
%\documentclass{amsart}
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\newcommand{\UpdateDate}{April 29, 2002}
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\title{Covering Property Axiom CPA$_{\rm cube}$ and its consequences}
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\ignore{
%%%%%%%%%%
\author{Krzysztof Ciesielski}
\address{Department of Mathematics, West Virginia University, 
        Morgantown, WV 26506-6310, USA}
\email{K\_Cies@math.wvu.edu; web page: {\tt http://www.math.wvu.edu/\~{}kcies}}
\thanks{The work of the first author was partially supported by 
        NATO Grant PST.CLG.977652.}
\author{Janusz Pawlikowski}
\address{Department of Mathematics, University of  Wroc\l aw,
         pl. Grunwaldzki 2/4, 50-384 Wroc\l aw, Poland}
\email{pawlikow@math.uni.wroc.pl}
\thanks{The author wishes to thank West Virginia University for its hospitality
   in years 1998-2001, where most of the results presented here were obtained.}
\subjclass{Primary 03E35; Secondary 03E17, 26A03}
\date{December 8, 2001}
%\dedicatory{This paper is dedicated to our authors.}
\keywords{Continuous images, 
perfectly meager, universally null, cofinality of null ideal,
uniform convergence, Martin's axiom}
%%%%%%%%%%%%
}
%\ignore{
\date{}
\pagestyle{myheadings}
\markboth{K.~Ciesielski and J.~Pawlikowski
\ \ \ \ \ \UpdateDate
}{Covering Property Axiom CPA$_{\rm cube}$ and its consequences 
\ \ \ \UpdateDate}
\author{
Krzysztof Ciesielski%
\thanks{AMS classification numbers: Primary 03E35; Secondary 
03E17, 26A03. \endgraf
\ \ Key words and phrases: continuous images, 
perfectly meager, universally null, cofinality of null ideal,
uniform convergence, Martin's axiom. \endgraf 
\ \ The work of the first author was partially supported by 
NATO Grant PST.CLG.977652.
}\\
{\footnotesize Department of Mathematics,}
{\footnotesize West Virginia University,} \\
{\footnotesize Morgantown, WV 26506-6310, USA}\\
{\footnotesize e-mail: K\_Cies@math.wvu.edu}; %\\
{\footnotesize web page: {\tt http://www.math.wvu.edu/\~{}kcies}}
\and
Janusz Pawlikowski\thanks{
%The work of the second author was partially supported by 
%??? %KBN 
%Grant ??? %2 P03A 031~14.
%\newline
The second
author wishes to thank West Virginia University for its hospitality
during 1998--2001, where the results presented here were obtained. 
}\\
{\footnotesize Department of Mathematics,}
{\footnotesize University of  Wroc\l aw,} \\
{\footnotesize pl. Grunwaldzki 2/4, 50-384 Wroc\l aw, Poland;} %\\
{\footnotesize e-mail: pawlikow@math.uni.wroc.pl}\\
}
%}

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\newcommand{\Implies}{\Longrightarrow}
\newcommand{\SoIC}{{s_0^{\rm prism}}}
\newcommand{\SoST}{{s_0^{\rm cube}}}
\newcommand{\psm}{{\rm CPA}}
\newcommand{\psmC}{\mbox{{\rm CPA$_{\rm cube}$}}}
\newcommand{\psmP}{\mbox{{\rm CPA$_{\rm prism}$}}}
\newcommand{\psmCsec}{\mbox{{\rm CPA$_{\rm cube}^{\rm sec}$}}}
\newcommand{\psmPsec}{\mbox{{\rm CPA$_{\rm prism}^{\rm sec}$}}}
\newcommand{\psmPLUS}{{\rm CPA$_{\rm cube}^+$}}
\newcommand{\psmPrPLUS}{{\rm CPA$_{\rm prism}^+$}}
\newcommand{\psmPrGame}{{\rm CPA$_{\rm prism}^{\rm game}$}}
\newcommand{\psmCGame}{{\rm CPA$_{\rm cube}^{\rm game}$}}
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\newcommand{\Ccube}{{\mathcal C}_{\rm cube}}
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% characteristic function
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\def\cof{{\rm cof}}
\def\cl{{\rm cl}}
\def\diam{{\rm diam}}
\def\dist{{\rm dist}}
\def\inter{{\rm int}}
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\def\continuum{{\mathfrak c}}
\def\co{\continuum}
\def\Cantor{{\mathfrak C}}
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\def\ra{\rangle}

\def\dom{{\rm dom}}
\def\range{{\rm range}}

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\begin{document}
 
\maketitle



\begin{abstract}
In the paper we formulate a Covering Property Axiom \psmC,
which holds in the iterated perfect set model, 
and show that it implies easily the following facts. 
\begin{itemize}
\item[(a)]
For every $S\subset\real$ of cardinality continuum there exists a uniformly
continuous function $g\colon\real\to\real$ with $g[S]=[0,1]$. 

\item[(b)] If $S\subset\real$ is either perfectly meager or universally null
then $S$ has cardinality less than~$\continuum$. 

\item[(c)] $\cf(\NN)=\omega_1<\continuum$, i.e., 
the cofinality of the measure ideal $\NN$ is $\omega_1$.

\item[(d)] For every 
uniformly bounded sequence 
$\la f_n\in\real^\real\ra_{n<\omega}$ of Borel 
functions there are the sequences:
$\la P_\xi\subset\real\colon\xi<\omega_1\ra$ of compact sets
and $\la W_\xi\in[\omega]^\omega\colon\xi<\omega_1\ra$
such that $\real=\bigcup_{\xi<\omega_1}P_\xi$ and for every
$\xi<\omega_1$:
\begin{quote}
$\la f_n\restriction P_\xi\ra_{n\in W_\xi}$ is a monotone uniformly 
convergent sequence of uniformly continuous functions.
\end{quote}


\item[(e)] {\sc Total failure of Martin's Axiom:}
$\continuum>\omega_1$ and 
for every non-trivial ccc forcing $\mathP$ there
exists $\omega_1$-many dense sets in $\mathP$ such 
that no filter intersects all of them.
\end{itemize}
\end{abstract}

%\newpage

\section{Axiom \psmC\ and other preliminaries} 
Our set theoretic terminology is standard and follows 
that of~\cite{CiBook}. 
In particular, $|X|$ stands for the cardinality of a set $X$
and $\continuum=|\real|$. 
A Cantor set $2^\omega$ will be denoted by a symbol $\Cantor$. 
We use the term {\em Polish space}\/
for a complete separable metric space {\tt without
isolated points}. 

