\documentstyle[amsfonts,amssymb]{amsart}
%\def \L{\char 147}
%\def \l{\char 149}
\title{Marczewski fields and ideals}
\author{M. Balcerzak,
A. Bartoszewicz, J. Rzepecka, S. Wro\'nski}
\address{Institute of Mathematics, {\L}\'od\'z Technical University,
al. Politechniki 11, I-2, 90-924 {\L}\'od\'z, Poland\\
and Faculty of Mathematics, {\L}\'od\'z University, ul. Banacha 22,
90-238 {\L}\'od\'z}
\email{mbalce@@krysia.uni.lodz.pl}
%\author{Artur Bartoszewicz}
\address{Institute of Mathematics, {\L}\'od\'z Technical University,
al. Politechniki 11, I-2, 90-924 {\L}\'od\'z, Poland}
\email{arturbar@@ck-sg.p.lodz.pl}
%\author{Joanna Rzepecka}
\address{Institute of Mathematics, {\L}\'od\'z Technical University,
al. Politechniki 11, I-2, 90-924 {\L}\'od\'z, Poland}
%\email{@@ck-sg.p.lodz.pl}
%\author{Stanis{\l}aw Wro\'nski}
\address{Institute of Mathematics, {\L}\'od\'z Technical University,
al. Politechniki 11, I-2, 90-924 {\L}\'od\'z, Poland}
\email{wronskis@@ck-sg.p.lodz.pl}
\keywords{Marczewski sets, field of sets, Baire category, Lebesgue
measure, density topology.\\
The first author
was partially supported by NSF Cooperative
Research Grant INT-9600548 and its Polish part financed by KBN.
The main results of the paper were presented during the 13th Spring
Miniconference in Real Analysis in Auburn (March 5-6, 1999)}
\subjclass{28A05, 03E75, 06E25, 54A10, 54E52}

%\date{ }

\newcommand{\hw}{\hspace*{1cm} }
\newcommand{\hs}{\hspace*{0.5cm} }
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\newtheorem {tw} {Theorem}[section]
\newtheorem {lem} {Lemma}[section]
\newtheorem {pr} {Proposition}[section]
\newtheorem {wn} {Corollary}[section]
\newtheorem {rem} {Remark}[section]
\newtheorem {ex} {Example}[section]
\newtheorem{prb}{Problem}[section]
\newcommand{\mf}{Marczewski field }
\newcommand{\tf}{topological field }
\newcommand{\tfs}{topological fields }
\newcommand{\mc}{mutually coinitial }
\newcommand{\sodf}{$S({\cal F})$ }
\newcommand{\soodf}{$S^0({\cal F})$ }
\newcommand{\Sf}{S({\cal F}) }
\newcommand{\Sof}{S^0({\cal F}) }
\newcommand{\sh}{S({\cal H})}
\newcommand{\Sh}{S({\cal H}) }
\newcommand{\f}{{\cal F} }
\newcommand{\bb}{{\cal B}}
\newcommand{\uv}{U\cap V }
\newcommand{\uvw}{U\cap V\cap W }
\newcommand{\sumawzf}{\bigcup\limits_{W\in \F} }
\newcommand{\zr}{\subsetneq}
%{\stackrel{\subset}{{\scriptstyle \neq}} }

\newcommand{\eps}{\varepsilon}
\newcommand{\real}{{\Bbb R}}
\newcommand{\rat}{{\Bbb Q}}
\newcommand{\F}{{\cal F} }
\newcommand{\A}{{\cal A} }
\newcommand{\C}{{\cal C} }
\newcommand{\G}{{\cal G} }
\newcommand{\B}{{\cal B} }
\newcommand{\D}{{\cal D} }
\newcommand{\I}{{\cal I} }
\newcommand{\M}{{\cal M} }
\newcommand{\N}{{\cal N} }
\newcommand{\Hp}{{\cal H} }
\newcommand{\Pp}{{\cal P} }

