unital Banach algebra.
\prooflet{(\HR, C.2.)}
%4A6J

\leader{4A6F}{Proposition} If $U$ is a Banach algebra and
$\phi:U\to\RoverC$ is a multiplicative linear functional, then
$|\phi(u)|\le\|u\|$ for every $u\in U$.

\proof{\Quer\ Otherwise, there is a $u$ such that $|\phi(u)|>\|u\|$;
set $v=\Bover1{\phi(u)}u$, so that $\phi(v)=1$ and $\|v\|<1$.   Since
$\|v^n\|\le\|v\|^n$ for every $n\ge 1$, 
$w=\sum_{n\in\Bbb N\setminus\{0\}}v^n$ is
defined in $U$ (4A4Ie), and $w=vw+v$ (because $u\mapsto vu$ is a continuous
linear operator, so we can use 4A4Bh to see that
$vw=\sum_{n\in\Bbb N\setminus\{0\}}v^{n+1}$).   But this means that
$\phi(w)=\phi(v)\phi(w)+\phi(v)=\phi(w)+1$, which is impossible.\ \Bang
}%end of proof of 4A6F

\leader{4A6G}{Definition} Let $U$ be a normed algebra and $u\in U$.

\spheader 4A6Ga For any $u\in U$, $\lim_{n\to\infty}\|u^n\|^{1/n}$
is defined and equal to $\inf_{n\ge 1}\|u^n\|^{1/n}$.
\prooflet{(\HR, C.4.)}

\spheader 4A6Gb This common value is\cmmnt{ called} the {\bf spectral
radius} of $u$.

\leader{4A6H}{Theorem} If $U$ is a unital Banach algebra,
then the set $R$ of invertible elements is open, and
$u\mapsto u^{-1}$ is a continuous function from $R$ to itself.   
If $v\in U$ and $\|v-e\|<1$, then $v\in R$ and
$\|v^{-1}-e\|\le\Bover{\|v-e\|}{1-\|v-e\|}$.
\prooflet{(\Folland, 1.4;  \HR, C.8 \& C.10;  \Rudin, 10.7 \& 10.12.)}

\vleader{48pt}{4A6I}{Theorem} Let $U$ be a complex unital Banach algebra and
$u\in U$.   Write $r$ for the spectral radius of $u$.

(a) If $\zeta\in\Bbb C$ and $|\zeta|>r$ then $\zeta e-u$ is invertible.

(b) There is a $\zeta$ such that $|\zeta|=r$ and $\zeta e-u$ is not
invertible.
%4A6K 4A6J

\proof{ \Folland, 1.8;  \HR, C.24;  \Rudin, 1.13.}
%mt4abits

\leader{4A6J}{Theorem} Let $U$ be a commutative complex unital Banach
algebra, and $u\in U$.   Then for any $\zeta\in\Bbb C$ the following are
equiveridical:

(i) there is a non-zero multiplicative linear functional
$\phi:U\to\Bbb C$ such that $\phi(u)=\zeta$;

(ii) $\zeta e-u$ is not invertible.
%4A6K

\proof{ \Folland, 1.13;  \HR, C.20;  \Rudin, 11.5.}

\leader{4A6K}{Corollary} Let $U$ be a commutative complex Banach
algebra and $u\in U$.   Then its spectral radius $r(u)$ is
$\max\{|\phi(u)|:\phi$ is a multiplicative linear functional on $U\}$.
\prooflet{(\Folland, 1.13;  \HR, C.20;  \Rudin, 11.9.)}

\leader{4A6L}{Exponentiation} Let $U$ be a unital Banach algebra.
For any $u\in U$,\cmmnt{ $\sum_{k=0}^{\infty}\|\Bover1{k!}u^k\|
\penalty-100\le\sum_{k=0}^{\infty}\Bover1{k!}\|u\|^k$ is finite, so}

\Centerline{$\exp(u)=\sum_{k\in\Bbb N}\Bover1{k!}u^k$}

\noindent is defined in $U$\cmmnt{ (4A4Ie)}.   \cmmnt{(In this
formula, interpret $u^0$ as $e$ for every $u$.)}
%4A6M, 4A6N

\leader{4A6M}{Lemma} Let $U$ be a unital Banach algebra.

(a) If $u$, $v\in U$ and $\max(\|u\|,\|v\|)\le\gamma$ then
$\|\exp(u)-\exp(v)-u+v\|\le\|u-v\|(\exp\gamma-1)$.   So if
$\max(\|u\|,\|v\|)\le\bover23$ and $\exp(u)=\exp(v)$ then $u=v$.
%4A6K

(b) If $\|u-e\|\le\bover16$ then there is a $v$ such that $\exp(v)=u$
and $\|v\|\le 2\|u-e\|$.
%4A6N

(c) If $u$, $v\in U$ and $uv=vu$ then $\exp(u+v)=\exp(u)\exp(v)$.
%4A6N

\proof{{\bf (a)} Note first that if $k\ge 1$ then

$$\eqalignno{\|u^k-v^k\|
&=\|\sum_{i=0}^{k-1}u^{k-i}v^i-u^{k-i-1}v^{i+1}\|
=\|\sum_{i=0}^{k-1}u^{k-i-1}(u-v)v^i\|\cr
&\le\sum_{i=0}^{k-1}\|u\|^{k-i-1}\|u-v\|\|v\|^i
\le\sum_{i=0}^{k-1}\gamma^{k-1}\|u-v\|
=k\gamma^{k-1}\|u-v\|.\cr}$$

\noindent So

$$\eqalign{\|\exp(u)-\exp(v)-u+v\|
&=\|\sum_{k\in\Bbb N\setminus\{0,1\}}\Bover1{k!}(u^k-v^k)\|
\le\sum_{k=2}^{\infty}\Bover1{k!}\|u^k-v^k\|\cr
&\le\sum_{k=2}^{\infty}\Bover{k}{k!}\gamma^{k-1}\|u-v\|
=\|u-v\|(\exp\gamma-1).\cr}$$

\noindent Now if $\exp(u)=\exp(v)$ and $\gamma\le\bover23$,
$0\le\exp\gamma-1<1$ so $\|u-v\|=0$ and $u=v$.

\medskip

{\bf (b)} Set $\gamma=\|u-e\|$.   Define $\sequencen{v_n}$ in $U$ by
setting $v_0=0$, $v_{n+1}=v_n+u-\exp(v_n)$ for $n\in\Bbb N$.   Then

\Centerline{$\|v_{n+1}-v_n\|=\|u-\exp(v_n)\|\le 2^{-n}\gamma$,
\quad$\|v_n\|\le 2(1-2^{-n})\gamma$}

\noindent for every $n\in\Bbb N$.   \Prf\ Induce on $n$.   The induction
starts with $\|v_0\|=0$ and $\|u-\exp(v_0)\|=\|u-e\|=\gamma$.   Given
that $\|v_n\|\le 2(1-2^{-n})\gamma$ and
$\|u-\exp(v_n)\|\le 2^{-n}\gamma$, then

\Centerline{$\|v_{n+1}\|
\le\|v_n\|+\|u-\exp(v_n)\|
\le 2(1-2^{-n})\gamma+2^{-n}\gamma
=2(1-2^{-n-1})\gamma$.}

\noindent Now $\max(\|v_{n+1}\|,\|v_n\|)\le 2\gamma\le\bover13$, so

$$\eqalign{\|u-\exp(v_{n+1})\|
&=\|v_{n+1}-v_n+\exp(v_n)-\exp(v_{n+1})\|
\le\|v_{n+1}-v_n\|(\exp\Bover13-1)\cr
&\le\Bover12\|v_{n+1}-v_n\|
=\Bover12\|u-\exp(v_n)\|
\le 2^{-n-1}\gamma,\cr}$$

\noindent and the induction continues.\ \Qed

Since $\sum_{n=0}^{\infty}\|v_{n+1}-v_n\|$ is finite,
$v=\lim_{n\to\infty}v_n$ is defined in $U$, and
$\|v\|=\lim_{n\to\infty}\|v_n\|\le 2\gamma$.   Accordingly

$$\eqalign{\|\exp(v)-\exp(v_n)\|
&\le\|v-v_n\|+\|\exp(v)-\exp(v_n)-v+v_n\|\cr
&\le\|v-v_n\|(1+\exp\Bover13-1)
\to 0\cr}$$

\noindent as $n\to\infty$, and $\exp(v)=\lim_{n\to\infty}\exp(v_n)=u$.

