On nice equivalence relations on ${}^\lambda 2$
Saharon Shelah \thanks {\null\newline I would like to thank Alice Leonhardt for the beautiful typing \null\newline This research was supported by The Israel Science Foundation founded by the Israel Academy of Sciences and Humanities. Publication 724.}\endthanks
724.tex
%%% History
%%%  Written 99/11
%%%  First Typed - 2000/Feb/12
%%%  previous version - 00/Apr/18
%%%  Latest Revision - 00/June/6
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% processed by citealice (August 1999) on Fri Jun  9 08:54:46 CDT 2000
% To: Saharon Shelah <shelah@math.huji.ac.il>, Andrzej Roslanowski <roslanow@e-math.ams.org>
% Subject: Sh724
% Date: Tue, 6 Jun 2000 14:53:38 -0400 (EDT)
% From: Alice Leonhardt <leonhard@math.rutgers.edu>
% Mime-Version: 1.0
% Content-Description: revisions
% X-sliced-and-diced-by: 'savemail' 0.3, Feb 1999

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% (was f396)
\title{On nice equivalence relations on ${}^\lambda 2$} \endtitle
\author {Saharon Shelah \thanks {\null\newline I would like to thank 
Alice Leonhardt for the beautiful typing \null\newline
This research was supported by The Israel Science Foundation
founded by the Israel Academy of Sciences and Humanities. Publication
724.}\endthanks} \endauthor    

\affil{Institute of Mathematics\\
 The Hebrew University\\
 Jerusalem, Israel
 \medskip
 Rutgers University\\
 Mathematics Department\\
 New Brunswick, NJ  USA} \endaffil
\endtopmatter
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% lines of the file, your name, along with an electronic address at which
% others who might use the file may reach you, and (3) you remove each
% occurrence of Oren's name and electronic address from this file.  These
% restrictions help ensure that all standard versions of these macros are
% identical, and that Oren doesn't get deluged with inappropriate e-mail.
%
% This file, btxmac.tex, contains TeX macros that allow BibTeX, a
% bibliography program that was originally designed for use with LaTeX,
% to work with plain TeX.  Please report any bugs (outright goofs,
% improvable macros, misfeatures, or unclear documentation) to Oren
% Patashnik (opbibtex@cs.stanford.edu).  These macros will become frozen
% shortly after BibTeX version 1.00 is released.
%
% AMS-TEX WARNING: We tried very hard, for version .99i of these macros,
% to make them compatible with AmS-TeX.  We succeeded to the extent
% that, if you use one of the standard bibliography styles, you probably
% won't notice any problems with version 0.99i of btxmac.tex.  But
% ultimately we failed, in that the inherent incompatibilities between
% plain TeX and AmS-TeX kept making these macros break, for certain
% inputs or certain styles.  Examples:  (1) AmS-TeX treats at-signs as
% special, in ways that plain TeX and LaTeX don't, so that, for example,
% you can't have any `@' characters in an argument to the \cite command,
% the way you can in TeX or LaTeX; (2) AmS-TeX decided that plain TeX's
% and LaTeX's macron-accent control sequence `\=' should be undefined;
% so you'll need to define `\=' to be `\B' to get the xampl.bib example
% suggested below to work with AmS-TeX; (3) AmS-TeX redefines the tie
% character `~' of plain TeX, and AmS-TeX's `amsppt' style redefines
% plain TeX's `\nobreak' macro, so that if you use an author-date style
% like `apalike' and you have a multiple-author reference for which the
% author-date style automatically produces a citation in the text like
% `(Jones et~al., 1992)' you will throw AmS-TeX's `amsppt' style into
% an infinite loop, exceeding its input stack size.  In practice, such
% incompatibilities surface infrequently; but it is now clear to us that
% it's not worth the effort (perhaps it's not even possible) to make the
% btxmac.tex macros robust when used both with plain TeX and Ams-TeX.
% If the BibTeX/AmS-TeX results attainable with the current btxmac.tex
% macros are sufficient, fine.  But if there's a demand for more robust
% BibTeX/AmS-TeX behavior, then someone who's very familiar with the
% AmS-TeX package should probably make an amsbtxmc.tex version of the
% macros (remembering to follow the copyright restrictions above).
% Until then, if you're an AmS-TeX user, or a LaTeX or plain TeX user
% sharing files with an AmS-TeX user, beware.
% END OF AMS-TEX WARNING.
%
% To use these macros you should be familiar with how BibTeX interacts
% with LaTeX, since BibTeX's interaction with TeX is very similar; that
% interaction is explained in the LaTeX manual.  It also helps to
% have read "BibTeXing", the documentation that accompanies BibTeX.
% Then, if you want, you should redefine any of the macros that begin
% with `\bbl' or `\biblabel' or `\print' that you need to get formatting
% different from the default (the default settings are designed to
% accompany a bibliography style like BibTeX's standard style `plain').
% The macros you might want to change are described briefly a few
% paragraphs hence.  [To get started without reading any documentation,
% try running the nine-line .tex file below through TeX and BibTeX.
% Remember the general scheme: Running (La)TeX writes information on
% the .aux (auxiliary) file; then running BibTeX reads information from
% the .aux, .bst (style), and .bib (database) files, and writes
% information (the bibliography) on a .bbl file; then running (La)TeX
% incorporates the bibliography; then running (La)TeX once more fixes
% the remaining forward references into the bibliography.  Thus, to get
% everything incorporated into your output, you'll have to run (La)TeX,
% BibTeX, (La)TeX, (La)TeX.  (Standup, sitdown, fight, fight, fight.)]
%
% These macros can stand alone or they can be \input into a macro
% package, like Eplain, that is sufficiently compatible with plain TeX.
% To use these macros to format the 0.99 version of the xampl.bib file
% that's distributed with BibTeX (that version of the file has no
% self-identification), you'll need to define \mbox, which is a LaTeX
% command, to be \hbox, as in the example below.
%
% Here's a nine-line plain TeX file for trying out btxmac.tex; of course
% you'll have to remove the comment characters at the beginning of each
% line, and, depending on your system, you might have to take steps so
% that BibTeX can "see" the files xampl.bib and plain.bst (BibTeX will
% give you two empty-field warning messages that you should ignore).
%
%     \def\mbox#1{\leavevmode\hbox{#1}}
%     \input btxmac
%     \noindent This cites Aamport's gnominious article~\cite{article-full}.
%     \medskip
%     \leftline{\bf References}
%     \nocite{*}   % put all database entries into the reference list
%     \bibliography{xampl}   % specify the database files; here, just xampl.bib
%     \bibliographystyle{plain}   % specify plain.bst as the style file
%     \bye
%
%
%   HISTORY
%
% Karl Berry wrote the original version of these macros in 1989 and
% 1990, for use in his `Eplain' package.  Oren Patashnik modified them
% slightly in July 1990, as part of the official BibTeX distribution.
%
%    1-Aug-90  Version 0.99a, not released to the general public.
%   14-Aug-90  0.99b, first general release.
%   26-Aug-90  0.99c, made \@undefinedmessage work with other macro packages.
%    6-Sep-90  0.99d, allowed for general formatting of bibliography labels,
%                     for general formatting of (in-text) citations, and for
%                     changing certain catcodes while reading the .aux file.
%   14-Nov-90  0.99e, changed the way \@setletters works, made some \new...'s
%                     non-outer, and changed the way Eplain reads this file.
%   12-Dec-90  0.99f, made \@resetnumerals change the `,' and `.' catcodes; and
%                     added \biblabelextrahang, \@getoptionalarg, and \bblsc.
%   11-Mar-91  0.99g, made a few minor changes required by the way Eplain reads
%                     this file, but no functional changes.
%   24-Apr-91  0.99h, inhibited the reading and writing of the .aux file if it
%                     isn't used or if the \noauxfile macro is defined, and
%                     removed some .aux-file-opening detritus; printed the
%                     cite-key of undefined citations in \tt font; changed the
%                     catcode of `_' inside \cite; and called \@resetnumerals
%                     from inside a group.
%   29-Feb-92  0.99i, made these macros semi-compatible with AmS-TeX; removed
%                     \@resetnumerals, \@setletters, \@tokstostring, and
%                     friends; changed the way \cite handles catcodes; changed
%                     \@getoptionalarg, and had \bibitem and \newcommand use
%                     it; added \@futurenonspacelet and (to facilitate the use
%                     of multiple reference lists) \bblfilebasename; changed
%                     \biblabelprint to use the new macros \biblabelprecontents
%                     and \biblabelpostcontents, and to, by default, right-
%                     justify numeric labels; and renamed \biblabelextrahang to
%                     the more descriptive \biblabelextraspace.
%   14-Mar-92  0.99j, made 0.99i's use of `\\' local to btxmac.tex.
%
%
% The LaTeX-related commands defined in this file include (a) the four
% commands that a user types (\bibliography, \bibliographystyle, \cite,
% and \nocite); (b) the three commands that BibTeX looks for in the .aux
% file (\bibdata, \bibstyle, and \citation---there is a fourth command
% that BibTeX looks for, but that command is related to LaTeX's \include
% facility, so these macros ignore that command); and (c) a LaTeX
% command (\newcommand) that's written by one of the four standard
% bibliography styles (alpha).  The definitions here are not exactly the
% same as the corresponding LaTeX definitions (those eight LaTeX
% definitions depend on a significant fraction of LaTeX itself).  But
% the only substantial differences are with \newcommand, which here,
% without complaining, lets you redefine a preexisting control sequence
% (in LaTeX, \newcommand won't let you redefine a preexisting command),
% and which here doesn't make the control sequences it defines \long (in
% LaTeX, that happens automatically); there may also be other minor
% differences.  To summarize: Unless you know what you're doing, you
% shouldn't define any control sequences with these eight names:
%
% \bibdata
% \bibliography
% \bibliographystyle
% \bibstyle
% \citation
% \cite
% \newcommand
% \nocite
%
% There are three other commands written by one or more of the four
% standard (plain, abbrv, alpha, unsrt) or four semistandard (acm,
% apalike, ieeetr, siam) bibliography styles; they take effect only
% within the bibliography, and are redefinable, as explained later:
%
% \em
% \newblock
% \sc
%
% There's one control sequence you might want to use (but not redefine)
% in redefining \biblabelprint:
%
% \biblabelwidth
%
% There are fifteen other control sequences (explained later in more detail)
% that the macros of this file will use if you define them---you should
% define them after the \input btxmac command but before the \bibliography
% command.  The first six begin with `\bbl' and affect fonts, spacing,
% perhaps other characteristics of the bibliography, and which .bbl files
% get read; the next five begin with `\biblabel' and determine how labels
% are formatted in the bibliography; and the last four begin with `\print'
% and determine how the in-text citations are formatted:
%
% \bblem
% \bblfilebasename
% \bblhook
% \bblnewblock
% \bblrm
% \bblsc
% \biblabelcontents
% \biblabelprecontents
% \biblabelprint
% \biblabelpostcontents
% \biblabelextraspace
% \printbetweencitations
% \printcitefinish
% \printcitenote
% \printcitestart
%
% If it's defined before the \input btxmac command, the control sequence
% below inhibits the reading and writing of the .aux file(s), and the
% issuing of related warning messages.  Any definition will do.  This
% feature might help when you're working on draft stages of a document:
%
% \noauxfile
%
%
% Here's another control sequence (it's described later) that you
% probably won't want to redefine unless you are writing another macro
% package; if you do redefine it, however, do it before the \input btxmac
% command (and notice that it has an `@' in its name):
%
% \@undefinedmessage
%
% Any other control sequence in this file that might conflict with
% something you've defined will have an `@' in its name, so such conflicts
% are unlikely; but if you're worried about a specific control sequence
% name, do a text search of this file to look for it.
%
%
% So to start things off we turn `@' into a letter (category code 11),
% keeping track of the old category code for future restoration.
% (Simply resetting it to 12 when we leave these macros is
% insufficient.)  The use of `\cite' as a temporary control sequence is
% a kludge, but it's a reasonably simple way to accomplish what we need
% without possibly overwriting something (without an `@' in its name)
% that might already be defined.
%
\edef\cite{\the\catcode`@}%
\catcode`@ = 11
\let\@oldatcatcode = \cite
\chardef\@letter = 11
\chardef\@other = 12
%
%
% Next come some things that will be useful later.
%
% Make an outer definition into an inner one (due to Chris Thompson).
% The arguments should be the control sequence to be defined, and the
% new of the \outer control sequence, as characters; the control
% sequence #1 is defined to be just the same as \csname#2\endcsname, but
% not \outer.  For example, \@innerdef\innernewcount{newcount} would
% define \innernewcount to be a non-outer version of \newcount.
%
\def\@innerdef#1#2{\edef#1{\expandafter\noexpand\csname #2\endcsname}}%
%
% We use \@innerdef to make some of our allocations local, because
% Eplain includes our code inside a conditional.  We put @'s in the
% names to minimize the (already small) chance of conflicts.
%
\@innerdef\@innernewcount{newcount}%
\@innerdef\@innernewdimen{newdimen}%
\@innerdef\@innernewif{newif}%
\@innerdef\@innernewwrite{newwrite}%
%
%
% Swallow one parameter.
%
\def\@gobble#1{}%
%
%
% Use TeX 3.0's \inputlineno to get the line number, for better error
% messages, but if we're using an old version of TeX, don't do anything.
%
\ifx\inputlineno\@undefined
   \let\@linenumber = \empty % Pre-3.0.
\else
   \def\@linenumber{\the\inputlineno:\space}%
\fi
%
%
% The following macro \@futurenonspacelet (from the TeXbook) behaves
% essentially like \futurelet except that it discards any implicit or
% explicit space tokens that intervene before a nonspace is scanned:
%
\def\@futurenonspacelet#1{\def\cs{#1}%
   \afterassignment\@stepone\let\@nexttoken=
}%
\begingroup % The grouping here avoids stepping on an outside use of `\\'.
\def\\{\global\let\@stoken= }%
\\ % now \@stoken is a space token (\\ is a control symbol, so that
   % space after it is seen).
\endgroup
\def\@stepone{\expandafter\futurelet\cs\@steptwo}%
\def\@steptwo{\expandafter\ifx\cs\@stoken\let\@@next=\@stepthree
   \else\let\@@next=\@nexttoken\fi \@@next}%
\def\@stepthree{\afterassignment\@stepone\let\@@next= }%
%
%
% \@getoptionalarg\CS gets an optional argument from the input, enclosed
% in brackets, then expands \CS.  We set \@optionalarg to \empty if we
% don't find one, otherwise to the text of the argument.  This assumes
% the brackets don't have a funny category code.
%
\def\@getoptionalarg#1{%
   \let\@optionaltemp = #1%
   \let\@optionalnext = \relax
   \@futurenonspacelet\@optionalnext\@bracketcheck
}%
%
% The \expandafter's in this macro let us avoid the use of \aftergroup,
% which is somewhat more expensive.
%
\def\@bracketcheck{%
   \ifx [\@optionalnext
      \expandafter\@@getoptionalarg
   \else
      \let\@optionalarg = \empty
      % We can't do the \temp after the \fi, because then the \temp gets
      % in the way of reading the optional argument from the input, if
      % we do have one.
      \expandafter\@optionaltemp
   \fi
}%
%
\def\@@getoptionalarg[#1]{%
   \def\@optionalarg{#1}%
   \@optionaltemp
}%
%
%
% From LaTeX.
%
\def\@nnil{\@nil}%
\def\@fornoop#1\@@#2#3{}%
%
\def\@for#1:=#2\do#3{%
   \edef\@fortmp{#2}%
   \ifx\@fortmp\empty \else
      \expandafter\@forloop#2,\@nil,\@nil\@@#1{#3}%
   \fi
}%
%
\def\@forloop#1,#2,#3\@@#4#5{\def#4{#1}\ifx #4\@nnil \else
       #5\def#4{#2}\ifx #4\@nnil \else#5\@iforloop #3\@@#4{#5}\fi\fi
}%
%
\def\@iforloop#1,#2\@@#3#4{\def#3{#1}\ifx #3\@nnil
       \let\@nextwhile=\@fornoop \else
      #4\relax\let\@nextwhile=\@iforloop\fi\@nextwhile#2\@@#3{#4}%
}%
%
%
% This macro tests if a file \jobname.#1 exists, and sets \if@fileexists
% appropriately.  If an optional argument is given, it is used as the
% root part of the filename instead of \jobname.
%
\@innernewif\if@fileexists
%
\def\@testfileexistence{\@getoptionalarg\@finishtestfileexistence}%
\def\@finishtestfileexistence#1{%
   \begingroup
      \def\extension{#1}%
      \immediate\openin0 =
         \ifx\@optionalarg\empty\jobname\else\@optionalarg\fi
         \ifx\extension\empty \else .#1\fi
         \space
      \ifeof 0
         \global\@fileexistsfalse
      \else
         \global\@fileexiststrue
      \fi
      \immediate\closein0
   \endgroup
}%
%
%
%% [[[start of BibTeX-specific stuff]]]
%
% Now come the four main LaTeX commands and their associated .aux
% commands.  Just as in LaTeX, \bibliographystyle defines the BibTeX
% style name (.bst file, that is), and \bibliography defines the
% database (.bib) file(s).  The corresponding .aux-file commands are
% \bibstyle and \bibdata, which are there only for BibTeX's (but not
% LaTeX's) use.
%
\def\bibliographystyle#1{%
   \@readauxfile
   \@writeaux{\string\bibstyle{#1}}%
}%
\let\bibstyle = \@gobble
%
% As well as writing the \bibdata command to tell BibTeX which .bib
% files to read, we read the .bbl file that BibTeX (or a person,
% conceivably) has produced.  We use \bblfilebasename as the root of the
% filename to read; this defaults to \jobname.
%
\let\bblfilebasename = \jobname
\def\bibliography#1{%
   \@readauxfile
   \@writeaux{\string\bibdata{#1}}%
   \@testfileexistence[\bblfilebasename]{bbl}%
   \if@fileexists
      % We just output a non-discardable item (the `whatsit' with the
      % \bibdata command).  This means that the glue that will be
      % inserted next (\parskip or \baselineskip, most likely) will be a
      % legal breakpoint.  Most likely, this is after some kind of
      % heading, where we don't want to allow a page break.  So:
      \nobreak
      \@readbblfile
   \fi
}%
\let\bibdata = \@gobble
%
% The \nocite{label,label,...} command writes its argument to \@auxfile,
% unless instructed not to, but produces no text in the document.  Both
% the \nocite and \cite commands produce \citation commands in the .aux file.
%
\def\nocite#1{%
   \@readauxfile
   \@writeaux{\string\citation{#1}}%
}%
%
\@innernewif\if@notfirstcitation
%
% \cite[note]{label,label,...} produces the citations for the labels as
% well.  If the optional argument `note' is present, it's added after
% the labels.  Since \cite calls \nocite to do its .aux-file writing,
% \cite doesn't need to call \@readauxfile (\nocite does).
%
\def\cite{\@getoptionalarg\@cite}%
%
% Typeset the citations for the labels in #1, followed by the note, if
% it exists.  To change the citation's format in the text, redefine one
% or more `\print...' macros, whose defaults appear later in this file.
%
\def\@cite#1{%
   % Remember the optional argument, in case one of the macros we call
   % below ends up looking for an optional argument itself.  For
   % example, if a \cite[note] triggers reading the .aux file, then the
   % [note] would be clobbered, since \@testfileexistence looks for an
   % optional arg.
   \let\@citenotetext = \@optionalarg
   % Start printing the text, beginning with a left bracket by default.
   \printcitestart
   % It's complicated, but because \nocite puts a `whatsit' onto the list,
   % \nocite should follow \printcitestart.  It's conceivable, but very
   % unlikely, that this `whatsit' will cause a problem (glue that doesn't
   % disappear when you want it to is the most likely symptom), requiring
   % a change either to \printcitestart or to the label that the .bst file
   % produces.
   \nocite{#1}%
   \@notfirstcitationfalse
   \@for \@citation :=#1\do
   {%
      \expandafter\@onecitation\@citation\@@
   }%
   \ifx\empty\@citenotetext\else
      \printcitenote{\@citenotetext}%
   \fi
   \printcitefinish
}%
%
\def\@onecitation#1\@@{%
   \if@notfirstcitation
      \printbetweencitations
   \fi
   %
   \expandafter \ifx \csname\@citelabel{#1}\endcsname \relax
      \if@citewarning
         \message{\@linenumber Undefined citation `#1'.}%
      \fi
      % Give it a dummy definition:
      \expandafter\gdef\csname\@citelabel{#1}\endcsname{%
% Change: marginal remark added, goldstrn@math.huji.ac.il, 
% goldstern@tuwien.ac.at, May 1996 mg
%  !!! change !!!
\strut
\vadjust{\vskip-\dp\strutbox
\vbox to 0pt{\vss\parindent0cm \leftskip=\hsize 
\advance\leftskip3mm
\advance\hsize 4cm\strut\openup-4pt 
\rightskip 0cm plus 1cm minus 0.5cm ?  #1 ?\strut}}
         {\tt
            \escapechar = -1
            \nobreak\hskip0pt
            \expandafter\string\csname#1\endcsname
            \nobreak\hskip0pt
         }%
      }%
   \fi
   % Now produce the text, whether it was undefined or not.
   \csname\@citelabel{#1}\endcsname
   \@notfirstcitationtrue
}%
%
% Given a label `foo', the macro `\b@foo' is supposed to
% hold the text that should be produced.
%
\def\@citelabel#1{b@#1}%
%
% So, how does a citation label get defined?  When we read the .bbl file
% (below), a \bibitem writes out a \@citedef command.  And when we read
% the \@citedef, we define \@citelabel{#1}, where #1 is the user's
% label.
%
\def\@citedef#1#2{\expandafter\gdef\csname\@citelabel{#1}\endcsname{#2}}%
%
%
% Reading the .bbl file also produces the typeset bibliography.  Please
% notice, however, that we do not produce the title for the references
% (e.g., `References'), as LaTeX does.  The formatting and spacing of
% that title, whether it should go into the headline, and so on, are all
% things determined by your format.  We cannot know those things in
% advance.  If you wish, you can define \bblhook to produce the title.
% Or just do it before the \bibliography command.
%
\def\@readbblfile{%
   % Define a counter to tell us which item number we are on, unless
   % we've already defined it (because the document has more than one
   % bibliography).
   \ifx\@itemnum\@undefined
      \@innernewcount\@itemnum
   \fi
   %
   \begingroup
      \def\begin##1##2{%
         % ##1 is just `thebibliography'.
         % ##2 is the widest label.
         % We set (new dimen) \biblabelwidth based on the widest label
         \setbox0 = \hbox{\biblabelcontents{##2}}%
         \biblabelwidth = \wd0
      }%
      \def\end##1{}% ##1 is `thebibliography' again.
      %
      % Here we have two possibilities:
      % \bibitem[typesetlabel]{citationlabel}
      % \bibitem{citationlabel}
      % If we have the second of these, the citations are numbered, starting
      % from one; we use our own count register \@itemnum for this.
      %
      \@itemnum = 0
      \def\bibitem{\@getoptionalarg\@bibitem}%
      \def\@bibitem{%
         \ifx\@optionalarg\empty
            \expandafter\@numberedbibitem
         \else
            \expandafter\@alphabibitem
         \fi
      }%
      \def\@alphabibitem##1{%
         % Need \xdef here for various reasons.
         \expandafter \xdef\csname\@citelabel{##1}\endcsname {\@optionalarg}%
         % Left-justify alpha labels, unless \biblabel{pre,post}contents
         % are already defined.
         \ifx\biblabelprecontents\@undefined
            \let\biblabelprecontents = \relax
         \fi
         \ifx\biblabelpostcontents\@undefined
            \let\biblabelpostcontents = \hss
         \fi
         \@finishbibitem{##1}%
      }%
      %
      \def\@numberedbibitem##1{%
         \advance\@itemnum by 1
         \expandafter \xdef\csname\@citelabel{##1}\endcsname{\number\@itemnum}%
         % Right-justify numeric labels, unless \biblabel{pre,post}contents
         % are already defined.
         \ifx\biblabelprecontents\@undefined
            \let\biblabelprecontents = \hss
         \fi
         \ifx\biblabelpostcontents\@undefined
            \let\biblabelpostcontents = \relax
         \fi
         \@finishbibitem{##1}%
      }%
      %
      \def\@finishbibitem##1{%
         \biblabelprint{\csname\@citelabel{##1}\endcsname}%
         \@writeaux{\string\@citedef{##1}{\csname\@citelabel{##1}\endcsname}}%
         \ignorespaces
      }%
      %
      % Do the printing (we're producing the bibliography, remember).
      %
      \let\em = \bblem
      \let\newblock = \bblnewblock
      \let\sc = \bblsc
      % Punctuation won't affect spacing;
      \frenchspacing
      % the penalties below are from LaTeX's [article,book,report].sty;
      \clubpenalty = 4000 \widowpenalty = 4000
      % the next two values come from LaTeX's \sloppy command;
      \tolerance = 10000 \hfuzz = .5pt
      \everypar = {\hangindent = \biblabelwidth
                      \advance\hangindent by \biblabelextraspace}%
      \bblrm
      % the \parskip is a guess at what looks good;
      \parskip = 1.5ex plus .5ex minus .5ex
      % and the space between label and text comes from LaTeX's \labelsep.
      \biblabelextraspace = .5em
      \bblhook
      %
      \input \bblfilebasename.bbl
   \endgroup
}%
%
% The widest label's width is useful for redefining \biblabelprint;
% you redefine \biblabelwidth, in effect, by redefining the
% \biblabelcontents macro that appears below.  And \biblabelextraspace,
% which is redefinable inside \bblhook, is added to \biblabelwidth to
% determine the amount of hanging indentation.
%
\@innernewdimen\biblabelwidth
\@innernewdimen\biblabelextraspace
%
% Now come the main macros that are related to the printing of the
% bibliography.  Since you might want to redefine them, they are given
% default definitions outside of \@readbblfile.
%
% The first one controls the printing of a bibliography entry's label.
% If you change it, make sure that it starts with something like
% \noindent or \indent or \leavevmode that puts TeX into horizontal mode
% (even if the label itself is empty); otherwise, the hanging
% indentation will get messed up in certain circumstances.
%
\def\biblabelprint#1{%
   \noindent
   \hbox to \biblabelwidth{%
      \biblabelprecontents
      \biblabelcontents{#1}%
      \biblabelpostcontents
   }%
   \kern\biblabelextraspace
}%
%
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\bn

