% Final minor changes made 1/18/02 (according to the referee's suggestions)

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\markboth{A. Bartoszewicz and K. Ciesielski}{MB-representations and topological algebras}

\title%[MB-representations and topological algebras]
{MB-representations and topological algebras}


\author{
Artur Bartoszewicz, Institute of Mathematics, {\L}\'od\'z Technical University,
Al. Politechniki 11, 90-924 {\L}\'od\'z, Poland  (arturbar@ck-sg.p.lodz.pl)
\and
Krzysztof Ciesielski%
\thanks{The second author was partially supported by NATO Grant PST.CLG.977652.
\endgraf
Papers authored or
co-authored by a Contributing Editor are managed
by a Managing Editor or one of the other Contributing Editors.
}%
,
Department of Mathematics, West Virginia
University, Morgantown, WV 26506-6310, USA
(K\_Cies@math.wvu.edu)
{\tt http://www.math.wvu.edu/\~{}kcies}
}





\date{January 18, 2002}

\MathReviews{Primary 54E52; Secondary 06E25, 03E35}
\keywords{Generalized Marczewski's sets, topology.}

\newcommand{\F}{{\mathcal F}}
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%[MB-representations and topological algebras]
%{MB-representations and topological algebras}
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%\author[A.Bartoszewicz, K.Ciesielski]{Artur Bartoszewicz, Krzysztof Ciesielski}
%\address{Technical University of Lodz,
%         Institute of Mathemathics,
%         Al. Politechniki 11,
%         90-924 Lodz,
%         Poland}


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\begin{document}
\maketitle
\begin{abstract}
For an algebra $\A$ and an ideal $\I$ of subsets of a set $X$ we consider pairs
$\la \A,\I\ra$ which have  the common inner Marczewski-Burstin
representation. The main goal of the paper is to 
investigate which inner Marczewski-Burstin representable algebras and pairs are topological.
\end{abstract}

\section{Introduction}
Let $X$ be a nonempty set and let  $\mathcal F$ be a nonempty family of nonempty subsets
of $X$.
Following the idea of Burstin and Marczewski we define:
$$ 
S(\F)=\{A \subset X\colon (\forall P \in\F) (\exists Q \in \F)
(Q \subset A \cap P \mbox{ or } Q \subset P \setminus A) \}
$$
and
$$ S_0(\mathcal F )= \{ A \subset X\colon  (\forall P \in \mathcal F)(\exists Q \in
\mathcal F ) (Q \subset P \setminus A ) \}.
$$
Then $S(\mathcal F)$ is an algebra of subsets of $X$ 
and $S_0(\mathcal F)$ is an ideal on $X$. 
(See \cite{BBRW}. In this paper family  $S_0(\F)$ is denoted by $S^0(\F)$.)
Burstin \cite{Bu}
showed that if we take as $\mathcal F$ the family of perfect subsets of $\mathbb R$ with
a positive Lebesgue measure then $S(\F)$ equals to the $\sigma$-algebra of
measurable sets and $S_0(\F)$ is the ideal of null sets. 
On the other hand, if $\mathcal F$ is the family of all perfect subsets of $\mathbb R$ then
$S(\F)$ and $S_0(\F)$ become Marczewski's $\sigma$-algebra and 
Marczewski's $\sigma$-ideal, which are closely related to a class of  
Sierpi\'nski functions~\cite{Ma}.

We say that an algebra $\A$ (an ideal $\I$) of subsets of $X$ 
{\em has a Marczewski-Burstin representation}\/ 
if there exists a nonempty family $\F$ of nonempty subsets of $X$ 
such that $\A = S(\F)$ ($\I = S_0(\F)$,   respectively).
If in addition $\F \subset\A$ then we say that $\A$ {\em is inner MB-representable}.
For $\I\subset\A$ we say that the pair $\la\A,\I\ra$
{\em is MB-representable}\/ provided $\la\A,\I\ra=\la S(\F),S_0(\F)\ra$
for some family  $\F$. 
If in addition $\F\subset\A$ then we say that $\la\A,\I\ra$
{\em is inner MB-representable}. 
MB-representations of algebras and ideals were recently considered in 
the papers~\cite{Re,BR,BET,BBRW,BBC}. 
In the first three of these papers a family $\F$ was always chosen from ``nice'' sets: Borel 
or perfect with respect to some topology; 
papers  \cite{BBRW,BBC} contain the
systematic studies of MB-representations for quite arbitrary families $\F$.