For a Polish space $X$, the symbol $\perf(X)$
will stand for the 
collection of all subsets of $X$ homeomorphic to a Cantor set $\Cantor$.
We will consider $\perf(X)$ as ordered by inclusion.
Thus, a family $\E\subset\perf(X)$ is {\em dense in\/ $\perf(X)$}
provided for every $P\in\perf(X)$ there exists a $Q\in\E$
such that $Q\subset P$. 

Axiom \psmC\ is of the form
\begin{quote}
$\continuum=\omega_2$ and 
if $\E\subset\perf(X)$ is {\em appropriately}\/ dense in $\perf(X)$
then $|X\setminus\bigcup\E_0|<\continuum$ for some 
$\E_0\in[\E]^{\leq\omega_1}$. 
\end{quote}
If the word ``appropriately'' in the above is ignored,
then it implies the following statement.
\begin{description}
\item[{\bf Na\"\i ve-$\cpa$:}]
If $\E$ is dense in $\perf(X)$
then $|X\setminus\bigcup\E|<\continuum$. 
\end{description}
It is a very good candidate for our axiom in the sense that
it implies all the properties we are interested in.
It has, however, one major flaw --- {\em it is false!}\/ 
This is the case since 
$S\subset X\setminus\bigcup\E$ for some dense set $\E$ in $\perf(X)$
provided 
\begin{quote}
for each $P\in\perf(X)$ 
there is a $Q\in\perf(X)$ such that $Q\subset P\setminus S$.
\end{quote}
This means that the family $\G$ of all sets of the form $X\setminus\bigcup\E$,
where $\E$ is dense in $\perf(X)$, coincides with the 
$\sigma$-ideal $s_0$ of Marczewski's sets,
since $\G$ is clearly hereditary. Thus we have 
\begin{equation}\label{eq0}
s_0=\left\{X\setminus\bigcup\E\colon \mbox{ $\E$ is dense in $\perf(X)$}\right\}.
\end{equation}
However, it is well known (see e.g. 
\cite[thm.~5.10]{AMillerSubsetsOfR}) that 
there are $s_0$-sets of cardinality~$\continuum$.
Thus, our Na\"\i ve-$\cpa$ ``axiom'' cannot be consistent with ZFC.


In order to formulate the real axiom \psmC\ we need the following 
terminology and notation. 
A subset $C$ of a product $\Cantor^\eta$ of the Cantor set is
said to be a {\em perfect cube}\/ 
if 
$C=\prod_{n\in\eta} C_n$, where $C_n\in\perf(\Cantor)$ for each $n$.
For a fixed Polish space $X$ 
let $\Fcube$\label{PageFcube}
stand for the family of all continuous 
injections from a perfect cube $C\subset\Cantor^\omega$ onto a set $P$ 
from $\perf(X)$. 
We consider each function $f\in\Fcube$ from $C$ onto $P$ 
as a coordinate system imposed on~$P$.
We say that $P\in\perf(X)$ is a {\em cube}\/ if
it is determined by an (implicitly given) 
witness function $f\in\Fcube$ onto $P$,
and $Q$ is a {\em subcube of a cube}\/ $P\in\perf(X)$
provided $Q=f[C]$, where $f\in\Fcube$ is a witness function for $P$
and $C\subset\dom(f)\subset\Cantor^\omega$ 
is a perfect cube.
Here and in what follows the symbol 
$\dom(f)$
stands for the domain of $f$. 


We say that a family $\E\subset\perf(X)$ is {\em cube dense}\/
in $\perf(X)$
provided every cube $P\in\perf(X)$ contains a subcube
$Q\in\E$. 
More formally, 
$\E\subset\perf(X)$ is cube dense provided 
\begin{equation}\label{eqDefFdense}
\forall f\in\Fcube\ \exists g\in\Fcube\ 
(g\subset f\ \&\ \range(g)\in\E). 
\end{equation}
It is easy to see that the notion of
cube density is a generalization of 
the notion of density as defined in the first paragraph of this section:
\begin{equation}\label{eqFcubeF}
\mbox{if $\E$ is cube dense in $\perf(X)$
then $\E$ is dense in $\perf(X)$.} 
\end{equation}
On the other hand, the converse implication is not true, as shown by the 
following simple example. 

\ex{exCubeDens}{Let $X=\Cantor\times\Cantor$ and let $\E$ be the family 
of all $P\in\perf(X)$ such that either
\begin{itemize}
\item all vertical sections $P_x=\{y\in\Cantor\colon \la x,y\ra\in P\}$
      of $P$ are countable, or 
\item all horizontal sections $P^y=\{x\in\Cantor\colon \la x,y\ra\in P\}$
      of $P$ are countable. 
\end{itemize}
Then $\E$ is dense in $\perf(X)$, but it is not cube dense in $\perf(X)$. 
}

\proof To see that $\E$ is dense in $\perf(X)$ let $R\in\perf(X)$.
We need to find a $P\subset R$ with $P\in\E$. 
Clearly at least one of the projections 
$\proj_0(R)$ or $\proj_1(R)$ is uncountable.
Assume that $\proj_0(R)$ is uncountable and let $p\colon \proj_0(R)\to\Cantor$
be a Borel function. (For example, if $p$ is defined by $p(x)=\min R_x$
then $p\subset R$ is Baire class~1.) So, there is a $Q\in\perf(\Cantor)$ such
that
$p\restriction Q$ is continuous. 
In particular, $p\restriction Q$ (identified with its graph) 
is a closed subset of $X=\Cantor\times\Cantor$. 
So $P=p\restriction Q\in\E$
is a subset of $R$.

To see that $\E$ is not $\Fcube$-dense in $\perf(X)$
it is enough to notice that $P=X=\Cantor\times\Cantor$ 
considered as a cube, 
where the second coordinate is identified with 
$\Cantor^{\omega\setminus\{0\}}$, has no subcube in $\E$. 
More formally, 
let $h$ be a homeomorphism from $\Cantor$ onto $\Cantor^{\omega\setminus\{0\}}$,
let $g\colon\Cantor\times\Cantor\to
\Cantor^\omega=\Cantor\times\Cantor^{\omega\setminus\{0\}}$ be 
given by $g(x,y)=\la x,h(y)\ra$, 
and let 
$f=g^{-1}\colon\Cantor^\omega\to\Cantor\times\Cantor$ be
the coordinate function 
making $\Cantor\times\Cantor=\range(f)$ a cube.
Then $\range(f)$ does not contain a subcube from $\E$.
\qed

With these notions in hand we are ready to formulate our 
axiom \psmC.\label{PageCPAcube}
\begin{description}
\item[{\bf \psmC:}] $\continuum=\omega_2$ and 
for every Polish space $X$ and every cube dense
family $\E\subset\perf(X)$ there is an $\E_0\subset\E$ 
such that $|\E_0|\leq\omega_1$ and $|X\setminus\bigcup\E_0|\leq\omega_1$. 
\end{description}

The proof that \psmC\ holds in the iterated perfect set model
can be found in~\cite{CP83} and~\cite{CPAbook}.