\medskip

{\bf (c)} Because $uv=vu$,
$(u+v)^m=\sum_{j=0}^m\Bover{m!}{j!(m-j)!}u^jv^{m-j}$ for every
$m\in\Bbb N$ (induce on $m$;  the point is that $uv^j=v^ju$ for every
$j\in\Bbb N$).   Next, $\sum_{j,k\in\Bbb N}\Bover1{j!k!}\|u\|^j\|v\|^k$
is finite.   So

$$\eqalignno{\exp(u+v)
&=\sum_{m\in\Bbb N}\Bover1{m!}(u+v)^m\cr
&=\sum_{m\in\Bbb N}\bigl(\sum_{j+k=m}
  \Bover1{j!k!}u^jv^k\bigr)
=\sum_{(j,k)\in\Bbb N\times\Bbb N}\Bover1{j!k!}u^jv^k\cr
\displaycause{using 4A4I(e-ii)}
&=\sum_{j\in\Bbb N}\bigl(\sum_{k\in\Bbb N}\Bover1{j!k!}u^jv^k)\cr
\displaycause{4A4I(e-ii) again}
&=\sum_{j\in\Bbb N}
  \bigl(\Bover1{j!}u^j\sum_{k\in\Bbb N}\Bover1{k!}v^k\bigr)\cr
\displaycause{by 4A4Bh, because $w\mapsto\Bover1{j!}u^jw$ is a
continuous linear operator for each $j$}
&=\sum_{j\in\Bbb N}\Bover1{j!}u^j\exp(v)
=\bigl(\sum_{j\in\Bbb N}\Bover1{j!}u^j\bigr)\exp(v)\cr
\displaycause{4A4Bh again}
&=\exp(u)\exp(v),\cr}$$

\noindent as claimed.
}%end of proof of 4A6M

\leader{4A6N}{Lemma} If $U$ is a unital Banach algebra, $u\in U$ and
$\|u^n-e\|\le\bover16$ for every $n\in\Bbb N$, then $u=e$.

\proof{ For every $n\in\Bbb N$ there is a $v_n\in U$ such
that $\exp(v_n)=u^{2^n}$ and $\|v_n\|\le\bover13$ (4A6Mb).   Then
$\exp(v_{n+1})=\exp(v_n)^2=\exp(2v_n)$ (4A6Mc),
$\|v_{n+1}\|\le\bover13$ and $\|2v_n\|\le\bover13$ so $v_{n+1}=2v_n$ for
every $n$ (4A6Ma).   Inducing on $n$, $v_n=2^nv_0$ for every $n$, so
that $\|v_0\|\le 2^{-n}\|v_n\|\to 0$ as $n\to\infty$, and $u=\exp(v_0)=e$.
}%end of proof of 4A6N

\leader{4A6O}{Proposition} Let $U$ be a normed algebra, and $U^*$, 
$U^{**}$ its dual and bidual as a normed space.   For a bounded linear
operator $T:U\to U$ let $T':U^*\to U^*$ be the adjoint of $T$ and
$T'':U^{**}\to U^{**}$ the adjoint of $T'$.

(a) We have bilinear maps, all of norm at most $1$,

$$\eqalign{(f,x)\mapsto f\frsmallcirc x&:U^*\times U\to U^*,\cr
(\phi,f)\mapsto \phi\frsmallcirc f&:U^{**}\times U^*\to U^*,\cr
(\phi,\psi)\mapsto \phi\frsmallcirc\psi&:U^{**}\times U^{**}\to U^{**}
\cr}$$

\noindent defined by the formulae

$$\eqalign{(f\frsmallcirc x)(y)&=f(xy),\cr
(\phi\frsmallcirc f)(x)&=\phi(f\frsmallcirc x),\cr
(\phi\frsmallcirc\psi)(f)&=\phi(\psi\frsmallcirc f)\cr}$$

\noindent for all $x$, $y\in U$, $f\in U^*$ and $\phi$, $\psi\in U^{**}$.

(b)(i) Suppose that $S:U\to U$ is a bounded linear operator such that
$S(xy)=(Sx)y$ for all $x$, $y\in U$.   Then
$S''(\phi\frsmallcirc\psi)=(S''\phi)\frsmallcirc\psi$ for all
$\phi$, $\psi\in U^{**}$.

\quad(ii) Suppose that $T:U\to U$ is a bounded linear operator such that
$T(xy)=x(Ty)$ for all $x$, $y\in U$.   Then 
$T''(\phi\frsmallcirc\psi)=\phi\frsmallcirc(T''\psi)$ for
all $\phi$, $\psi\in U^{**}$.

\proof{{\bf (a)} The calculations are elementary if we take them one at a
time.

\medskip

{\bf (b)(i)}\grheada
 $(S'f)\frsmallcirc x=f\frsmallcirc(Sx)$ for every $f\in U^*$
and $x\in U$.   \Prf\ 

\Centerline{$((S'f)\frsmallcirc x)(y)=(S'f)(xy)=f(S(xy))=f((Sx)y)
=(f\frsmallcirc(Sx))(y)$}

\noindent for every $y\in U$.\ \Qed

\medskip

\qquad\grheadb\ 
$\psi\frsmallcirc(S'f)=S'(\psi\frsmallcirc f)$ for every $f\in U^*$.   \Prf\

\Centerline{$(\psi\frsmallcirc(S'f))(x)
=\psi((S'f)\frsmallcirc x)=\psi(f\frsmallcirc(Sx))
=(\psi\frsmallcirc f)(Sx)=(S'(\psi\frsmallcirc f))(x)$}

\noindent for every $x\in U$.\ \Qed

\medskip

\qquad\grheadc\ So

$$\eqalign{(S''(\phi\frsmallcirc\psi))(f)
&=(\phi\frsmallcirc\psi)(S'f)
=\phi(\psi\frsmallcirc(S'f))\cr
&=\phi(S'(\psi\frsmallcirc f))
=(S''\phi)(\psi\frsmallcirc f)
=((S''\phi)\frsmallcirc\psi)(f)\cr}$$

\noindent for every $f\in U^*$, and 
$S''(\phi\frsmallcirc\psi)=(S''\phi)\frsmallcirc\psi$.

\medskip

\quad{\bf (ii)}\grheada\ $(T'f)\frsmallcirc x=T'(f\frsmallcirc x)$ for
every $f\in U^*$ and $x\in U$.   \Prf\

\Centerline{$((T'f)\frsmallcirc x)(y)
=(T'f)(xy)
=f(T(xy))
=f(x(Ty))
=(f\smallcirc x)(Ty)
=(T'(f\frsmallcirc x))(y)$}

\noindent for every $y\in U$.\ \Qed

\medskip

\qquad\grheadb\ $\psi\frsmallcirc(T'f)=(T''\psi)\frsmallcirc f$ for every 
$f\in U^*$.   \Prf\ 

$$\eqalign{(\psi\frsmallcirc(T'f))(x)
&=\psi((T'f)\frsmallcirc x)
=\psi(T'(f\frsmallcirc x))\cr
&=(T''\psi)(f\frsmallcirc x)
=((T''\psi)\frsmallcirc f)(x)\cr}$$

\noindent for every $x\in U$.\ \Qed

\medskip

\qquad\grheadc\ So

\Centerline{$(T''(\phi\frsmallcirc\psi))(f) 
=(\phi\frsmallcirc\psi)(T'f)
=\phi(\psi\frsmallcirc(T'f))
=\phi((T''\psi)\frsmallcirc f)
=(\phi\frsmallcirc(T''\psi))(f)$}

\noindent for every $f\in U'$, and
$T''(\phi\frsmallcirc\psi)=\phi\frsmallcirc(T''\psi)$.
}%end of proof of 4A6O

\cmmnt{\medskip

\noindent{\bf Remark} I must not abandon you at this point without telling
you that $\frsmallcirc:U^{**}\times U^{**}\to U^{**}$ is an 
{\bf Arens multiplication}, and that it is associative, so that
that $U^{**}$ is a Banach algebra.
}

%\exercises{\leader{4A6X}{Basic exercises (a)}

%\leader{4A6Y}{Further exercises (a)}
%}%end of exercises

%\endnotes{
%\Notesheader{4A6}

%}%end of notes

\discrpage

\frfilename{mt4A7.tex}
\versiondate{13.3.22}
\copyrightdate{2022}
     
\def\chaptername{Appendix}
\def\sectionname{Algebraic topology}
     
\newsection{4A7} A fundamental theorem about the topology of Euclidean space is
used in \S472.  \query (8.7.22) no idea what I was doing here

\leader{4A7A}{Definitions} Suppose that $X$ and $Y$ are topological spaces,
and $f:X\to Y$, $g:X\to Y$ are continuous functions.   

\spheader 4A7Aa A 
{\bf homotopy from $f$ to $g$}  is a continuous function $F:X\times[0,1]\to Y$
such that $F(x,0)=f(x)$ and $F(x,1)=g(x)$ for every $x\in X$.

\spheader 4A7Ab $f$ and $g$ are {\bf homotopic} if there is a homotopy from
$f$ to $g$.

\leader{4A7B}{Theorem} If $r\ge 1$ and $S_{r-1}=\partial B(\tbf 0.1)$ 
is the unit sphere
$\{x:\|x\|=1\}$ in $\BbbR^r$, then the identity function from $S_{r-1}$ to
itself is not homotopic to a constant function.