%\newcommand{\F1}{{\cal F}_1 }
\newcommand{\h}{Hashimoto }
\newcommand{\m}{Marczewski }
\newcommand{{\leb}}{Le\-bes\-gue }
\newcommand{\e}{Ellentuck }
\newcommand{\r}{Ramsey }
\newcommand{\fm}{Fund. Math. }
\newcommand{\rae}{Real Anal. Exchange }
\newcommand{\so}{$(s^0)$-sets }
\newcommand{\s}{$(s)$-sets }
\newcommand{\px}{{\cal P}(X)}
\newcommand{\zp}{\setminus \{ \emptyset \}}
\newcommand{\fft}{F\cap F_t}
\newcommand{\htn}{H_t^n}
\newcommand{\htm}{H_t^m}
\newcommand{\pert}{\mbox{Perf}(\tau)}
\newcommand{\spert}{S(\mbox{Perf}(\tau))}
\newcommand{\tm}{T^{\star }_{{\cal M}}}
\newcommand{\tn}{T^{\star }_{{\cal N}}}
\newcommand{\dn}{\D_{{\cal N}}}
\newcommand{\dm}{\D_{{\cal M}}}
\newcommand{\Dn}{\D_{{\cal N}}}
\newcommand{\Dm}{\D_{{\cal M}}}
\newcommand{\Tm}{T^{\star }_{{\cal M}}}
\newcommand{\Tn}{T^{\star }_{{\cal N}}}
\newcommand{\sopert}{S^0(\mbox{Perf}(\tau))}


\newcommand{\bd}{\begin{description}}
\newcommand{\ed}{\end{description}}
\newcommand{\be}{\begin{enumerate}}
\newcommand{\ee}{\end{enumerate}}
\newcommand{\bi}{\begin{itemize}}
\newcommand{\ei}{\end{itemize}}
\newcommand{\bp}{\begin{pf}}
\newcommand{\ep}{\end{pf}}

\begin{document}

\maketitle
\begin{abstract}
For an $X\neq\emptyset$
and a given family $\F\subset {\cal P}(X)\setminus \{ \emptyset \} $, we
consider the \mf \sodf which consists of sets $A\subset X$ such that each set
$U\in \F$ contains a set $V\in \F$ with $V\subset A$ or $V\cap A=\emptyset$.
We also study the respective ideal $\Sof$.
We show general properties of \sodf and certain representation theorems.
For instance we prove that the interval algebra in $[0,1)$ is a
Marczewski field. We are also interested in
situations where \sodf $=S(\tau\setminus \{ \emptyset \} )$ for a topology
$\tau $ on $X$.
We propose a general method which establishes \sodf and \soodf provided that
$\F$ is the family of perfect sets with respect to $\tau $, and $\tau$ is
a certain ideal topology on $\real$ connected with measure or category.
\end{abstract}

\section{General properties}
The notions of $(s)$-sets and $(s^0)$-sets are due to \m \cite{Sz}.
They have been investigated by many authors. (See \cite{Mi1}, \cite{Mi2},
 \cite{BrCo} and also \cite{Br2}, \cite{Co}, \cite{Wa}.)
The scheme defining \s and \so
was used for more general settings
in several publications (see e.g. \cite{Mo}, \cite{Bre},
\cite{Pa}, \cite{R}, \cite{BR}).
We observe that this scheme turns out interesting without any essential
restrictions on a generating family of sets.
Namely, let ${\cal F}$
be a family of nonempty subsets of a given set $X$. We put\\
$S(\F)=\{ A\subset X: (\forall\;U\in \F)\;(\exists V\in \F)\;
(V\subset U\cap A \;\vee \; V\subset U\setminus A)\},$\\
$S^0(\F)=\{ A\subset X: (\forall\;U\in \F)\;(\exists V\in \F)\;
V\subset U\setminus A\},$\\
$H(\F)=\{ A\subset X: (\forall\;B\subset A)\;B\in \F\}.$\\
Note that
$H(\F)$ is the maximal hereditary family contained in $\F$. In the
case when $\F$ consists of all perfect
subsets of a given Polish space, \sodf
and \soodf are exactly the families of classical \m \s and $(s^0)$-sets.

Our notation is standard. By $\px $ we denote the power set of $X$.
Throughout the paper $X\neq\emptyset$.
\begin{pr}\label{p1}
Let $\F \subset \px\zp$. Then we have
\bd
\item[(1)] $S({\cal F})$ is a field of sets,
\item[(2)] $\Sof \subset H(\Sf),$ and \soodf is an ideal of sets,
\item[(3)] $\F\cap \Sof =\emptyset $,
\item[(4)] $(\forall\;U\in \Sf\setminus\Sof)
(\exists V \in \F )V\subset U$,
\item[(5)] $\F\subset \Sf \Leftrightarrow (\forall U, V\in \F)
(\exists W\in \F) (W\subset \uv\; \vee \;W\subset U\setminus V)$,
\item[(6)] if $\{ x\}\in \F $ for all $x\in X$ then $\Sf=\px$ and
$\Sof =\{\emptyset \} $.
\ed
\end{pr}
\begin{pf}
(1) From the definition of \sodf it immediately follows that, if
$A\in \Sf $ then $X\setminus A\in \Sf$. Now, assume that $A,B\in \Sf $.
Let $C\in \F$. If there is a $D\in \F$
such that either $D\subset C\cap A$ or
$D\subset C\cap B$ then $D\subset C\cap (A\cup B)$.
If such $D$ does not exist,
there is a $D_1\in \F $ such that $D_1\subset C\setminus A$ and there is a
$D_2\in \F$ such that $D_2\subset D_1\setminus B$.
Thus $D_2\subset C\setminus
(A\cup B)$. Hence $A\cup B\in\Sf $.