It is also worth noticing that in order to check
that $\E$ is cube dense it is
enough to consider in condition (\ref{eqDefFdense})
only functions $f$ defined on the entire space $\Cantor^\omega$,
that is

\fact{factFcube}{ 
$\E\subset\perf(X)$ is cube dense if and only if
\begin{equation}\label{eqDefFdenseWeak}
\forall f\in\Fcube,\ 
\dom(f)=\Cantor^\omega,\ 
\exists g\in\Fcube\ 
(g\subset f\ \&\ \range(g)\in\E). 
\end{equation}
}
\proof
To see this, let $\Phi$ be
the family of all bijections 
$h=\la h_n\ra_{n<\omega}$ between perfect subcubes 
$\prod_{n\in\omega} D_n$ and $\prod_{n\in\omega} C_n$
of $\Cantor^\omega$ such that
each $h_n$ is a homeomorphism between $D_n$ and $C_n$. 
Then 
\begin{equation}
f\circ h\in\Fcube\ \mbox{ for every $f\in\Fcube$ and $h\in\Phi$
with $\range(h)\subset\dom(f)$}.
\end{equation}
Now take an arbitrary $f\colon C\to X$ from $\Fcube$
and choose an $h\in\Phi$
mapping $\Cantor^\omega$ onto $C$.
Then $\hat f=f\circ h\in\Fcube$ maps $\Cantor^\omega$ into $X$
and, using (\ref{eqDefFdenseWeak}), we can find 
$\hat g\in\Fcube$ such that $\hat g\subset\hat f$ and $\range(\hat g)\in\E$.
Then $g=f\restriction h[\dom(\hat g)]$
satisfies condition (\ref{eqDefFdense}). \qed


Next, let us consider\label{PageSoST}
\begin{eqnarray}
\SoST & = &
\left\{X\setminus\bigcup\E\colon 
\mbox{ $\E$ is cube dense in $\perf(X)$}\right\}\label{eq00}\\ 
& = &
\left\{S\subset X\colon \forall\ 
\mbox{cube $P\in\perf(X)$ $\exists$ 
subcube $Q\subset P\setminus S$}\right\}.\nonumber
\end{eqnarray}
Notice that

\fact{fact1}{$[X]^{<\continuum}\subset \SoST\subset s_0$ for every Polish space $X$.}

An easy proof of this Fact~\ref{fact1} can be found in~\cite{CPAbook}.
It can be also shown in ZFC that $\SoST$ forms a $\sigma$-ideal.
However, neither of these facts will be used in the sequel. 
On the other hand we will be interested in the following proposition. 

\prop{propBacic}{If \psmC\ holds then $\SoST=[X]^{\leq\omega_1}$.
%for every Polish space~$X$. 
}

\proof It is obvious that \psmC\ implies  $\SoST\subset[X]^{<\continuum}$. 
We will not be interested in the other inclusion, though it follows 
immediately from Fact~\ref{fact1}. \qed



\rem{remMiller}{$\SoST\neq[X]^{\leq\omega_1}$ 
in a model obtained by 
adding Sacks numbers side-by-side.
In particular \psmC\ is false in this model. 
}

\proof This follows from the fact that $\SoST=[X]^{\leq\omega_1}$
implies the property (A) (see Corollary~\ref{cor:forA}) 
while it is false in the model mentioned above,
as noticed by Miller in~\cite[p.~581]{Mi}.
(In this model the set $X$ of all Sacks generic numbers
cannot be mapped continuously onto $[0,1]$.) \qed




\section{Continuous images of sets of cardinality $\continuum$}

An important quality of the ideal $\SoST$, and so the power of 
the assumption $\SoST=[X]^{<\continuum}$, is well depicted 
by the following fact. 

\prop{prop1}{If $X$ is a Polish space  and $S\subset X$ does not belong to
$\SoST$ then there exist a $T\in[S]{^\continuum}$ and 
a uniformly continuous function $h$ from $T$ onto~$\Cantor$.}

\proof Take an $S$ as above and let 
$f\colon\Cantor^\omega\to X$ be a continuous injection 
such that 
$f[C]\cap S\neq\emptyset$
for every perfect cube $C$. %=\prod_{n<\omega}C_n$.
Let $g\colon\Cantor\to\Cantor$ be a continuous 
function such that $g^{-1}(y)$ is perfect for every $y\in\Cantor$. 
Then 
$h_0=g\circ\proj_0\circ f^{-1}\colon f[\Cantor^\omega]\to\Cantor$
is uniformly continuous. Moreover, if $T=S\cap f[\Cantor^\omega]$ then  
$h_0[T]=\Cantor$ since 
\[
T\cap h_0^{-1}(y)=
T\cap f[\proj_0^{-1}(g^{-1}(y))]=
S\cap f[g^{-1}(y)\times\Cantor\times\Cantor\times\cdots]\neq\emptyset
\]
for every $y\in\Cantor$. \qed

\cor{cor:forA}{Assume $\SoST=[X]^{<\continuum}$ for a Polish space $X$. If
$S\subset X$ has cardinality $\continuum$ then there exists 
a uniformly continuous function $f\colon X\to[0,1]$ such that $f[S]=[0,1]$.

In particular, \psmC\ implies property\/ {\rm (a)}.
}

\proof If $S$ is as above then, by \psmC, $S\notin\SoST$.
Thus, by Proposition~\ref{prop1}
there exists a uniformly continuous function $h$ from  a subset of $S$ onto 
$\Cantor$. Consider $\Cantor$ as a subset of $[0,1]$
and let $\hat h\colon X\to[0,1]$ be a uniformly continuous extension
of $h$. If $g\colon[0,1]\to[0,1]$ is continuous
and such that $g[\Cantor]=[0,1]$
then $f=g\circ\hat h$ is as desired. \qed

The fact that (a) holds in the iterated perfect set model was first proved 
by A.~Miller in~\cite{Mi}.