\proof{\query

}%end of proof of 4A7B                                                          
                                                                                
\leader{4A7C}{Corollary} If $r\ge 1$, $B(0,1)$ is the unit ball in $\BbbR^r$
and $h:B(\tbf 0,1)\to B(\tbf 0,1)$ is a continuous
function such that $h[S_{r-1}]=S_{r-1}$, then $h[B(\tbf 0,1)]=B(\tbf 0,1)$.

\proof{\query

}%end of proof of 4A7C

%\leader{4A7C}{}
%\leader{4A7D}{}
%\leader{4A7E}{}
%\leader{4A7F}{}
%\leader{4A7G}{}
%\leader{4A7H}{}
%\leader{4A7I}{}
%\leader{4A7J}{}
%\leader{4A7K}{}
%\leader{4A7L}{}
%\leader{4A7M}{}
%\leader{4A7N}{}
%\leader{4A7O}{}
%\leader{4A7P}{}
%\leader{4A7Q}{}
%\leader{4A7R}{}
%\leader{4A7S}{}
%\leader{4A7T}{}
%\leader{4A7U}{}
%\leader{4A7V}{}
%\leader{4A7W}{}

\discrpage
          
\leader{472G}{Theorem} ({\smc Bagnara Gennaioli Leccese \& Luongo p22})
Let $r\ge 1$ be an integer, $B\subseteq\BbbR^r$ a closed balls, $\rho_E$ 
the Euclidean metric on $\BbbR^r$, and $\rho$ a metric on $B$ 
inducing the usual topology on $B$.   Then there is a
$(\rho,\rho_E)$-Lipschitz surjection from $B$ onto itself.

\proof{{\bf (a)} For the time being (down to the end of (c) below), 
suppose that $B=\{x:x\in\BbbR^r$, $\|x\|\le 1\}$ is the ordinary unit ball of
$\Bbb R^r$.   Fix $i$ such that $1\le i\le r$ for the moment, and
write $\pi_i$ for the $i$th coordinate map from $\BbbR^r$ to $\Bbb R$.

\medskip

\quad{\bf (i)} For $\alpha>0$ and $x\in B$, set 

\Centerline{$f_i(\alpha,x)
=\inf_{z\in B}\pi_i(z)+\Bover1{\alpha}\rho(x,z)$;}

\noindent since $B$ is compact and $z\mapsto\pi_i(z)+\Bover1{\alpha}\rho(x,z)$
is continuous, the infimum is attained at $z_{x\alpha i}$ say.   
(We have two metrics in this theorem,
but only one topology on $B$, so `continuous' and `compact' in the last
sentence are unambiguous.)   

\medskip

\quad{\bf (ii)} Observe that

\Centerline{$-1=\inf_{z\in B}\pi_i(z)
\le f_i(\alpha,x)\le\pi_i(x)+\Bover1\alpha\rho(x,x)\le\pi_i(x)\le 1$}

\noindent for every $x\in B$.   

\medskip

\quad{\bf (iii)} If $0<\alpha\le\beta$ and $x\in B$ then

\Centerline{$f_i(\beta,x)
=\inf_{z\in B}\pi_i(z)+\Bover1{\beta}\rho(x,z)
\le\inf_{z\in B}\pi_i(z)+\Bover1{\alpha}\rho(x,z)=f_i(\alpha,x)$}

\noindent and 

$$\eqalign{f_i(\alpha,x)-f_i(\beta,x)&
\le\sup_{z\in B}\bigl((\pi_i(z)+\Bover1{\alpha}\rho(x,z))
  -(\pi_i(z)+\Bover1{\beta}\rho(x,z))\bigr)\cr&
=(\Bover1{\alpha}-\Bover1{\beta})\sup_{z\in B}\rho(x,z)
\le M(\Bover1{\alpha}-\Bover1{\beta})\cr}$$

\noindent where $M=\sup_{y,z\in B}\rho(y,z)$ is finite, again because
$B$ is compact.

\medskip

\quad{\bf (iv)} If $x$, $y\in B$ then

$$\eqalign{f_i(\alpha,y)&
=\inf_{z\in B}\pi_i(z)+\Bover1{\alpha}\rho(y,z)
\le\inf_{z\in B}\pi_i(z)+\Bover1{\alpha}(\rho(y,x)+\rho(x,z))\cr&
=\Bover1{\alpha}\rho(y,x)+\inf_{z\in B}\pi_i(z)+\Bover1{\alpha}\rho(x,z)
=\Bover1{\alpha}\rho(y,x)+f_i(\alpha,x)\cr}$$

\noindent and similarly
$f_i(\alpha,x)\le\Bover1{\alpha}\rho(x,y)+f_i(\alpha,y)$, so
$|f_i(\alpha,x)-f_i(\alpha,y)|\le\bover1{\alpha}\rho(x,y)$.

\medskip

\quad{\bf (v)}
For any $x\in B$, $\alpha\mapsto f_i(\alpha,x)$ is non-increasing and
bounded above, so $\lim_{\alpha\downarrow 0}f_i(\alpha,x)
=\sup_{\alpha>0}f_i(\alpha,x)$ is defined.
But now we have

\Centerline{$\pi_i(z_{x\alpha i})+\Bover1{\alpha}\rho(x,z_{x\alpha i})
=f_i(\alpha,x)\le\pi_i(x)$}

\noindent so $\rho(x,z_{x\alpha i})\le\alpha$ for every $\alpha>0$ and
$x=\lim_{\alpha\downarrow 0}z_{x\alpha i}$.   It follows that
$\pi_i(x)=\lim_{\alpha\downarrow 0}\pi_i(z_{x\alpha i})$ and

$$\eqalign{\pi_i(x)&
\ge\lim_{\alpha\downarrow 0}f_i(\alpha,x)
=\lim_{\alpha\downarrow 0}
  (\pi_i(z_{x\alpha i})+\Bover1{\alpha}\rho(x,z_{x\alpha i}))\cr&
=\pi_i(x)+\lim_{\alpha\downarrow 0}
  \Bover1{\alpha}\rho(x,z_{x\alpha i})
\ge\pi_i(x);\cr}$$

\noindent thus $\lim_{\alpha\downarrow 0}f_i(\alpha,x)=\pi_i(x)$.

As the real-valued
functions $x\mapsto\pi_i(x)-f_i(\alpha,x)$ are all continuous 
and $B$ is compact, Dini's theorem (436Ic) tells us that 

\Centerline{$\lim_{\alpha\downarrow 0}\sup_{x\in B}(\pi_i(x)-f_i(\alpha,x))
=\inf_{\alpha>0}\sup_{x\in B}(\pi_i(x)-f_i(\alpha,x))=0$.}

\medskip

\quad{\bf (vi)} Extend $f_i$ to the whole of $\coint{0,\infty}\times B$ by
setting $f_i(0,x)=\pi_i(x)$ for every $x\in B$.   Then $f_i$ is continuous.
\Prf\ Take $\alpha\ge 0$ and $x\in B$.   If $\alpha>0$, then

$$\eqalign{|f_i(\beta,y)-f_i(\alpha,x)|&
\le|f_i(\beta,y)-f_i(\alpha,y)|+|f_i(\alpha,y)-f_i(\alpha,x)|\cr&
\le M|\Bover1{\beta}-\Bover1{\alpha}|+\Bover1{\alpha}\rho(y,x)\cr}$$

\noindent whenever $\beta>0$ and $y\in B$, by (iii) and (iv) above.   So
$f_i$ is continuous at $(\alpha,x)$.   If $\alpha=0$, then

\Centerline{$\lim_{\beta\downarrow 0}\sup_{y\in B}|f_i(\beta,y)-\pi_i(y)|
=0=\sup_{y\in B}|f_i(0,y)-\pi_i(y)|$}

\noindent so

$$\eqalign{&\limsup_{(\beta,y)\to(\alpha,x)}|f_i(\beta,y)-f_i(\alpha,x)|\cr&
\mskip100mu\le\limsup_{(\beta,y)\to(\alpha,x)}|f_i(\beta,y)-\pi_i(y)|
   +\limsup_{(\beta,y)\to(\alpha,x)}|\pi_i(y)-\pi_i(x)|
=0\cr}$$

\noindent and again $f_i$ is continuous at $(\alpha,x)$.\ \Qed

\medskip

{\bf (b)} Define $f:\coint{0,\infty}\times B\to\BbbR^r$ by setting

\Centerline{$f(\alpha,x)=(f_1(\alpha,x),\ldots,f_r(\alpha,x))$}

\noindent for $x\in B$.   By (a-vi), $f$ is continuous, and if $\alpha>0$ then
by (a-iv)

\Centerline{$\|f(\alpha,x)-f(\alpha,y)\|
\le\sum_{i=1}^r|f_i(\alpha,x)-f_i(\alpha,y)|
\le\Bover{r}{\alpha}\rho(x,y)$}

\noindent for all $x$, $y\in B$ so $x\mapsto f(\alpha,x)$ is
$(\rho,\rho_E)$-Lipschitz.   By (a-v), 

$$\limsup_{\alpha\downarrow 0}\sup_{x\in B}\|f(\alpha,x)-x\|
\le\sum_{i=1}^r\limsup_{\alpha\downarrow 0}
   \sup_{x\in B}|\pi_i(x)-f_i(\alpha,x)|
=0$$

\noindent so there is a $\delta>0$ such that
$\|f(\alpha,x)-x\|\le\bover12$ whenever $x\in B$ and $0<\alpha\le\delta$.

