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\documentclass{rae}
\usepackage{amsmath,amsthm,amssymb}
%\coverauthor{Stewart Baldwin and Jack Brown}
%\covertitle{A Simple Proof that $(s)/(s^0)$ is a Complete Boolean Algebra}
\received{August 24, 1998}
\MathReviews{Primary 28A05: Secondary 03G05}
\keywords{Complete Boolean Algebra, Marczewski Measurable}
\firstpagenumber{1}
\markboth{Stewart Baldwin and Jack Brown}{$(s)/(s^0)$ is a Complete Boolean Algebra}
\author{Stewart Baldwin and Jack Brown, Department of Mathematics, Auburn University, AL
36849-5310,
e-mail: {\tt baldwsl@mail.auburn.edu} and {\tt brownj4@mail.auburn.edu}}
\title{A SIMPLE PROOF THAT $(s)/(s^0)$ IS A COMPLETE BOOLEAN ALGEBRA}
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\begin{document}
\maketitle
\begin{abstract}
Let $X$ be a complete separable metric space, let $(s)$ be the set
of all Marczewski \cite{sm} measurable subsets of $X$, and let $(s^0)$ be
the the
set of all Marczewski null subsets of $X$. It is already known that $(s)/(s^0)$
is a complete Boolean algebra, but the known proofs of this involve
complicated preliminaries. We present a simple proof that $(s)/(s^0)$
is a complete Boolean algebra.
\end{abstract}
\bigskip
Let $X$ and $Y$ be complete, separable, metric spaces. In \cite{s2}, Kenneth
Schilling constructed a $(s)$-set in $X \times Y$ which was not in the
$\sigma$-algebra generated by sets of the form $A \times B$, where
$A$ and $B$ are $(s)$-sets in $X$ and $Y$ respectively. The proof used
the fact that $(s)/(s^0)$ is a complete Boolean Algebra, but the two
proofs \cite{m},\cite{s1} of this fact which were described in \cite{s2}
both involve complicated
preliminaries (see \cite{s2} for more details). The present paper gives a
simple,
direct proof that $(s)/(s^0)$ is a complete Boolean Algebra.
First, we give the basic definitions.
\bigskip
\noindent
{\bf Definition 1.}
Let $X$ be a complete separable metric space. A set
$A \subseteq X$ is said to be {\em Marczewski measurable}, or an $(s)$-set
(written $A \in (s)$)
iff for every perfect set $P \subseteq X$, there is a perfect set
$Q \subseteq P$ such that either $Q \subseteq A$ or $Q \cap A = \emptyset$.
$A \subseteq X$ is said to be {\em Marczewski null}, or an $(s^0)$-set
(written $A \in (s^0)$)
iff for every perfect set $P \subseteq X$, there is a perfect set
$Q \subseteq P$ such that $Q \cap A = \emptyset$ (or, equivalently,
$A \in (s^0)$ iff $A \in (s)$ and $A$ does not contain a perfect subset).
Rather than introduce the basic definitions from Boolean Algebra, we
shall translate the statement that $(s)/(s^0)$ is a complete Boolean
Algebra into a statement which is more convenient for our purposes.
Thus, define an equivalence relation $\sim$ on $(s)$
by $Y \sim Z$ iff $Y \Delta Z \in (s^0)$, where
$Y \Delta Z = (Y \cup Z) \setminus (Y \cap Z)$ is the symmetric
difference of $Y$ and $Z$. Define the relation $\leq$ on $(s)$ (which we may
think of as ``almost'' a subset of, mod $(s^0)$), by $Y \leq Z$ iff
$Y \setminus Z \in (s^0)$, and note that $Y \sim Z$ iff $Y \leq Z$ and
$Z \leq Y$. We could form the Boolean Algebra $(s)/(s^0)$ by going to
equivalence classes, but it will be more convenient to avoid the
additional layer of notation which that would entail. The terms ``upper
bound'', ``lower bound'', ``least upper bound'', and ``greatest lower
bound'' (with respect to the relation $\leq$) are then defined in the
obvious way, with l.u.b.'s (and g.l.b.'s) not
necessarily being unique. However, it is easy to see that l.u.b.'s and
g.l.b.'s are unique modulo the equivalence relation $\sim$ (if they
exist). The statement that $(s)/(s^0)$ is a complete Boolean Algebra
is then easily seen to be
equivalent to the statement that ``Every subset of $(s)$ has a
least upper bound in $(s)$ with respect to the relation $\leq$.''
(This also implies the existence of g.l.b.'s.)
%\end{defi}
The proof given here was motivated by a similar result due to John Walsh
\cite{w}. If ${\cal A}$ is a $\sigma$-algebra of subsets of some set $X$, and
${\cal I} \subseteq {\cal A}$ is a $\sigma$-ideal, then we say that the
pair $({\cal A}, {\cal I})$ has the {\em hull property} iff whenever
$U \subseteq X$ there is a $V \in {\cal A}$ such that $U \subseteq V$,
and if $W \in {\cal A}$ such that $U \subseteq W$, then $V \setminus W
\in {\cal I}$. Walsh's result was that $((s),(s^0))$ has the hull
property. In fact, the argument given below turns out to be somewhat simpler
than Walsh's corresponding argument for the hull property. The reason
for this
is that, in the proof below, every element of ${\cal B}$ is an
$(s)$-set. In the hull property proof given in \cite{w}, $U$ is not
necessarily an
$(s)$-set, and the set corresponding to ${\cal C}$ in that proof is
necessarily more complicated.
\begin{thm}
Let $X$ be a complete separable metric space, and let
$(s)$ and $(s^0)$ be as above. Then $(s)/(s^0)$ is a complete Boolean algebra.