It is worth to note here that the function 
$f$ in Corollary~\ref{cor:forA} cannot be
required to be either monotone or 
in the class ``$D^1$''\label{PageD1}
of all functions having finite or infinite derivative at every point.
This follows immediately from the following proposition, 
since each function which is either monotone or ``$D^1$''
belongs to the Banach class\label{PageT2} 
\[
(T_2)=
\left\{f\in\C(\R)\colon \{y\in\R\colon |f^{-1}(y)|>\omega\}\in\NN\right\}.
\]
(See \cite{Foran} or \cite[p.~278]{Saks}.)


\prop{prop:AinZFC}{There exists, in ZFC, an $S\in[\real]^{\continuum}$
such that $[0,1]\not\subset f[S]$ for every $f\in(T_2)$.}

\proof Let $\{f_\xi\colon \xi<\continuum\}$
be an enumeration of all functions from $(T_2)$
whose range contains $[0,1]$. 
Construct by induction a sequence 
$\la\la s_\xi,y_\xi\ra\colon\xi<\continuum\ra$
such that for every $\xi<\continuum$
\begin{itemize}
\item[(i)] $y_\xi\in[0,1]\setminus f_\xi[\{s_\zeta\colon\zeta<\xi\}]$
and $|f^{-1}_\xi(y_\xi)|\leq\omega$.
\item[(ii)] $s_\xi\in\real\setminus
\left(\{s_\zeta\colon\zeta<\xi\}\cup
\bigcup_{\zeta\leq\xi}f^{-1}_\zeta(y_\zeta)\right)$.
\end{itemize}
Then the set $S=\{s_\xi\colon\xi<\continuum\}$ is as required
since $y_\xi\in [0,1]\setminus f_\xi[S]$ for every $\xi<\continuum$. 
\qed





\section{Perfectly meager and  universally null sets}

The fact that (b) holds in the iterated perfect set model was first proved 
by A.~Miller in~\cite{Mi}.



\thm{thm:PMandUN}{If $S\subset\real$ is either perfectly meager 
or universally null then $S\in \SoST$.

In particular, \psmC\ implies property\/ {\rm (b)}.
}

\proof Take an $S\subset\real$ which is either perfectly meager 
or universally null and let $f\colon\Cantor^\omega\to\real$
be a continuous injection.
Then $S\cap f[\Cantor^\omega]$ is either meager or null in 
$f[\Cantor^\omega]$.
Thus $G=\Cantor^\omega\setminus f^{-1}(S)$
is either comeager or of full measure in $\Cantor^\omega$.
Hence the theorem follows immediately from the following claim. \qed



\claim{claim1}{Consider $\Cantor^\omega$ with standard 
topology and standard product measure. 
If $G$ is a Borel subset of $\Cantor^\omega$ which is either 
of second category or of positive measure
then $G$ contains a perfect cube $\prod_{i<\omega}P_i$. 
}

The measure version of the claim is a variant 
the following theorem:
\begin{itemize}
\item[(m)] for every full measure
subset $H$ of $[0,1]\times[0,1]$ there are
a perfect set $P\subset[0,1]$ and a positive inner measure subset
$\hat H$ of $[0,1]$ such that $P\times \hat H\subset H$ 
\end{itemize}
which was proved by Eggleston \cite{Eg} and, independently, by 
Brodski\v\i~\cite{Br}. The category version of the claim 
is a consequence of 
the category version of (m):
\begin{itemize}
\item[(c)] for every Polish space $X$ and every comeager 
subset $G$ of $X\times X$ there are
a perfect set $P\subset X$ and a comeager subset
$\hat G$ of $X$ such that $P\times \hat G\subset G$.
\end{itemize}
This well known result can be found in~\cite[Exercise~19.3]{Ke}. 
(Its version for $\real^2$ is also proved, for example, 
in \cite[condition ($\star$), p. 416]{CW}.) 
For completeness, we will show here in detail how to deduce
the claim from (m) and (c). 

We will start the argument with a
simple fact, in which we will use the following notations.
If $X$ is a Polish space endowed with a Borel measure then
$\psi_0(X)$ will stand for the sentence
\begin{description}
\item[$\psi_0(X)$:] For every full measure
subset $H$ of $X\times X$ there are
a perfect set $P\subset X$ and a positive inner measure subset
$\hat H$ of $X$ such that $P\times\hat H\subset H$. 
\end{description}
Thus $\psi_0([0,1])$ is is a restatement of (m). 
We will also use the following 
seemingly stronger variants of $\psi_0(X)$.
\begin{description}
\item[$\psi_1(X)$:] For every full measure
subset $H$ of $X\times X$ there are a perfect set $P\subset X$ and 
a subset $\hat H$ of $X$ of full measure
such that $P\times\hat H\subset H$. 
\item[$\psi_2(X)$:] For a subset $H$ of $X\times X$ 
of positive inner measure 
there are a perfect set $P\subset X$ and 
a positive inner measure subset
$\hat H$ of $X$ such that $P\times\hat H\subset H$. 
\end{description}


\fact{factForClaim}{Let $n=1,2,3,\ldots$. 
\begin{itemize}
\item[{\rm (i)}] If $E\subset\real^n$ has a positive 
Lebesgue measure then
$\rational^n+E=\bigcup_{q\in\rational^n}(q+E)$ has a full
measure.

\item[{\rm (ii)}] $\psi_k(X)$ holds for all $k<3$ and 
$X\in\{[0,1],(0,1),\real,\Cantor\}$. 
\end{itemize}
}

\proof (i) Let $\lambda$ be the Lebesgue measure on $\real^n$ and
for $\e>0$ and $x\in\real^n$ let $B(x,\e)$ be an open ball
in $\real^n$ of radius $\e$ centered at $x$. 
By way of contradiction assume that there exists
a positive measure set 
%$A\subset \real^n\setminus(\rational^n+E)$.
$A\subset \real^n$ disjoint with $\rational^n+E$.
Let $a\in A$ and $x\in E$ be the Lebesgue density points 
of $A$ and $X$, respectively. 
Take an  $\e>0$ such that 
$\lambda(A\cap B(a,\e))>(1-4^{-n})\lambda(B(a,\e))$  
and
$\lambda(E\cap B(x,\e))>(1-4^{-n})\lambda(B(x,\e))$.  
Now, if $q\in\rational^n$ is such that 
$q+x\in B(a,\e/2)$ then 
$A\cap(q+E)\cap B(a,\e/2)\neq\emptyset$ 
since $B(a,\e/2)\subset B(a,\e)\cap B(q+x,\e)$
and thus 
$\lambda(A\cap(q+E)\cap B(a,\e/2))> 
\lambda(B(a,\e/2))-2 \cdot 4^{-n}\lambda(B(a,\e))\geq 0$.
Hence $A\cap(\rational^n+E)\neq\emptyset$ contradicting 
the choice of $A$. 