We say that the pair $\la\A,\I\ra$ (an algebra $\A$, or an ideal $\I$)  
{\em is topological}\/ provided  there exists a topology $\tau$ on $X$
such that $\la\A,\I\ra=\la S(\F),S_0(\F)\ra$
($\A=S(\F)$, or $\I=S_0(\F)$, respectively), where $\F=\tau\setminus\{\emptyset\}$.
It was noticed in~\cite[prop.~1.3]{BBRW} that 
$\I=S_0(\tau\setminus\{\emptyset\})$ is equal to the ideal $NWD(\tau)$
of $\tau$-nowhere dense sets, while $\A=S(\tau\setminus\{\emptyset\})$ is the algebra
of subsets of $X$ with nowhere dense boundary. 
Clearly every topological pair $\la\A,\I\ra$ 
is inner MB-representable. The main question 
we investigate in this note is whether the converse is also true, that is,
more precisely 
\begin{quotation}
{\em Which inner MB-representable pairs $\la\A,\I\ra$ are topological?}
\end{quotation}

%\fact{fact1}{
%$\mathcal F \subset S(\mathcal F ) \iff (\forall U,V \in \mathcal F )
%(\exists W \in \mathcal F ) (W \subset U \cap V \; or \; W \subset U \setminus V)$.}

We say that the families $\F_1$ and $\F_2$ of subsets of $X$ are {\em mutually
coinitial}\/ provided 
\[
( \forall U \in \mathcal F_1 ) (\exists V \in \mathcal F_2  )(V \subset U )
\ \ \mbox{ and }\ \ 
(\forall U \in \mathcal F_2 )(\exists V \in \mathcal F_1 )(V \subset U).
\]
We will need the following facts from~\cite{BBRW}.

\fact{fact2}{If families $\F_1$ and $\F_2$ are mutually coinitial then
%$\la S(\F_1),S_0(\F_1)\ra=\la S(\F_2),S_0(\F_2)\ra$.
$S(\F_1)=S(\F_2)$ and $S_0(\F_1) = S_0(\F_2)$.
}

\fact{fact3}{If $\F_1\subset S(\F_1)$, 
$\F_2 \subset S(\F_2)$ and $\la S(\F_1),S_0(\F_1)\ra=\la S(\F_2),S_0(\F_2)\ra$
then $\F_1$ and $\F_2$ are mutually coinitial.
}

Since topological algebras are always inner MB-representable, the problem:
{\em is a given pair $\la S(\F), S_0(\F)\ra $, with 
$\F\subset S(\F)$, topological?}
is equivalent to:
{\em is $\F$ mutually coinitial with some topology
(or with a base of some topology) on $X$?}\/
If we consider only an inner MB-representable algebra $S(\F)$, the problem
{\em is $S(\F)$ topological}\/ cannot be formulated in these terms:
the ideals $S_0(\F)$ and $S_0(\tau \setminus \{\emptyset\})$ can be quite
different and so $\F$ and $\tau\setminus\{\emptyset\}$ need not be mutually coinitial.
On the other hand, any ideal $\mathcal I$ is the ideal of nowhere dense sets in some
topology (see~\cite{CJ}), so in our terms any ideal of sets is topological.


\section{The results}

We use the standard set theoretic notation as in~\cite{Ci}.
  
\thm{thm1}{Let $|X|=\kappa\geq\omega$ and $\I$ be a proper ideal of
subsets of $X$ such that $\I\subset [X]^{<\kappa}$. 
If \/ $\bigcup\I= X$ then the pair $\la\P(X),\I\ra$ is inner MB-representable but is not
topological.
}


\proof   To see that $\la\P(X),\I\ra$ is inner MB-representable put 
$\F=\P(X)\setminus\I$ and notice that 
$S_0(\F)=\I$ and $S(\F)=\P(X)$. (It is true for any proper ideal~$I$.)