\end{thm}
\begin{proof}[{\sc Proof.}]
Let ${\cal B} \subseteq (s)$,
and let ${\cal P}$ be the set of all perfect subsets of $X$.
Let ${\cal C}=\{C \in {\cal P}$: For some $B \in {\cal B},
C \subseteq B\}$,
and ${\cal D}=\{D \in {\cal P}$: For every $B \in {\cal B},
D \cap B \in (s^0)\}$.
Observe that, if $C \in {\cal C}$ and $D \in {\cal D}$, then $C \cap D$
is a closed set which cannot contain a perfect subset, and is therefore
countable.
Without loss of generality, we may assume that ${\cal C}$ and ${\cal D}$
are both nonempty, since otherwise either $X$ or $\emptyset$ would be a
least upper bound of ${\cal B}$. Thus ${\cal C}$ and ${\cal D}$ both
have cardinality ${\c}$ (the cardinality of the continuum), and we
may enumerate ${\cal C}=\{C_{\alpha}: \alpha < {\c}\}$ and
${\cal D}=\{D_{\alpha}: \alpha < {\c}\}$.
Let $L=\bigcup_{\alpha < {\c}}(C_{\alpha} \setminus \bigcup_{\beta <
\alpha}D_{\beta})$, and we claim that $L$ is the desired least upper
bound. First, to show that $L \in (s)$, let $P \in {\cal P}$. There are
then two cases.
Case 1: $P \in {\cal D}$.
Then $P=D_{\alpha}$ for some $\alpha < {\c}$.
Thus $P \cap L = D_{\alpha} \cap L \subseteq
D_{\alpha} \cap \bigcup_{\beta \leq \alpha}C_{\beta}$, which has
cardinality less than ${\c}$. Thus, there is a perfect set
$Q \subseteq P$ such that $Q \cap L = \emptyset$. Thus $L \in (s)$.
Case 2: $P \notin {\cal D}$.
Then there is a $B \in {\cal B}$ such that $P \cap B \notin (s^0)$,
and if we pick a perfect subset $P'$ of $P \cap B$,
then $P' \in {\cal C}$. Then $P' = C_{\alpha}$ for some
$\alpha < {\c}$.
Then $C_{\alpha} \setminus \bigcup_{\beta < \alpha}D_{\beta}
\subseteq L$, and since $C_{\alpha} \cap \bigcup_{\beta < \alpha}D_{\beta}$
has cardinality less than ${\c}$, $L \cap P'$ contains a perfect
subset, and $L \in (s)$.
To see that $L$ is an upper bound of ${\cal B}$, let $B \in {\cal B}$.
Let $P$ be a perfect subset of $B$. Then $P=C_{\alpha}$ for some
$\alpha < {\c}$. Since $C_{\alpha} \cap \bigcup_{\beta <
\alpha}D_{\beta}$ has cardinality less than ${\c}$, we see that
$P \cap L \neq \emptyset$, and thus, since $P$ was arbitrary,
$B \setminus L$ does not contain a perfect subset. Thus, since
$B \setminus L \in (s)$, $B \setminus L \in (s^0)$, and $B \leq L$.
Thus, $L$ is an upper bound of ${\cal B}$.
Now, let $U \in (s)$ be any upper bound of ${\cal B}$. We need to show that
$L \leq U$. Let $P$ be any perfect subset of $L$. Then $P$ cannot be
equal to $D_{\alpha}$ for any $\alpha < {\c}$, for the construction of
$L$ guarantees that $L$ intersects $D_{\alpha}$ in fewer than ${\c}$
points. Thus, $P \notin {\cal D}$, so there is a $B \in {\cal B}$ such
that $P \cap B \notin (s^0)$. Thus, since $P \cap B \in (s)$,
$P \cap B$ must contain a perfect subset $Q$. Since $U$ is an upper
bound of ${\cal B}$, $B \leq U$, and therefore
$Q \cap U \neq \emptyset$. Thus $P \cap U \neq \emptyset$, and since $P$
was an arbitrary perfect subset of $L$, $L \setminus U \notin (s^0)$.
Thus, since $L \setminus U \in (s)$, $L \leq U$, and we are done.
\end{proof}
The similarity between the above proof and Walsh's proof for the hull
property motivates the following obvious question, for which there
appears to be no obvious answer:
\bigskip
{\bf Question:} If ${\cal A}$ is a $\sigma$-algebra of subsets of some
set $X$, and ${\cal I} \subseteq {\cal A}$ is a $\sigma$-ideal, then
does either of the statements ``${\cal A}/{\cal I}$ is a complete Boolean
Algebra'' and ``$({\cal A},{\cal I})$ has the hull property'' imply the
other?
\bigskip
\begin{thebibliography}{999999}
\bibitem{m}John Morgan II, {\it Point Set Theory}, Marcel Dekker, New York,
1990.
\bibitem{w} J. T. Walsh, ``Marczewski Sets, Measure, and the Baire Property,
II,'' {\it Proc. of the Amer. Math. Soc.}
{\bf 106} (1989), 1027-1030.
\bibitem{s1} K. Schilling, ``Some category bases which are equivalent to
topologies", {\it Real Analysis Exchange} {\bf 14} (1988-89), 210-214.
\bibitem{s2} \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_,
``A Tale of Two (s)-ities,'' to appear in {\it Real Analysis Exchange}.
\bibitem{sm} E. Szprilrajn (Marczewski), ``
Sur une classe de fonction de M. Sierpinski et la classe correspondante
d'ensembles", {\it Fund. Math.} {\bf 24} (1935), 17-34.
\end{thebibliography}
\end{document}
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{A Simple Proof that $(s)/(s^0)$ is a Complete Boolean Algebra}
{Stewart Baldwin and Jack Brown}
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