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%Date: Thu, 7 May 1998 14:22:44 EDT
%From: "Dr. Kenneth Schilling"
%% Abstract requested 5/22/98
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% last edit:jch 22/10/98
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% Inroads Section from Brown
\documentclass{rae}
\usepackage{amsmath,amsthm,amssymb}
%\coverauthor{Kenneth Schilling}
%\covertitle{A Tale of Two $(s)$-ities}
\received{March 19, 1998}
\MathReviews{Primary: 28A05}
\firstpagenumber{1}
\markboth{Kenneth Schilling}{A Tale of Two $(s)$-ities}
\author{Kenneth Schilling, Mathematics Department,
University of Michigan-Flint}
\title{A TALE OF TWO $(s)$-ITIES}
%%%%%%Put Author's Definitions Below Here %%%%%%%%%%%
\def\pf{\noindent{\sc Proof.} }
\newcommand{\allconsecutive}
{\newtheorem{theor}{Theorem}
\newtheorem{pr}[theor]{Proposition}
\newtheorem{lem}[theor]{Lemma}
\newtheorem{co}[theor]{Corollary}
}
\newcounter{ite}
\def\begown#1{\medskip\noindent{\bf #1 }}
\def\endown{\par\rm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\maketitle
\begin{abstract}
In the product $X\times Y$ of two uncountable complete
separable metric spaces, not every $(s)$-set belongs to the
$\sigma$-algebra generated by the products of $(s)$-sets in $X$ with
$(s)$-sets in~$Y$. The construction makes use of the fact that the
Boolean algebra $(s)/(s_0)$ is complete.
\end{abstract}
\bigskip
\allconsecutive
\begin{center}
{\small It was the best of $\times$, it was the worst of $\times$. }
\end{center}
Let $X$ and $Y$ be complete, separable metric spaces.
Jack Brown posed the
following question: Does every $(s)$-set in $X\times Y$
belong to
the $\sigma$-algebra generated by all sets of the form $A\times B$,
where $A\subset X$ and $B\subset Y$ are\break $(s)$-sets?
This question is particularly natural in the light of
the recent result of Elalaoui-Talibi \cite{1}
that the graph of an $(s)$-measurable function
$f:X\rightarrow Y$ does belong to this $\sigma$-algebra.
However, the answer to Brown's question is ``no.''
The construction
and proof are motivated by
John Morgan's theory of Category Bases, as we discuss below.
Thanks to Jack Brown for suggesting the problem and drastically
simplifying
the solution.
\bigskip
A set $M$ in a complete separable metric space is said to be have
property
$(s)$ if every perfect set has a perfect subset which is either a
subset of $M$
or is disjoint from $M$. The set $M$ has property $(s_0)$ if every
perfect
set has a perfect subset which is disjoint from $M$. These notions
were
introduced by Szpilrajn-Marczewski in~\cite{7},
where it is proved, among many other things,
that the class of sets having property $(s)$ is a $\sigma$-algebra,
and the
class of sets having property $(s_0)$ is a $\sigma$-ideal.
Now let $Z$ be a complete, separable metric space. The Boolean
algebra ${\cal
S}(Z)$ of $(s)$-sets modulo the $(s_0)$-sets is defined in the usual
manner;
for each set $M\subset Z$ with property $(s)$, $\left[ M\right]$,
the equivalence class
of $M$ `mod $(s_0)$,' is an element of ${\cal S}(Z)$.
It is clear from the facts cited above
that ${\cal S}(Z)$ is a countably complete Boolean
algebra. In fact, more is true;
$$ {\cal S}(Z) \mbox{ is a complete Boolean
algebra.} \eqno\mbox{(cBa)} $$
There are two proofs of (cBa) in the
literature. In \cite{6}, it is proved that there is a topology $\cal T$
on $Z$
such that the $\cal T$-Baire property on $Z$ coincides with property
$(s)$,
and the class of $\cal T$-meager (= first category) sets in $Z$
coincides with
the class of sets with property $(s_0)$.