To see that $\la\P(X),\I\ra$ is not topological suppose,
by way of contradiction, that for some topology $\tau$
we have $\I=S_0(\tau \setminus\{\emptyset\})$
and $\P(X)=S(\tau \setminus \{\emptyset\})$.
Consider a family $\{A_{\alpha}\colon\alpha < \kappa\} \subset\P(X)$
of disjoint sets such that $|A_{\alpha}|=\kappa$ for each $\alpha < \kappa$.
For every $\alpha<\kappa$ the interior 
$\inter(A_{\alpha})$ of $A_\alpha$ is nonempty since the boundary of
$A_{\alpha}$ belongs to $\mathcal I$ and has the cardinality less then $\kappa$.
Moreover $|\inter(A_{\alpha})|= \kappa $.
For each $\alpha<\kappa$ choose an $x_{\alpha}\in\inter(A_{\alpha})$.
Then $A = \{ x_{\alpha}\colon \alpha < \kappa\} $ has cardinality $\kappa$,
so $\inter(A)\neq \emptyset$.
Pick $x_{\alpha _{0}} \in\inter(A)$.
Then $\{x_{\alpha _0}\} =\inter(A_{\alpha _0})\cap\inter(A)\in \tau $.
 But $\{x_{\alpha _0}\} \in \I =NWD(\tau)$, a contradiction. \qed


\rem{rem1}{The condition $\bigcup\I = X$ in Theorem~\ref{thm1} is essential.
For example, if $x_0\in X$ and $\I=\{\emptyset,\{x_0 \}\}$ then 
the pair $\la\P(X),\I\ra$ is made topological by a topology 
$\tau=\{A\subset X\colon x_0 \notin A\}\cup\{X\}$.
}

For a family $\G$ of sets we let 
$i(\G)\stackrel{\rm def}{=}\left\{\bigcap\G_0\colon \G_0\in[\G]^{<\omega}\right\}$.


\thm{th2}{
Let $\kappa$ be an infinite cardinal and $\F$ be a family of nonempty subsets of $X$ 
such that $\F \subset S(\F)$,  $|\F|\leq\kappa$, and 
\begin{itemize}
\item $S_0(\F)$  contains all sets 
$\bigcup\J$ where $\J\in[i(\F)\cap  S_0(\F)]^{< \kappa}$.
\end{itemize}
Then the pair $\la S(\F),S_0(\F)\ra$ is topological.
}

\proof Recall that, by \cite[prop.~1.1(3)]{BBRW}, we have $\F\cap S_0(\F)=\emptyset$. 
Let
$\F=\{P_{\alpha}\colon \alpha<\kappa\}$. 
For every $\alpha<\kappa$ put
\[
Z_\alpha=\bigcup\big(S_0(\F)\cap i(\{P_\xi\colon \xi\leq\alpha\})\big)
\ \ \ \mbox{ and }\ \ \ 
Q_\alpha=P_\alpha\setminus Z_\alpha.
\]
Note that by our assumptions we have $Z_\alpha\in S_0(\F)$,
so $Q_\alpha\in S(\F)\setminus S_0(\F)$. 

Let $\tau$ be a topology on $X$ generated by 
$\B=i(\{Q_{\alpha}\colon \alpha<\kappa\})$.
By Fact~\ref{fact2} it is enough to show that
families $\F$ and $\B\setminus\{\emptyset\}$ are mutually coinitial. 

Clearly for every $P_\alpha\in\F$ we have $Q_{\alpha} \subset P_{\alpha}$
and $Q_{\alpha}\in \B\setminus\{\emptyset\}$.
So, $\B\setminus\{\emptyset\}$ is coinitial with $\F$. 
To see that $\F$ is coinitial with $\B\setminus\{\emptyset\}$
take $Q\in\B\setminus\{\emptyset\}$.
Since $Q\in S(\F)$ it is enough to show that $Q\notin S_0(\F)$
(as for every $A\in S(\F)\setminus S_0(\F)$
there are $P,P'\in\F$ with $P\subset A\cap P')$.
Let $\alpha_1<\cdots<\alpha_n<\kappa$
be such that
\[
Q=\bigcap_{i=1}^n Q_{\alpha_i}=\bigcap_{i=1}^n (P_{\alpha_i}\setminus Z_{\alpha_i})
=\bigcap_{i=1}^n P_{\alpha_i}\setminus \bigcup_{i=1}^n Z_{\alpha_i}
\]
Since $\bigcap_{i=1}^n P_{\alpha_i}\in i(\{P_\xi\colon \xi\leq\alpha_n\})$,
it cannot belong to $S_0(\F)$, as otherwise we would have
$\bigcap_{i=1}^n P_{\alpha_i}\subset Z_{\alpha_n}$
contradicting our assumption that $Q\neq\emptyset$. 
Thus $\bigcap_{i=1}^n P_{\alpha_i}\in S(\F)\setminus S_0(\F)$
and $\bigcup_{i=1}^n Z_{\alpha_i}\in S_0(\F)$, leading to 
$Q\in S(\F)\setminus S_0(\F)$. 
\qed 