\medskip

{\bf (c)} For $x\in\BbbR^r$ let $h(x)\in B$ be such that
$\|x-h(x)\|=\min_{z\in B}\|x-z\|$ (3A5Md);  then $\|h(x)-h(y)\|\le\|x-y\|$ for
all $x$, $y\in\BbbR^r$ (3A5Me\Latereditions), while $\|h(x)\|=1$ whenever
$\|x\|\ge 1$.   So if we set
$F(t,x)=h(2f(\delta t,x))$ for $t\in[0,1]$ and $x\in B$, $F$ will be continuous
and $x\mapsto F(1,x):B\to B$ will be $(\rho,\rho_E)$-Lipschitz.

Moreover, $\|f(\delta t,x)-x\|\le\bover12$ for every $x\in B$, by the choice of
$\delta$. so if $x\in S_{r-1}$ we shall have
$\|f(\delta t,x)\|\ge\bover12$ and $\|F(t,x)\|=1$ for every $t\in[0,1]$.
So $F\restr[0,1]\times S_{r-1}$ is a homotopy between $g_0$ and 
$g\restr S_{r-1}$ where

\Centerline{$g_0(x)=F(0,x)=h(2f(0,x))=h(2x)=x$}

\noindent for $x\in S_{r-1}$ and $g(x)=F(1,x)$ for $x\in B$.   But this means
that $g\restr S_{r-1}$ is homotopic to the identity on $S_{r-1}$.   By 4A7C,
$g[B]=B$, while $g$ is $(\rho,\rho_E)$-Lipschitz.

\medskip

{\bf (d)} For a general closed ball $B\subseteq\BbbR^r$, if $B$ is a singleton
then the result is trivial.   If $B=\{x:x\in\BbbR^r$, $\|x-z\|\le\delta\}$ where
$z\in\BbbR^r$ and $\delta>0$, then $(\BbbR^r,\rho_E,B)$ and 
$(\BbbR^r,\rho _E,B(\tbf 0,1))$ are lipeomorphic in the sense that there is
a bijection $\phi:\BbbR^r\to\BbbR^r$ such that $\phi$ and $\phi^{-1}$ are
$(\rho_E,\rho_E)$-Lipschitz and $\phi[B(\tbf 0,1)]=B$.   
(Take $\phi(x)=z+\delta x$ for $x\in\BbbR^r$.)   Now, given a metric
$\rho$ on $B$ inducing its topology, set
$\tilde\rho(x,y)=\rho(\phi(x),\phi(y))$ for $x$, $y\in B(\tbf 0,1)$;  
then $\tilde\rho$ is a metric on $B(\tbf 0,1)$ inducing its topology, and
$\phi^{-1}:B\to B(\tbf 0,1)$ is $(1,\rho,\tilde\rho)$-Lipschitz.   By
(a)-(c), there is a $(\tilde\rho,\rho_E)$-Lipschitz surjection
$\tilde g:B(\tbf 0,1)\to B(\tbf 0,1)$.   But now
$g=\phi\tilde g\phi^{-1}$ is a $(\rho,\rho_E)$-Lipschitz surjection from $B$ 
onto itself. 
}%end of proof of 472G

\leader{472H}{Corollary} Let $r\ge 1$ be an integer, $\rho$ a metric on
$\BbbR^r$ inducing the usual topology on $\BbbR^r$, and $\mu_{Hr}^{(\rho)}$ the
corresponding $r$-dimensional Hausdorff measure on $\BbbR^r$.    Then
$\mu_{Hr}^{(\rho)}$ is strictly positive.

\proof{ If $G\subseteq\BbbR^r$ is a non-empty open set, it includes a
non-trivial closed ball $B$ say.   By 472G, there is a surjection
$g:B\to B$ which is $(\rho,\rho_E)$-Lipschitz where $\rho_E$ is the usual
metric on $\BbbR^r$;  let $\gamma>0$ be such that $g$ is
$(\gamma,\rho,\rho_E)$-Lipschitz.   As the $r$-dimensional Hausdorff measure
$\mu_{Hr}^{(\rho_E)}$ is just a multiple of Lebesgue measure (264I), 

\Centerline{$0<\mu_{Hr}^{(\rho_E)}B\le\gamma^r\mu_{Hr}^{(\rho)}B$}

\noindent (471J) and 

\Centerline{$\gamma^r\mu_{Hr}^{(\rho)}G\ge\gamma^r\mu_{Hr}^{(\rho)}B>0$.}

}%end of proof of 472H

%\leader{472I}{}
%\leader{472J}{}
%\leader{472K}{}
%\leader{472L}{}
%\leader{472M}{}
%\leader{472N}{}
%\leader{472O}{}
%\leader{472P}{}
%\leader{472Q}{}
%\leader{472R}{}
%\leader{472S}{}
%\leader{472T}{}
%\leader{472U}{}
%\leader{472V}{}
%\leader{472W}{}
     

\frfilename{mt4aconc.tex}
\versiondate{10.1.17}
\copyrightdate{2017}
     
\def\chaptername{Appendix}
\def\sectionname{Concordance}
     
\centerline{\bf Concordance for Appendix}
     
\bigskip
     
I list here the section and paragraph numbers which have 
(to my knowledge) appeared in print in references to this volume, 
and which have since been changed.

\leader{4{}A2Jf}{Uniformities on completely regular spaces} 4{}A2Jf,
referred to in the 2009 edition of Volume 5, has been moved to 4A2Jg.

\leader{4{}A3Q}{Baire property and cylindrical $\sigma$-algebras}
4{}A3Q-4{}A3T and 4{}A3V, referred to in the 2008 and 2015 editions of Volume
5, are now 4A3R-4A3U %4A3R 4A3S 4A3T 4A3U
and 4A3W.

\leader{4{}A4B}{Bounded sets in linear topological spaces} 4{}A4Bg,
referred to in the 2008 edition of Volume 5, has been moved to 3A5Nb.

\frnewpage
\references{
\frfilename{mt4r.tex}
\versiondate{9.2.13}
\copyrightdate{2002}

% 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}

\Loadcyrillic

\Centerline{\bf References for Volume 4}


\medskip%A

Aldaz J.M.\ [95] `On compactness and Loeb measures', Proc.\ Amer.\
Math.\ Soc.\ 123 (1995) 173-175.
\cmmnt{[439Xa.]}

Aldaz J.M.\ [97] `Borel extensions of Baire measures', Fund.\ Math.\ 154
(1997) 275-293.
\cmmnt{[434Yn.]}

Aldaz J.M.\ \& Render H.\ [00] `Borel measure extensions of measures
defined on sub-$\sigma$-algebras', Advances in Math.\ 150 (2000)
233-263.
\cmmnt{[432D.]}

Alexandroff P.S.\ \& Urysohn P.S.\ [1929] {\it  M\'emoire sur les espaces
topologiques compacts},
Verhandelingen Akad.\ Wetensch.\ Amsterdam 14 (1929) 1-96.
\cmmnt{[419L.]}

Anderson I.\ [87] {\it Combinatorics of Finite Sets.}  Oxford U.P.,
1987.
\cmmnt{[4A4N.]}

Andretta A.\ \& Camerlo R.\ [13] `The descriptive set theory of the
Lebesgue density theorem', Advances in Math.\ 234 (2013).
%{\tt http://arxiv.org/abs/1105.3355}.
\cmmnt{[475Yg.]}

Arkhangel'skii A.V.\ [92] {\it Topological Function Spaces.}  Kluwer,
1992.
\cmmnt{[\S462 {\it intro.}, \S467 {\it notes\/}.]}

\leaveitout{
Armitage D.H.\ \& Gardiner S.J.\ [01] {\it Classical Potential Theory.}
Springer, 2001.
478Yd}

Aronov B., Basu S., Pach J.\ \& Sharir M.\ [03] (eds.)
{\it Discrete and Computational Geometry;  the Goodman-Pollack
Festschrift,}  Springer, 2003.