\head {\S0 Introduction} \endhead  \resetall \sectno=0
 % \resetall 
\bigskip

The main question here is the possible generalization of the following
theorem on ``simple" equivalence relation on ${}^\omega 2$ to higher
cardinals.
\proclaim{\stag{0.1} Theorem}  1) Assume that
\mr
\item "{$(a)$}"  $E$ is a Borel 2-place relation on ${}^\omega 2$
\sn
\item "{$(b)$}"  $E$ is an equivalence relation
\sn
\item "{$(c)$}"  if $\eta,\nu \in {}^\omega 2$ and
$(\exists!n)(\eta(n) \ne \nu(n))$, \ub{then} $\eta,\nu$ are not
$E$-equivalent.
\ermn
\ub{Then} there is a perfect subset of ${}^\omega 2$ of pairwise non
$E$-equivalent members. \nl
2) Instead of ``$E$ is Borel", ``$E$ is analytic (or even a Borel
combination of analytic relations)" is enough. \nl
3) If $E$ is a $\pi^1_2$ relation which is an equivalence relation
satisfying clauses (b) + (c) in $\bold V^{\text{Cohen}}$, \ub{then} the
conclusion of (1) holds.
\endproclaim
\bn
See \cite{Sh:273}, it was used to prove a result on the homotopy
group: if $X$ is Hausdorff metric, compact, separable, arc-connected,
locally arc-connected and the homotopy group is not finitely generated
then it has cardinality continuum; the proof of \scite{0.1} used
forcing in \cite{Sh:273}, see \cite{PaSr98} for eliminating the
forcing. 

We may restrict $E$ to be like $r_p(\text{Ext}(G,\Bbb Z))$ or just
closer to group theory as in Grossberg Shelah \cite{GrSh:302}, 
\cite{GrSh:302a}, Mekler Roslanowski Shelah \cite{MRSh:314},
\cite{Sh:664}.  In \S5 we say somewhat more. \nl
We here continue \cite{Sh:664} but do not rely on it.

Turning to ${}^\lambda 2$ the problem split according to the character
of $\lambda$ and the ``simplicity" of $E$.  If $E$ is $\pi^1_1$ and
$\lambda = \lambda^{< \lambda}$ and $\lambda \ge \beth_\omega$ (or
just $(Dl)_\lambda$ holds) a generalization holds.  If $E$ is $\Sigma^1_1,\lambda =
\lambda^{< \lambda}$ the generalization in general fails; all this in
\S1.  Now if $\lambda$ is singular, strong limit for
simplicity, it is natural to consider ${}^{\text{cf}(\lambda)}\lambda$
instead of ${}^\lambda 2$.  If $\lambda$ has uncountable cofinality we
get strong negative results in \S2.  If $\lambda$ has countable cofinality,
and is the limit of ``somewhat large cardinals", e.g. measurable
cardinals, (but $\lambda = \aleph_\omega$ may be O.K., i.e. consistently) the
generalization holds (in \S3), but if the universe is closed to $L$ (e.g. in
$L$ there is no weakly compact or just no Erdos cardinal) then we get
negative results (see \S4).  Note that theorems of the form ``if $E$ has many
equivalence classes it has continuum" do not generalize well, see
\cite{ShVs:719} even for weakly compact.  
\bigskip

\definition{\stag{0.2} Definition}  For a cardinal $\lambda$ and let
${\Cal B}_\lambda$ be ${}^\lambda 2$ (or ${}^\lambda \lambda$ or
${}^{\text{cf}(\lambda)} \lambda)$. \nl
1)  For a logic ${\Cal L}$ we say that $E$ is a ${\Cal L}$-\ub{nice},
say 2-place for simplicity, relation on ${\Cal B}$ \ub{if} there is
a model $M$ with universe $\lambda$ and finite vocabulary $\tau$, and
unary function symbols $F_1,F_2 \notin \tau$ (denoting possibly partial
unary functions), such that letting $\tau^+ = \tau \cup \{F_1,F_2\}$, for some
sentence $\psi = \psi(F_1,F_2)$ in $L(\tau)$ we have
\mr
\item "{$\circledcirc$}"  for any $\eta_1,\eta_2 \in {\Cal B}$ letting
$M_{\eta_1,\eta_2} = (M,\eta_1,\eta_2)$ be the $\tau^+$-model
expanding $M$ with $F^{M_{\eta_1,\eta_2}}_\ell = \eta_\ell$ we have
\nl
$\eta_1 E \eta_2 \Leftrightarrow (M,\eta_1,\eta_2) \models \psi$. \nl
We may write $M \models \psi[\eta_1,\eta_2]$ and
$\psi[\eta_1,\eta_2,M]$ or $\psi(x,y,M)$ or write $a \subseteq
\lambda$ coding $M$ instead of $M$.
\ermn
2) $E$ is a $\pi^1_1$-relation on ${\Cal B}$ means that above we allow
$\psi$ to be of the form $(\forall X)\varphi,\varphi$ first order or
even inductive logic (i.e. we have variables on sets and are allowed
to form the first fix point); $X$
vary on sets.  Similarly $\Sigma^1_1,\pi^1_2$, projective; writing
nice means ${\Cal L}$ is first order + definition by induction.  We
may write nice$_{\Cal B},\Sigma^1_1[{\Cal B}]$ etc, and may replace
${\Cal B}$ by $\lambda$ if ${\Cal B} = {\Cal B}_\lambda$.  We write
very nice for ${\Cal L}$-nice, ${\Cal L}$ first order.
\enddefinition
\bn
\ub{Notation}:

$(\forall^* i < \delta)$ means ``for every large enough $i < \delta$".

$J^{\text{bd}}_\delta$ is the ideal of bounded subsets of $\delta$.
\bigskip

\definition{\stag{0.3} Definition}  Let $(D \ell)_\lambda$ means that $\lambda$
is regular, uncountable and there is $\bar{\Cal P} = \langle {\Cal
P}_\alpha:\alpha < \lambda \rangle$ such that ${\Cal P}_\alpha$ is a
family of $< \lambda$ subsets of $\alpha$ and for every $X \subseteq
\lambda$ the set $\{\delta < \lambda:X \cap \delta \in {\Cal
P}_\delta\}$ is stationary; hence $\lambda = \lambda^{< \lambda}$.  
[By \cite{Sh:460}, $\lambda = \lambda^{< \lambda} 
\ge \beth_\omega \Rightarrow (D \ell)_\lambda$ and (by Kunen)
$\lambda = \mu^+ \Rightarrow (D \ell)_\lambda \equiv \diamondsuit_\lambda$].
\enddefinition
\bigskip

\definition{\stag{0.4} Definition}  ${\Cal Q} \subseteq {}^\lambda 2$
is called perfect or $\lambda$-perfect if:
\mr
\item "{$(a)$}"  $A \ne \emptyset$
\sn
\item "{$(b)$}"  if $\eta \in A$ then $\{\ell g(\eta \cap \nu):\nu \in
A \backslash \{\eta\}\} \subseteq \lambda$ is unbounded
\sn
\item "{$(c)$}"  $\{\eta \restriction \zeta:\eta \in A,\zeta \le
\lambda\}$ is closed under the union of $\triangleleft$-increasing sequences.
\ermn
Equivalently, ${\Cal Q} = \langle \rho_\eta:\eta \in {}^\lambda 2
\rangle$ such that
\mr
\item "{$(a)'$}"  $\rho_\eta \in {}^\lambda 2$
\sn
\item "{$(b)'$}"  $\eta_1 \ne \eta_2 \in {}^\lambda 2 \Rightarrow
\rho_{\eta_2} \ne \rho_{\eta_2}$
\sn
\item "{$(c)'$}"  if $\eta_0,\eta_1,\eta_2 \in {}^\lambda 2$ are
distinct and if $\eta_1 \cap \eta_2 \triangleleft \eta_1 \cap \eta_0$
(so $\eta_1 \cap \eta_2 \ne \eta_1 \cap \eta_0)$ then $\rho_{\eta_1}
\cap \rho_{\eta_2} \triangleleft \rho_{\eta_1} \cap \rho_{\eta_0}$ and
$\rho_{\eta_1} (\ell g(\rho_{\eta_1} \cap \rho_{\eta_2})) =
\eta_1(\ell g(\eta_1 \cap \eta_2))$.
\endroster
\enddefinition   
\bn
Note
\demo{\stag{0.5} Fact}  1) If $\lambda = \lambda^{< \kappa}$ and $E$
is ${\Cal L}$(induction)-nice \ub{then} $E$ is $L_{\lambda^+,\kappa}$-nice.
\enddemo
\newpage

\head {\S1} \endhead  \resetall \sectno=1
 % \resetall 
\bigskip

We here continue \cite[\S2]{Sh:664}, the theorem and most proofs can be
read without it.  The claims below generalize \cite{Sh:273}.
\proclaim{\stag{1.1} Claim}  Assume
\mn
$\boxtimes_1(a) \quad \lambda = \lambda^{< \lambda}$ and 
$\lambda \ge \beth_\omega$ or just
$(Dl)_\lambda$ (see \scite{0.3})
\smallskip

$(b) \quad E$ is a nice 2-place relation on $\lambda$
\smallskip

$(c)(\alpha) \quad  E$ is a equivalence relation on ${}^\lambda 2$
\smallskip

\hskip17pt $(\beta) \quad$   if $\eta,\nu \in {}^\lambda 2$ and
$(\exists!\alpha < \lambda)(\eta(\alpha) \ne \nu(\alpha))$ then $\neg
\eta E \nu$.
\mn
\ub{Then} $E$ has $2^\lambda$ equivalence classes, moreover a perfect
set of pairwise non $E$-equivalent members of ${}^\lambda 2$.
\endproclaim
\bigskip

\demo{Proof}  Note that
\mr
\item "{$\circledast$}"  of $\Bbb P$ is a $\lambda$-complete forcing
(or just $\lambda$-strategically complete) then $\Vdash_{\Bbb P}$
``clauses (c), $(\alpha), (\beta)$ are still true".  \nl
So we can apply \scite{1.1A} below.  \hfill$\square_{\scite{1.1}}$
\endroster
\enddemo
\bn
A relative is
\proclaim{\stag{1.1A} Claim}  Assume
\mn
$\boxtimes_2(a),(c) \quad$   as in $\boxtimes_1$
\smallskip

$(b)$ \hskip23pt   is $\pi^1_1[\lambda]$ 2-place relation on
${}^\lambda 2$, say defined by $(\exists z)\varphi(x,y,z,\bar a)$ \nl

\hskip40pt see Definition \scite{0.2}  
\sn

$(c)^+=(c)^+_{\text{Cohen}} \quad$   if $\Bbb P = ({}^{\lambda >}
2,\triangleleft)$, i.e. $\lambda$-Cohen then in $V^{\Bbb P}$ \nl

\hskip80pt clauses (c) still hold.
\endproclaim
\bigskip

\demo{Proof}  \ub{Stage A}:  Let $(\eta_0,\eta_1) \in {}^\lambda 2
\times {}^\lambda 2$ be generic over $\bold V$ for the forcing $\Bbb Q =
({}^{\lambda>} 2) \times ({}^{\lambda >}2)$.  Now do we have $\bold
V[\eta_0,\eta_1] \models ``\eta_0 E \eta_1"$?  If so, then for some
$(p_0,p_1) \in ({}^{\lambda >} 2) \times ({}^{\lambda >} 2)$ we have
$(p_0,p_1) \Vdash_{\Bbb Q} ``{\underset\tilde {}\to \eta_0} 
E {\underset\tilde {}\to \eta_1}"$, let
$\alpha < \lambda$ be $> \ell g(p_0),\ell g(p_1)$ and by clause
$(c)^+ (\beta)$ in $\bold V[\eta_0,\eta_1]$
we can find $\eta'_1 \in {}^\lambda 2$ such that $\eta^1_1
\restriction \alpha = \eta_1 \restriction \alpha$, and for some $\beta
\in (\alpha,\lambda),\eta'_1 \restriction [\beta,\lambda) = \eta_1
\restriction [\beta,\lambda)$, (here $\beta = \alpha +1$ is O.K.) and
$\bold V[\eta_0,\eta_1] \models \neg \eta'_1 E \eta_1$.