(ii) First note that 
$\psi_k(\real)\Equi\psi_k((0,1))\Equi\psi_k([0,1])
\Equi\psi_k(\Cantor)$ for every $k<3$. 
This is justified by the fact that for the mappings
$f\colon (0,1)\to\real$ given by $f(x)=\cot(x\pi)$,
the identity mapping $id\colon (0,1)\to[0,1]$,
and $d\colon\Cantor\to[0,1]$ given by 
$d(x)=\sum_{i<\omega}\frac{x(i)}{2^{i+1}}$, 
the image and the preimage of measure zero set 
(full measure set) is of measure zero (full measure).

Since, by (m),
$\psi_0([0,1])$ is true, we have also that 
$\psi_0(X)$ holds also for $X\in\{(0,1),\real,\Cantor\}$.
To finish the proof it is enough to show that 
$\psi_0(\real)$ implies $\psi_1(\real)$
and $\psi_2(\real)$.

To prove $\psi_1(\real)$ let 
$H$ be a full measure subset of $\real\times\real$
and let us define $H_0=\bigcap_{q\in\rational}(\la 0,q\ra+H)$.
Then $H_0$ is still of full measure so, by $\psi_0(\real)$,
there are perfect set $P\subset\real$ and a positive inner measure
subset $\hat H_0$ of $\real$ such that $P\times\hat H_0\subset H_0$. 
Thus, for every $q\in\rational$ we also have
$P\times(q+\hat H_0)\subset \la 0,q\ra+H_0=H_0$.
Let $\hat H=\bigcup_{q\in\rational}(q+\hat H_0)$. 
Then $P\times\hat H\subset H_0\subset H$ and, by (i),
$\hat H$ has full measure. So, $\psi_1(\real)$ is proved. 

To prove $\psi_2(\real)$ let 
$H\subset\real\times\real$ be of positive inner measure.
Decreasing $H$, if necessary, we can assume that $H$ is compact. 
Let $H_0=\rational^2+H$. Then, by (i), 
$H_0$ is of full measure so, by $\psi_0(\real)$,
there are perfect set $P_0\subset\real$ and a positive inner measure
subset $\hat H_0$ of $\real$ such that $P_0\times\hat H_0\subset H_0$. 
Once again, decreasing $P_0$ and $\hat H_0$ if necessary, 
we can assume that they are homeomorphic to $\Cantor$
and that no relatively open subset of $\hat H_0$
has measure zero. 
Since 
$P_0\times\hat H_0\subset\bigcup_{q\in\rational^2}(q+H)$ 
is covered by countably many compact sets 
$(P_0\times\hat H_0)\cap(q+H)$ with $q\in\rational^2$,
there is a $q=\la q_0,q_1\ra\in\rational^2$ such that
$(P_0\times\hat H_0)\cap(q+H)$
has a non-empty interior in $P_0\times\hat H_0$. 
Let $U$ and $V$ be non-empty clopen subsets of $P_0$ 
and $\hat H_0$, respectively, such that
$U\times V\subset (P_0\times\hat H_0)\cap(q+H)
\subset \la q_0,q_1\ra+H$.
Then $U$ and $V$ are perfect and $V$ has positive measure.
Let $P=-q_0+U$ and $\hat H=-q_1+V$.
Then $P\times\hat H=(-q_0+U)\times(-q_1+V)=
-\la q_0,q_1\ra+(U\times V)\subset H$,
so $\psi_2(\real)$ holds. 
\qed

\smallskip 

\noindent{\sc Proof of Claim~\ref{claim1}.} 
Since the natural homeomorphism between 
$\Cantor$ and $\Cantor^{\omega\setminus\{0\}}$ preserves 
product measure, we can identify
$\Cantor^\omega=\Cantor\times\Cantor^{\omega\setminus\{0\}}$ 
with $\Cantor\times\Cantor$ considered 
with its usual topology and its usual product measure.
With this identification the result follows 
easily, by induction
on coordinates, from the following fact.
\begin{itemize}
\item[($\bullet$)] For every Borel subset $H$ of $\Cantor\times\Cantor$
which is of second category (of positive measure) 
there are a perfect set $P\subset\Cantor$ and a second category 
(positive measure) subset $\hat H$ of $\Cantor$ 
such that $P\times \hat H\subset H$. 
\end{itemize}

The measure version of ($\bullet$) is a restatement of
$\psi_2(\Cantor)$, which was proved in Fact~\ref{factForClaim}(ii).
To see the category version of ($\bullet$) 
let $H$ be a Borel subset of $\Cantor\times\Cantor$
of second category. Then there are clopen subsets 
$U$ and $V$ of $\Cantor$ such that 
$H_0=H\cap(U\times V)$ is comeager in $U\times V$.
Since $U$ and $V$ are homeomorphic to $\Cantor$, we can 
apply (c) to $H_0$ and $U\times V$ we can find
a perfect set $P\subset U$ and a comeager Borel subset
$\hat H$ of $V$ 
such that $P\times\hat H\subset H_0\subset H$, finishing the proof.
\qed



\section{$\cf(\NN)=\omega_1<\continuum$}


Next we show that 
\psmC\ implies that $\cof(\NN)=\omega_1$.
So, under \psmC, all cardinals from Cicho\'n's diagram 
(see e.g. \cite{BJ}) are equal to~$\omega_1$. 
The fact that this holds in the iterated perfect set model has been 
well known. 

Let $\C_H$\label{PageCsubH} 
be the family of all subsets $\prod_{n<\omega}T_n$
of $\omega^\omega$ such that $T_n\in[\omega]^{\leq n+1}$ for all $n<\omega$.
We will use the following characterization. 