Thus (cBa) follows from the classical Birkhoff-Ulam
theorem (see \cite[p.~75]{5}): In any topological space, ``Baire property
modulo
meager'' is a complete Boolean algebra. A second proof, using the
general
notion of
`category bases,' is found in \cite[p.~37]{4}.
We now use this result to show that the Fubini property fails
spectacularly
for the properties $(s)$ and $(s_0)$. Let $X$ and $Y$ be complete
separable
metric spaces, and consider the space $X\times Y$.
To fix notation, for every $M\subset X\times Y$, $x\in X$ and $y\in
Y$, the
cross-sections of $M$ are defined by $M_x=M\cap (\{x\}\times Y)$ and
$M^y=M\cap (X\times \{y\})$.
\begin{pr} \label{null-F} Let $\left[ \hat{M} \right] =
\Sigma_{y\in Y} \left[ X\times \{y\} \right]$.
(This infinite join is computed in the complete Boolean algebra
${\cal
S}(X\times Y)$). Then $\hat{M}$ has property $(s)$, and
\begin{list}{(\roman{ite})}{\usecounter{ite}}
\item for all $y\in Y$, the set $(X\times \{y\}) \setminus \hat{M}^y$
has
property $(s_0)$, and
\item for all $x\in X$, $\hat{M}_x$ has property $(s_0)$.
\end{list}
\end{pr}
In other words, in the sense of properties $(s)$ and $(s_0)$,
the cross-sections $\hat{M}^y$
are all nearly full subsets of $(X\times \{y\})$, while
the cross-sections $\hat{M}_x$
are all negligible subsets of $\{x\}\times Y$. (The
best of $\times$, the worst of $\times$??)
\pf To prove \ref{null-F}(i), fix $y\in Y$.
Note that by definition of join, $\left[ \hat{M}\right]
\geq \left[ X\times \{y\} \right]$,
so $\left[ \hat{M}^y \right] = \left[ \hat{M} \cap (X\times
\{y\})
\right]
= \left[ X\times \{y\} \right]$. To prove (ii), fix $x\in X$.
Note that for all
$y\in Y$, the set $(\{x\}\times Y) \cap (X\times \{y\})$ is a
singleton and
therefore has property $(s_0)$. Now $\left[ \hat{M}_x \right]
= \left[ (\{x\}\times Y)\cap \hat{M} \right] =
\Sigma_{y\in Y} \left[ (\{x\}\times Y) \cap (X\times \{y\})
\right]$,
and (ii) follows.\qed
\medskip
Another characterization of $\hat{M}$ is
\begin{list}{(\roman{ite})}{\usecounter{ite}}
\item For every horizontal cross-section $H$ of $X\times Y$,
$H\setminus \hat{M}$ has property $(s_0)$, and
\item For any $N \subset X\times Y$, if for every horizontal
cross-section $H$ of $X\times Y$,
$H\setminus N$ has property $(s_0)$, then $ \hat{M}\setminus N$ has
$(s_0)$.
\end{list}
Let $\sigma(s(X)\times s(Y))$ denote the smallest $\sigma$-algebra
containing all sets $A\times B$ where $A\subset X$ and $B\subset Y$
have property $(s)$.
Our main goal here is to prove that the set $(s)$-set $\hat{M}$
does not belong to $\sigma(s(X)\times s(Y))$.
To this
end, we introduce two new properties, $(s^2)$ and $(s_0^2)$.
Define a {\em perfect rectangle} to be a set of the form $P\times
Q$,
where $P\subset X$ and $Q\subset Y$ are perfect sets.
We say that $M\subset X\times Y$ has property $(s^2)$
if every perfect rectangle has a subset which is a perfect rectangle
and is either a subset of $M$ or is disjoint from $M$.
We say that $M$ has property $(s^2_0)$
if every perfect rectangle has a subset which is a perfect rectangle
and is disjoint from~$M$.
\begin{pr}\label{Brown}
\leavevmode
\begin{list}{(\roman{ite})}{\usecounter{ite}\topsep0pt\itemsep0.2truecm\parsep0pt
\labelwidth0.8truecm
\leftmargin1truecm
\labelsep0.2truecm
}
\item If $A\subset X$ and $B\subset Y$ have property $(s)$,
then $A\times B$ has property $(s^2)$.\par
\item The class of $(s_0^2)$-sets is a $\sigma$-ideal on $X\times Y$.