So $\bold V[\eta_0,\eta_1] \models ``\neg \eta_0 E \eta'_1"$ (again as in
$\bold V[\eta_0,\eta_1],E$ is an equivalence relation by clause
$(c)^+$ and $\bold V[\eta_0,\eta_1] \models ``\eta_0 E \eta_1"$).  
Also $(\eta_0,\eta'_1)$ is generic over $\bold V$ for
$({}^{\lambda >} 2) \times ({}^{\lambda >}2)$ with $(p_0,p_1)$ in the
generic set and $\bold V[\eta_0,\eta_1] = \bold V[\eta_0,\eta'_1]$ so
we get a contradiction to $(p_0,p_1) \Vdash ``\neg {\underset\tilde
{}\to \eta_0} \in {\underset\tilde {}\to \eta_1}"$.  Hence
\mr
\item "{$\circledast_1$}"  $\Vdash_{({}^{\lambda>} 2) \times ({}^{\lambda
>} 2)} ``\neg({\underset\tilde {}\to \eta_0} E
{\underset\tilde {}\to \eta_1})"$.
\endroster
\bn
\ub{Stage B}:

Let $\chi$ be large enough and let $N \prec ({\Cal H}(\chi),\in)$ be
such that $\|N\| = \lambda,N^{< \lambda} \subseteq N$ and the
definition of $E$ belongs to $N$.  Note that
\mr
\item "{$\circledast_2$}"  if $(\eta_0,\eta_1) \in ({}^\lambda 2)
\times ({}^\lambda 2)$ (and is in $\bold V$) and $N[\eta_0,\eta_1]
\models ``\neg \eta_0 E \eta_1"$, \ub{then} $\neg \eta_0 E
\eta_1$. \nl
[Why?  As $E$ is $\pi^1_1$, in $N[\eta_0,\eta_1]$ there is a witness
$\in {}^\lambda 2$ for failure, and it also witnesses in $\bold V$
that $\neg \eta_0 E \eta_1$.]
\endroster
\enddemo
\bn
Clearly to finish proving \scite{1.1}, it suffices to prove
\proclaim{\stag{1.2} Subclaim}   1) Assume $\lambda = \lambda^{< \lambda}$
and $(Dl)_\lambda$.

If ${\Cal H}(\lambda) \subseteq N,N^{<\lambda} \subseteq N,\|N\| =
\lambda$ and $N \models ZFC^-$, \ub{then} there is a perfect ${\Cal Q}
\subseteq {}^{\lambda >}2$ such that for any $\eta_0 \ne \eta_1$ from
${\Cal Q}$ the pair $(\eta_0,\eta_1)$ is generic over $N$ for
$[({}^\lambda 2) \times ({}^{\lambda >} 2)]^N$. \nl
2) Assume $\lambda = \lambda^{< \lambda}$ and there is a tree with
$\lambda$ levels each of cardinality $< \lambda$ and $2^\lambda$ \,
$\lambda$-branches.  \ub{Then} for some $X \subseteq {}^\lambda 2,|X|
= 2^\lambda$ and $\eta_0 \ne \eta_1 \in X \Rightarrow (\eta_0,\eta_1)$
generic over $N$ for $({}^{\lambda >} 2) \times ({}^{\lambda >} 2)$.
\endproclaim
\bigskip

\demo{Proof}  1) Let $\langle {\Cal P}_\alpha:\alpha < \lambda \rangle$
be such that ${\Cal P}_\alpha \subseteq {\Cal P}(\alpha),|{\Cal
P}_\alpha| < \lambda$, and for every $X \subseteq \lambda$ the set
$\{\alpha:X \cap \alpha \in {\Cal P}_\alpha\}$ is stationary.  So by
coding we can find ${\Cal P}'_\alpha \subseteq
\{(\eta_0,\eta_1):\eta_0,\eta_1 \in {}^\alpha 2\}$ of cardinality $<
\lambda$ such that for every $\eta_0,\eta_1 \in {}^\lambda 2$ the set
$\{\alpha < \lambda:(\eta_0 \restriction \alpha,\eta_1 \restriction
\alpha) \in {\Cal P}'_\alpha\}$ is stationary.  Lastly, let $\langle {\Cal
I}_\alpha:\alpha < \lambda \rangle$ list the dense open subsets of
$({}^{\lambda >} 2) \times ({}^{\lambda >} 2)$ which belongs to $N$.
Now we define by induction on $\alpha < \lambda,\langle \rho_\eta:\eta
\in {}^\alpha 2 \rangle$ such that:
\mr
\item "{$(a)$}"  $\rho_\eta \in {}^{\lambda >}2$
\sn
\item "{$(b)$}"  $\beta < \ell g(\eta) \Rightarrow \rho_{\eta
\restriction \beta} \triangleleft \rho_\eta$
\sn
\item "{$(c)$}"  $\rho_\eta \char 94 \langle \ell \rangle
\triangleleft \rho_{\eta \char 94 \langle \ell \rangle}$
\sn
\item "{$(d)$}"  if $(\eta_0,\eta_1) \in {\Cal P}'_\alpha,\ell_0 <
2,\ell_1 < 2$ then
$(\rho_{\eta_0 \char 94 \langle \ell_0 \rangle},
\rho_{\eta_1 \char 94 \langle \ell_1 \rangle}) \in \dbca_{\beta \le
\alpha} {\Cal I}_\beta$.
\ermn
There is no problem to carry the definition (using 
$|{\Cal P}'_\alpha| < \lambda = \text{ cf}(\lambda)$) and $\{\dbcu_{\alpha <
\lambda} \rho_{\eta \restriction \alpha}:\eta\in {}^\lambda 2\}$ is a
perfect set as required. \nl
2) Easier as we can assume that such a tree belongs to $N$.  
\hfill$\square_{\scite{1.2}},\square_{\scite{1.1A}}$
\enddemo
\bigskip

\proclaim{\stag{1.3} Claim}  1) In \scite{1.1}, \scite{1.1A} we can weaken clause
$(\beta)$ (in $(c), (c)^+$, call it $(c)^-,(c)^\pm$ respectively) to
\mr
\item "{$(\beta)^-$}"  if $\eta \in {}^\lambda 2,\alpha < \lambda$
then for some $\beta \in (\alpha,\lambda)$ and $\rho \in
{}^{[\alpha,\beta)} 2$ we have \nl
$\neg \eta E((\eta \restriction \alpha)
\char 94 \rho \char 94 \eta \restriction [\beta,\lambda))$.
\ermn
2) In \scite{1.1}, \scite{1.1A} and in \scite{1.3}(1) for any $\varepsilon^* \le
\lambda$ we can replace $E$ by $\langle E_\varepsilon:\varepsilon <
\varepsilon^* \rangle$, each $E_\varepsilon$ satisfying clauses (b)
and $(c),(c)^+,(c)^-,(c)^\pm$ there respectively and in the conclusion:
\mr
\item "{$(*)$}"  there is a $\lambda$-perfect ${\Cal Q}$ such that
{\roster
\itemitem{ $(\alpha)$ }  ${\Cal Q} = \langle \rho_\eta:\eta \in
{}^\lambda 2 \rangle$ and
\sn
\itemitem{ $(\beta)$ }  if $\eta_1 \ne \eta_2$ are from ${}^\lambda 2$
then $\rho_{\eta_1} \ne \rho_{\eta_2}$ and $\varepsilon < \varepsilon^*
\Rightarrow \neg(\rho_{\eta_1} E_\beta \rho_{\eta_2})$
\sn
\itemitem{ $(\gamma)$ }   for $\eta \in {}^\lambda 2$ the set $\{\ell
g(\rho_\eta \cap \rho_\nu):\nu \in {}^\lambda 2 \backslash \{\eta\}\}$
is a closed unbounded subset of $\lambda$.
\endroster}
\ermn
3) In \scite{1.1A}, \scite{1.3}(1),(2) we can weaken
$(c)^+$ or $(c)^\pm$ to
\mr
\item "{$(*)$}"  for a stationary set of $N \in [{\Cal
H}(\lambda^+)]^\lambda$ there is (in $\bold V$) \, $\eta \in
{}^\lambda 2$ which is Cohen over $N$ such that $\pi^1_1[\lambda]$
sentences are absolute from $N[\eta]$ to $\bold V$ (for
$\Sigma^1_1[\lambda]$ sentences this is necessarily true and clause
(c) (or (c)$^-$) holds.
\endroster
\endproclaim
\bigskip

\demo{Proof}  The same as the proof of \scite{1.1}.
\enddemo
\bn
Now we would like not to restrict ourselves to $\pi^1_1[\lambda]$
equivalence relations.
\proclaim{\stag{1.4} Claim}  1) Assume
\mr
\item "{$(a)$}"  $\lambda = \lambda^{< \lambda},\mu \le 2^\lambda$
\sn
\item "{$(b)$}"  $E$ is a $\pi^1_2[\lambda]$ 2-place relation on
${}^\lambda 2$, say definable by
$(\forall z_1)(\exists z_2)\varphi(x,y,z_1,z_2,a)$
{\roster
\itemitem{ $(c)(\alpha)$ }  $E$ is an equivalence relation on ${}^\lambda
2$
\sn
\itemitem{$(\beta)$ }  if $\eta,\nu \in {}^\lambda 2$ and $(\exists!
\alpha < \lambda)(\eta(\alpha)) \ne \nu(\alpha))$ then $\neg(\eta E
\nu)$
\endroster}
\item "{$(c)^+$}"  if $\eta \in {}^\lambda 2$ is generic over $\bold
V$ for $({}^{\lambda >} 2,\triangleleft)$, i.e. is Cohen \ub{then} in
$\bold V[\eta]$, clause (c) still
holds
\nl
(note that for $\rho_1,\rho_2 \in ({}^\lambda 2)^{\bold V}$ anyhow
$\bold V \models ``\rho_1 E \rho_2" \Leftrightarrow \bold V[\eta] \models
``\rho_1 E \rho_2"$)
\sn
\item "{$(d)$}"  for every $A \subseteq \lambda$ and $\chi > 2^\lambda$
there are $N,\langle \rho_\varepsilon:\varepsilon < \mu \rangle$ such
that
{\roster
\itemitem{ $(i)$ }  $N \prec ({\Cal H}(\chi),\in),N^{< \lambda} 
\subseteq N,\|N\| = \lambda,A \in N$
\sn
\itemitem{ $(ii)$ }  $\rho_\varepsilon \in {}^\lambda 2$ and $[\varepsilon <
\zeta \Rightarrow \rho_\varepsilon \ne \rho_\varepsilon]$
\sn
\itemitem{ $(iii)$ }  for $\varepsilon \ne
\zeta,(\rho_\varepsilon,\rho_\zeta)$ is generic over $N$ for
$({}^{\lambda >} 2 \times {}^{\lambda >} 2)$
\sn
\itemitem{ $(iv)$ }  $\pi^1_1[\lambda]$ formulas are preserved from
$N[\rho_\varepsilon,\rho_\zeta]$ to $\bold V$ for $\varepsilon < \zeta
< \mu$.
\endroster}
\ermn
\ub{Then} $E$ has $\ge \mu$ equivalence classes. \nl
2) We can replace $\ge \mu$ by ``perfect" in the conclusion if in (c),
$\{\rho_\varepsilon:\varepsilon < \mu\} \subseteq {}^\lambda 2$ is
perfect [see \scite{0.4}]. \nl
3) We can replace ${}^{\lambda >} 2$ by ${\Cal T} \subseteq
{}^{\lambda >} 2$ subtree such that forcing with ${\Cal T}$ adds no
bounded subset to $\lambda$.
\endproclaim
\bigskip

\demo{Proof}  By \cite[2.2t]{Sh:664}.
\enddemo
\bigskip

\definition{\stag{1.5} Definition}  Clause (d) of \scite{1.4} is called
``$\lambda$ is $[\lambda,\mu)$-w.c.a." (as in \cite[2.1t]{Sh:664}'s notation).
\enddefinition
\bigskip

\proclaim{\stag{1.6} Claim}  We can strengthen \scite{1.4} just as
\scite{1.3} strenghthens \scite{1.1}.
\endproclaim
\bn
We may wonder when does clause (d) of \scite{1.4} holds.
\proclaim{\stag{1.7} Claim}  1) Assume
\mr
\widestnumber\item{$(iii)$}
\item "{$(i)$}"  $\lambda = \lambda^{< \lambda}$ in $\bold V$
\sn
\item "{$(ii)$}"  $\Bbb P$ is a forcing notion
\sn
\item "{$(iii)$}"  $\langle {\underset\tilde {}\to
\eta_\varepsilon}:\varepsilon < \mu \rangle$ is a sequence of $\Bbb
P$-names,
\sn
\item "{$(iv)$}"  $\Vdash_{\Bbb P} ``{\underset\tilde {}\to
\eta_\zeta} \ne {\underset\tilde {}\to \eta_\varepsilon} \in
{}^\lambda 2$ for $\varepsilon < \zeta < \mu"$
\sn
\item "{$(v)$}"  if $A \subseteq \lambda,p \in \Bbb P,\chi$ large
enough then there are $N \prec ({\Cal H}(\chi),\in),\|N\| =
\lambda,N^{< \lambda} \subseteq N,\{A,p\} \in N$ and $q$ such
that $p \le q \in \Bbb P,q$ is $(N,\Bbb P)$-generic, $q \Vdash ``({}^{\lambda >}
2)^{V^{\Bbb P}} \subseteq N[{\underset\tilde {}\to G_{\Bbb P}}]"$ and
$\Bbb P' \lessdot \Bbb P$ such that $q \Vdash_{\Bbb P} ``\text{for
some } u \in [\mu]^\mu$, for every $\varepsilon \ne \zeta$ from 
$u,({\underset\tilde {}\to \eta_\varepsilon},{\underset\tilde {}\to
\eta_\zeta})$ is generic over $N[{\underset\tilde {}\to G_{{\Bbb
P}'}}]$ for $({}^{\lambda >} 2 \times {}^{\lambda >} 2)^{V^{\Bbb P}}$
and the forcing $\Bbb P/(\Bbb P + {\underset\tilde {}\to
\eta_\varepsilon} + {\underset\tilde {}\to \eta_\zeta})$ is
$\lambda$-complete (or at least $\lambda$-strategically complete).
\ermn
\ub{Then} $\lambda$ is $(\lambda,\mu)$-c.w.a. (see \scite{1.5}) in the universe $\bold
V^{\Bbb P}$.
\endproclaim
\bigskip

\demo{Proof}  Straightforward.  
\enddemo
\newpage

\head {\S2 Singulars of uncountable cofinality} \endhead  \resetall \sectno=2
 % \resetall 
\bigskip

In this section we show that usually generalization of \scite{0.1}
probably fail badly for ${}^{\text{cf}(\lambda)} \lambda,\lambda$
singular of uncountable cofinality.
\bigskip

\proclaim{\stag{2.0} Claim}  1) For every $\lambda$ strong limit
singular we can find a very nice equivalence relation $E$ on
${}^\lambda 2$ such that:
\mr
\item "{$(a)$}"  $E$ has exactly two equivalence classes
\sn
\item "{$(b)$}"  if $\eta,\nu \in {}^\lambda 2$ and $(\exists
!i)(\eta(i) \ne \nu(i))$, \ub{then} $\neg(\eta E \nu)$.
\ermn
2) For every finite abelian group $H$ and $a_\alpha \in H$ for $\alpha
< \lambda$, we can find a very nice equivalence relation $E$ on
${}^\lambda 2$ with $|H|$-equivalence classes such that the
$E$-equivalence classes can be listed as $\langle X_b:b \in H \rangle$
and
\mr
\item "{$(c)$}"  if $\eta_0,\eta_1 \in {}^\lambda 2,\eta_0(i) =
0,\eta_1(i) = 1,(\forall j)[j \ne i \Rightarrow \eta(j) =
\nu(j)]$ and $\eta_\ell \in X_{b_\ell}$ then $H \models ``b_1 - b_0
= a_i"$.
\endroster
\endproclaim
\bigskip

\remark{\stag{d.5} Remark}  1) In part (2) we can look at ${}^\lambda H$ getting a
similar result.
\endremark
\bigskip