Asanov M.O.\ \& Velichko N.V.\ [81] `Compact sets in $C_p(X)$', Comm.\
Math.\ Helv.\ 22 (1981) 255-266.
\cmmnt{[462Ya.]}

Austin T.\ [10a] `On the norm convergence of nonconventional ergodic
averages', Ergodic Theory and Dynamical Systems 30 (2010) 321-338.
%({\tt http://{\penalty-100}arxiv.org/abs/{\penalty-100}0805.0320}).
\cmmnt{[497M.]}

Austin T.\ [10b] `Deducing the multidimensional Szemer\'edi theorem from
the infinitary hypergraph removal lemma',
J.\ d'Analyse Math.\ 111 (2010) 131-150.
%({\tt http://arxiv.org{\bsp}abs/0808.2267}).
\cmmnt{[497M.]}

\leaveitout{
Avil\'es A., Plebanek G.\ \& Rodr\'{\i}guez J.\ [p11] `Measurability in
$C(2^{\kappa})$ and Kunen cardinals', 2011.
%material on  C(\{0,1\}^{\kappa}) ;  Baire and Borel \sigma-algebras
%in weak and pointwise topologies
}

\leaveitout{Axler S. \& Ramey W.\ [95]
`Harmonic polynomials and Dirichlet-type problems',
Proc.\ Amer.\  Math.\ Soc.\ 123 (1995) 3765-3773.
http://www.axler.net/HarmonicPolynomials.html
478Yd}

\medskip%B

Balogh Z.\ [96] `A small Dowker space in ZFC', Proc.\ Amer.\ Math.\
Soc.\ 124 (1996) 2555-2560.
\cmmnt{[439O.]}

Banach S.\ [1923] `Sur le probl\`eme de la mesure',
Fundamenta Math.\ 4 (1923) 7-33.
\cmmnt{[449O.]}

Becker H.\ \& Kechris A.S.\ [96] {\it The descriptive set theory of
Polish group actions.}  Cambridge U.P., 1996 (London Math.\ Soc.\
Lecture Note Series 232).  \cmmnt{[424H, 448P, \S448 {\it notes\/}.]}

Bellow A.\ {\it see} Ionescu Tulcea A.

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Springer, 1976 (Lecture Notes in Mathematics 541).
%4r

Bergelson V.\ [85] `Sets of recurrence of $\Bbb Z^m$-actions and properties of
sets of differences in $\Bbb Z^m$',
J.\ London Math.\ Soc.\ (2) 31 (1985) 295-304.
%http://people.dm.unipi.it/dinasso/ULTRA2015/BUMN/Bergelson%20-%20Sets%20of%20recurrence%20of%20Zm-actions%20and%20properties%20of%20sets%20of%20differences%20in%20Zm%20(1985).pdf
\cmmnt{[491Yc.]}

Bergman G.M.\ [06] `Generating infinite symmetric groups', Bull.\ London
Math.\ Soc.\ 38 (2006) 429-440.
%http://math.berkeley.edu/~gbergman/papers/
%({\tt http:/{\bsp}arxiv.org{\bsp}abs{\bsp}math{\bsp}0401304v2}).
\cmmnt{[\S494 {\it notes}.]}

Billingsley P.\ [99] {\it Convergence of Probability Measures.}  Wiley,
1999.
\cmmnt{[\S4A3 {\it notes}.]}
%second edition ISBN 0-471-19745-9

Blackwell D.\ [56] `On a class of probability spaces',
pp.\ 1-6 in {\smc Neyman 56}.
\cmmnt{[419K, 452P.]}

Bochner S.\ [1933] `Monotone Funktionen, Stieltjessche Integrale, und
harmonische Analyse', Math.\ Ann.\ 108 (1933) 378-410.
\cmmnt{[445N.]}

\indexheader{Bogachev}
Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.
\cmmnt{[437Q, 437Xs, 437Yo, 437Yu, 457K.]}

Bollob\'as B.\ [79] {\it Graph Theory.}   Springer, 1979.
\cmmnt{[4A1G, 4A4N.]}

Bongiorno B., Piazza L.di \ \& Preiss D.\ [00]
`A constructive minimal integral which includes Lebesgue integrable
functions and derivatives', J.\ London Math.\ Soc.\ (2) 62 (2000) 117-126.
\cmmnt{[483Yi.]}

Borodulin-Nadzieja P.\ \& Plebanek G.\ [05] `On compactness of measures on
Polish spaces', Illinois J.\ Math.\ 49 (2005) 531-545.
\cmmnt{[451L.]}

Bourbaki N.\ [65] {\it Int\'egration,} chaps.~1-4.   Hermann, 1965
(Actualit\'es Scientifiques et Industrielles 1175).
\cmmnt{[\S416 {\it notes\/}, \S436 {\it notes\/}.]}

Bourbaki N.\ [66] {\it Espaces Vectoriels Topologiques}, chaps.~1-2,
$2^{\text{e}}$ \'ed.
Hermann, 1966 (Actualit\'es Scientifiques et Industrielles 1189).
\cmmnt{[4A2J.]}
%Store A QA 37.B6 no 7792 0070 5

Bourbaki N.\ [69] {\it Int\'egration,} chap.~9.   Hermann, 1969
(Actualit\'es Scientifiques et Industrielles 1343).
\cmmnt{[\S436 {\it notes.}]}

Bourbaki N.\ [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[\S461 {\it notes\/}, \S4A4.]}

\indexheader{Bourgain}
Bourgain J., Fremlin D.H.\ \& Talagrand M.\ [78] `Pointwise compact
sets of Baire-measurable functions', Amer.\ J.\ Math.\ 100 (1978)
845-886.
\cmmnt{[462Ye.]}

\indexheader{Brodski\v{\i}}
Brodski\v{\i} M.L.\ [1949] `On some properties of sets of positive measure',
Uspehi Matem.\ Nauk (N.S.) 4 (1949) 136-138.
\cmmnt{[498B.]}

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Math.\ Systems Theory 4 (1970) 243-248.
\cmmnt{[449D.]}

Burke D.K.\ \& Pol R.\ [05] `Note on separate continuity and the Namioka
property', Top.\ Appl.\ 152 (2005) 258-268.
\cmmnt{[\S463 {\it notes.}]}

Burke M.R., Macheras N.D.\ \& Strauss W.\ [p21] `The strong marginal lifting
problem for hyperstonian spaces', preprint (2021).
%bms5s.pdf

\medskip%C

Carrington D.C.\ [72] `The generalised Riemann-complete integral', PhD
thesis, Cambridge, 1972.
\cmmnt{[481L.]}

\v{C}ech E.\ [66] {\it Topological Spaces.}  Wiley, 1966.
\cmmnt{[\S4A2, 4A3S.]}

Chacon R.V. [69]
`Weakly mixing transformations which are not strongly mixing',
Proc.\ Amer.\ Math.\ Soc.\ 22 (1969) 559-562.
\cmmnt{[494F.]}

\indexheader{Choquet}
Choquet G.\ [55] `Theory of capacities', Ann.\ Inst.\ Fourier (Grenoble)
5 (1955) 131-295.
\cmmnt{[432K.]}

\indexheader{Chung}
Chung K.L.\ [95] {\it Green, Brown and Probability.}   World Scientific,
1995.
% 978-9810225339  HA33.4.C5  479Xt

\indexheader{Ciesielski}
Ciesielski K.\ \& Pawlikowski J.\ [03]
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Fundamenta Math. 176 (2003) 63-75.
\cmmnt{[498C.]}

Cs\'asz\'ar \'A.\ [78] {\it General Topology.}  Adam Hilger, 1978.
\cmmnt{[\S4A2, \S4A5.]}

\medskip%D

Davies R.O.\ [70] `Increasing sequences of sets and Hausdorff measure', Proc.\ London Math.\ Soc.\ (3) 20 (1970) 222-236.
\cmmnt{[471G.]}

Davies R.O.\ [71] `Measures not approximable or not specifiable by means
of balls', Mathematika 18 (1971) 157-160.
\cmmnt{[\S466 {\it notes\/}.]}

Davis W.J., Ghoussoub N.\ \& Lindenstrauss J.\ [81] `A lattice-renorming
theorem and applications to vector-valued processes', Trans.\ Amer.\
Math.\ Soc.\ 263 (1981) 531-540.
\cmmnt{[\S467 {\it notes\/}.]}

\indexheader{de Caux}
de Caux P.\ [76] `A collectionwise normal, weakly $\theta$-refinable
Dowker space which is neither irreducible nor realcompact', Topology
Proc.\ 1 (1976) 67-77.
\cmmnt{[439O.]}

Dellacherie C.\ [80] `Un cours sur les ensembles analytiques', pp.\
183-316 in {\smc Rogers 80}.
\cmmnt{[\S432 {\it notes\/}.]}

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$(\ell_{\infty}/c_0)$,weak) is not a Radon space', Proc.\ Amer.\ Math.\
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\cmmnt{[466I.]}

Deville R., Godefroy G.\ \& Zizler V.\ [93] {\it Smoothness and
Renormings in Banach Spaces.}  Pitman, 1993.
\cmmnt{[\S467 {\it intro.}, 467Ye, \S467 {\it notes\/}.]}

Diaconis, P.\ \& Freedman, D.\ [80] `Finite exchangeable sequences.'
Ann. of Probability 8 (1980) 745-764.
\cmmnt{[459Xd.] }