\rem{remRef1}{It was pointed to us by the referee that a very similar result 
(with almost identical proof) was proved earlier by Schilling in~\cite[thm.~3]{S}.
More precisely, Schilling considers the $\sigma$-ideals $S^\sigma_0(\F)$ generated by $S_0(\F)$
(which he denotes by $\M(\F)$), 
defines $S^\sigma(\F)$ as
\[
\{A \subset X\colon (\forall P \in\F) (\exists Q \in \F,\ Q\subset P)
(Q \cap A \in S^\sigma_0(\F) \mbox{ or } Q \setminus A \in S^\sigma_0(\F))\}
\]
%\[
%S^\sigma(\F)=\{A \subset X\colon (\forall P \in\F) (\exists Q \in \F)
%(Q \cap A \in S^\sigma_0(\F) \mbox{ or } Q \setminus A \in S^\sigma_0(\F))\}
%\]
(which he denotes by $\B(\F)$\footnote{If $\la X,\tau\ra$ is a topological space
then $\B(\tau\setminus\{\emptyset\})=S^\sigma(\tau\setminus\{\emptyset\})$
is the family of all subsets of $X$ with the Baire property.}), and
proves that if $\la X,\F\ra$ is a category base,
$\kappa=|\F|$, and the condition $\bullet$ holds for $S^\sigma_0(\F)$ in place of $S_0(\F)$ 
then there exists a topology $\tau$ on $X$ such that
$S^\sigma(\F)=S^\sigma(\tau\setminus\{\emptyset\})$  and $S^\sigma_0(\F)$ 
is equal to the $\sigma$-ideal $\M(\tau)$ of meager subsets of $\la X,\tau\ra$.

It is not difficult to see that our result implies Schilling's theorem 
since, by Fact~\ref{fact3}, 
$\la S(\F),S_0(\F)\ra=\la S(\tau\setminus\{\emptyset\}),S_0(\tau\setminus\{\emptyset\})\ra$
implies that $\F$ and $\tau\setminus\{\emptyset\}$ are mutually coinitial
so $S^\sigma(\F)=S^\sigma(\tau\setminus\{\emptyset\})$
and $S^\sigma_0(\F)$ is equal to the 
$\sigma$-ideal generated by $S_0(\tau\setminus\{\emptyset\})=NWD(\tau)$,
that is, $S^\sigma_0(\F)=\M(\tau)$.

The relation between both results is the most straightforward when
$S_0(\F)$ is a $\sigma$-ideal, since then we have
$S^\sigma_0(\F)=S_0(\F)=NWD(\tau)=\M(\tau)$ and 
$S^\sigma(\F)=S(\F)=S(\tau\setminus\{\emptyset\})=S^\sigma(\tau\setminus\{\emptyset\})$.

We also should point here that our condition $\bullet$ implies that 
$\la X,\F\cup\{X\}\ra$ forms a category base. 
}




Applying Theorem~\ref{th2} to $\kappa$ equal to continuum $\continuum$ and 
the family $\F$ of perfect subsets of the real line we obtain immediately 
the following corollary. 

\cor{cor1}{The pair $\la S, S_0\ra$ of the classical Marczewski sets is topological.}

The fact that $S=S^\sigma(\tau\setminus\{\emptyset\})$ 
(which is equal to $S(\tau\setminus\{\emptyset\})$) 
was first proved by Aniszczyk~\cite{A} under the additional set-theoretical assumption 
that the ideal $S_0$ is continuum additive. 
Schilling~\cite{S} noticed that 
there is a topology $\tau$ on the real line for which
$\la S, S_0\ra=\la S^\sigma(\tau\setminus\{\emptyset\}),
S^\sigma_0(\tau\setminus\{\emptyset\})\ra$
which, as we noticed in Remark~\ref{remRef1},
is equal to $\la S(\tau\setminus\{\emptyset\}),S_0(\tau\setminus\{\emptyset\})\ra$.