\demo{Proof}  Use \cite[\S2]{Sh:664}.
\enddemo
\bigskip

\proclaim{\stag{2.1} Claim}  Assume
\mr
\item "{$(a)$}"  $\lambda > \kappa = \text{ cf}(\lambda) > \aleph_0$
\sn
\item "{$(b)$}"  $2^\kappa + \lambda^{< \kappa} = \lambda$.
\ermn
\ub{Then} there is $E$ such that
\mr
\item "{$(\alpha)$}"  $E$ is an equivalence relation on ${}^\kappa
\lambda$
\sn
\item "{$(\beta)$}"  $E$ is very nice (see Definition \scite{0.2})
\sn
\item "{$(\gamma)$}"  if $\eta_1,\eta_2 \in {}^\kappa \lambda$ and
$(\forall^* i < \kappa)(\eta_1(i) = \eta_2(i))$ then $\eta_1 E \eta_2
\Leftrightarrow \eta_1 = \eta_2$
\sn
\item "{$(\delta)$}"  $E$ has exactly $\lambda$ equivalence classes.
\endroster
\endproclaim 
\bigskip

\demo{\stag{2.1A} Observation}  In \scite{2.1}, and in the rest of this
section: (of course, we have to translate the results we leave it as an
exercise to the reader). \nl
1) We can restrict ourselves to $\dsize \prod_{i <
\kappa} \lambda_i,\lambda_i < \lambda = \dsize \sum_{j < \kappa}
\lambda_j$, see the proof. \nl
2) We can consider ${}^\kappa \lambda$ as a subset of ${}^\lambda 2$,
in fact a very nice one:

we identify $\eta \in {}^\kappa \lambda$ with $\nu_\eta \in {}^\kappa
2$ when $\nu_\eta(i) = 1 \Leftrightarrow i \in
\{\text{pr}(\zeta,\eta(\zeta)):\zeta < \kappa\}$ for any choice of a
pairing function pr, in fact, any one to one pr:$\kappa \times \lambda
\rightarrow \lambda$ is O.K. \nl
3) If $\lambda$ is strong limit we can identify $\dsize
\prod_{i < \kappa} \lambda_i$ with ${}^\lambda 2$ as follows:
\wilog \, $\lambda_i = 2^{\mu_i}$ with $\mu_i$ increasing, let
$\langle g^i_\varepsilon:\varepsilon < {}^{\mu_i} 2 \rangle$ list the
functions from $[\dbcu_{j < i} \mu_j,\mu_i)$ to $\{0,1\}$ and we
identify $\eta \in \dsize \prod_{i < \kappa} \lambda_i$ with $\dbcu_{i
< \kappa} g^i_{\eta(i)} \in {}^\lambda 2$. \nl
4) We can identify any $\dsize \prod_{i < \kappa} \lambda_i$ with
${}^\kappa \lambda$ when $\lambda_i \le \lambda = \dsize \sum_{i <
\kappa} \lambda_i = \text{ lim sup}\langle \lambda_i:i < \kappa
\rangle$ by identifying $\eta \in \dsize \prod_{i < \kappa} \lambda_i$
with $\nu_\eta \in {}^\kappa \lambda$ such that $1 +
\nu_\eta(\varepsilon) = \eta(\zeta)$ when $\eta(\zeta) > 0,\varepsilon
= \text{ otp}\{\xi < \zeta:\eta(\xi) > 0\}$.
\enddemo
\bigskip

\demo{Proof}  We choose $\bar \lambda = \langle \lambda_i:i < \kappa
\rangle$, nondecreasing $i < j \Rightarrow \lambda_i \le \lambda_j$
with limit $\lambda$, (e.g. $\lambda_i = \lambda$) let $\mu_j = \dsize \prod_{i < j}
\lambda_i$ so $\mu_j \le \lambda$ and let $\bar f^i = \langle
f^i_\alpha:\alpha < \mu_i \rangle$ list $\dsize \prod_{j < i}
\lambda_j$ or just a set of representatives of $\dsize \prod_{j < i}
\lambda_j/J^{bd}_i$.
\nl
For every $\eta \in {}^\kappa \lambda$ let
\mr
\item "{$(a)$}"  for limit $i < \kappa$ let $\alpha_i(\eta) = \text{ Min}\{\alpha:\eta
\restriction i = f^i_\alpha \text{ mod } J^{bd}_i\}$ 
\sn
\item "{$(b)$}"  for $\varepsilon < \kappa$ let
$B_\varepsilon(\eta) = \{i:i < \kappa$ is a limit
ordinal, $\varepsilon < i$ and $f^i_{\alpha_i(\eta)}(\varepsilon) =
\eta(\varepsilon)\}$ \nl
and lastly
\sn
\item "{$(c)$}"  $A(\eta) = \{\varepsilon <
\kappa:B_\varepsilon(\eta)$ is not stationary$\}$.  
\ermn
Now we define two binary relations $E_0,E_1$ on ${}^\kappa \lambda$:
\mr
\item "{$(d)$}"  $\eta_1 E_0 \eta_2$ \ub{iff} for every $\varepsilon <
\kappa$ we have $B_\varepsilon(\eta_1) = B_\varepsilon(\eta_2)$ mod
${\Cal D}_\kappa$, where ${\Cal D}_\kappa$ is the club filter on $\kappa$
\sn
\item "{$(e)$}" $\eta_1 E_1 \eta_2$ \ub{iff} $\eta_1 E_0 \eta_2 \and
\eta_1 \restriction A(\eta_1) = \eta_2 \restriction A(\eta_2)$.
\ermn
Clearly
\mr
\item "{$(\alpha)$}"  $E_0$ is an equivalence relation on ${}^\kappa
\lambda$ with $\le 2^\kappa < \lambda$ classes
\sn
\item "{$(\beta)$}"  $E_1$ is an equivalence relation on ${}^\kappa
\lambda$, refining $E_0$
\sn
\item "{$(\gamma)$}"  $E_0,E_1$ are very nice
\sn
\item "{$(\delta)$}"  if $\eta_1,\eta_2 \in {}^\kappa \lambda$ and
$\eta_1 E_0 \eta_2$ then $A(\eta_1) = A(\eta_2)$
\sn
\item "{$(\varepsilon)$}"  for $\eta \in {}^\kappa \lambda,A(\eta_1)$
is a bounded subset of $\kappa$ \nl
[why?  otherwise let $C = \{\delta < \kappa:\delta = \sup(A(\eta_1)
\cap \delta)\}$, it is a club of $\kappa$, and for each $i \in C$ there
is $j_i < i$ such that $\eta \restriction [j_i,i) =
f^i_{\alpha_i(\eta)} \restriction [j_i,i)$, clearly $j_i$ exists by the
defintion of $\alpha_i(\eta)$.  By Fodor lemma, for some $j(*) <
\kappa$ the set $S_{j(*)} = \{i \in C:j_i = j(*)\}$ is stationary, now
choose $\varepsilon \in A(\eta) \backslash j(*)$, so clearly
$B_\varepsilon(\eta)$ includes $S_{j(*)}$ hence is a stationry subset
of $\kappa$ hence by the definition of $A(\eta)$ clearly $\varepsilon$
does not belong to $A(\eta)$, contradiction.]
\nl
So clearly
\sn
\item "{$(\zeta)$}"  $E_1$ has $\le ({}^\kappa \lambda/E_0) + \Sigma\{
\dsize \prod_{j < i} \lambda_j:i <\kappa\} \le \lambda$ equivalence
class.
\nl
Now
\item "{$(\eta)$}"  if $\eta_1,\eta_2 \in {}^\kappa \lambda$ and
$\eta_1 = \eta_2 \text{ mod } J^{bd}_\kappa$ \ub{then} for every limit
$i < \kappa$ large enough we have $\alpha_i(\eta_1) =
\alpha_i(\eta_2)$ \nl
[why?  let $i^* = \sup\{j+1:\eta_1(j) \ne \eta_2(j)\}$ so by the
assumption, if $i$ limit $\and i \in (i^*,\kappa)$ then $\eta_1
\restriction i = \eta_1 \restriction i$ mod $J^{bd}_i$ hence
$\alpha_i(\eta_1) = \alpha_i(\eta_2)$ by the definition of
$\alpha_i(-)$, which is the desired conclusion of $(\eta)$.]
\sn
\item "{$(\theta)$}"  if $\eta_1,\eta_2 \in {}^\kappa \lambda$ and
$\eta_1 = \eta_2$ mod $J^{bd}_\kappa$ \ub{then} $\eta_1 E_1 \eta_2
\Leftrightarrow \eta_1 = \eta_2$ \nl
[why?  if $\eta_1 = \eta_2$ clearly $\eta_1 E_1 \eta_2$; so assume
$\eta_1 E_1 \eta_2$ and we shall show that $\eta_1 = \eta_2$,
i.e. $\varepsilon < \kappa \Rightarrow \eta_1(\varepsilon) =
\eta_2(\varepsilon)$.  By the definition of $E_1$ we have $\eta_1 E_0
\eta_2$ hence by clause $(\delta)$ we have $A(\eta_1) = A(\eta_2)$,
call it $A$.  If $\varepsilon \in A$, by the definition of $E_1$ we have
$\eta_1 \restriction A = \eta_2 \restriction A$ hence
$\eta_1(\varepsilon) = \eta_2(\varepsilon)$.  So assume $\varepsilon \in
\kappa \backslash A$, first we can find $j^* < \kappa$ such that for every
limit $i \in (j^*,\kappa)$ we have $\alpha_i(\eta_1) =
\alpha_i(\eta_2)$, it exists by clause $(\eta)$.  Second, then
$B_\varepsilon(\eta_1),B_\varepsilon(\eta_2)$ are stationary (as
$\varepsilon \notin A(\eta_\ell)$) and equal modulo ${\Cal D}_\kappa$
(as $\eta_1 E_0 \eta_2$); so we can find $i \in B_\varepsilon(\eta_1)
\cap B_\varepsilon(\eta_2)$ which satisfy $i > j^*$.  Now $\eta_1(\varepsilon)
= f^i_{\alpha_i(\eta_1)}(\varepsilon)$ by the definition of
$B_\varepsilon(\eta_1)$ as $i \in B_\varepsilon(\eta_1)$ and
$\alpha_i(\eta_1) = \alpha_i(\eta_2)$ as $i > j^*$ and
$f^i_{\alpha_i(\eta_2)}(\varepsilon) = \eta_2(\varepsilon)$ by the
definition of $B_\varepsilon(\eta_2)$ as $i \in
B_\varepsilon(\eta_1)$; together $\eta_1(\varepsilon) =
\eta_2(\varepsilon)$.  So we have completed the proof that $\varepsilon
< \kappa \Rightarrow \eta_1(\varepsilon) = \eta_2(\varepsilon)$ thus
proving $\eta_1 = \eta_2$ as required.]
\sn
\item "{$(\iota)$}"  $E_1$ has $\ge \lambda_i$ equivalence classes for
any $i < \kappa$ \nl
[why? let $\eta^* \in {}^\kappa \lambda$ and for $\alpha < \lambda_i$
let $\eta^*_\alpha \in {}^\kappa \lambda$ be defined by
$\eta^*_\alpha(\varepsilon)$ is $\alpha$ if $\varepsilon = i$ and is
$\eta^*(\varepsilon)$ otherwise.  By clause $(\theta)$ we have $\alpha
< \beta < \lambda_i \Rightarrow \neg \eta^*_\alpha E_1 \eta^*_\beta$,
hence $|{}^\kappa \lambda/E_1| \ge \lambda_i$.]
\sn
\item "{$(\kappa)$}"  $E_1$ has exactly $\lambda$ equivalence classes
\nl
[why?  by clause $(\iota), E_1$ has $\ge \sup\{\lambda_i:i < \kappa\}
= \lambda$ equivalence classes and by clause $(\zeta)$ has $\le
\lambda$ equivalence classes.]  \hfill$\square_{\scite{2.1}}$
\endroster
\enddemo
\bigskip

\proclaim{\stag{2.2} Claim}  Assume
\mr
\item "{$(a)$}"  $\lambda > \kappa = \text{ cf}(\lambda) > \aleph_0$
\sn
\item "{$(b)$}"  $2^\kappa + \lambda^{< \kappa} = \lambda$
\sn
\item "{$(c)$}"  $\lambda \le \theta \le \lambda^\kappa$.
\ermn
\ub{Then} there is $E$ such that
\mr
\item "{$(\alpha)$}"  $E$ is an equivalence relation on ${}^\kappa
\lambda$
\sn
\item "{$(\beta)$}"  $E$ is very nice
\sn
\item "{$(\gamma)$}"  if $\eta_1,\eta_2 \in {}^\kappa \lambda$ and
$\eta_1 = \eta_2$ mod $J^{bd}_\kappa$ then $\eta_1 E \eta_2
\Leftrightarrow \eta_1 = \eta_2$
\sn
\item "{$(\delta)$}"  $E$ has $\theta$ equivalence classes.
\endroster
\endproclaim
\bigskip

\demo{Proof}  We can find a tree ${\Cal T} \subseteq {}^\kappa
\lambda$ with $\lambda$ nodes and exactly $\theta$ \,
$\kappa$-branches (\cite{Sh:262}); we can easily manage that $\eta \ne
\nu \in \text{ lim}_\kappa({\Cal T}) \Rightarrow (\exists^\kappa i <
\kappa)(\eta(i)) \ne \nu(i))$.  We proceed as in the proof of
\scite{2.1}, but in the definition of $E_1$ we add

$$
\eta_1 \in \text{ lim}_\kappa(T) \equiv \eta_2 \in \text{
lim}_\kappa({\Cal T}) \and (\eta_1 \in \text{ lim}_\kappa({\Cal T})
\rightarrow \eta_1 = \eta_2).
$$
${{}}$  \hfill$\square_{\scite{2.2}}$
\enddemo
\bigskip


\proclaim{\stag{2.3} Claim}  In Claim \scite{2.1} we replace clause
$(\gamma)$ by
\mr
\item "{$(\gamma)_1$}"  for every $\eta^* \in {}^\kappa \lambda$,
the set $\{\eta \in {}^\kappa \lambda:\eta = \eta^* \text{ mod }
J^{bd}_\kappa\}$ is a set of representatives for the family of
$E$-equivalence classes.
\endroster
\endproclaim
\bigskip

\demo{Proof}  Let $\bar \lambda$ be as there.

Let $K_i$ be an additive group, with set of elements $\lambda_i$ and
zero $0_{K_i}$.  Let $<^*$ be a well ordering of ${}^\kappa({\Cal P}(\kappa))$.

For every $\eta \in \dsize \prod_{i < \kappa} \lambda_i$ let

$$
\Xi_\eta = \{\langle B_\varepsilon(\nu):\varepsilon < \kappa
\rangle:\nu \in \dsize \prod_{i < \kappa} \lambda_i \text{ and } 
\nu = \eta \text{ mod } J^{bd}_\kappa\}
$$
\mn
so it is a nonempty subset of ${}^\kappa({\Cal P}(\kappa))$ and let
$\bar B^*_\eta$ be its $<^*$-first member and let $\Theta_\eta = \{\nu \in
\dsize \prod_{i < \kappa} \lambda_i:B_\varepsilon(\nu) = B^*_{\eta,\varepsilon}$ for every
$\varepsilon < \kappa$ and $\nu = \eta$ mod $J^{\text{bd}}_\kappa\}$.
\sn
Now note
\mr
\item "{$(*)_0$}"  $\Theta_\eta \ne \emptyset$ so $\bar B^*_\eta$ is
well defined for $\eta \in \dsize \prod_{i < \kappa} \lambda_i$
\sn
\item "{$(*)_1$}"  if $\nu \in \Theta_\eta$ \ub{then} for every limit $i <
\kappa$ large enough we have $\alpha_i(\nu) = \alpha_i(\eta)$
\sn
\item "{$(*)_2$}"  if $\nu_1,\nu_2 \in \Theta$ and $\varepsilon <
\kappa$, \ub{then} for every limit $i$ large enough we have:
$f^i_{\alpha_i(\nu_1)}(\varepsilon) =
f^i_{\alpha_i(\nu_2)}(\varepsilon)$.
\ermn
Now for $\eta \in \dsize \prod_{i < \kappa} \lambda_i$ we define
$\rho_\eta \in \dsize \prod_{i < \kappa} \lambda_i$ by

$$
\align
\rho_\eta(\varepsilon) \text{ is}: &f^i_{\alpha_i(\eta)}(\varepsilon)
\text{ for every } i \in B^*_\varepsilon \\
  &\text{ large enough if } B^*_{\eta,\varepsilon} \text{ is stationary}
\endalign
$$

$$
0_{K_i} \text{ if } B^*_\varepsilon \text{ is not stationary}.
$$
\mn
It is easy to see that
\mr
\item "{$(*)_3$}"  if $\nu \in \kappa_\lambda$ then $\rho_\eta(\varepsilon) =
\eta(\varepsilon)$ for every $\varepsilon < \kappa$ large enough hence
\sn
\item "{$(*)_4$}"  $\rho_\eta = \eta$ mod $J^{bd}_\kappa$
\sn
\item "{$(*)_5$}"  if $\eta_1,\eta_2 \in \dsize \prod_{i < \kappa}
\lambda_i$ and $\eta_1 = \eta_2 \text{ mod } J^{bd}_\kappa$ \ub{then}
$\rho_{\eta_1} = \rho_{\eta_2}$.
\ermn
Lastly, we define the equivalence relation $E$:

for $\eta_1,\eta_2 \in \dsize \prod_{i < \kappa} \lambda_i$ we define:

$$
\eta_1 E \eta_2 \text{ \ub{iff} (for every } i < \kappa \text{ we
have } K_i \models ``\eta_1(i)-\rho_{\eta_1}(i) =
\eta_2(i)-\rho_{\eta_2}(i)").
$$ 
\mn
Now check.  \hfill$\square_{\scite{2.3}}$
\enddemo
\bn
We may like to weaken the cardinal arithmetic assumptions.
\proclaim{\stag{2.4} Claim}  Assume
\mr
\item "{$(a)$}"  $\lambda > \kappa = \text{ cf}(\lambda) > \aleph_0$
\sn
\item "{$(b)$}"  $\kappa^{\aleph_0} < \lambda = \lambda^{\aleph_0}$.
\ermn
\ub{Then} the results \scite{2.1}, \scite{2.3} and \scite{2.2} (if
there is a tree ${\Cal T}$ with $\lambda$ nodes and $\kappa$-branches)
holds \ub{if} we replace the ideal $J^{bd}_\kappa$ by the ideal
$[\kappa]^{< \aleph_0}$.
\endproclaim
\bigskip

\demo{Proof}  Without loss of generality $\lambda_i >
\kappa^{\aleph_0},\langle \lambda_i:i < \kappa \rangle$ as in \scite{2.1}.
Let $\langle D_i:i < \kappa^{\aleph_0} \rangle$ list the
subsets of $\kappa$ of order type $\omega$ and let $\bar f^i = \langle
f^i_\alpha:\alpha < \dsize \prod_{j \in D_i} \lambda_j \rangle$ list
$\dsize \prod_{j \in D_i} \lambda_j$.  For $\eta \in \dsize
\prod_{\varepsilon < \kappa} \lambda_\varepsilon$ let $\alpha_i(\eta)
= \text{ Min}\{\alpha:\eta \restriction D_i = f^i_\alpha \text{ mod
} J^{bd}_{D_i}\}$.  With those choices the proofs are similar, we
write below the proof of the parallel of \scite{2.1}.