\prop{propBarto}{{\rm (Bartoszy\'nski~\cite[thm. 2.3.9]{BJ})}
\[
\cof(\NN)=
\min\left\{|\F|\colon \F\subset\C_H\ \&\ \bigcup\F=\omega^\omega\right\}.
\]
}

\lem{lemCofN}{The family 
$\C_H^*=\{X\subset\omega^\omega\colon \ X\subset T\ \mbox{ for some }\ T\in\C_H\}$ is 
$\Fcube$-dense in $\perf(\omega^\omega)$.}

\proof Let $f\colon\Cantor^\omega\to\omega^\omega$ be a continuous 
function. 
By (\ref{eqDefFdenseWeak})
it is enough to find a perfect cube $C$ in $\Cantor^\omega$
such that $f[C]\in\C_H^*$. 

Construct, by induction on $n<\omega$,
the families 
$\{E^i_s\colon s\in 2^n\ \&\ i<\omega\}$ of non-empty clopen 
subsets of $\Cantor$
such that for every $n<\omega$ and $s,t\in 2^n$
\begin{itemize}
\item[(i)]  $E^i_s=E^i_t$ for every $n\leq i<\omega$;
\item[(ii)] $E^i_{s\hat{\ }0}$ and $E^i_{s\hat{\ }1}$
            are disjoint subsets of $E^i_s$ for every $i<n+1$;
\item[(iii)] for every $\la s_i\in 2^n\colon i<\omega\ra$ 
\[
f(x_0)\restriction 2^{(n+1)^2}=f(x_1)\restriction 2^{(n+1)^2}
\ \ \ \mbox{ for every }\ \ \ x_0,x_1\in\prod_{i<\omega}E_{s_i}.
\]
\end{itemize}
For each $i<\omega$ the fusion of 
$\{E^i_s\colon s\in 2^{<\omega}\}$
will give us the $i$-th coordinate set of the desired perfect cube $C$. 

Condition (iii) can be ensured by uniform continuity of $f$. Indeed,
let $\delta>0$ be such that
$f(x_0)\restriction 2^{(n+1)^2}=f(x_1)\restriction 2^{(n+1)^2}$
for every $x_0,x_1\in\Cantor^\omega$ of distance less than $\delta$.
Then it is enough to choose 
$\{E^i_s\colon s\in 2^n\ \&\ i<\omega\}$ such that
(i) and (ii) are satisfied and every set 
$\prod_{i<\omega}E_{s_i}$ from (iii) has diameter less than $\delta$.
This finishes the construction.

Next for every $i,n<\omega$ let 
$E^i_n=\bigcup\{E^i_s\colon s\in 2^n\}$
and $E_n=\prod_{i<\omega}E^i_n$.
Then $C=\bigcap_{n<\omega}E_n=\prod_{i<\omega}\left(\bigcap_{n<\omega}E^i_n\right)$
is a perfect cube, since 
$\bigcap_{n<\omega}E^i_n\in\perf(\Cantor)$ for every $i<\omega$. 
Thus, to finish the proof it is enough to show that 
$f[C]\in\C_H^*$.

So, for every $k<\omega$ let $n<\omega$ be 
such that $2^{n^2}\leq k+1<2^{(n+1)^2}$, 
put  
\[
T_k=\{f(x)(k)\colon x\in E_n\}
=\left\{f(x)(k)\colon 
x\in \prod_{i<\omega}E_{s_i}
\ \mbox{ for some } \la s_i\in 2^n\colon i<\omega\ra\right\},
\]
and notice that $T_k$ has at most $2^{n^2}\leq k+1$ elements. 
Indeed, by (iii), the set 
$\left\{f(x)(k)\colon x\in \prod_{i<\omega}E_{s_i}\right\}$
has precisely one element for every $\la s_i\in 2^n\colon i<\omega\ra$ 
while (i) implies that 
$\left\{\prod_{i<\omega}E_{s_i}\colon \la s_i\in 2^n\colon i<\omega\ra\right\}$
has $2^{n^2}$ elements. 
Therefore $\prod_{k<\omega}T_k\in\C_H$. 

To finish the proof it is enough to notice that 
$f[C]\subset\prod_{k<\omega}T_k$.
\qed

\cor{corCofN}{If \psmC\ holds then $\cof(\NN)=\omega_1$.
}

\proof By \psmC\ and Lemma~\ref{lemCofN} there exists 
an $\F\in[\C_H]^{\leq\omega_1}$ such that 
$|\omega^\omega\setminus \bigcup\F|\leq\omega_1$.
This and Proposition~\ref{propBarto} imply $\cof(\NN)=\omega_1$. \qed

\section{Pointwise convergent of subsequences of real-valued
functions}

A sequence $\la f_n\ra_{n<\omega}$ of real-valued functions is 
{\em uniformly bounded}\/
provided there exists an $r\in\real$ such that $\range(f_n)\subset[-r,r]$
for every $n$. 
In 1932 Mazurkiewicz~\cite{Maz} proved the following variant
of Egorov's theorem.

\begin{quote}
%{\bf Mazurkiewicz' Theorem} 
{\it For every uniformly bounded sequence $\la f_n\ra_{n<\omega}$
of real-valued continuous functions 
defined on a Polish space $X$ 
there exists a 
subsequence which is uniformly convergent on some perfect set $P$.}
\end{quote}

The main result of this section is the following theorem. 


\thm{thmSelNEW}{If \psmC\ holds then
\begin{itemize}
\item[($*$)]  
for every Polish space $X$ and 
uniformly bounded sequence $\la f_n\colon X\to \real\ra_{n<\omega}$ of Borel 
measurable functions there are the sequences:
$\la P_\xi\colon\xi<\omega_1\ra$ of compact subsets of $X$
and $\la W_\xi\in[\omega]^\omega\colon\xi<\omega_1\ra$
such that $X=\bigcup_{\xi<\omega_1}P_\xi$ and for every $\xi<\omega_1$:
\begin{quote}
$\la f_n\restriction P_\xi\ra_{n\in W_\xi}$ is a monotone uniformly 
convergent sequence of uniformly continuous functions.
\end{quote}

\end{itemize}
}
Theorem~\ref{thmSelNEW} is a variant of~\cite[theorem~2]{CP82}
and its corollary, according to which condition 
($*$) for continuous functions $f_n$ can be deduced from
the assumptions that
{\em $\cf(\NN)=\omega_1$ and 
there exists a selective $\omega_1$-generated ultrafilter on $\omega$}.

\medskip

\proof We first note that the family $\E$ of all $P\in\perf(X)$ 
for which there exists a $W\in[\omega]^\omega$ such that 
\[
\mbox{the sequence $\la f_n\restriction P\ra_{n\in W}$ 
is monotone and uniformly convergent}
\]
is $\Fcube$-dense in $\perf(X)$.