Similarly, we show that \soodf is an ideal (condition (2)).
The remaining statements, except for (4), can be checked directly without
troubles.

(4) Suppose that there is a $U_0\in \Sf\setminus\Sof $ such that
$V\setminus U_0\neq\emptyset$ for each $V\in\F$.
Since $U_0\notin\Sof $, there
is a $V_0\in \F $ such that $W\cap U_0\neq\emptyset $ for each $W\in \F$,
$W\subset V_0$. Since $U_0\in \Sf $, there is a $W_0\in\F$ such that
$W_0\subset V_0\cap U_0$. But then $W_0\setminus U_0=\emptyset$,
a contradiction.
\end{pf}

\begin{wn}\label{c1}
If $\F\subset\px $ is a field of sets, then
$$\F\zp \subset S(\F\zp)\setminus
S^0(\F\zp).$$
\end{wn}

\begin{pf}
 Use Proposition \ref{p1} (3) and (5).
\end{pf}


Note that the classical \s and \so form a
$\sigma $-field and a $\sigma $-ideal,
respectively (the proof of $\sigma$-additivity for \so is
based on the fusion lemma; see \cite{Sz}).
Additionally, in that case $H(\Sf)=\Sof$ \cite[3.1]{Sz}.
On the other hand, there are families
$\F$ for which $H((S(\F ))\neq S^0(\F )$ \cite[Cor. 1.10]{R}.
Observe that there are cases when $S(\F )$ forms a
$\sigma$-field but $S^0(\F )$ is not a $\sigma$-ideal. That
happens if $X$ is an infinite set and $\F$ stands for
the family of all infinite subsets of $X$;
then $S(\F )={\cal P}(X)$ and $S^0(\F )$ consists of all
finite subsets of $X$. In \cite[Lemma 2]{Pa} it was proved
that if $\F\subset S(\F )$ and $S^0(\F )$ is $\sigma$-additive
then $S(\F )$ is a $\sigma$-field.

The operation $S$ can be iterated.
For a family $\F\subset \px\zp $ we define
$ S_0(\F)=\F$ and
$ S_{\alpha}(\F)=S(\cup_{\gamma <\alpha}S_{\gamma}(\F)\zp )$
for any ordinal $\alpha >0$. Of course we may consider only
$\alpha\le 2^{2^{\kappa}}$ where $\kappa$ is the cardinality of $X$.
The families $S_\alpha (\F)$, $\alpha >0$, are fields and
from Corollary \ref{c1} it follows that
$S_\gamma (\F)\subset S_\alpha (\F)$ for any ordinals $\gamma ,\alpha$ with
$0<\gamma <\alpha$. In our future studies, we plan to establish
the maximal number of different fields that can be obtained in
a sequence of type $\langle S_\alpha (\F)\colon\alpha >0\rangle$.
In the former version of the paper we claimed incorrectly that
this number is $2$. The referee has observed that it is at least $3$.
(See Remark \ref{r1}.)

We say that two families $\F_1, \F_2\subset \px\zp $ are {\em \mc}
if
$$ (\forall \;U\in \F_1)\;(\exists \;V\in \F_2)\;V\subset U$$
$$ \mbox{ and }(\forall \;U\in \F_2)\;(\exists \;V\in \F_1)\;V\subset U.$$

\begin{pr}\label{p3}
Let $ \F_1,\F_2\subset \px\zp$.
\bd
\item[(i)]If $\F_1,\F_2$ are \mc then $S(\F_1)=S(\F_2)$ and
$S^0(\F_1)=S^0(\F_2)$.
\item[(ii)] Assume that $\F_1\subset S(\F_1)$ and $\F_2\subset S(\F_2)$.
If $S(\F_1)=S(\F_2)$ and
$S^0(\F_1)=S^0(\F_2)$ then $\F_1,\F_2$ are mutually coinitial.
\ed
\end{pr}