Define a two place relation $E^*$ on $\dsize \prod_{\varepsilon <
\kappa} \lambda_\varepsilon:\eta E \nu$ iff $(\forall i <
\kappa^{\aleph_0})(\alpha_i(\eta) = \alpha_i(\nu))$. Clearly $E^*$ is
a very nice equivalence relation and $\eta E^* \nu$ iff $(\eta,\nu
\in \dsize \prod_{\varepsilon < \kappa} \lambda_\varepsilon$ and
$\{\varepsilon:\eta(\varepsilon) \ne \nu(\varepsilon)\}$ is finite).
For $\eta \in \dsize \prod_{\varepsilon < \kappa} \lambda_\varepsilon$
let $A_{\eta,\varepsilon} = \{f^i_{\alpha_i(\eta)}(\varepsilon):i <
\kappa^{\aleph_0}$ and $\varepsilon \in D_i\}$ and let
$\gamma_{\eta,\varepsilon}$ be $0$ if $\eta(\varepsilon) \in
A_{\eta,\varepsilon}$ and $\eta(\varepsilon)-\sup(A_{\eta,\varepsilon}
\cap \eta(\varepsilon))$ otherwise stipulating $\alpha - \beta = 0$ if
$\alpha \le \beta$.  Lastly, $\eta E \nu$ iff
$(\forall \varepsilon)(\gamma_{\eta,\varepsilon} \equiv
\gamma_{\nu,\varepsilon})$.
\hfill$\square_{\scite{2.4}}$
\enddemo
\bigskip

\proclaim{\stag{2.5} Claim}  1) If 
$2^\kappa < \lambda = \lambda^{\aleph_0},\aleph_0 < \kappa =
\text{ cf}(\lambda) < \lambda$, \ub{then} we can find $E$ as in
\scite{2.1}$(\alpha),(\beta)$ (but not $(\gamma)$) and
\mr
\item "{$(\gamma)^*$}"  if $\eta \in {}^\kappa \lambda$ and $i <
\kappa$ then $X_{\alpha,i} = \{\nu \in {}^\kappa \lambda:(\forall j)(j
< \kappa \and j \ne i \rightarrow \nu(j) = \eta(j)\}$ is a set of
representations for $E$.
\ermn
2)  If $2^\kappa \le \lambda = \lambda^{\aleph_0},\aleph_0 < \kappa =
\text{ cf}(\lambda) < \lambda,1 \le \theta \le \lambda$, \ub{then} we
can find $E$ as in \scite{2.1}$(\alpha),(\beta)$ and
\mr
\item "{$(\gamma)^*$}"  if $\eta \in {}^\kappa \lambda$ and $i <
\kappa$ then $X_{\alpha,i}$ contains a set of representations
\sn
\item "{$(\delta)^*$}"   $E$ has $\theta$ equivalence classes.
\endroster
\endproclaim
\bigskip

\demo{Proof}  1)  We let $K$ be an abelian group with universe
$\lambda$ and in the proof of \scite{2.3},
define $\rho_\eta$ such that $\eta = \rho_\eta$ mod
$[\kappa]^{< \aleph_0}$ and let $a_\eta = \{i < \kappa:\eta(i) \ne
\rho_\eta(i)\} \in [\kappa]^{< \aleph_0}$ and define $E$ by: $\eta E
\nu$ iff $K \models \dsize \sum_{i \in a_\eta} \eta(i) -
\rho_\eta(i) = \dsize \sum_{i \in a_\nu} \nu(i) - \rho_\nu(i)$. \nl
2) Similar but we use equality in $K/K_1,K_1$ a subgroup of $K$ such that
$(K:K_1) = \theta$.  \hfill$\square_{\scite{3.5}}$
\enddemo
\bigskip 

\remark{\stag{2.6} Concluding Remark}  Instead $\langle J^{bd}_{D_i}:i
< \kappa^{\aleph_0} \rangle$ we can use $\langle (D_i,J_i):i < i^{\bar
\lambda} \rangle,D_i \subseteq \kappa,J_i$ an ideal on $D_i$ such that
$|\dsize \prod_{\varepsilon \in D_i} \lambda_\varepsilon/J_i| \le
\lambda,I = \{D \subseteq \kappa$: for every $i < i^*$ we have $D \cap
D_i \in J_i\}$ is included in $J^{bd}_\kappa$. \nl
Have not pursued this.
\endremark
\newpage

\head {\S3 Countable cofinality: positive results} \endhead  \resetall \sectno=3
 % \resetall 
\bigskip

We first phrase sufficient conditions which related to large
cardinals.  Then we prove that they suffice.  The proof of \scite{3.1}
is later in this section.
\proclaim{\stag{3.1} Lemma} Assume
\mr
\item "{$(a)$}"  $\lambda$ is strong limit of cofinality $\aleph_0$
\sn
\item "{$(b)$}"  $\lambda$ is a limit of measurables, or just 
\sn
\item "{$(b)^-$}"  for every $\theta < \lambda$ for some $\mu,\chi$
satisfying $\theta \le \mu \le \chi < \lambda$, 
there is a $(\chi,\mu,\theta)$-witness (see Definition \scite{3.2} below) 
\sn
\item "{$(c)$}"  $E$ is a nice equivalence relation on ${}^\omega
\lambda$ (or has enough absoluteness, as proved in \scite{3.11})
\sn
\item "{$(d)$}"  if $\eta,\nu \in {}^\omega \lambda$ and
$(\exists!n)(\eta(n) \ne \nu(n))$ then $\neg(\eta E \nu)$.
\ermn
\ub{Then} $E$ has $2^\lambda$ equivalence classes, moreover if
$\lambda_n < \lambda_{n+1} < \lambda = \dsize \Sigma_{n < \omega} \,
\lambda_n$ \ub{then} there is a subtree of
${}^{\omega >} \lambda$ isomorphic to $\dbcu_m \dsize \prod_{n < m}
\lambda_n$, with the $\omega$-branches pairwise non $E$-equivalent
(even somewhat more, see \scite{3.14}). 
\endproclaim
\bigskip

\definition{\stag{3.2} Definition}  1) We say $(\Bbb Q,I^{s_1,s_2}_2)$ is a
$(\lambda,\mu,\theta)$-witness if ($\lambda \ge \mu \ge \theta$ and):
\mr
\item "{$(a)$}"  $\Bbb Q$ is a $\theta$-complete forcing notion
\sn
\item "{$(b)$}"  $s_1$ is a function from $\Bbb Q$ to ${\Cal P}(\lambda)
\backslash \{\emptyset\}$
\sn
\item "{$(c)$}"  $s_2$ is a function from $\Bbb Q$ to $\{A:A \subseteq
\{(\alpha,\beta):\alpha < \beta < \lambda\}\}$
\sn
\item "{$(d)$}"  if $\Bbb Q \models p \le q$ then $s_\ell(q) \subseteq
s_\ell(p)$ for $\ell = 1,2$
\sn
\item "{$(e)$}"  $(\alpha,\beta) \in s_2(p) \Rightarrow
\{\alpha,\beta\} \subseteq s_1(p)$
\sn
\item "{$(f)$}"  for every $p \in \Bbb Q$ there is $q$ such that $p \le q
\in \Bbb Q$ and \nl
$(\forall \beta)(\exists \alpha,\gamma)[\beta \in s_1(q) \rightarrow
(\alpha,\beta) \in s_2(p) \and (\beta,\gamma) \in s_2(p)]$
\sn
\item "{$(g)$}"  if $p \in \Bbb Q$ and $A \subseteq \lambda \times
\lambda$, \ub{then} for some $q$ we have $p \le q \in \Bbb Q$ and $s_2(p)
\subseteq A \vee s_2(p) \cap A = \emptyset$
\sn
\item "{$(h)$}"  if $p \in \Bbb Q$ then for some $Y \in [\lambda]^\mu$
for every $\alpha < \beta$ from $Y$ we have $(\alpha,\beta) \in s_2(p)$. 
\ermn
2) We say $(\Bbb Q,s_1,s_2)$ is a $(\lambda,\mu,\theta,\varrho)$-witness if we
can strengthen clause $(g)$ to
\mr
\item "{$(g)^+$}"  if $f:{}^2 \lambda \rightarrow \varrho$ 
and $p \in \Bbb Q$ \ub{then} for some $q$ we have $p \le q \in
\Bbb Q$ and $f \restriction s_2(q)$ is constant.
\ermn
3) We call $(\Bbb Q,s_1,s_2)$ uniform if $\lambda =: \cup\{s_1(p):p
\in \Bbb Q\}$ is a cardinal and for every $p \in \Bbb Q$ and $\alpha < \lambda$ for some
$q$ we have $p \le q \in \Bbb Q$ and $s_1(q) \cap \alpha =
\emptyset$. \nl
4) We replace $\varrho$ by $< \varrho$ if in clause $(g)^+$, Rang$(f)$
is a subset of $\varrho$ of cardinality $< \varrho$.
\enddefinition
\bigskip

\definition{\stag{3.3} Definition}  1) We say that $(\Bbb Q,\bar s)$ is a
$(\lambda,\mu,\theta,\varrho;n)$-witness if $\lambda \ge \mu \ge
\theta,\lambda \ge \varrho$ and $\bar s = \langle s_m:m =
1,\dotsc,n \rangle$ and
\mr
\item "{$(a)$}"  $\Bbb Q$ is a $\theta$-complete forcing
\sn
\item "{$(b)$}"  $s_m$ is a function from $\Bbb Q$ to ${\Cal P}(\{\bar
\alpha:\bar \alpha = \langle \alpha_\ell:\ell < m \rangle,\alpha_\ell
< \alpha_{\ell +1}$ for $\ell < m-1\})$
\sn
\item "{$(c)$}"  if $\Bbb Q \models p \le q$ and $m \in
\{1,\dotsc,n\}$ then $s_m(p) \subseteq s_m(q)$
\sn
\item "{$(d)$}"  if $\langle \alpha_\ell:\ell < m+1 \rangle \in s_{m+1}(p)$
and $k < m+1$ then \nl
$\langle \alpha_\ell:\ell < k \rangle \char 94 \langle
\alpha_\ell:\ell = k+1,\dotsc,m \rangle \in s_m(p)$
\sn
\item "{$(e)$}"  for every $m \in \{1,\dotsc,n-1\},k<m$ and $p \in
\Bbb Q$ there is $q$ satisfying $p \le q \in \Bbb Q$ and $(\forall \bar \alpha \in
s_m(q))(\exists \bar \beta \in s_{m+1}(p))[\bar \alpha = (\bar \beta
\restriction k) \char 94 (\bar \beta \restriction [k+1,m))]$
\sn
\item "{$(f)^+$}"  if $m \in \{1,\dotsc,n\}$ and $f:{}^m \lambda
\rightarrow \varrho$ and $p \in \Bbb Q$ \ub{then} for some $q$ we have
$p \le q \in \Bbb Q$ and $f \restriction s_m(q)$ is constant
\sn
\item "{$(g)$}"  if $p \in \Bbb Q$ then for some $Y \in [\lambda]^\mu$
every increasing $\bar \alpha \in {}^n Y$ belongs to $s_n(p)$.
\ermn
2) $(\Bbb Q,\bar s)$ is a $(\lambda,\theta,\varrho;\omega)$-witness 
is defined similarly (using $s_m$ for $m \in [1,\omega)$). \nl
3) If $\varrho = 2$ we may omit it, also in Definition \scite{3.2}.
Also ``uniform" and ``$< \varrho$" means as in Definition \scite{3.2}. 
\enddefinition
\bigskip

\proclaim{\stag{3.4} Claim}  1) If $(\Bbb Q,s_1,s_2)$ is a
$(\lambda,\mu,\theta;n)$-witness and $\varrho < \theta,n < \omega$,
\ub{then} $(\Bbb Q,s_1,s_2)$ is a
$(\lambda,\mu,\theta,2^\varrho;n)$-witness. \nl
2) If ${\Cal D}$ is a normal ultrafilter on $\lambda$ so $\lambda$ is
measurable, $\Bbb Q = ({\Cal
D},\supseteq),s_1(A) = A,s_2(A) = \{(\alpha,\beta):\alpha < \beta$ are
from $A\}$, \ub{then} $(Q,s_1,s_2)$ is a uniform
$(\lambda,\lambda,\lambda)$-witness. \nl
3) If in (2), $s_m(A) = \{\bar \alpha:\bar \alpha = \langle
\alpha_\ell:\ell < m \rangle$ is increasing, $\alpha_\ell \in A\},\bar
s = \langle s_{m+1}:m < n \rangle$ and $n \le \omega$ \ub{then} $(\Bbb Q,\bar s)$ is a
$(\lambda,\lambda,\lambda,< \lambda;n)$-witness. \nl
4) If there is a $(\lambda,\mu,\theta;n)$-witness and $2^{< \theta} \le
\lambda$, \ub{then} there is such $(\Bbb Q,\bar s)$ with $|\Bbb Q| \le
2^\lambda$. \nl
5) Definition \scite{3.2}(1) is the case $n=2$ of Definition
\scite{3.3}(1).
\endproclaim
\bigskip

\demo{Proof}  Easy.
\enddemo
\bigskip

\proclaim{\stag{3.5} Claim}  1) Assume
\mr
\item "{$(a)$}"  $2 \le n < \omega,\lambda = \beth_{n-1}(\theta)^+$
\sn
\item "{$(b)$}"  $\theta$ is a compact cardinal or just a $\lambda$-compact cardinal
\sn
\item "{$(c)$}"  $\mu = \mu^{< \mu} < \theta$
\sn
\item "{$(d)$}"  $\Bbb P = \text{ Levy}(\mu,< \theta)$.
\ermn
\ub{Then} in $V^{\Bbb P}$, there is a $(\lambda,\mu,\theta;n)$-witness
$(\Bbb Q,\bar s)$. \nl
2) If there are $\lambda_n$ for $n < \omega,\lambda_n < \lambda_{n+1}$
and $\lambda_n$ is
$2^{(2^{\lambda_n})^+}$-compact, \ub{then} for some set forcing $\Bbb P$ in
$V^{\Bbb P}$ the cardinal $\lambda = \beth_\omega = \aleph_\omega$ is
dichotomically good (see Definition \scite{3.6A} below).
\endproclaim
\bigskip

\demo{Proof}  By \cite{Sh:124}.
\enddemo
\bn
Toward proving Lemma \scite{3.1} assume (till the end of this section) that
\demo{\stag{3.6} Hypothesis}  $\lambda = \dsize \sum \lambda_n,\dsize
\sum_{\ell < n} 2^{\lambda_\ell} < \theta_n \le \lambda_n$ and $(\Bbb
P_n,s_{n,1},s_{n,2})$ is a $(\lambda_n,\mu^+_n,\theta_n)$-witness and
for simplicity $\mu_n < \mu_{n+1},\lambda = \dsize \sum_n \mu_n$.
\enddemo
\bigskip

\definition{\stag{3.6A} Definition}  We call $\lambda$ dichotomically
good if there are $\lambda_n,\mu_n,\theta_n,\Bbb P_n,s_{n,1},s_{n,2}$
as in \scite{3.6}.
\enddefinition
\bigskip