\item The class of $(s^2)$-sets is a $\sigma$-algebra on $X\times
Y$.
\end{list}
\end{pr}
\pf For (i), let $P\times Q$ be a perfect rectangle. If $P\cap A$ and $Q
\cap B$
contain perfect sets $P',Q'$, respectively, then $(P\times Q)\cap
(A\times
B)\supset P'\times Q'$. On the other hand, if $P\setminus A$
contains the
perfect set $P'$ (or $Q \setminus B$ contains the perfect set $Q'$),
then
$(P\times Q)\setminus (A\times B) \supset (P'\times Q)$ (or
$(P\times
Q)\setminus (A\times B) \supset (P\times Q')$, respectively).
For (ii), we use a standard ``dyadic schema'' argument.
Let $A_1, A_2, \ldots$ have property $(s^2_0)$, and let $P\times Q$
be a perfect rectangle. We build recursively two dyadic schemata
$\{P_{i_1,\ldots,i_n}\}$ and $\{Q_{i_1,\ldots,i_n}\}$
of perfect sets, such that, for all binary sequences
$ (i_1,\ldots,i_n)$,
\begin{align*}
P_{i_1,\ldots,i_{n}} & \subset P_{i_1,\ldots,i_{n-1}}\subset X , \
\ \
Q_{i_1,\ldots,i_{n}}\subset Q_{i_1,\ldots,i_{n-1}}\subset Y ,
\tag1\\
& P_{i_1,\ldots,i_{n}} \mbox{ and } Q_{i_1,\ldots,i_{n}} \mbox{
have diameter }<1/n, \\
P_{i_1,\ldots,i_{n},0} & \cap P_{i_1,\ldots,i_{n},1} =
Q_{i_1,\ldots,i_{n},0} \cap Q_{i_1,\ldots,i_{n},1} = \emptyset , \mbox{ and }
\\
& (P_{i_1,\ldots,i_{n}}\times Q_{i_1,\ldots,i_{n}})
\cap A_n = \emptyset. \end{align*}
Indeed, let $P_{\emptyset}=P$, $Q_{\emptyset}=Q$. Now suppose that
the perfect sets
$P_{i_1,\ldots,i_{n-1}}$ and $Q_{i_1,\ldots,i_{n-1}}$ have been
constructed
so as to satisfy (1). By hypothesis, there exist $P^*\subset
P_{i_1,\ldots,i_{n-1}}$ and
$Q^*\subset Q_{i_1,\ldots,i_{n-1}}$ such that $(P^*\times Q^*)\cap
A_n =
\emptyset$. Finally, choose $P_{i_1,\ldots,i_{n-1},0}$ and
$P_{i_1,\ldots,i_{n-
1},1}$ to be disjoint perfect subsets of $P^*$ each of diameter $<
1/n$, and
$Q_{i_1,\ldots,i_{n-1},0}$ and $Q_{i_1,\ldots,i_{n-1},1}$ to be
disjoint
perfect subsets of $Q^*$ each of diameter $< 1/n$. Then
$P_{i_1,\ldots,i_{n-1},i_n}$ and $Q_{i_1,\ldots,i_{n-1},i_n}$
satisfy (1), as well.
Let ${\displaystyle P' = \bigcap_n \bigcup_{i_1,\ldots,i_n}
P_{i_1,\ldots,i_{n}} }$ and ${\displaystyle Q'= \bigcap_n
\bigcup_{i_1,\ldots,i_n}
Q_{i_1,\ldots,i_{n}} }$. It is now a routine matter to verify that
$(P'\times Q')\subset (P\times Q)$ is a perfect rectangle that does
not
intersect~$\bigcup_n A_n$.