\bp
 (i) is evident.\\
(ii) Let $U\in \F_1$. Then $U\not\in S^0(\F_1)$ by Proposition \ref{p1}(3).
Hence $U\in S(\F_1)\setminus S^0(\F_1)=S(\F_2)\setminus S^0(\F_2)$ and by
Proposition \ref{p1}(4) there is a $V\in \F_2$ such that $V\subset U$.
Analogously, for each $U\in \F_2$
there is a $V\in \F_1$ such that $V\subset U$.
\ep

Note that an idea similar to that contained in Proposition \ref{p3}
was used in \cite[Th. 1, p. 23]{Mo}.
The referee has asked whether the converse of (i) is true.
The answer is ``no''which follows from
Remark \ref{remi} in the next section.

Now, consider a field
$\Sigma $ (respectively, an ideal $\I$) of subsets of
$X$. We say that:
\begin{itemize}
\item $\Sigma $ (respectively, $\I$) is a {\em \tf} (respectively,
a {\em topological ideal}) if there is a topology
$\tau $ on $X$ such that $\Sigma $
consists of all sets with $\tau $-nowhere dense boundary (respectively, $\I$
consists of $\tau$-nowhere dense sets). (Cf. \cite[\S 8.V]{Ku}.)
We thus write $\Sigma=\Sigma (\tau )$ and
$\I=NWD(\tau )$.
\item $\Sigma $ (respectively, $\I$) is a {\em \mf} (respectively,
a {\em \m ideal}) if there is a family
${\cal F}\subset\px\zp$ such that $\Sigma=\Sf$
(respectively, $\I=\Sof$).
\end{itemize}

Note that papers \cite{BET} and \cite{BBC} use different terminology
for Marczewski fields: the authors of \cite{BET} say that
``${\cal F}$ is a basis for a Marczewski-Burstin-like characterization
of $\Sigma$'', and in \cite{BBC}, $\Sigma$ is called ``Marczewski-Burstin
representable''.

An easy connection between the above
notions is contained in the following
\begin{pr}\label{p4}
(Cf. \cite{BR}.) If $\tau $ is a topology on $X$
then $S(\tau\zp)=\Sigma(\tau)$
and $S^0(\tau \zp)=NWD(\tau)$.
Consequently, every topological field (ideal)
is a Marczewski field (ideal).
\end{pr}

From Proposition \ref{p3}(i) we derive

\begin{pr}\label{p1.5}
If a family $\F\subset \px\zp$ is \mc with a topological base on
$X$ then the field \sodf is topological.
\end{pr}

In Section 2 we shall show that the interval algebra is
a Marczewski field. This will imply that
there are Marczewski fields that are not topological fields.
A Boolean-theoretical characterization of \tfs was given in
\cite{Wr1}. Article \cite{CJ} was devoted to
extensive studies of topological
ideals; the authors considered also an additional requirement stating that
an ideal consists of meager sets in some topology. In Section 2 we
discuss some connections between Marczewski fields, topological fields
and category bases (introduced by John Morgan II, see \cite{Mo}).

The class of Marczewski fields seems to be rich.
From \cite{Bu} it follows that the Lebesgue measurable sets
in $\real$ form a Marczewski field. (Note that paper \cite{Bu} is
much earlier than \cite{Sz}.) Also the sets with the Baire
property in $\real$ constitute a Marczewski field \cite{Br1}, \cite{BET}.
When we started to prepare our paper, it was not even known
whether there exists a non-Marczewski field of subsets of $\real$.
Now, our knowledge is wider. Namely, the forthcoming paper \cite{BBC}
contains a construction of a non-Marczewski field on $\real$
provided $2^\omega =\omega_1$ and $2^{\omega_1}=\omega_2$.
Another result of \cite{BBC} states that
$2^\omega =\omega_1$ and $2^{\omega_1}=\omega_2$ imply
that the Borel subsets of $\real$ form a Marczewski
field.