\definition{\stag{3.7} Definition}  1) We define the forcing notion
$\Bbb Q_1$
\mr
\item "{$(a)$}"  $\Bbb Q_1 = \bigl\{p:p = (\eta,\bar A) = (\eta^p,\bar A^p) \text{ such
that letting } n(p) = \ell g(\eta)$ \nl
\hskip80pt $\text{we have } n^p < \omega,
\eta^p \in \dsize \prod_{\ell < n[p]} \lambda_\ell$ and \nl
\hskip80pt $\bar A^p = \langle A^p_\ell:\ell \in [n(p),\omega) \rangle \text{ and } 
A^p_\ell \in \Bbb P_\ell \bigr\}$
\sn
\item "{$(b)$}"  $p \le_{{\Bbb Q}_1} q$ iff $\eta^p \trianglelefteq
\eta^q,n(p) \le \eta(q),\ell \in [n(q),\omega) \Rightarrow \Bbb P_\ell
\models ``A^p_\ell \le A^q_\ell"$ and $n(p) \le \ell < n(q)
\Rightarrow \eta^q(\ell) \in s_1(A^p_\ell)$
\sn
\item "{$(c)$}"  We define the $\Bbb Q_1$-name $\underset\tilde {}\to
\eta$ by: $\underset\tilde {}\to \eta[G] = \cup\{\eta^p:
p \in {\underset\tilde {}\to G_{{\Bbb Q}_1}}\}$
\sn
\item "{$(d)$}"  We define
{\roster
\itemitem{$ (\alpha)$ }   $p \le^{{\Bbb Q}_1}_{pr} q$ iff $p
\le_{{\Bbb Q}_1} q \and n(p) = n(q)$
\sn
\itemitem{ $(\beta)$ }  $p \le^{{\Bbb Q}_1}_{apr} q$ iff
$p \le_{{\Bbb Q}_1} q \and \dsize \bigwedge_{\ell \ge n(q)} A^q_\ell =
A^p_\ell$
\sn
\itemitem{ $(\gamma)$ }  $p \le^{{\Bbb Q}_1}_{pr,n} q$ iff $p
\le^{{\Bbb Q}_1}_{pr} q$ and $\bar A^p \restriction [n(p),n(q)) = \bar
A^q \restriction [n(p),n(q))$.
\endroster}
\ermn
2) We define the forcing notion $\Bbb Q_2$ by: 
\mr
\item "{$(a)$}" $\Bbb Q_2 = \bigl\{p:p = (\eta_0,\eta_1,\bar A) =
(\eta^p_0,\eta^p_1,\bar A^p) \text{ where for some } n(p) < \omega
\text{ we have}$:
\nl
\hskip40pt  $\eta^p_0,\eta^p_1 \in \dsize \prod_{\ell < n[p]}
\lambda_\ell,\bar A^p = \langle A^p_\ell:\ell \in [n(p),\omega)
\rangle$ and $A^p_\ell \in \Bbb P_\ell \bigr\}$
\sn
\item "{$(b)$}"  $p \le_{{\Bbb Q}_2} q$ iff
{\roster
\itemitem{ $(i)$ }   $n(p) \le n(q)$ 
\sn
\itemitem{ $(ii)$ }  $\eta^p_\ell \trianglelefteq \eta^q_\ell$ for
$\ell=0,1$ 
\sn
\itemitem{ $(iii)$ }  $A^q_\ell \subseteq A^p_\ell$ for $\ell \in
[n(q),\omega)$ 
\sn
\itemitem{ $(iv)$ }  the pair 
$(\eta^p_0(\ell),\eta^p_1(\ell))$ is from $s_2(A^p_\ell)$ for $\ell
\in [n(p),n(q))$
\endroster}
\item "{$(c)$}"  we define
\footnote{why don't we ask $(\forall \ell <
n^p)(\eta^p_0(\ell) < \eta^p_1(\ell))$? to be able to construct the
perfect set, but, of course, $p \Vdash_{{\Bbb Q}_2} ``{\underset\tilde
{}\to \eta_0}(\ell) < {\underset\tilde {}\to \eta_1}(\ell)$ for $\ell
\in [n(p),\omega)$"} the $\Bbb Q_2$-name ${\underset\tilde {}\to
\eta_\ell}$ (for $\ell =0,1)$ by 
${\underset\tilde {}\to \eta_\ell}[G] = \cup\{\eta^p_\ell:p \in
{\underset\tilde {}\to G_{{\Bbb Q}_2}}\}$
\sn
\item "{$(d)$}"  we define
{\roster
\itemitem{ $(\alpha)$ }   $p \le^{{\Bbb Q}_2}_{pr} q$ iff $p
\le_{{\Bbb Q}_1} q \and n(p) = n(q)$ and
\sn
\itemitem{$ (\beta)$ }   $p \le^{{\Bbb Q}_2}_{apr} q$
iff $p \le_{{\Bbb Q}_2} q \and \dsize \bigwedge_{\ell \ge n(q)}
A^q_\ell = A^p_\ell$ and
\sn
\itemitem{ $(\gamma)$ }  $p \le^{{\Bbb Q}_2}_{pr,n} q$ iff
$p \le^{{\Bbb Q}_2}_{pr} q$ and $\bar A^p \restriction [n(p),n) \equiv
\bar A^q \restriction [n(p),n)$.
\endroster}
\ermn
3) If for a fix $k < \omega$, we have 
$(\Bbb P_n,\bar s^n)$ is a $(\lambda_n,\mu_n,\theta_n;k)$-witness for $n <
\omega$ \ub{then} we can define $\Bbb Q_\ell$ for $\ell = 1,\dotsc,k$ naturally. \nl
4) If $(\Bbb P_n,\bar s^n)$ is a $(\lambda_n,\mu_n,\theta_n;n)$-witness for $n <
\omega$ \ub{then} we can define $\Bbb Q = \{(\eta_0,\bar A):n < \omega,\eta(\ell) \in
{}^\ell(\lambda_\ell)\}$ is increasing, $\bar A = \langle A_\ell:\ell \in
[n,\omega),A_\ell \in {\Bbb P}_\ell\}$ with the natural order. \nl
We shall not pursue (3), (4).
\enddefinition
\bigskip

\demo{\stag{3.9} Fact}  Let $\ell \in \{1,2\}$. \nl
1) If $p \le_{{\Bbb Q}_\ell} q$ \ub{then} 
for some $q$ we have $p \le^{{\Bbb Q}_\ell}_{pr,n(q)} q \le^{{\Bbb Q}_\ell}_{apr}
r$. \nl
2) If $\bar p = \langle p_i:i < \alpha \rangle$ is $\le^{{\Bbb
Q}_\ell}_{pr}$-increasing, $\alpha < \theta_{n(p_0)}$, \ub{then}
$\bar p$ has a $\le^{{\Bbb Q}_\ell}_{pr}$-upper bound. \nl
3) If $\underset\tilde {}\to \tau$ is a ${\Bbb Q}_\ell$-name of an
ordinal, $p \in {\Bbb Q}_\ell$, \ub{then} for some $q$ and $n$ we
have:
\mr
\item "{$(a)$}"  $p \le_{pr} q$
\sn
\item "{$(b)$}"  if $q \le_{apr} r$ and $n(r) \ge n$, \ub{then} $r$ forces a
value to $\underset\tilde {}\to \tau$.
\ermn
4) In (3), if $\Vdash ``\underset\tilde {}\to \tau < \omega$ or just
$< \alpha^* < \theta_{n(p)}"$ then \wilog \, $n = n(p)$.
\enddemo
\bigskip

\demo{Proof}  Easy.
\enddemo
\bigskip

\proclaim{\stag{3.10} Claim}  1) $\underset\tilde {}\to \eta$ is
generic for ${\Bbb Q}_1$. \nl
2) The pair $({\underset\tilde {}\to \eta_0},{\underset\tilde {}\to \eta_1}), 
{\underset\tilde {}\to \eta_1}$ is generic for $\Bbb Q_2$.
\endproclaim
\bigskip

\proclaim{\stag{3.11} Claim}  Forcing by $\Bbb Q_2$ preserve ``$E$ is a
nice (see Definition \scite{0.2}(2)) equivalence relation on $\dsize
\prod_{n < \omega} \lambda_n$ satisfying clause (d) of \scite{3.1}.
\endproclaim
\bigskip

\demo{Proof}  Assume toward contradiction that $p^* \Vdash_{{\Bbb
Q}_2} ``{\underset\tilde {}\to \nu_0},{\underset\tilde {}\to
\nu_1},{\underset\tilde {}\to \nu_2} \in \dsize \prod_{\ell < \omega}
\lambda_\ell$ form a counterexample that is: ${\underset\tilde {}\to
\nu_0} E {\underset\tilde {}\to \nu_1} \wedge 
{\underset\tilde {}\to \nu_1} E {\underset\tilde {}\to \nu_2} 
\wedge \neg \nu_0 E {\underset\tilde {}\to \nu_1}$ or $\neg
{\underset\tilde {}\to \nu_0} E {\underset\tilde {}\to \nu_0}$ or
${\underset\tilde {}\to \nu_0} E {\underset\tilde {}\to \nu_1} \wedge
\neg {\underset\tilde {}\to \nu_1} E {\underset\tilde {}\to \nu_0}$ or
${\underset\tilde {}\to \nu_0} E {\underset\tilde {}\to \nu_1}
\wedge (\exists !n)({\underset\tilde {}\to \nu_0}(n) \ne 
{\underset\tilde {}\to \nu_1}(n))"$. \nl
Choose $\chi$ large enough and $\bar N = \langle N_n:n < \omega
\rangle,N$ such that:
\mr
\item "{$\circledast^\chi_{\bar N}(i)$}"  $N_n
\prec_{L_{\lambda^+_n,\lambda^+_n}} ({\Cal H}(\chi),\in)$ and $\|N_n\|
= 2^{\lambda_n}$ and $\{p^*,E,{\underset\tilde {}\to
\nu_0},{\underset\tilde {}\to \nu_1},
{\underset\tilde {}\to \nu_2},N_0,\dotsc,N_{n-1}\}$ belong to $N_n$
\sn
\item "{$(ii)$}"  $N = \dbcu_{n < \omega} N_n$ so $N \prec ({\Cal
H}(\chi),\in)$.
\ermn
Now we choose $p_n$ by induction on $n < \omega$ such that:
\mr
\item "{$(*)(i)$}"  $p_0=p$,
\sn
\item "{$(ii)$}"  $p_n \in N_n,n(p_n) \ge n$
\sn
\item "{$(iii)$}"  $p_n \le p_{n+1}$
\sn
\item "{$(iv)$}"  if $\underset\tilde {}\to \tau \in N_n$ is a $\Bbb
Q_2$-name of an ordinal then for some $k_n(\underset\tilde {}\to \tau)
> n+1$ we have: if $p_{n+1} \le q$ and $n(q) \ge k_n({\underset\tilde {}\to
\tau_n})$ then $q$ forces a value to $\underset\tilde {}\to \tau$.
\ermn
This is possible by \scite{3.9}(2),(3).  Now let $G = \{q:q \in N \cap
\Bbb Q_2$ and $q \le p_n$ (or just $p_n \Vdash ``q \in \underset\tilde
{}\to G")$ for some $n\}$, it is a subset of $\Bbb Q^N_2$ generic over $N$.
(Why?  If $N \models ``{\Cal I} \subseteq \Bbb Q_2$ is dense" then ${\Cal I}
\subseteq \Bbb Q_2$ is dense and there is ${\Cal I}' \subseteq {\Cal
I}$, a maximal antichain of $\Bbb Q_2$ which belongs to $N$ hence to
some $N_n$; there is $g \in N_n$, a one to one function from ${\Cal
I}'$ onto $|{\Cal I}'|$, so it define a $\Bbb Q_2$-name
$\underset\tilde {}\to \tau,
\underset\tilde {}\to \tau(G) = \gamma \Leftrightarrow (\forall q)(q
\in {\Cal I}' \cap G \rightarrow f(q) = \gamma) \Leftrightarrow
(\exists q)(q \in {\Cal I}' \cap G \and f(q) = \gamma)$, 
so $k_n(\underset\tilde {}\to \tau) < \omega$ is well 
defined and so $p_{k_n(\underset\tilde {}\to \tau)}$
forces a value to $\underset\tilde {}\to \tau$ hence forces $q \in
\underset\tilde {}\to G$ for some $q \in {\Cal I}' \subseteq {\Cal
I}$, hence $q \in G$ so $G \cap {\Cal I} \ne \emptyset$ as required).
Now by straightforward absoluteness argument, ${\underset\tilde {}\to
\nu_0}[G],
{\underset\tilde {}\to \nu_1}[G],{\underset\tilde {}\to \nu_2}[G] \in
\dsize \prod_{\ell < \omega} \lambda_\ell$ give contradiction to an
assumption.  \hfill$\square_{\scite{3.11}}$
\enddemo
\bn
In fact
\demo{\stag{3.11A} Observation}  Assume
\mr
\item "{$(a)$}"  $\lambda^*$ is strong limit of cofinality
$\aleph_0,\lambda^* = \dsize \sum_{n < \omega} \lambda^*_n,\lambda^*_n <
\lambda^*_{n+1}$
\sn
\item "{$(b)$}"
{\roster
\itemitem{ $(i)$ }   $\Bbb Q$ is a forcing notion
\sn
\itemitem{ $(ii)$ }   $\le_{pr}$ is included in $\le_{\Bbb Q}$ and
\sn
\itemitem{ $(iii)$ }  $\bold n:\Bbb Q \rightarrow \omega$ is such
that, for each $n$ the set ${\Cal I}_n = \{p \in \Bbb Q:\bold n(p) \ge
n\}$ is a dense subset of $\Bbb Q$
\sn
\itemitem{ $(iv)$ }  for $p \in \Bbb Q,\{q \in \Bbb Q:p \le_{pr} q\}$ is
$\lambda^*_{n(p)}$-complete and
\sn
\itemitem{ $(v)$ }   $\Bbb Q$ has pure decidability for $\Bbb
Q$-names of truth values
\sn
\itemitem{ $(vi)$ }  if $p \in \Bbb Q$ and $\underset\tilde {}\to
\tau$ is a $\Bbb Q$-name of an ordinal, \ub{then} there are $m <
\omega$ and $q$ satisfying: $p \le_{pr} q$ and $q \le r \wedge m \le
n(r) \Rightarrow$ ($r$ forces a value to $\underset\tilde {}\to \tau$)
\endroster}
\item "{$(c)$}"  $N,\langle N_n:n < \omega \rangle$ as above, 
$\{\Bbb Q,\le,\le_{pr}\} \in N_0$.
\ermn
\ub{Then} there is $G \subseteq \Bbb Q^N$ generic over $N$ hence
${\Cal H}(\lambda)^{N[G]} = {\Cal H}(\lambda)$.
\enddemo
\bigskip

\demo{Proof}  Should be clear.
\enddemo
\bigskip

\definition{\stag{3.12} Definition/Claim}  Assume that $F$ is a permutation of
$(\dsize \prod_{\ell < n(*)} \lambda_\ell) \times (\dsize \prod_{\ell
< n(*)} \lambda_\ell)$. \nl
1) Let $\Bbb Q^{\ge n(*)}_2 = \{p \in \Bbb P_2:n(p) \ge n(*)\}$ and let
$\hat F$ be the following function from $\Bbb Q^{\ge n(*)}_2$ to $\Bbb
Q^{\ge n(*)}_2$

$$
\align
\hat F(p) = q \text{ iff } &n(q) = n(p) \\
  &(\eta^q_0 \restriction n(*),\eta^q_1 \restriction n(*)) =
F((\eta^p_0 \restriction n(*),\eta^p_1 \restriction n(*))) \\
  &\eta^q_0 \restriction [n(*),n(p)) = \eta^p_0 \restriction
[n(*),n(p)) \\
  &\eta^q_1 \restriction [n(*),n(p)) = \eta^q_1 \restriction
[n(*),n(p)) \\
  &\bar A^q = \bar A^p.
\endalign
$$
\mn
2) For $p \in \Bbb Q^{\ge n(*)}_2,\hat F(p)$ is well defined $\in \Bbb
Q^{\ge n(*)}_2$. \nl
3) $\hat F$ is a permutation of $\Bbb Q^{\ge n(*)}_2$ preserving
$\le,\le_{pr},\le_{pr,n},\le_{apr}$ (and their negations), and $F \mapsto
\hat F$ is a group homomorphism. \nl
4) If $G \subseteq \Bbb Q_2$ is generic over $\bold V$ then
$\hat{\bold F}(G) = \{r \in \Bbb Q_2$: for some $q \in G \cap \Bbb
Q^{\ge n(*)}_2$ we have $r \le \hat F(q)\}$ is a generic over $\bold V$ and
$\bold V[\hat F(G)] = \bold V[G]$ and even $N[\hat F(G)] = N[G]$ if
say $N \prec ({\Cal H}(\chi),\in),\Bbb Q_2 \in N,F \in N,\lambda
\subseteq N$.
\enddefinition
\bigskip

\demo{Proof}  Easy.
\enddemo
\bigskip
 
\proclaim{\stag{3.13} Claim}

$\Vdash_{{\Bbb Q}_2} ``\neg {\underset\tilde {}\to \eta_0} E
{\underset\tilde {}\to \eta_1}"$.
\endproclaim
\bigskip

\demo{Proof}  If not, let $p \in \Bbb Q_2$ be such that $p \Vdash ``{\underset\tilde
{}\to \eta_0} E {\underset\tilde {}\to \eta_1}"$.
Now by clause (f) of Definition \scite{3.2}(1), we can find $p_1$ such
that:
\mr
\item "{$(i)$}"  $\Bbb Q_2 \models p \le_{pr} p_1$
\sn
\item "{$(ii)$}"  if $n(p) \le n < \omega$ and $\beta \in s_1(A^{p_1}_n)$
then for some $\alpha,\gamma$ we have $(\alpha,\beta),(\beta,\gamma)
\in s_2(A^p_n)$.
\ermn
Let $G_1 \subseteq \Bbb Q_2$ be generic over $\bold V$ such that $p_1 \in G_1$ and
let $\eta_\ell = {\underset\tilde {}\to \eta_\ell}[G_1]$ so $\bold V[G_1]
\models \eta_0 E \eta_1$.  By \scite{3.11} in $\bold V[G_1],E$ is still an
equivalence relation satisfying clause (d) of \scite{3.1} and
$\eta_1(n) \in s_1(A^{p_1}_n)$.  We can
find $\alpha < \lambda_{n(p)}$ such that $\alpha^* < \eta_1(n^*)$ and
$(\alpha^*,\eta_1(n^*)) \in s_2(A^p_{n^*})$.  Let us define $\eta'_0
\in \dsize \prod_{n < \omega} \lambda_n$ by $\eta'_0(n)$ is $\alpha^*$
if $n=n^*$ and $\eta_0(n)$ otherwise. \nl
Now the pairs $(\eta_0 \restriction (n(*)+1),\eta_1 \restriction
(n(*)+1))$ and $(\eta'_0 \restriction (n(*))+1),\eta_1 \restriction
(n(*)+1))$ are from $(\dsize \prod_{n \le n(*)} \lambda_n) \times
(\dsize \prod_{n \le n(*)} \lambda_n)$, so there is a permutation $F$
of this set interchanging those two pairs and is the identity
otherwise.  Let $G_2 = \hat F(G_1)$.  Now by \scite{3.12}
\mr
\item "{$(*)_1$}"  $G_2$ is a generic subset of $\Bbb Q_2$ over $\bold
V$
\sn
\item "{$(*)_2$}"  $\bold V[G_2] = \bold V[G_1]$
\sn
\item "{$(*)_3$}"  ${\underset\tilde {}\to \eta_0}[G_2] = \eta'_0,
{\underset\tilde {}\to \eta_1}[G_2] = \eta_1$.
\ermn
By \scite{3.11} we have
\mr 
\item "{$(*)_4$}"  $\bold V[G_\ell] \models \neg \eta_0 E \eta'_0$.
\ermn
As $p \le p_1 \in G_1$, by the choice of $p$ clearly
\mr
\item "{$(*)_5$}"  $\bold V[G_1] \models ``\eta_0 E \eta_1"$.
\ermn
By the choice of $p_1$ and $(\alpha,\eta_1(n^*))$ clearly $p \le \hat
F(p_1) \in G_2$ so
\mr
\item "{$(*)_6$}"  $\bold V[G_2] \models ``{\underset\tilde {}\to \eta'_0}[G_2]E
{\underset\tilde {}\to \eta_1}[G_2]"$ hence $\bold V[G_\ell] \models ``\eta'_0 E
\eta_1"$.
\ermn
Now $(*)_4 + (*)_5 + (*)_6$ contradict \scite{3.11}.
\hfill$\square_{\scite{3.13}}$
\enddemo
\bigskip