Indeed, let $g\in\Fcube$, $g\colon\Cantor^\omega\to X$,  
and consider the functions $h_n=f_n\circ g$.
Since $h=\la h_n\colon n<\omega\ra\colon\Cantor^\omega\to\real^\omega$
is Borel measurable, there is a dense $G_\delta$ subset 
$G$ of $\Cantor^\omega$ such that $h\restriction G$ is continuous.
So, we can find a perfect cube $C\subset G\subset\Cantor^\omega$,
and for this $C$ function $h\restriction C$ is continuous. 
Thus, identifying the coordinate spaces of 
$C$ with $\Cantor$, without loss of generality we can assume that 
$C=\Cantor^\omega$, that is, that each function 
$h_n\colon\Cantor^\omega\to\real$ is continuous.  
Now, by \cite[thm. 6.9]{TF}, there is a perfect cube $C$ in $\Cantor^\omega$
and a $W\in[\omega]^\omega$ such that 
the sequence $\la h_n\restriction C\ra_{n\in W}$ 
is monotone and uniformly convergent.%
\footnote{Actually \cite[thm. 6.9]{TF} is stated for 
functions defined on $[0,1]^\omega$. 
However, the proof presented there for 
works also for functions defined on $\Cantor^\omega$.}
So $P=g[C]$ is in $\E$. 

Now, by \psmC, there exists an 
$\E_0\in[\E]^{\leq\omega_1}$ such that $|X\setminus\bigcup\E_0|\leq\omega_1$.
Then $\{P_\xi\colon\xi<\omega_1\}=
\E_0\cup\left\{\{x\}\colon x\in X\setminus\bigcup\E_0\right\}$ is as desired:
if $P_\xi\in\E_0$ then the existence of an appropriate $W_\xi$
follows from the definition of $\E$. If $P_\xi$ is a singleton, then
the existence of $W_\xi$ follows from 
%Ramsey theorem. 
the fact that every sequence of reals has a monotone subsequence.
\qed

\section{Total failure of Martin's Axiom}

In this section we prove that \psmC\ implies
the total failure of Martin's Axiom, that is, 
the property that 
\begin{quote}
for every non-trivial ccc forcing $\mathP$ there
exists $\omega_1$-many dense sets in $\mathP$ such 
that no filter intersects all of them. 
\end{quote}
The consistency of this fact with
$\continuum>\omega_1$ was first proved by Baumgartner~\cite{Baum}
in a model obtained by adding Sacks reals side-by-side. 
The topological and boolean algebraic formulations
of the theorem follow immediately from
the following proposition. 



\prop{pr:AntiMart}{The following conditions are equivalent. 
\begin{itemize}
\item[{\rm (a)}] For every non-trivial ccc forcing $\mathP$ there
exists $\omega_1$-many dense sets in $\mathP$ such 
that no filter intersects all of them. 
\item[{\rm (b)}] Every compact ccc topological space without isolated
points is a union of $\omega_1$ nowhere dense sets.
\item[{\rm (c)}] For every atomless ccc complete Boolean algebra $B$
there exists $\omega_1$-many dense sets in $B$ such 
that no filter intersects all of them. 
\item[{\rm (d)}] For every atomless ccc complete Boolean algebra $B$
there exists $\omega_1$-many maximal antichains in $B$ such 
that no filter intersects all of them. 
\item[{\rm (e)}] For every countably generated atomless 
ccc complete Boolean algebra $B$
there exists $\omega_1$-many 
maximal antichains in $B$ such 
that no filter intersects all of them. 
\end{itemize}
}

\proof The equivalence of the conditions (a), (b), (c), and (d)
is well known. In particular, equivalence (a)--(c) 
is explicitly given in \cite[thm. 0.1]{Baum}. 
Clearly (d) implies (e). 
The remaining implication, 
(e)$\Implies$(d), is a version of 
the theorem from~\cite[p.~158]{MS}.
However, it is expressed there in a bit different
language, so we include here its proof. 

So, let 
$\la B,\vee,\wedge,{\mathbf 0},{\mathbf 1}\ra$ be 
an atomless ccc complete Boolean algebra.
For every $\sigma\in 2^{<\omega_1}$ define, by induction on the 
length $\dom(\sigma)$ of a sequence $\sigma$, a $b_\sigma\in B$ 
such that the following conditions are satisfied. 

\begin{itemize}
\item $b_\emptyset={\mathbf 1}$. 
\item $b_\sigma$ is a disjoint union of
$b_{\sigma\hat{\ }0}$ and $b_{\sigma\hat{\ }1}$. 
\item If $b_\sigma>{\mathbf 0}$ then
$b_{\sigma\hat{\ }0}>{\mathbf 0}$ and $b_{\sigma\hat{\ }1}>{\mathbf 0}$.

\item If $\lambda=\dom(\sigma)$ 
is a limit ordinal then
$b_\sigma=\bigwedge_{\xi<\lambda}b_{\sigma\restriction\xi}$.
\end{itemize}
%Note that, because of ccc, 
%for every $s\in 2^{\omega_1}$ there is an $\alpha<\omega_1$
%such that $b_{s\restriction\alpha}={\mathbf 0}$.

Let $T=\{s\in 2^{<\omega_1}\colon b_s>{\mathbf 0}\}$. 
Then $T$ is a subtree of $2^{<\omega_1}$;
its levels determine antichains in $B$, so they are countable. 

First assume that $T$ has a countable height. 
Then $T$ itself is countable. Let $B_0$ be the smallest 
complete subalgebra of $B$ containing $\{b_\sigma\colon \sigma\in T\}$
and notice that $B_0$ is atomless.  
Indeed, if there were an atom $a$ in $B_0$
then $S=\{\sigma\in T\colon a\leq b_\sigma\}$ would be a branch 
in $T$ so that $\delta=\bigcup S$ would belong to
$2^{<\omega_1}$. Since $b_\delta\geq a>{\mathbf 0}$, we would also
have $\delta\in T$.
But then 
$a\leq b_\delta=b_{\delta\hat{\ }0}\vee b_{\delta\hat{\ }1}$ 
so  that either 
$\delta\hat{\ }0$ or $\delta\hat{\ }1$ belongs to $S$, which is impossible. 