For any filter ${\cal F}$ of the
algebra $\A=\px $ we denote
$-\F=\{ X\setminus E:\;E\in\F\} $ and
$\A_\F=\F\cup -\F$.
%Then $\A_\F$ forms a subalgebra of $\A$.

\begin{pr}\label{pl1}
 For any filter $\F$ of $\A$ we have $\Sf =\A_\F$ and $S^0(\F )=-\F$.
\end{pr}
\begin{pf}
We easily check that $\F\subset\Sf$ and $-\F\subset S^0(\F )$.
Since $\A_\F$ is the smallest field containing $\F$, we
have $\A_\F\subset \Sf$. To show the reverse inclusion consider a
$U\in \Sf$. Since $X\in \F $, we can find a $V\in \F $ such that
either $V\subset U $ or $V\subset X\setminus U$. Hence either $U\in \F$ or
$X\setminus U\in \F$ which means that $U\in \A_\F$. Thus $\Sf\subset\A_\F$.
It can be similarly shown that $S^0(\F )\subset -\F$.
\end{pf}

\begin{pr}
For a set $X$ of cardinality $| X|=\kappa$ there are $2^{2^\kappa}$
nonisomorphic \m fields on $X$ containing all singletons.
\end{pr}
\bp
 We follow the argument given in \cite{F}. Let $\Phi$ consist of all
filters in $\A=\px$ which are intersections of two free ultrafilters.
Then $| \Phi |= 2^{2^\kappa}$ and $\A_{\F_1}\neq\A_{\F_2}$ for
any distinct $\F_1,\F_2\in \Phi$.
Additionally, $\{ x\}\in \A_\F$ for any $x\in X$ and $\F\in \Phi$.
Thus, by Proposition \ref{pl1}, there are $2^{2^{\kappa}}$ \m
fields on $X$ containing all singletons.
Any isomorphism between subalgebras of $\px$ containing all singletons is
induced by a bijection from $X$ to $X$. Hence each isomorphism class of such
subalgebras has at most $2^\kappa$ elements. Finally,
observe that if $h$ is a
bijection of $X$ onto $X$ and $\F\in \Phi $ then
$$\{ h[U]:\;U\in \Sf\}=S(\{h[V]: \;V\in\F\})\mbox{ and }
\{ h[V]:\;V\in\F\}\in \Phi.$$
Thus there are $2^{2^\kappa}$ different classes of isomorphic \m fields on
$X$ containing all singletons.
\ep

\section{Marczewski fields, topological fields and category bases}
Let $X=[0,1)$. The family of all finite unions of half-open
intervals $[a,b)$ (where $0\le a< b\le 1$) form a field of
subsets of $X$. It is called the {\em interval algebra} of $X$
\cite[1.11]{K}.
\begin{tw}\label{tw2.3}
The interval algebra ${\cal A}$ of $X=[0,1)$ is a Marczewski
field.
\end{tw}
\begin{pf}
Let $\rat$ stand for the set of all rationals and let
$\frak c$ denote the cardinality of $\Bbb R$. Consider the equivalence
relation $x\sim y\iff x-y\in\rat$. Let $F\colon [0,1]\to\real /\sim$
be a one-to-one function such that $x\notin F(x)$ for $x\in [0,1]$.
(Note that $F$ can be easily constructed by transfinite induction.
Indeed, arrange all points of $[0,1]$ into a one-to-one sequence
$x_\gamma$, $\gamma <{\frak c}$, and consider an $\alpha <{\frak c}$.
If the values $F(x_\gamma )$ for $\gamma <\alpha$ have been defined,
we pick $x\in [0,1]\setminus\bigcup_{\gamma <\alpha}[F(x_\gamma )]
\setminus [x_\alpha ]$ and put $F(x_\alpha )=[x]$ where $[x]$ denotes
the respective equivalence class.)

For $x\in [0,1]$ let\\
$\F_r(x)=\{ ([x,x+\eps )\setminus F(x))\cap X:\eps >0\}$,\\
$\F_l(x)=\{ ((x-\eps ,x)\setminus F(x))\cap X:\eps >0\}$,\\
$\F(x)=\F_l(x)\cup\F_r(x)$.\\
Note that $\F_r(1)=\F_l(0)=\emptyset$, otherwise $\F_r(x)$ and $\F_l(x)$
are nonempty. Finally, let
$\F=\bigcup\limits_{x\in [0,1]}\F(x)$.

From the definitions of $\F$ and $S(\F )$ it easily follows that
$[a,b)\in S(\F )$ for any $a,b$ with $0\le a<b\le 1$. Since $S(\F )$ is
a field of sets, we have ${\cal A}\subset S(\F )$.

\bigskip
\noindent
{\bf Claim 1.}
Let $x\in [0,1]$ and $k\in\{ r,l\}$. If $U\in\F_k(x)$ then for each
$y\in [0,1]$ and for each $V\in\F(y)$ such that $V\subset U$, we have
$y=x$, and moreover $V\in\F_k(x)$.