\proclaim{\stag{3.14} Claim}  1) Fix $\chi > \lambda$ large enough and
choose $N_n \prec_{L_{\lambda_n,\lambda_n}} ({\Cal H}(\chi),\in)$ such
that $\|N_n\| = 2^{\lambda_n},\{E,p_n:n < \omega\},\{N_\ell:\ell <
n\}$ belongs to $N_n$, and let $N = \dbcu_{n < \omega} N_n$; (certainly
can be done).  \ub{Then} we can find $\langle \rho_\nu:\nu \in \dsize \prod_{\ell <
n} \mu_\ell,n < \omega \rangle$ and $N,\langle N_n:n < \omega
\rangle$
\mr
\item "{$(\alpha)$}"  $\rho_\nu \in \dsize \prod_{\ell < \ell g(\nu)} \lambda_\ell$ 
\sn
\item "{$(\beta)$}"  $\nu_1 \triangleleft \nu_2 \Rightarrow
\rho_{\nu_1} \triangleleft \rho_{\nu_2}$
\sn
\item "{$(\gamma)$}"  if $\nu_1,\nu_2 \in \dsize \prod_{\ell <n}
\lambda_\ell$ and $m \le k < n$ and $\nu_2(m) < \nu_2(m)$ then
$\eta_{\nu_1}(k) < \eta_{\nu_2}(m)$
\sn
\item "{$(\delta)$}"  if $\nu \in \dsize \prod_{i < \omega}
\lambda_\ell$ then $\rho_\nu =: \dbcu_{n < \omega} \rho_{\nu
\restriction n}$ is generic for $(N,\Bbb Q_1)$ 
\sn
\item "{$(\varepsilon)$}"  if $\nu_0,\nu_1 \in \dsize \prod_{\ell < \omega}
\lambda_\ell,\nu_0 <_{lex} \nu_1$ then $(\rho_{\nu_0},\rho_{\nu_1})$
is generic for $(N,\Bbb Q_2)$ hence
\sn
\item "{$(\zeta)$}" if $\nu_0 \ne \nu_1 \in \dsize \prod_{\ell <
\omega} \lambda_\ell$ then $\neg \eta_{\nu_0} E \eta_{\nu_1}$.
\ermn
2) Moreover, for some $p \in \Bbb Q_2,n(p) = 0$ and non-principal
ultrafitler $D$ on $\omega$ we have
\mr
\item "{$(*)$}"  if $\eta,\nu \in \dsize \prod_{n < \omega}
s_1(A^p_n)$ and $\eta/D \ne \nu/D$ then $\neg \eta E \nu$.
\endroster
\endproclaim
\bigskip

\demo{Proof}   Let $M_0 \prec_{L_{\aleph_1,\aleph_1}} N_0,\|M_0\| =
2^{\aleph_0}$, let
$N = \dbcu_{n < \omega} N_n$. \nl
As above we choose $p_n$ by induction on $n$ such that:
\mr
\widestnumber\item{$(iii)$}
\item "{$(i)$}"  $p_n \in \Bbb Q_n$
\sn
\item "{$(ii)$}"  $p_n \in N_n$
\sn
\item "{$(iii)$}"  $n(p_0) = 0$
\sn
\item "{$(iv)$}"  $p_n \le_{pr} p_{n+1}$
\sn
\item "{$(v)$}"  for every $\Bbb Q_2$-name of an ordinal
$\underset\tilde {}\to \tau \in N_n$, for some $k_n(\underset\tilde
{}\to \tau) \in [n,\omega)$ we have: if $\Bbb Q_2 \models ``p_{n+1} \le q"$
and $n(q) \ge k_n(\underset\tilde {}\to \tau)$ then $q$ forces a value to
$\underset\tilde {}\to \tau$
\sn
\item "{$(vi)$}"  if $\underset\tilde {}\to \tau \in M_0$ is a $\Bbb
Q_2$-name of a natural number then $p_0$ forces a value to it.
\ermn
We can find $p_\omega \in \Bbb Q_2$ such that $n < \omega \Rightarrow
p_n \le_{pr} p_\omega$ and we can find $p^*$ such that $p_\omega \le
p^*$ and $(\forall n)(\forall \beta)(\exists \alpha,\gamma)[\beta \in
s_1(A^{p^*}_n) \rightarrow (\alpha,\beta),(\beta,\gamma) \in
s_2(A^{p_\omega}_n)]$ and we shall show that $p^*$ is as required.
Now clearly
\mr
\item "{$\boxtimes$}"  if $\eta_0,\eta_1 \in \dsize \prod_{n < \omega}
\lambda_n$ and $(\forall \ell < 2)(\forall n < \omega)(\eta_\ell(n) \in
s_1(A^{p^*}_n)$ and for every $n < \omega$ large enough
$(\eta_0(n),\eta_1(n)) \in s_2(A^{p^*}_n)$ \ub{then}
{\roster
\itemitem{ $(a)$ }  for some subset $G$ of $\Bbb Q^N_2$ generic over $N$
we have ${\underset\tilde {}\to \eta_0}[G] = \eta_0,{\underset\tilde
{}\to \eta_1}[G] = \eta_1$
\sn
\itemitem{ $(b)$ }  $\neg \eta_0 E \eta_1$ \nl
[why? by \scite{3.12}.]
\endroster}
\ermn
This suffices for part (1): by clause (b) of Definition \scite{3.2}(1),
we can find $Y_n \subseteq \lambda_n$ of
cardinality $\mu^+_n$ (really otp$(Y_n) = \mu_n \times
\mu_{n-1} \times \ldots \times \mu_0$ is enough) such that for
any $\alpha < \beta$ from $Y,(\alpha,\beta) \in s_2(A^p_n)$.  Now we
can choose by induction on $n,\langle \rho_\nu:\nu \in \dsize
\prod_{\ell < n} \mu_\ell \rangle$ as required in
$(\alpha),(\beta),(\gamma)$ of \scite{3.14}(1), they are as required.
\sn
For $B \subseteq \omega$ let ${\underset\tilde {}\to \eta_B}$ be the
following $\Bbb Q_2$-name:

$$
\eta_B(n) \text{ is } {\underset\tilde {}\to \eta_0}(n) \text{ if } n
\in B \text{ and is } {\underset\tilde {}\to \eta_1}(n) \text{ if } n
\in \omega \backslash B.
$$
\mn
Clearly ${\underset\tilde {}\to \eta_B}$ is a $\Bbb Q_2$-name of a
member of ${}^\omega \lambda$ and ${\underset\tilde {}\to \eta_B} \in
M_0$ hence for $B_1,B_2 \subseteq \omega$ the following $\Bbb Q_2$-name of a
truth value, Truth Value$({\underset\tilde {}\to \eta_{B_1}} E
{\underset\tilde {}\to \eta_{B_2}})$, is decided by $p_0$, say it is
$\bold t(B_1,B_2)$.
\sn
Define a two place relation $E'$ on ${\Cal P}(\omega):B_2 E' B_2$ iff
$\bold t(B_1,B_2) =$ truth. \nl
Let $J = \{B \subseteq \omega:\bold t(\emptyset,B) =$ truth$\}$. \nl
Clearly
\mr
\item "{$(*)_0$}"  $E'$ is an equivalence relation on ${\Cal
P}(\omega)$
\sn
\item "{$(*)_1$}"  $\omega \notin J$, even $[n,\omega) \notin J$.
\ermn
Let $\alpha^1_n < \alpha^2_n < \alpha^3_n < \alpha^4_n$ be from $Y_n$
for $n < \omega$ and for $h \in {}^\omega \{1,2,3,4\}$ let $\nu_h \in \dsize
\prod_{n < \omega} \lambda_n$ be $\nu_h(n) = \alpha^{h(n)}_n$.  Easily
\mr
\item "{$(*)_2$}"  if $B_1,B_2,B_3,B_4 \subseteq \omega$ and $B_1
\backslash B_2 = B_3 \backslash B_4,B_2 \backslash B_1 = B_4
\backslash B_3$ then $\bold t(B_1,B_2) = \bold t(B_3,B_4)$ that is $B_1 E' B_2 \equiv
B_3 E B'_4$ 
\sn
\item "{$(*)_3$}"  if $B_1,B_2,B_3,B_4 \subseteq \omega$ and $B_2
\triangle B_2 = B_3 \triangle B_4$ then $B_1 E' B_2 \equiv B_3
E' B_1$ \nl
[why? this is the meaning of $(*)_2$]
\sn
\item "{$(*)_4$}"  for $B_1,B_2 \subseteq \omega,B_1 E' B_2$ iff $B_1
\triangle B_2 \in J$ \nl
[why?  use $(*)_2$ with $B_1,B_2,B_1 \triangle B_2,\emptyset$ here
standing for $B_1,B_2,B_3$ by there] 
\sn
\item "{$(*)_5$}"  $B_1 \subseteq B_2 \in J \Rightarrow B_2 \in J$ \nl
[why?  we can find $A_1,A-2,A_3 \subseteq \omega$ such that $(A_1
\triangle A_2) = B_2 = (A_2 \triangle A_3)$ and $B_1 = A_1 \triangle
A_3$ by the if part of $(*)_4$ we have $\bold t(A_1,A_2) =$ truth,
$\bold t(A_2,A_3) =$ truth, by the definition of $\bold t$ we have
$\bold t(A_1,A_3) =$ truth, hence by $(*)_4$ the only if part we have
$B_1 = A_1 \triangle A_3 \in J$ as required] 
\sn
\item "{$(*)_6$}"  $J$ is an ideal \nl
[why?  by $(*)_5$ it is enough to show that for disjoint $B_1,B_2 \in
J$ also $B = B_1 \cup B_2 \in J$.  Now we have $B \triangle B_1 = B_2$
hence by $(*)_4$ we have $B E' B_1$ and $B_1 \triangle \emptyset = B_1
\in J$ hence $B_1 E' \emptyset$ hence $B E' \emptyset$; but by $(*)_4$
this implies $B \in J$ as required.] 
\ermn
So by $(*)_6$ there is an ultrafilter $D$ on $\omega$ disjoint to $J$,
and by $(*)_1$ it is non-principal, so we have proved also part (2).
\hfill$\square_{\scite{3.14}} \quad \square_{\scite{3.1}}$
\enddemo
\newpage

\head {\S4 The countable cofinality case: negative results} \endhead  \resetall \sectno=4
 % \resetall 
\bigskip

\proclaim{\stag{d.1} Claim}  Assume
\mr
\item "{$(a)$}"  $\lambda > \text{ cf}(\lambda) = \aleph_0$ 
\sn
\item "{$(b)$}"   $(\forall \alpha < \lambda)[|\alpha|^{\aleph_0} < \lambda]$
\sn
\item "{$(c)$}"   there is an algebra ${\frak B}$ with universe
$\lambda$, no infinite free subset and with $< \lambda$ functions.
\ermn
\ub{Then} there is $E$ such that
\mr
\item "{$(\alpha)$}" $E$ is an equivalence relation on ${}^\omega \lambda$
\sn
\item "{$(\beta)$}"  $E$ is very nice (see Definition \scite{0.2})
\sn
\item "{$(\gamma)$}"  if $\eta,\nu \in {}^\omega \lambda$ and $\eta
=^* \nu$ (i.e. $(\exists^{< \aleph_0} n)(\eta(n) \ne \nu(n))$ then
$\eta E \nu \Leftrightarrow \eta = \nu$
\sn
\item "{$(\delta)$}"  $E$ has $\lambda$ equivalence classes.
\endroster
\endproclaim 
\bigskip

\remark{\stag{d.2} Remark}  1) We can replace ${}^\omega \lambda$ by the set of
increasing $\omega$-sequences or by $\dsize \prod_{n < \omega}
\lambda_n$ when $\lambda_n < \lambda_{n+1} < \lambda = \dsize \sum_{m
< \omega} \lambda_m$ or by $\{A \subseteq \lambda:(\forall
n)(\exists!\alpha)(\alpha \in A \and \dsize \sum_{\ell < n}
\lambda_\ell \le \alpha < \lambda_n)$. \nl
2)  We can omit clause (b) if we weaken clause $(\gamma)$.  We can 
imitate \scite{3.3}, \scite{3.4}, \scite{3.5}.
\endremark 
\bigskip

\demo{Proof}  Without loss of generality ${\frak B}$ has $\aleph_0$
function and the individual constants $\{\alpha:\alpha < \lambda_0\}$
where $\lambda_0 < \lambda$.  

We define $E_0$ on ${}^\omega \lambda$ by

$$
\align
\eta E_0 \nu \text{ \ub{iff}}: &\text{ if } n <
\omega,\sigma(x_1,\dotsc,x_{n-1}) \text{ a } {\frak B} \text{-term,
and} \\
  &k,k_1,\dotsc,k_n < \omega \text{ then} \\
  &\eta(k) = \sigma(\eta(k_1),\eta(k_2),\dotsc,\eta(k_n)) \Leftrightarrow
\nu(k) = \sigma(\nu(k_1),\nu(k_2),\dotsc,\nu(k_n)).
\endalign
$$
\mn
So $E_0$ is an equivalence relation with $\le \lambda^{\aleph_0}_0 <
\lambda$ equivalence classes.  For $\eta \in {}^\omega \lambda$ let
$A(\eta) = \{k:\text{ for some } k^* < \omega$ there are no $n <
\omega,k_1,\dotsc,k_n \in [k^*,\omega)$ and ${\frak B}$-term
$\sigma(x_1,\dotsc,x_n)$ such that $\eta(k) =
\sigma(\eta(k_1),\dotsc,\eta(k_n))\}$. \nl
Lastly, we define $E_1$ by

$$
\eta E_1 \nu \text{ iff } \eta E_0 \nu \and \eta \restriction A(\eta)
= \nu \restriction A(\nu).  
$$ 
\mn
The rest is as in \S3.  \hfill$\square_{\scite{d.1}}$
\enddemo
\bigskip

\proclaim{\stag{d.3} Claim}  1) In \scite{d.1} we can demand
\mr
\item "{$(\delta)$}"  for each $\eta \in {}^\omega
\lambda,\eta/J^{bd}_\omega$ is a set of representatives of $E$.
\ermn
2) We can weaken in \scite{d.1} assumption (b) to
\mr
\item "{$(b)'$}"  $(\aleph_0 + |\tau({\frak B})|)^{\aleph_0} < \lambda$.
\ermn
3) If in \scite {d.1} we change clause $(\gamma)$ in the conclusion to
$(\gamma)^-$ below, we can omit clause (b) of the assumption
\mr
\item "{$(\gamma)^-$}"  for every equivalence class $X$ of $E$ for
every $\eta \in {}^\omega \lambda$ and for some
$\alpha,\eta_{n,\alpha} \in X$ where $\eta_{\alpha,n} \in {}^\omega
\lambda$ is: $\eta_{\alpha,n}(\ell) = \alpha$ if $\ell = n$ and
$\eta_{\alpha,n}(\ell) = \eta(\ell)$ otherwise.
\endroster
\endproclaim
\bigskip

\demo{Proof}  1) We imitate \scite{3.3} only letting $\Xi_\eta =
\bigl\{ \{ \langle k,k_1,\dotsc,k_n,\sigma \rangle:\nu(k) =
\sigma(k_1,\dotsc,k_n)\}:\nu \in {}^\omega \lambda,\nu/J^{bd}_\omega =
\eta/J^{bd}_\omega \bigr\}$. \nl
2) The same proof. \nl
3) For $\eta \in {}^\omega \lambda$ let $\bold n(\eta) < \omega$ be
the minimal $n \in [\bold n(\eta),\omega) \Rightarrow c \ell_{\frak
B}\{\eta(\ell):\ell \in [n,\omega)\} = c \ell_{\frak
B}\{\eta(\ell):\ell \in [\bold n(\eta),\omega)\}$.  Let $K$ be an
additive group with universe $\lambda,K_1$ a subgroup, $|K_1| =
\lambda,(K:K_1) = \lambda$ and $\eta E \nu$ iff $\dsize \sum_{n <
\bold n(\eta)} \eta(n) = \dsize \sum_{n < n(\nu)} \nu(\eta)$ mod
$K_1$. \nl
${{}}$  \hfill$\square_{\scite{d.3}}$
\enddemo
\bn
\ub{Question}:  What about having $\sigma \in (\lambda,2^\lambda)$
equivalence classes?
\newpage

\head {\S5 On $r_p(\text{Ext}(G,\Bbb Z)$} \endhead  \resetall \sectno=5
 % \resetall 
\bigskip

\definition{\stag{e.1} Definition}  For an abelian group $G$ and prime
$p$ let $r_p(G)$ be the rank of $G/pG$ as a vector space over $\Bbb
Z/p \Bbb Z$.