We also get

\cor{cor2}{Assume the Continuum Hypothesis.
If $\emptyset\notin\F\in[\P(X)]^{\leq\continuum}$ is such that 
$S_0(\F)$ is a $\sigma$-ideal and $\F\subset S(\F)$ then 
the pair $\la S(\F), S_0(\F)\ra$ is topological.
}

For the rest of this note we will assume that $X$ is a set
of cardinality $\kappa\geq\omega$. We say that 
a family $\F_0\subset[X]^\kappa$ is
{\em almost disjoint}\/ provided 
$|F_1\cap F_2|<\kappa$ for every distinct $F_1,F_2\in\F_0$. 
It is a basic fact that there exist
an almost disjoint family $\F_0\subset[X]^\kappa$ of cardinality greater than $\kappa$. 


Notice the following simple fact.

\fact{factA}{If $\F_0\subset[X]^\kappa$ is almost disjoint and 
$\F=\{F\triangle A\colon F\in\F_0\ \&\ A\in[X]^{<\kappa}\}$ then 
$$
\F\subset
S(\F)=\{A\colon (\forall F\in\F)(|F \setminus A|<\kappa\ \mbox{ or }\ |F \cap A|< \kappa)\}
$$
and $[X]^{<\kappa}\subset S_0(\F)=\{A\colon  (\forall F\in\F)(|F\cap A|< \kappa)\}$.

Moreover, $S_0(\F)=[X]^{<\kappa}$ if and only if 
$\F_0$ is maximal almost disjoint.
}


\thm{th3}{
Let $\F=\{F\triangle A\colon F\in\F_0\ \&\ A\in[X]^{<\kappa}\}$, where 
$\F_0\subset[X]^\kappa$ is almost disjoint.
\begin{itemize}
\item[(a)] If $\kappa$ is a regular cardinal and $|\F_0|\leq\kappa$
    then the pair $\la S(\F),S_0(\F)\ra$ is topological.

\item[(b)] If $|\F_0|> \kappa$ then the algebra $S(\F)$ is not topological.
\end{itemize}
}


\proof (a) Let $X=\{x_\alpha\colon \alpha<\kappa\}$ and put
\[\F_1=\{F\setminus\{x_\xi\colon\xi<\alpha\}\colon F\in\F_0\ \&\ \alpha<\kappa\}.
\]
Regularity of $\kappa$ implies that families $\F$ and $\F_1$
are mutually coinitial. 
So, by Fact~\ref{fact2}, we have 
$\la S(\F),S_0(\F)\ra=\la S(\F_1),S_0(\F_1)\ra$.
Clearly$|\F_1|\leq\kappa$ and $\F_1\subset \F\subset S(\F)\subset S(\F_1)$. 

Since regularity of $\kappa$ implies also that $S_0(\F)$ is $\kappa$-additive
(i.e., union if less than $\kappa$-many sets from $S_0(\F)$ 
belongs to $S_0(\F)$), 
condition $\bullet$ from Theorem~\ref{th2} holds and so 
$\la S(\F_1),S_0(\F_1)\ra$ is topological. 



(b) By way of contradiction suppose that there exists a topology $\tau$ on $X$
such that $S(\F)=S(\tau_0)$, where 
$\tau_0=\tau\setminus\{\emptyset\}$.

Note that for every $F\in\F$ we have 
$F\in S(\F)\setminus S_0(\F)=S(\tau_0)\setminus NWD(\tau)$,
so
\begin{equation}\label{eq2}
\inter_\tau(F)\neq\emptyset \ \mbox{ for every } F\in\F.
\end{equation}
Also, if $F_0,F_1\in\F_0$ are different then 
\[
\inter_\tau(F_0)\cap\inter_\tau(F_1)\subset F_0\cap F_1\in [X]^{<\kappa}
\subset S_0(\F)=S_0(\tau_0)=NWD(\tau).
\]
So, $\{\inter_\tau(F)\colon F\in\F_0\}$ is the family
of non-empty pairwise disjoint subsets of $X$ of cardinality $|\F_0|>|X|$, which is impossible. 
\qed

\rem{rem2}{Notice that if $\kappa$ has uncountable cofinality,
$\F_0\subset[X]^\kappa$ is maximal almost disjoint,
and $\F$ is as in Fact~\ref{factA} then 
the algebra $\cal A$ generated by
the family $\cal F$ (i.e., the closure of $\cal F$  
under finite unions, finite intersections and 
complements in $X$)
is not inner MB-representable. 
This follows immediately from \cite[thm.~13]{BBC}.
}


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\end{thebibliography}


\end{document}