Indeed, suppose that $y\neq x$. Let $U=I\setminus F(x)$ and
$V=J\setminus F(y)$ where $I$ and $J$ are the respective intervals.
From $V\subset U$ and the density of $F(x)$ it follows that $\emptyset\neq
J\cap F(x)\subset J\cap F(y)$ which contradicts the disjointness of
$F(x)$ and $F(y)$. Thus $y=x$ and so, $V\in\F_m(x)$ for some $m\in\{r,l\}$.
However, $m=k$ since otherwise $U\cap V=\emptyset$.

\bigskip

We have already observed that
${\cal A}\subset S(\F )$.
To prove that $S(\F )\subset{\cal A}$ fix an
$A\in S(\F )\setminus\{\emptyset\}$.

\bigskip
\noindent
{\bf Claim 2.} For each $x\in A$ there exists an $\eps >0$ such that
$[x,x+\eps )\subset A$. For each $x\in X\setminus A$ there exists an
$\eps >0$ such that $[x,x+\eps )\cap A=\emptyset$.

The latter assertion follows from the former applied to $X\setminus A$.
To show the former assertion, suppose $x\in A$ and $[x,x+\eps )\setminus A
\neq\emptyset$ for each $\eps >0$. Consider a $U\in\F_r(x)$. Since
$A\in S(\F )$, there is a $V\in\F$ such that either $V\subset U\cap A$ or
$V\subset U\setminus A$. By Claim 1 we have $V\in\F_r(x)$. Since
$x\in A\cap V$, we infer that $V\subset U\cap A$. Let
$V=([x,x+\eps )\setminus F(x))\cap X$ where $\eps >0$. We may assume
that $x+\eps\le 1$. By our supposition, pick a $y\in (x,x+\eps )\setminus A$.
Let $\tilde{V}=[y,x+\eps )\setminus F(y)$. Then $\tilde{V}\in\F_r(y)$
and since $A\in S(\F )$, there is a $W\in\F$ such that either $W\subset
\tilde{V}\cap A$ or $W\subset\tilde{V}\setminus A$. Again, by Claim 1,
we have $W\in\F_r(y)$, so we may assume that $W=[y,y+\eps_1 )\setminus F(y)$
where $y+\eps_1\le x+\eps$. Since $y\notin A$, we have $W\cap A=\emptyset$.
The set $[y,y+\eps_1 )\setminus (F(x)\cup F(y))$ is nonempty (uncountable)
contained in $[x,x+\eps )\setminus F(x)=V\subset A$ and simultaneously in
$[y,y+\eps_1 )\setminus F(y)=W\subset X\setminus A$. Contradiction.

\bigskip
\noindent
{\bf Claim 3.} For each $x\in (0,1]$ there exists an $\eps >0$
such that either $(x-\eps,x)\subset A$ or $(x-\eps,x)\cap A=\emptyset$.

To show the claim, suppose that there exists an $x\in (0,1]$ such that
$(x-\eps ,x)\setminus A\neq\emptyset$ and $(x-\eps ,x)\cap A\neq\emptyset$
for each $\eps >0$.
Let $U\in\F_l(x)$. Since $A\in S(\F )$, there is a $V\in\F$ such that
either $V\subset U\cap A$ or $V\subset U\setminus A$. By Claim 1 we have
$V\in\F_l(x)$ and we may assume that $V=(x-\eps ,x)\setminus F(x)$ where
$x-\eps \ge 0$. If $V\subset A$, by our supposition we can pick
$y\in (x-\eps,x)\setminus A$. By Claim 2 there is an $\eps_1>0$ such that
$[y,y+\eps_1)\cap A=\emptyset$ and we may assume that $y+\eps_1\le x$.
On the other hand, $\emptyset\neq [y,y+\eps_1)\setminus F(x)\subset
(x-\eps,x)\setminus F(x)\subset V\subset A$, a contradiction.
If $V\cap A=\emptyset$, by our supposition we can pick
$y\in (x-\eps ,x)\cap A$. By Claim 2 there is an $\eps_1>0$ such that
$[y,y +\eps_1)\subset A$ and we may assume that $y+\eps_1\le x$.
On the other hand, $\emptyset\neq [y,y+\eps_1)\setminus F(x)\subset
(x-\eps ,x)\setminus F(x)=V\subset X\setminus A$, a contradiction.

\bigskip
From Claim 2 it follows that each connected component $I$ of $A$ is
a nondegenerate interval with $b=\sup I\notin I$. Denote $a=\inf I$ and
observe that $a\in I$. Indeed, suppose that $a\notin I$. We know that
$[a,b)\in S(\F )$. Thus $A\cap [a,b)=(a,b)\in S(\F )$ and
consequently $[a,b)\setminus (a,b)=\{ a\}\in S(\F )$ which contradicts
Claim 2.