Instead using a definition of the abelian group Ext$(G,\Bbb Z)$, for
$G$ torsion free abelian group we quote (see \cite{Fu}).
\enddefinition
\bigskip

\proclaim{\stag{e.2} Claim}  For a torsion free abelian group $G$ and
prime $p,r_p(\text{\rm Ext\/}(g,\Bbb Z))$ is the rank of Hom$(G,\Bbb
Z/pZ)/(\text{Hom}(G,\Bbb Z)/p \Bbb Z)$ where Hom$(G,\Bbb Z/p \Bbb Z)$ is
the abelian group of homomorphisms from $G$ to $\Bbb Z/p \Bbb Z$,
Hom$(\Bbb Z,G)/p \Bbb Z$ is the abelian group of homomorphism $h$ from $G$
to $\Bbb Z/p \Bbb Z$ such that for some homomorphism $g$ from $G$ to
$\Bbb Z$ we have $x \in G \Rightarrow g(x)/\Bbb Z \in h(x)$.
\endproclaim
\bn
More generally (see \cite{Sh:664})
\definition{\stag{e.3} Definition}  1) We say ${\Cal Y} = (\bar A,\bar
K,\bar G,\bar D,\bar g^*)$ is a $\lambda$-system if
\mr
\item "{$(A)$}"  $\bar A = \langle A_i:i \le \lambda \rangle$ is an
increasing sequence of sets, $A = A_\lambda = \cup \{A_i:i < \lambda\}$
\sn
\item "{$(B)$}"  $\bar K = \langle K_t:t \in A \rangle$ is a sequence
of finite groups
\sn
\item "{$(C)$}"  $\bar G = \langle G_i:i \le \lambda \rangle$ is a
sequence of groups, $G_i \subseteq \dsize \prod_{t \in A_i} K_t$, each
$G_i$ is closed and $i < j \le \lambda \Rightarrow G_i = \{g
\restriction A_i:g \in G_j\}$ and \nl
$G_\lambda = \{g \in \dsize \prod_{t \in A_\lambda} K_t:(\forall i <
\lambda)(g \restriction A_i \in G_i)\}$ 
\sn
\item "{$(D)$}"  $\bar D = \langle D_\delta:\delta \le \lambda$ (a
limit ordinal)$\rangle,D_\delta$  an ultrafilter on $\delta$ such that
$\alpha < \delta \Rightarrow [\alpha,\delta) \in D_\delta$
\sn
\item "{$(E)$}"  $\bar g^* = \langle g^*_i:i < \lambda \rangle,g^*_i
\in G_\lambda$ and $g^*_i \restriction A_i = e_{G_i} = \langle
e_{K_t}:t \in A_i \rangle$.
\ermn
Of course, formally we should write $A^{\Cal Y}_i,K^{\Cal Y}_t,G^{\Cal
Y}_\delta,D^{\Cal Y}_\delta,g^{\Cal Y}_i$, eta., if clear from the
context we shall not write this. \nl
2) Let ${\Cal Y}^-$ be the same omitting $D_\lambda$ and we call it a
lean $\lambda$-system.
\enddefinition
\bigskip

%pgs8-11: \nl
We can deduce Sageev Shelah \cite{SgSh:148} result (if $|G| = \lambda$
is weakly compact $(> \aleph_0)$ and $p$ is prime, \ub{then}
$r_p(\text{Ext}(G,\Bbb Z)) \ge \lambda \Rightarrow
r_p(\text{Ext}(G,\Bbb Z)) = 2^\lambda)$.  For this note
\proclaim{\stag{e.8} Claim}  1)  Assume
\mr
\item "{$(a)$}"  ${\Cal Y}$ is a $\lambda$-system
\sn
\item "{$(b)$}"  $\bar H = \langle H_i:i < \lambda \rangle$ is a
sequence of groups, $\bar \pi = \langle \pi_{i,j}:i < j < \lambda
\rangle$, \nl
$\pi_{i,j} \in \text{ Hom}(H_j,H_i)$, commuting
\sn
\item "{$(c)$}"  $\bar h = \langle h_i:i < \lambda \rangle,h_i \in
\text{ Hom}(H_i,G^{\Cal Y}_i)$, and $i < j < \lambda \and x \in
H_j \Rightarrow (h_j(x)) \restriction A_i = h_i(\pi_{i,j}(x))$
\sn
\item "{$(d)$}"  $H_\lambda,\pi_{i,\lambda} \, (i < \lambda)$ form
the inverse limit of $\langle H_i,\pi_{i,j}:i < j < \lambda \rangle$,
and $h = h_\lambda$ the inverse limit of $\langle h_i:i < \lambda
\rangle$
\sn
\item "{$(e)$}"  $E_h$ is the following 2-place relation on $G_\lambda:f_1 E f_2
\Leftrightarrow f_1 f^{-1}_2 \in \text{ Rang}(h)$.
\ermn
\ub{Then}
\mr
\item "{$(\alpha)$}"  $h \in \text{ Hom}(H_\lambda,G_\lambda)$
\sn
\item "{$(\beta)$}"   if $(\forall i < \lambda)(|A_i| \le \lambda \and
|H_i| \le \lambda)$, then $E_h$ is a $\Sigma^1_1[\lambda]$-equivalence
relation
\sn
\item "{$(\gamma)$}"  if $(\forall i < \lambda)(|A_i| < \lambda \and
|H_i| < \lambda)$ and $\lambda$ is weakly compact uncountable,
\ub{then} the 2-place relation $E = E_h$ on $G_\lambda$ defined by $f_1
E f_2 \Leftrightarrow f_1 f^{-1}_2 \in \text{ Rang}(h)$ is very nice
\sn
\item "{$(\delta)$}"  under $(\gamma)$'s assumptions, if
$(G:\text{Rang}(h)) \ge \lambda$ then $(G:\text{Rang}(h)) = 2^\lambda$.
\ermn
2) If for $\varepsilon < \varepsilon(*) \le \lambda$ we have $\langle
H^\varepsilon_i:i < \lambda \rangle,\langle \pi^\varepsilon_{i,j}:i <
j < \lambda \rangle,\langle h^\varepsilon_i:i \le \lambda \rangle$ as
above, and for every $\alpha < \lambda$ there are $f^\alpha_i \in
G_\lambda$ (for $i < \alpha$) such that $\neg(f^\alpha_i
E_{h^\varepsilon_\lambda} f^\alpha_j)$ for $i < j < \alpha$, \ub{then}
there are $f_i \in G$ for $i < 2^\lambda$ such that $i < j < 2^\lambda
\and \varepsilon < \varepsilon^* \Rightarrow \neg(f_i E_{h_\varepsilon} f_j)$.
\endproclaim 
\bigskip

\demo{Proof}  Straight.  (The main point is in clause $(\gamma)$ of
\scite{e.8}(1), the point is that if $f \in G_\lambda \backslash
\text{ Rang}(h_\lambda)$ then for some $i < \lambda$ we have $\pi_{i,\lambda}(f)
\in G_i \backslash \text{ Rang}(h_i)$ by the tree property of
$\lambda$).  \hfill$\square_{\scite{e.8}}$
\enddemo
\bigskip

\proclaim{\stag{g.12} Claim}  Assume
\mr
\item "{$(A)(a)$}"  $\lambda$ is a strong limit cardinal and $\theta$
is a compact cardinal $< \lambda$
\sn
\item "{$(b)$}"  $K_i$ is a group for $i < \lambda$
\sn
\item "{$(c)$}"  $I$ is a directed partial order, $t \in I \Rightarrow A(t)
\subseteq \lambda$ and $\dbcu_{t \in I} A(t) = \lambda$
\sn
\item "{$(d)$}"  for $t \in I,G_t$ is a subgroup of $\dsize \prod_{i
\in u} \{K_i:i \in A(t)\}$
\sn
\item "{$(e)$}"  for $s \le t$ from $I$ we have $A(s) \subseteq A(t)$ and
$f \in G_t \Rightarrow f \restriction A(s) \in G_s$
\sn
\item "{$(f)$}"  $G_\infty$ is the inverse limit
\sn
\item "{$(B)(a)$}"  $\varepsilon(*) \le \lambda$
\sn
\item "{$(b)$}"  for $\varepsilon < \varepsilon(*),\langle
H^\varepsilon_u,\pi^\varepsilon_{u,w}:u \le w$ from $I\rangle$
is an inversely directed system of groups
\sn
\item "{$(c)$}"  $h^\varepsilon_u \in \text{
Hom}(H^\varepsilon_u,G_u)$ for $u \in I,\varepsilon < \varepsilon(*)$
\sn
\item "{$(d)$}"
$H^\varepsilon_\infty,h^\varepsilon,h^\varepsilon_{\infty,u}$ are the
limit of the inverse system
\sn
\item "{$(e)$}"  $E_\varepsilon$ is the equivalence relation on
$G_\infty:f E_\varepsilon g 
\Leftrightarrow fg^{-1} \in \text{ Rang}(h^\varepsilon_\infty)$
\endroster
\sn
$(C) \qquad$  for every $\mu < \lambda$ we can find $\langle
f^\mu_\alpha:\alpha < \mu \rangle$ from $G_\infty$ such that \nl

\hskip30pt $\varepsilon < \mu \cap \varepsilon(*) \and \alpha < \beta \Rightarrow
\neg(f^\mu_\alpha E_\varepsilon f^\mu_\beta)$
\sn
$(D) \qquad \theta$ is $> \underset {i < \lambda}{}\to \sup|K_i| +
\underset {t \in I}{}\to \sup|A(t)|$ and also 
$\sup_{t \in I,\varepsilon < \varepsilon(*)} |H^\varepsilon_t|$.
\mn
\ub{Then} there are $f_\alpha \in G$ for $\alpha < 2^\lambda$ such
that $\varepsilon < \varepsilon(*) \and \alpha < \beta = 2^\lambda
\Rightarrow \neg(f_\alpha E_\varepsilon f_\beta)$.
\endproclaim
\bigskip

\demo{Proof}  Let $\kappa = \text{ cf}(\lambda),\langle \lambda_i:i <
\kappa \rangle$ be increasing with limit $\lambda$.  We can choose by
induction on $i < \kappa,I_i,A_i$ such that
\mr
\item "{$(\alpha)$}"  $A_i \subseteq \lambda,|A_i| < \lambda,j < i
\Rightarrow A_j \subseteq A_i,\lambda_i \subseteq A_i$
\sn
\item "{$(\beta)$}"  $I_i \subseteq I$ is directed, $|I_i| < \lambda,j
< i \Rightarrow I_j \subseteq I_i$ and $t \in I_i \Rightarrow A(t) \subseteq
A_i$
\sn
\item "{$(\gamma)$}"  if we restrict ourselves to $A_i,I_i$, there is
$\langle f^i_\alpha:\alpha < \mu_i \rangle$, such that $f^i_\alpha \in
G^{I_i}_\infty = \text{ Lim}_{I_i} \langle G_u,f_{u,w}:u \le w$ from
$I_i\rangle$ and $\varepsilon < \mu_i \cap \varepsilon(*) \and \alpha
< \mu_i \Rightarrow \neg(f^i_\alpha E^{I_i} f^i_\beta)$.
\ermn
Now we can apply \cite[\S3]{Sh:664}.  \hfill$\square_{\scite{g.12}}$
\enddemo
\bigskip

\proclaim{\stag{g.13} Claim}   Assume
\mr
\item "{$(a)$}"   $\lambda > \text{ cf}(\lambda) = \kappa$, and
$\kappa$ is a measurable cardinal, say $D$ a normal ultrafilter on $\kappa$
\sn
\item "{$(b)$}"  $G$ is a torsion free abelian group
\sn
\item "{$(c)$}"  $|G| = \lambda$
\sn
\item "{$(d)$}"  $p$ is a prime number.
\ermn
1) If $r_p(\text{Ext}(G,\Bbb Z)) \ge \lambda$ and $\lambda =
\lambda^{< \kappa}$ then $r_p(\text{Ext}(G,\Bbb Z))
\ge \lambda^\kappa$. \nl
2) Let $\lambda = \dsize \sum_{i < \kappa} \lambda_i$; then
$r_p(\text{Ext}(G,\Bbb Z))$ is $\dsize \prod_{i < \kappa} \mu_i/D$ for
some $\mu_i \le 2^{\lambda_i}$.
\endproclaim
\bigskip


\demo{Proof}  1) Let $\langle G_i:i < \kappa \rangle$ be an increasing
sequence of pure subgroups of $G$ with union $G$ satisfying $i < \kappa \Rightarrow
|G_i| < \lambda$.  Now
\mr
\item "{$(*)$}"  if $g \in \text{ Hom}(G,\Bbb Z/p \Bbb Z)$ and $i <
\kappa \Rightarrow g \restriction G_i \in \text{ Hom}(G,\Bbb Z)/p \Bbb
Z$ then $g \in \text{ Hom}(G,\Bbb Z)/p \Bbb Z$. \nl
Why?  Let $g \restriction G_i = h_i/p \Bbb Z$ where $h_i \in \text{
Hom}(G,\Bbb Z)$ and let $h$ a function from $G$ to $\Bbb Z$ be defined
as $h(x)=n \Leftrightarrow \{i < \kappa:h_i(x)=n\} \in D$.  Clearly $h
\in \text{ Hom}(G,\Bbb Z)$ and $g = h/p \Bbb Z$, as required.]
\ermn
The rest should be clear. \nl
2) As in part (1), letting $\mu_i = r_p(\text{Ext}(G_i,\Bbb Z))$.
\hfill$\square_{\scite{g.13}}$
\enddemo
\bigskip

\demo{\stag{g.14} Conclusion}  If $\lambda$ is a strong limit and above a compact
cardinal and $G$ is a torsion free abelian group and $p$ is a prime
then $r_p(\text{Ext}(G,\Bbb Z)) \ge \lambda \Rightarrow
r_p(\text{Ext}(G,\Bbb Z)) = 2^\lambda$.
\enddemo
\bigskip

\demo{Proof}  By \scite{g.13}.
\enddemo
\bigskip

\remark{\stag{g.15} Remark} So for $\lambda$ strong limit singular the problem of
the existence of $G$ such that $|G| = \lambda,r_p(\text{Ext}(G,\Bbb
Z)) = \lambda$ is not similar to the problem of the existence of $M$
such that $\|M\| = \lambda$, nu$(M) = \lambda$. 
\endremark

%\newpage

\shlhetal

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\newpage
    
REFERENCES.  
\bibliographystyle{lit-plain}
\bibliography{lista,listb,listx,listf,liste}

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\bye




724.bbl
\def\germ{\frak} \def\scr{\cal}
  \ifx\documentclass\undefinedcs\def\rm{\fam0\tenrm}\fi%f**k-amstex!
  \def\defaultdefine#1#2{\expandafter\ifx\csname#1\endcsname\relax
  \expandafter\def\csname#1\endcsname{#2}\fi} \defaultdefine{Bbb}{\bf}
  \defaultdefine{frak}{\bf} \defaultdefine{mathbb}{\bf}
  \defaultdefine{mathcal}{\cal}
  \defaultdefine{beth}{BETH}\defaultdefine{cal}{\bf} \def\bbfI{{\Bbb I}}
  \def\mbox{\hbox} \def\text{\hbox} \def\om{\omega} \def\Cal#1{{\bf #1}}
  \def\pcf{pcf} \defaultdefine{cf}{cf} \defaultdefine{reals}{{\Bbb R}}
  \defaultdefine{real}{{\Bbb R}} \def\restriction{{|}} \def\club{CLUB}
  \def\w{\omega} \def\exist{\exists} \def\se{{\germ se}} \def\bb{{\bf b}}
  \def\equivalence{\equiv} \let\lt< \let\gt> \def\cite#1{[#1]}
  \def\implies{\Rightarrow}
\begin{thebibliography}{GrSh 302a}
\makeatletter \renewcommand{\@biblabel}[1]{[#1]} \makeatother
\par\noindent [References of the form {\tt math.XX/$\cdots$} refer to the
{\tt xxx.lanl.gov} archive]  \par

\bibitem[Fu]{Fu}Laszlo Fuchs.
\newblock {\em {Infinite Abelan Groups}}, volume I, II.
\newblock Academic Press, New York, 1970, 1973.

\bibitem[GrSh 302]{GrSh:302}Rami Grossberg and Saharon Shelah.
\newblock {On the structure of ${\rm Ext}_ p(G,{\Bbb Z})$}.
\newblock {\em {Journal of Algebra}}, {\bf 121}:117--128, 1989.
\newblock See also [GrSh:302a] below.

\bibitem[GrSh 302a]{GrSh:302a}Rami Grossberg and Saharon Shelah.
\newblock {On cardinalities in quotients of inverse limits of groups}.
\newblock {\em {Mathematica Japonica}}, {\bf 47}(2):189--197, 1998.
 {\tt math.LO/9911225}

\bibitem[MRSh 314]{MRSh:314}Alan~H. Mekler, Andrzej Ros{\l}anowski, and Saharon
  Shelah.
\newblock {On the $p$-rank of Ext}.
\newblock {\em {Israel Journal of Mathematics}}, {\bf 112}:327--356, 1999.
 {\tt math.LO/9806165}

\bibitem[PaSr98]{PaSr98}N.~Pandey and S.~M. Srivastava.
\newblock {A Powerless Proof of a Result of Shelah on Fundamental Groups}.
\newblock {\em Preprint}.

\bibitem[SgSh 148]{SgSh:148}Gershon Sageev and Saharon Shelah.
\newblock {Weak compactness and the structure of {\rm Ext}$(A,\,{\bold Z})$}.
\newblock In {\em {Abelian group theory (Oberwolfach, 1981)}}, volume 874 of
  {\em {Lecture Notes in Mathematics}}, pages 87--92. {Springer, Berlin-New
  York}, 1981.
\newblock {ed. Goebel, R. and Walker, A.E.}

\bibitem[Sh 664]{Sh:664}Saharon Shelah.
\newblock {Strong dichotomy of cardinality}.
\newblock {\em {Results in Mathematics}}, {\bf {accepted}}.
 {\tt math.LO/9807183}

\bibitem[Sh 124]{Sh:124}Saharon Shelah.
\newblock {$\aleph _{\omega }$ may have a strong partition relation}.
\newblock {\em {Israel Journal of Mathematics}}, {\bf 38}:283--288, 1981.

\bibitem[Sh 273]{Sh:273}Saharon Shelah.
\newblock {Can the fundamental (homotopy) group of a space be the rationals?}
\newblock {\em {Proceedings of the American Mathematical Society}}, {\bf
  103}:627--632, 1988.

\bibitem[Sh 262]{Sh:262}Saharon Shelah.
\newblock {The number of pairwise non-elementarily-embeddable models}.
\newblock {\em {The Journal of Symbolic Logic}}, {\bf 54}:1431--1455, 1989.

\bibitem[Sh 460]{Sh:460}Saharon Shelah.
\newblock {The Generalized Continuum Hypothesis revisited}.
\newblock {\em {Israel Journal of Mathematics}}, {\bf 116}:285--321, 2000.
 {\tt math.LO/9809200}

\bibitem[ShVs 719]{ShVs:719}Saharon Shelah and Pauli V\"ais\"anen.
\newblock {On equivalence relations $\Sigma^1_1$--definable over $H(\kappa)$}.
\newblock {\em Fundamenta Mathematicae}, {\bf submitted}.
 {\tt math.LO/9911231}

\end{thebibliography}