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Bull.\ London Math. Soc.\ 40 (2008) 463-472.
%{\tt http://web.mac.com/chollan1{\bsp}iWeb/Site{\bsp}publications.html}.
\cmmnt{[494Q.]}

Dubins L.E.\ \& Freedman D.A.\ [79]
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Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 48 (1979) 115-132.
\cmmnt{[459K.]}

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\cmmnt{[\S4A4.]}

Du Plessis, N.\ [70] {\it Introduction to Potential Theory.}  Oliver \&
Boyd, 1970.
%used for \S478

\medskip%E

Eggleston H.G.\ [54] `Two measure properties of Cartesian product sets',
Quarterly J.\ Math.\ (2) 5 (1954) 108-115.
\cmmnt{[498B.]}

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\cmmnt{[449J.]}

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[4A1A.]}

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\S435 {\it notes\/}, 462L, \S4A2, 4A4B.]}

Evans L.C.\ \& Gariepy R.F.\ [92] {\it Measure Theory and Fine
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\cmmnt{[Chap.\ 47 {\it intro.}, \S473 {\it intro.}, \S473 {\it notes\/},
\S474 {\it notes\/}, \S475 {\it notes\/}.]}

\medskip%F

Falconer K.J.\ [90] {\it Fractal Geometry.}  Wiley, 1990.
\cmmnt{[\S471 {\it intro.}]}

Federer H.\ [69] {\it Geometric Measure Theory.}
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\cmmnt{[\S441 {\it notes\/}, 442Ya,
\S443 {\it notes\/}, Chap.\ 47 {\it intro.}, 471Yk,
\S471 {\it notes\/}, \S472 {\it notes\/}, \S475 {\it notes\/}.]}

Fernique X.\ [97] {\it Fonctions al\'eatoires gaussiennes,
vecteurs al\'eatoires gaussiens}, CRM Publications, 1997.
\cmmnt{[\S456 {\it notes\/}.]}

Folland G.B.\ [95] {\it A Course in Abstract Harmonic Analysis.}   CRC
Press, 1995
\cmmnt{[\S4A5, \S4A6.]}

\indexheader{F{\o}lner}
F{\o}lner E.\ [55] `On groups with full Banach mean value',
Math.\ Scand.\ 3 (1955) 243-254.
\cmmnt{[\S449 {\it notes\/}.]}

\indexheader{Ford}
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\cmmnt{[4A4N.]}

Frankl P.\ \& R\"odl V.\ [02] `Extremal problems on set systems',
Random Structures Algorithms 20 (2002) 131-164.
\cmmnt{[497L.]}

\indexheader{Fremlin}
Fremlin D.H.\ [74] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[\S462 {\it notes\/}.]}

Fremlin D.H.\ [75a] `Pointwise compact sets of measurable functions',
Manuscripta Math.\ 15 (1975) 219-242.
\cmmnt{[463K.]}

Fremlin D.H.\ [75b] `Topological measure spaces:  two counter-examples',
Math.\ Proc.\ Cambridge Phil.\ Soc.\ 78 (1975) 95-106.
\cmmnt{[419C, 419D.]}

Fremlin D.H.\ [76] `Products of Radon measures:  a counterexample',
Canadian Math.\ Bull.\ 19 (1976) 285-289.
\cmmnt{[419E.]}

Fremlin D.H.\ [81] `Measurable functions and almost continuous
functions', Manuscripta Math.\ 33 (1981) 387-405.
\cmmnt{[451T.]}

Fremlin D.H.\ [84] {\it Consequences of Martin's Axiom.}
Cambridge U.P., 1984.
\cmmnt{[\S434 {\it notes\/}, 439Yd, 4A2E.]}

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304
in {\smc Judah 93}.
\cmmnt{[\S438 {\it notes\/}.]}

Fremlin D.H.\ [95] `The generalized McShane integral',
Illinois J.\ Math.\ 39 (1995) 39-67.
\cmmnt{[481N.]}

Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces',
Fundamenta Math.\ 165 (2000) 67-94.
\cmmnt{[451V.]}

Fremlin D.H.\ [n05] `Strictly localizable Borel measures', note of 7.5.05
({\tt http:/{\bsp}www.\penalty-100essex.ac.uk{\bsp}maths/people{\bsp}fremlin/preprints.htm}).
\cmmnt{[434Yr.]}

Fremlin D.H.\ [n15] `The Sorgenfrey line is not a Prokhorov space',
note of 5.8.15
({\tt http:/{\bsp}www.\penalty-100essex.ac.uk{\bsp}maths/people{\bsp}fremlin/preprints.htm}).
\cmmnt{[439Yk.]}

Fremlin D.H.\ \& Grekas S.\ [95] `Products of completion regular
measures', Fund.\ Math.\ 147 (1995) 27-37.
\cmmnt{[434Q.]}

Fremlin D.H.\ \& Talagrand M.\ [78] `On the representation of increasing
linear functionals on Riesz spaces by measures',
Mathematika 25 (1978) 213-215.
\cmmnt{[439I.]}

Fremlin D.H.\ \& Talagrand M.\ [79] `A decomposition theorem for
additive set functions and applications to Pettis integral and ergodic
means', Math.\ Zeitschrift 168 (1979) 117-142.
\cmmnt{[\S464.]}  %{\it 464C}, 464I, 464M, 464N, \S464 {\it notes\/}

Fristedt B.\ \& Gray L.\ [97] {\it A Modern Approach to Probability
Theory.}  Birkh\"auser, 1997.
\cmmnt{[\S455 {\it notes\/}, \S495 {\it notes\/}.]}
%HA33.4.F7

Frol{\'\i}k Z.\ [61] `On analytic spaces', Bull.\ Acad.\ Polon.\ Sci.\
%8 (1961) 747-750.
9 (1961) 721-726.   %Paulo Monteiro, email 8.9.22
\cmmnt{[422F.]}

Frol\'\i k Z.\ [88] (ed) {\it General Topology and its Relations to
Modern Analysis and Algebra VI} (Proc.\ Sixth Prague Topological
Symposium), Heldermann, 1988.

Furstenberg H.\ [77]
`Ergodic behavior of diagonal measures and a theorem of Szemer\'edi
on arithmetic progressions', J. d'Analyse Math.\ 31 (1977), 204-256.
\cmmnt{[\S497 {\it intro.}, \S497 {\it notes\/}.]}

Furstenberg H.\ [81]
{\it Recurrence in Ergodic Theory and Combinatorial Number Theory.}
Princeton U.P., 1981.
\cmmnt{[\S497 {\it intro.}, 497N.]}
% QA614

Furstenberg H.\ \& Katznelson Y.\ [85]
`An ergodic Szemer\'edi theorem for IP-systems and combinatorial theory',
J.\ d'Analyse Math.\ 45 (1985) 117-168.
\cmmnt{[\S497 {\it intro.}]}

\medskip%G

Gaal S.A.\ [64] {\it Point Set Topology.}  Academic, 1964.
\cmmnt{[\S4A2.]}

Gardner R.J.\ [75] `The regularity of Borel measures and Borel
measure-compactness', Proc.\ London Math.\ Soc.\ (3) 30 (1975) 95-113.
\cmmnt{[438M.]}

Gardner R.J.\ \& Pfeffer W.F.\ [84] `Borel measures', pp. 961-1043 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[434G.]}

Giordano T.\ \& Pestov V.\ [02] `Some extremely amenable groups',
C.R.A.S.\ Paris (S\'er.\ I) 334 (2002) 273-278.
\cmmnt{[494I.]}

Glasner S.\ [98] `On minimal actions of Polish groups', Topology and its Applications 85 (1998) 119-125.
\cmmnt{[\S493 {\it notes\/}.]}

Gordon R.A.\ [94] {\it The Integrals of Lebesgue, Denjoy, Perron and
Henstock.}   Amer.\ Math.\ Soc., 1994 (Graduate Studies in Mathematics
4).
\cmmnt{[481Q, \S483 {\it notes\/}.]}

Greenleaf F.P.\ [69] {\it Invariant Means on Topological Groups and Their
Applications.}   van Nostrand, 1969.
\cmmnt{[\S449 {\it notes.}]}

Grothendieck A. [92] {\it Topological Vector Spaces.}  Gordon \& Breach,
1992.
\cmmnt{[\S462 {\it notes\/}.]}

\indexheader{Grzegorek}
Grzegorek E.\ [81] `On some results of Darst and Sierpi\'nski concerning
universal null and universally negligible sets', Bull.\ Acad.\ Polon.\
Sci.\ (Math.) 29 (1981) 1-5.
\cmmnt{[439F, \S439 {\it notes\/}.]}

\medskip%H

\indexheader{Halmos}
Halmos P.R.\ [1944] `In general a measure-preserving transformation is
mixing', Annals of Math.\ 45 (1944) 786-792.
\cmmnt{[494E.]}