For (iii), it is clear from the symmetry of the definition that the
complement
of an $(s^2)$-set is another $(s^2)$-set. To show that $(s^2)$ is
closed
under countable unions,
let $A_1, A_2, \ldots$ have property $(s^2)$, and let $P\times Q$ be
a
perfect rectangle. If for some $n$ $A_n\cap (P\times Q)$ has a subset
which is
a perfect rectangle, then so does $(\bigcup_n A_n) \cap (P\times Q)$.
If not,
it is not hard to show that for all $n$, $A_n\cap (P\times Q)$ has
property
$(s^2_0)$, and so by part (ii) $(\bigcup_n A_n)\cap (P\times Q)$ has
property
$(s^2_0)$. Thus there exists a perfect rectangle $P'\times Q' \subset
(P\times
Q) \setminus (\bigcup_n A_n)$, and we are done.
\begin{co} The set $\hat{M}$ has property $(s)$, but is not an
element of
the\break $\sigma$-algebra $\sigma(s(X)\times s(Y))$.
\end{co}
Indeed, by \ref{null-F}(i) no perfect rectangle misses $\hat{M}$,
and by \ref{null-F}(ii), no perfect rectangle is a subset of
$\hat{M}$.
Thus $\hat{M}$ does not have property $(s^2)$. However, \ref{Brown}~implies
that the class of $(s^2)$-sets contains the $\sigma$-algebra
in question. We are done.\qed
\begown{Remark.}
The construction of the set $\hat{M}$ depends on the
result (cBa), whose proof belongs to John Morgan's theory of
Category Bases. (An introduction to the theory may be found in~\cite{2},
or
in~\cite{4}. The proof of (cBa) is found only in~\cite{4}.) A {\em category
base}
is a pair $(X,{\cal C})$ in which $\cal C$ is a class of subsets of~$X$,
satisfying axioms that guarantee that the classes of $\cal C$-Baire
property
sets and $\cal C$-meager sets (which are defined in the natural way)
are well-behaved.
For example, every topological space $(X,{\cal C})$ is a category
base,
and the classes
of $\cal C$-Baire property sets and $\cal C$-meager sets are
exactly
as usual.
The fact (cBa) and therefore the construction of $\hat{M}$ depend on
the
following facts
(also proved in~\cite{2} and~\cite{4}):
If $Z$ is a complete separable metric space and $\cal P$ is
the class of perfect sets in $Z$, then $(Z,{\cal P})$ is a category
base, the $\cal P$-Baire property coincides with property $(s)$,
and the $\cal P$-meager sets are exactly the sets with property
$(s_0)$.
We implicitly considered here another category base here. If $X$ and
$Y$
are complete separable metric spaces and ${\cal P}^2$ the class
of perfect rectangles in $X\times Y$, then $(X\times Y,{\cal P}^2)$
is a category base. (This fact is proved in~\cite{3}.) The present
Proposition~\ref{Brown} shows, in the language of category bases, that
property $(s^2)$ coincides with the ${\cal P}^2$-Baire property,
and the sets with property $(s^2_0)$ are exactly
the ${\cal P}^2$-meager sets.
\begin{thebibliography}{9}
\bibitem{1}
H.~Elalaoui-Talibi, {\em A remark on graphs of Marczewski
measurable
functions}, to appear in Real Analysis Exchange.
\bibitem{2} J. Morgan, {\em Baire category from an abstract viewpoint},
Fund. Math. {\bf 94} (1977), 13--23.
\bibitem{3} J. Morgan, {\em On product bases}, Pac. J.
Math.~{\bf99(1)} (1982), 105--126.
\bibitem{4} J. Morgan, {\em Point set theory}, M. Dekker, New York, 1990.
\bibitem{5} R. Sikorski, {\em Boolean algebras}, Springer-Verlag, N.Y.,
1969.
%?%
\bibitem{6} K. Schilling, {\em Some category bases which are equivalent
to topologies}, Real Analysis Exchange {\bf 14(1)}, 210--214.
\bibitem{7} E. Szprilrajn, {\em Sur une classe de fonction de M. Sierpinski
et la classe correspondante d'ensembles}, Fund. Math. {\bf 24} (1935),
17--34.
\end{thebibliography}
\end{document}