Thus, $B_0$ is a complete, countably generated, atomless
subalgebra of $B$. So, by (e), there exists a family $\A$ of 
$\omega_1$-many maximal antichains in $B_0$
with no filter in $B_0$ intersecting all of them. 
But then each $A\in\A$ is also a maximal antichain in $B$
and no filter in $B$ would intersect all of them.
So, we have (d).

Next, assume that $T$ has height $\omega_1$ 
and for every $\alpha<\omega_1$ let
\[
T_\alpha=\{\sigma\in T\colon\dom(\sigma)=\alpha\}
\]
be the $\alpha$-th level of $T$. Also let 
$b_\alpha=\bigvee_{\sigma\in T_\alpha}b_{\sigma}$.
Notice that $b_\alpha=b_{\alpha+1}$ for every $\alpha<\omega_1$.
On the other hand, it may happen that 
$b_\lambda>\bigwedge_{\alpha<\lambda}b_\alpha$
for some limit $\lambda<\omega_1$;
however, this may happen only countably many times,
since $B$ is ccc. Thus, there is an $\alpha<\omega_1$
such that $b_\beta=b_\alpha$ for every $\alpha<\beta<\omega_1$. 

Now, let $B_0$ be the smallest 
complete subalgebra of $B$ below ${\mathbf 1}\setminus b_\alpha$
containing $\{b_\sigma\setminus b_\alpha\colon \sigma\in T\}$.
Then $B_0$ is countably generated and, as before, it can be shown that 
$B_0$ is atomless. Thus, there exists 
a family $\A_0$ of 
$\omega_1$-many maximal antichains in $B_0$
with no filter in $B_0$ intersecting all of them. 
Then no filter in $B$ containing ${\mathbf 1}\setminus b_\alpha$
intersects every $A\in\A_0$.
But for every $\alpha<\beta<\omega_1$
the set $A^\beta=\{b_\sigma\colon \sigma\in T_\beta\}$
is a maximal antichain in $B$ below $b_\alpha$. Therefore,
$\A_1=\{A^\beta\colon \alpha<\beta<\omega_1\}$
is an uncountable family of maximal antichains in $B$ below $b_\alpha$
with no filter in $B$ containing $b_\alpha$
intersecting every $A\in\A_1$.
Then it is easy to see that 
the family $\A=\{A_0\cup A_1\colon a_0\in\A_0\ \&\ A_1\in\A_1\}$
is a family of $\omega_1$-many maximal antichains in $B$
with no filter in $B$ intersecting all of them. This proves (d).
\qed 








\thm{th:AntiMart}{\psmC\ implies the total failure of 
Martin's Axiom.
}


\proof Let $\A$ be a countably generated atomless 
ccc complete Boolean algebra
and let $\{A_n\colon n<\omega\}$ generate $\A$.
By Proposition~\ref{pr:AntiMart} it is enough to 
show that $\A$ contains $\omega_1$-many 
maximal antichains such 
that no filter in $\A$ intersects all of them. 

Next let $\B$ be the $\sigma$-algebra of 
Borel subsets of $\Cantor=2^\omega$.
Recall that it is a free countably generated
$\sigma$-algebra, with free generators 
$B_i=\{s\in\Cantor\colon s(i)=0\}$.
Define 
$h_0\colon \{B_n\colon n<\omega\}\to\{A_n\colon n<\omega\}$
by $h_0(B_n)=A_n$ for all $n<\omega$. 
Then $h_0$ can be uniquely extended 
to a $\sigma$-homomorphism $h\colon\B\to\A$ 
between $\sigma$-algebras $\B$ and $\A$. 
(See e.g. \cite[34.1 p. 117]{Sik}.)
Let $\I=\{B\in\B\colon h[B]={\mathbf 0}\}$. 
Then $\I$ is a $\sigma$-ideal in $\B$ 
and the quotient algebra $\B/\I$ is isomorphic to $\A$.
(Compare also Loomis-Sikorski theorem in \cite[p. 117]{Sik} or \cite{Loom}.)
In particular, $\I$ contains all singletons and is ccc,
since $\A$ is atomless and ccc. 

It follows that we need only to consider 
complete Boolean algebras of the form 
$\B/\I$, where $\I$ is some ccc $\sigma$-ideal 
of Borel sets containing all singletons.
To prove that such an algebra has $\omega_1$ maximal 
antichains as desired, it is enough to
prove that
\begin{itemize}
\item[($*$)] $\Cantor$ is a union of $\omega_1$ perfect sets 
$\{N_\xi\colon\xi<\omega_1\}$ which belong to $\I$.
\end{itemize}
Indeed, assume that ($*$) holds and 
for every $\xi<\omega_1$ let $\D_\xi^*$ be a family
of all $B\in\B\setminus\I$ with closures $\cl(B)$ 
disjoint from $N_\xi$. Then $\D_\xi=\{B/\I\colon B\in\D_\xi^*\}$
is dense in $\B/\I$, since $\Cantor\setminus N_\xi$ is $\sigma$-compact
and $\B/\I$ is a $\sigma$-algebra. 
Let $\A_\xi^*\subset\D_\xi^*$ be such that
$\A_\xi=\{B/\I\colon B\in\A_\xi^*\}$ is a maximal antichain
in $\B/\I$. It is enough to show that no filter intersects
all $\A_\xi$'s. But if there were a filter $\F$ in $\B/\I$
intersecting all $\A_\xi$'s
then for every $\xi<\omega_1$ there would exist
a $B_\xi\in\A_\xi^*$ with $B_\xi/\I\in\F\cap\A_\xi$.
Thus, the set $\bigcap_{\xi<\omega_1}\cl(B_\xi)$ would be
non-empty, despite the fact that it is disjoint from
$\bigcup_{\xi<\omega_1}N_\xi=\Cantor$. 

To finish the proof it is enough to show that ($*$) follows from $\psmC$.
But this follows immediately from the fact that 
any cube $P$ in $\Cantor$ contains a subcube
$Q\in\I$ as any cube $P$ can be partitioned into $\continuum$-many
disjoint subcubes and, by the ccc property of $\I$,
only countably many of them can be outside~$\I$. 
\qed 

The authors like to thank professor Anastasis Kamburelis
for many helpful comments 
concerning the results presented here. 
We are especially grateful him for pointing out 
a gap in an earlier version of the proof of Theorem~\ref{th:AntiMart}.


\bigskip

Other consequences of \psmC\ 
can be found in \cite{CP83}, \cite{FJordan2}, 
\cite{Nowik:CompRamsey}, and in the monograph in preparation~\cite{CPAbook}. 





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\end{document}