Halmos P.R.\ [50] {\it Measure Theory.}   Springer, 1974.
\cmmnt{[\S441 {\it notes\/}, \S442 {\it notes\/}, 443M, \S443 {\it notes\/}.]}

Halmos P.R.\ [56] {\it Lectures on ergodic theory.}  Chelsea, 1956.
\cmmnt{[494A.]}

Hansell R.W.\ [01] `Descriptive sets and the topology of non-separable
Banach spaces', Serdica Math.\ J.\ 27 (2001) 1-66.
\cmmnt{[466D.]}

Hart J.E.\ \& Kunen K.\ [99] `Orthogonal continuous functions',
Real Analysis Exchange 25 (1999) 653-659.
\cmmnt{[416Yh.]}

Hartman S.\ \& Mycielski J.\ [58] `On the imbedding of topological groups into connected topological groups', Colloq.\ Math.\ 5 (1958) 167-169.
\cmmnt{[493Ya.]}

Haydon R.\ [74] `On compactness in spaces of measures and
measure-compact spaces', Proc.\ London Math.\ Soc.\ (3) 29 (1974) 1-6.
\cmmnt{[438J.]}

Henry J.P.\ [69] `Prolongement des mesures de Radon', Ann.\ Inst.\
Fourier 19 (1969) 237-247.
\cmmnt{[416N.]}

Henstock R.\ [63] {\it Theory of Integration.}  Butterworths, 1963.
\cmmnt{[Chap.\ 48 {\it intro.}, 481J.]}

Henstock R.\ [91] {\it The General Theory of Integration.}   Oxford
U.P., 1991.
\cmmnt{[\S481 {\it notes\/}.]}

Herer W.\ \& Christensen J.P.R.\ [75] `On the existence of pathological submeasures and the construction of exotic topological groups', Math.\ Ann.\ 213 (1975) 203-210.
\cmmnt{[\S493 {\it notes\/}.]}

Herglotz G.\ [1911] `\"Uber Potenzreihen mit positivem reellen Teil im
Einheitskreis', Leipziger Berichte 63 (1911) 501-511.
\cmmnt{[445N.]}

Hewitt E.\ \& Ross K.A.\ [63] {\it Abstract Harmonic Analysis I.}
Springer, 1963 and 1979.
\cmmnt{[\S442 {\it notes\/}, \S444 {\it intro.}, \S444 {\it notes\/},
\S4A5, \S4A6.]}

Hewitt E.\ \& Savage L.J.\ [55]
`Symmetric measures on Cartesian products',
Trans.\ Amer.\ Math.\ Soc.\ 80 (1955) 470-501.
\cmmnt{[459Xe.]}

%Holick\'y P.\ \& Spurn\'y J.\ [03] `Perfect images of absolute Souslin
%and absolute Borel Tychonoff spaces', Topology and its Appl.\ 131 (2003)
%281-294.
%\cmmnt{[434Yr.]}

\indexheader{Howroyd}
Howroyd J.D.\ [95] `On dimension and on the existence of sets of finite
positive Hausdorff measure', Proc.\ London Math.\ Soc.\ (3) 70 (1995)
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\cmmnt{[471R, 471S.]}

\medskip%I

\indexheader{Ionescu Tulcea}
Ionescu Tulcea A.\ [73] `On pointwise convergence, compactness and
equicontinuity I', Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 26
(1973) 197-205.
\cmmnt{[463C.]}

Ionescu Tulcea A.\ [74] `On pointwise convergence, compactness, and
equicontinuity II', Advances in Math.\ 12 (1974) 171-177.
\cmmnt{[463G.]}

Ionescu Tulcea A.\ \& Ionescu Tulcea C.\ [67] `On the existence of a
lifting commuting with the left translations of an arbitrary locally
compact group', pp.\ 63-97 in {\smc LeCam \& Neyman 67}.
\cmmnt{[\S447 {\it intro.}, 447I, \S447 {\it notes\/}.]}

Ionescu Tulcea A.\ \& Ionescu Tulcea C.\ [69] {\it Topics in the Theory
of Lifting.}  Springer, 1969.
\cmmnt{[\S453 {\it intro.}]}

\indexheader{Ismail}
Ismail M.\ \& Nyikos P.J.\ [80] `On spaces in which countably compact
sets are closed, and hereditary properties', Topology and its Appl.\ 11
(1980) 281-292.
\cmmnt{[\S434 {\it notes\/}.]}

\medskip%J

%Jackson S.\ \& Mauldin R.D.\ [99] `On the $\sigma$-class generated by
%open balls', Math.\ Proc.\ Cambridge Phil.\ Soc.\ 127 (1999) 99-108.

Jameson G.J.O.\ [74] {\it Topology and Normed Spaces.}  Chapman \& Hall,
1974.
\cmmnt{[\S4A4.]}

\indexheader{Jayne}
Jayne J.E.\ [76] `Structure of analytic Hausdorff spaces', Mathematika
23 (1976) 208-211.
\cmmnt{[422Yc.]}

Jayne J.E.\ \& Rogers C.A.\ [95] `Radon measures on Banach spaces with
their weak topologies', Serdica Math.\ J.\ 21 (1995) 283-334.
\cmmnt{[466H, \S466 {\it notes\/}.]}

\indexheader{Jech}
Jech T.\ [78] {\it Set Theory.}  Academic, 1978.
\cmmnt{[423S, \S4A1.]}

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2003.
%ISBN 3540440852
\cmmnt{[\S4A1.]}

Johnson R.A.\ [82] `Products of two Borel measures', Trans.\ Amer.\
Math.\ Soc.\ 269 (1982) 611-625.
\cmmnt{[434R.]}

Judah H.\ [93] (ed.) {\it Proceedings of the Bar-Ilan Conference on Set
Theory and the Reals, 1991.} Amer.\ Math.\ Soc.\ (Israel Mathematical
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Juh\'asz I., Kunen K.\ \& Rudin M.E.\ [76] `Two more hereditarily
separable non-Lindel\"of spaces', Canadian J.\ Math.\ 29 (1976)
998-1005.
\cmmnt{[\S437 {\it notes\/}, 439K, \S439 {\it notes\/}.]}

Just W.\ \& Weese M.\ [96] {\it Discovering Modern Set Theory I.}  Amer.\ Math.\ Soc., 1996 (Graduate Studies in Mathematics 8).
\cmmnt{[4A1A.]}

Just W.\ \& Weese M.\ [97] {\it Discovering Modern Set Theory II.}  Amer.\ Math.\ Soc., 1997 (Graduate Studies in Mathematics 18).
\cmmnt{[\S4A1.]}

\medskip%K

Kampen, E.R.van [1935] `Locally bicompact abelian groups and their character groups',
Ann.\ of Math.\ (2) 36 (1935) 448-463.
\cmmnt{[445U.]}

\indexheader{Kaplansky}
Kaplansky I.\ [71] {\it Lie Algebras and Locally Compact Groups.}
Univ.\ of Chicago Press, 1971.
\cmmnt{[\S446 {\it notes.}]}

Kechris A.S.\ [95] {\it Classical Descriptive Set Theory.}  Springer,
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\cmmnt{[Chap.\ 42 {\it intro.}, 423T, \S423 {\it notes\/},
\S448 {\it notes\/}, \S4A2, 4A3S, 4A3T.]}

%Kechris A.S.\ [p09] {\it Global Aspects of Ergodic Group Actions.}
%Amer.\ Math.\ Soc.\

Kellerer H.G.\ [84] `Duality theorems for marginal problems', Zeitschrift
f\"ur Wahrscheinlichkeitstheorie verw.\ Gebiete 67 (1984) 399-432.
\cmmnt{[457M.]}

Kelley J.L.\ [55] {\it General Topology.} Van Nostrand, 1955.
\cmmnt{[438R.]}

Kelley J.L.\ \& Srinivasan T.P.\ [71] `Pre-measures on lattices of sets',
Math.\ Annalen 190 (1971) 233-241.
\cmmnt{[413Xr.]}

Kittrell J.\ \& Tsankov T.\ [09] `Topological properties of full groups',
Ergodic Theory and Dynamical Systems 2009 (doi:10.1017/S0143385709000078).
\cmmnt{[494O, \S494 {\it notes}.]}

Kolmogorov A.N.\ [1933] {\it Grundbegriffe der
Wahrscheinlichkeitsrechnung.}  Springer, 1933;  translated as {\it
Foundations of Probability Theory,}  Chelsea, 1950.
\cmmnt{[454D.]}

K\"othe G.\ [69] {\it Topological Vector Spaces I.}   Springer, 1969.
\cmmnt{[\S462 {\it notes\/}, \S4A4, 4A6C.]}

Koumoullis G.\ \& Prikry K.\ [83] `The Ramsey property and measurable
selectors', J.\ London Math.\ Soc.\ (2) 28 (1983) 203-210.
\cmmnt{[451T.]}