From the above we infer that $A$ is a union of at most countable family of
pairwise disjoint intervals of type $[a,b)$. This family however cannot be
infinite. Indeed, suppose that $A=\bigcup_{n=1}^\infty [a_n,b_n)$ with
$[a_n,b_n)\subset X$, $n\ge 1$, pairwise disjoint. Pick a strictly monotonic
subsequence $(a_{k_n})$ of $(a_n)$. If $a_{k_n}\searrow x$, we apply
Claim 2 to $x$ and we obtain a contradiction. If $a_{k_n}\nearrow x$, we
apply Claim 3 to $x$ and we obtain a contradiction.

Thus we have proved that $A\in{\cal A}$. Consequently,
$S(\F )\subset{\cal A}$.
\end{pf}

\begin{rem}\label{remi}
{\em Observe that in the above construction, we can choose, for $i=1,2$,
one-to-one functions $F^{(i)}\colon [0,1]\to\real /\sim$ with
disjoint ranges, and such that $x\notin F^{(i)}(x)$ for each $x\in [0,1]$.
Then ${\cal A}=S(\F^{(1)})=S(\F^{(2)})$ where $\F^{(i)}$ $(i=1,2)$ is
associated with $F^{(i)}$ as in the proof of Theorem \ref{tw2.3}.
Since $H({\cal A})=\{\emptyset\}$, we have
$S^0(\F^{(1)})=S^0(\F^{(2)})=\{\emptyset\}$ by Proposition \ref{p1}(2).
However, the argument for Claim 1 shows that $\F^{(1)}$ and $\F^{(2)}$
are not mutually coinitial. Thus the converse of
(i) in Proposition \ref{p3} is false.}
\end{rem}

\begin{rem}\label{r1}
{\em Since ${\cal A}$ and the family of nonempty open sets in $[0,1)$
are mutually coinitial, the field $\Sigma =
S({\cal A}\setminus\{\emptyset\})$ consists of all sets in $[0,1)$ with
nowhere dense boundary, and $S(\Sigma )={\cal P}([0,1))$ by
Proposition \ref{p1}(6). So we have $3$ different fields obtained by
the iteration of $S(\cdot )$.}
\end{rem}

\begin{wn}\label{wn2.1}
There exists an \mf which is not a topological field.
\end{wn}
\bp This follows from Theorem \ref{tw2.3} since every topological field
has an atom \cite{Wr1} and the algebra ${\cal A}$ has no atoms.
\ep

Although the class of topological subfields of ${\cal P}(X)$ is smaller
than the class of Marczewski fields, the former can be used to get
the following representation result:
\begin{tw}\label{tw2.4}
Every field $\Sigma $ of subsets of $X$ is
equal to the intersection of all \tfs
containing $\Sigma $.
\end{tw}
\bp
If $\Sigma =\px $, the assertion is obvious.
Assume that $\Sigma\neq \px$. It suffices to show that
for each $A\notin\Sigma$ there is a
\tf $\Sigma_A\supset\Sigma$ with $A\notin\Sigma_A$.
So, let $A\notin\Sigma$.
By \cite[Lemma 2]{Wr2} we find an ultrafilter
$\F_A$ of the field $\Sigma $
such that no subset of $A$ is in $\F_A$ and no
subset of $X\setminus A$ is in
$\F_A$. Thus $A\notin S(\F_A)$. Put $\Sigma_A =S(\F_A) $.
Observe that $\Sigma_A$
is a \tf since $\F_A$ forms a topological base, and thus
Proposition \ref{p1.5} can be used.
Because $\F_A$ is an ultrafilter of $\Sigma $,
we have $\Sigma\subset S(\F_A)$.
\ep


Recall \cite{Mo} that a pair
$(X,\C )$ is said to be a {\em category base} if
$\C $ stands for a family of subsets of a nonempty set $X$, and
nonempty sets in $\C $,
called {\em regions}, satisfy the following axioms:
\bd
\item[$1^0$]$\bigcup \C=X,$
\item[$2^0$]Let $A$ be a region and $\D$ -- a nonempty family of
disjoint regions
with $|\D |<|\C |$. Then
\bi
\item if $A\cap (\bigcup \D)$ contains  a region then there is a region
$D\in \D$ such that $A\cap D$ contains a region;
\item if $A\cap (\bigcup \D)$ contains  no region then there is a region
$B\subset A\setminus \bigcup \D$.
\ei
\ed