Kuipers L.\ \& Niederreiter H.\ [74] {\it Uniform Distribution of
Sequences.}   Wiley, 1974.
\cmmnt{[491Z.]}

Kunen K.\ [80] {\it Set Theory.}  North-Holland, 1980.
\cmmnt{[\S4A1, 4A2E.]}

Kunen K.\ \& Vaughan J.E.\ [84] (eds.) {\it Handbook of Set-Theoretic
Topology.}   North-Holland, 1984.
%4r

Kuratowski K.\  [66] {\it Topology,} vol.\ I.   Academic, 1966.
\cmmnt{[423T, 462Ye, \S4A2, 4A3S.]}

Kurzweil J.\ [57] `Generalized ordinary differential equations and
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\cmmnt{[Chap.\ 48 {\it intro.}]}

Kwiatowska A. \& Solecki S.\ [11] `Spatial models of Boolean actions and
groups of isometries', Ergodic Theory Dynam.\ Systems 31 (2011) 405-421.
%{\tt http://www.math.uiuc.edu/\penalty-100$^{\sim}$ssolecki/papers.html}.
\cmmnt{[\S448 {\it notes\/}.]}

\medskip%L

Laczkovich M.\ [02] `Paradoxes in measure theory', pp.\ 83-123 in
{\smc Pap 02}.
\cmmnt{[\S449 {\it notes\/}.]}

LeCam L.M.\ \& Neyman J.\ [67] (eds.) {\it Proc.\ Fifth Berkeley
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Levy A.\ [79] {\it Basic Set Theory.}  Springer, 1979.
\cmmnt{[\S4A1.]}

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\cmmnt{[\S495 {\it notes\/}.]}

\indexheader{Losert}
Losert V.\ [79] `A measure space without the strong lifting property',
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\cmmnt{[453N, \S453 {\it notes\/}.]}

\indexheader{Luke\v{s}}
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{\it Fine topology methods
in real analysis and potential theory,} Springer, 1986 (Lecture Notes in
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\cmmnt{[414Ye.]}

\medskip%M

\indexheader{Mackey}
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\cmmnt{[448S.]}

Marczewski E.\ [53] `On compact measures', Fundamenta Math.\ 40 (1953)
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\cmmnt{[413T, Chap.\ 45 {\it intro.}, \S451 {\it intro.}, 451J.]}

Marczewski E.\ \& Ryll-Nardzewski C.\ [53] `Remarks on the compactness
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\cmmnt{[454C.]}

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\cmmnt{[435C.]}

Mattila P.\ [95] {\it Geometry of Sets and Measures in Euclidean
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\cmmnt{[Chap.\ 47 {\it intro.}, \S471 {\it intro.}, \S472 {\it notes\/},
\S475 {\it notes\/}.]}
%QA311.M2 9520 6456 1

Mauldin R.D. \& Stone A.H.\ [81] `Realization of maps', Michigan Math.\
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\cmmnt{[418T.]}

Maurey B.\ [79] `Construction des suites sym\'etriques', C.R.A.S.\ (Paris) 288 (1979) 679-681.
\cmmnt{[492H.]}
  % one m in sym\'etrique

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Monthly 80 (1973) 349-359.
\cmmnt{[481M.]}

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\cmmnt{[\S495 {\it notes\/}.]}

Milman V.D.\ \& Schechtman G.\ [86]
{\it Asymptotic Theory of Finite Dimensional Normed Spaces.}
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\cmmnt{[492G, \S492 {\it notes\/}.]}

\leaveitout{Mohammed D.\ \& Saadi K.\ [p09#]
`Probl\`emes de mesurabilit\'es sur
l'espace des fonctions continues sur un compact', 2009.
\cmmnt{[466Y?.]}
%refereeing for Canad Math Bull;  rejected
}

Molt\'o A., Orihuela J., Troyanski S.\ \& Valdivia M.\ [09] {\it A
Non-linear Transfer Technique for Renorming.}   Springer, 2009 (Lecture
Notes in Mathematics 1951).
\cmmnt{[\S467 {\it notes.}]}

Montgomery D.\ \& Zippin L.\ [55] {\it Topological Transformation
Groups.}   Interscience, 1955.
\cmmnt{[\S446 {\it notes.}]}

\indexheader{Moran}
Moran W.\ [68] `The additivity of measures on completely regular
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\cmmnt{[439P.]}

Moran W.\ [69] `Measures and mappings on topological spaces', Proc.\
London Math.\ Soc.\ (3) 19 (1969) 493-508.
\cmmnt{[435Xk.]}

Moran W.\ [70] `Measures on metacompact spaces', Proc.\ London Math.\
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\cmmnt{[438J.]}

\indexheader{Moschovakis}
Moschovakis Y.N.\ [80] {\it Descriptive Set Theory.}   Springer, 1980.
\cmmnt{[Chap.\ 42 {\it intro.}, \S423 {\it notes\/}.]}

Muldowney P.\ [87] {\it A General Theory of Integration in Function
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\cmmnt{[481P, \S482 {\it notes\/}.]}

Mushtari D.H.\ [96] {\it Probabilities and Topologies on Linear Spaces.}
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%ISBN 5-900975-01-0

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\cmmnt{[451U.]}

\medskip%N

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\cmmnt{[497J.]}

%Naimark M.A. [70] {\it Normed Rings.}  Wolters-Noordhoff, 1970.


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\medskip%O
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\cmmnt{[434Yq.]}

\medskip%P

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\cmmnt{[452I, 452S, 452Ye, \S452 {\it notes\/}.]}

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\cmmnt{[452R.]}

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\cmmnt{[449K, \S449 {\it notes\/}.]}
% QA 403.P2 store A

Perron O.\ [1914] `\"Uber den Integralbegriff', Sitzungsberichte der
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\cmmnt{[\S493 {\it notes\/}.]}

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\cmmnt{[484Xb.]}

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\cmmnt{[484B.]}

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\cmmnt{[\S461 {\it notes\/}.]}

\indexheader{Plewik}
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\cmmnt{[443Yl.]}

\indexheader{Pontryagin}
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\indexheader{Preiss}
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Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 27 (1973) 109-116.
\cmmnt{[439S.]}
%ILL 31.7.15

Preiss D.\ \& Ti\v{s}er J.\ [91]
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\cmmnt{[462B, 462C.]}

\medskip%Q\medskip%R

Radon J.\ [1913] `Theorie und Anwendungen der absolut additivien
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\cmmnt{[\S416 {\it notes\/}.]}

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\cmmnt{[\S451 {\it notes\/}.]}

\indexheader{Rao B.V.}
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\cmmnt{[419F.]}

Rao B.V. \& Srivastava S.M.\ [94] `Borel isomorphism theorem',
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\cmmnt{[424C.]}

\indexheader{Rao M.}
Rao M.\ [77] {\it Brownian motion and classical potential theory.}
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{\tt http://www1.essex.ac.uk/maths/people{\bsp}fremlin{\bsp}rao.htm}.
%used for \S477

\indexheader{Ressel}
Ressel P.\ [77] `Some continuity and measurability results on spaces of
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\cmmnt{[417C, 437M.]}

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\cmmnt{[\S494 {\it notes\/}.]}

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\cmmnt{[\S471 {\it notes\/}.]}

Rogers C.A.\ [80] (ed.) {\it Analytic Sets.}   Academic, 1980.
\cmmnt{[Chap.\ 42 {\it intro.}, \S422 {\it notes\/}, \S423 {\it notes\/}.]}

Rogers L.C.J.\ \& Williams D.\ [94] {\it Diffusions, Markov Processes
and Martingales,} vol.\ I.   Wiley, 1994.
\cmmnt{[\S455 {\it notes\/}.]}
%HA33.4.W5

Rokhlin V.A.\ [1948] `A ``general'' measure-preserving transformation is not
mixing', Doklady Akademii Nauk SSSR 60 (1948) 349-351.
\cmmnt{[494E.]}

Rosendal C.\ [09]
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\medskip%S

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\indexheader{Sierpi\'nski}
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\indexheader{Szemer\'edi}
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\cmmnt{[497L.]}
\indexheader{Talagrand} %? before \medskip to act correctly

\medskip%T

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466I, \S466 {\it notes\/}.]}

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%465Xm

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\cmmnt{[\S4A4.]}

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\cmmnt{[437J.]}
%used in chap 41

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\cmmnt{[425D, 425Ya.]}

\medskip%U

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\medskip%V

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% `Ta{sh}ken{t}skogo instituta' taken from ref in Fund Math

\medskip%W

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\cmmnt{[\S434 {\it notes\/}.]}

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\indexheader{Wheeler}
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  % printer's error "2" for "1" at head of paper refers to part, not
%volume
\cmmnt{[\S435 {\it notes\/}, \S437 {\it notes\/}.]}

%\medskip%X\medskip%Y\medskip%Z

%Zindulka O.\ [p98] `Killing residual measures', preprint, 1998.

\frnewpage
}%end of references


%---------------------------------------------------------------%

\end