A set $E\subset X$ is called {\em singular} if
$E\in S^0(\C\setminus\{\emptyset\})$.

\begin{tw}\label{tw2.2}
Assume that for a family
$\F\subset\px\zp $ we have $X\in\F\subset\Sf $ and
$\bigcup (\Sof \cap \{ \uv :\;V\in \F\} )\in \Sof$ for each $U\in\F$.
Then $(X,\F)$ is
a category base whose ideal of singular sets equals $S^0(\F)$.
\end{tw}
\bp
 It is enough to check condition $2^0$ defining a category base.
Let $F,F_t\in \F$ for $t\in T$ (where $T$ is an arbitrary set of indices).
Assume that $F\cap \bigcup\limits_{t\in T}F_t$
contains a set from $\F$. Then there exists  a $t_0\in T$ such that
$F\cap F_{t_0}$ contains a set from $\F$.
Indeed, suppose that it is not the
case. Since $\F\subset\Sf $, we have $F\cap F_t\in \Sf$ for each $t\in T$.
Hence by Proposition \ref{p1}(4) we get $F\cap F_t\in \Sof $. Thus, by
assumption, we have $\bigcup\limits_{t\in T}(F\cap F_t)\in \Sof $ which
yields a contradiction (cf. Proposition \ref{p1}(3)).
Assume now that $F\cap \bigcup\limits_{t\in T}F_t$
contains no set from $\F$.
Then every set $\fft $, $t\in T$, contains no set from $\F$.
As before we infer that
$\fft\in\Sf$ for $t\in T$ and moreover $\fft\in\Sof$. By
assumption we have
$\bigcup \limits_{t\in T}(\fft )\in \Sof $ and consequently,
$$ F\setminus\bigcup\limits_{t\in T}(\fft )\in \Sf \setminus\Sof ,$$
since $F\in \Sf\setminus\Sof$. Hence, by Proposition \ref{p1}(4),
there is a set
from ${\cal F}$ contained in $F\setminus \bigcup\limits_{t\in T}F_t$.
\ep


\section{Marczewski fields and perfect sets}
Let $(X,\tau )$ be a topological space.
By a {\it $\tau $-perfect set} we mean
a nonempty $\tau$-closed set without isolated points.
Let $\pert$ stand for the
family of all $\tau $-perfect subsets of $X$.
As it was mentioned in Section 1,
if $X$ is a Polish space with topology $\tau$
then $\spert$ and $\sopert$
are exactly the classical families of \m \s and $(s^0)$-sets.
In \cite{R}, studies of $\spert$ and
$\sopert$ for other topological
spaces were initiated. (See also \cite{BR}.)
In this section we propose a general method
which enables us to reprove some
results of \cite{R} and \cite{BR}, and to show new applications.

If $\tau $ is a given topology on $X$
and we want to characterize $\spert$,
we shall use the following scheme:
\bd
\item[$1^0$] we conjecture that $\spert=\Sigma $
where $\Sigma$ is a known field of sets,
\item[$2^0$] we know that $\Sigma =\Sf $ for some family $\F\subset \px\zp$;
\item[$3^0$] in aim to confirm our conjecture $1^0$, it is enough (by
Proposition \ref{p3}(i)) to check that
$\pert$ and $\f$ are mutually coinitial.
\ed

A similar method works for $\sopert$.
The above scheme will be illustrated by
examples dealing with some ``ideal topologies'' on $\real$.

Let $\tau $ be a topology on $X$, and let $\I \subset \px$ be a
$\sigma $-ideal
containing all singletons.
The family
$${\cal B}_\I^\star =\{ U\setminus A:\; U\in \tau\; \&\; A\in\I\}$$
forms a base for a topology $\tau_\I^\star$, on $X$,
stronger than $\tau $, which will be called the
{\em \h topology} associated with $\tau$ and $\I$.
If $\tau$ is second countable (or even hereditary Lindel\"{o}f) then
$\tau_\I^\star ={\cal B}_\I^\star$. (See \cite{H}, \cite{JH},
\cite{LMZ}.) The following property is well known.

\begin{lem}\label{l3.1}
(Cf. \cite{BR}). Let $\I$ be a $\sigma $-ideal of subsets of
a separable metric space $X$ and let $\I$
contain all singletons. A set $F\subset X$ is
$\tau_{\I}^{\star}$-perfect if and only if
$F$ is $\tau $-perfect and $U\cap F\notin \I$ for each $U\in \tau$ with
$U\cap F\neq\emptyset$.
\end{lem}

By $\M$ and $\N$ we denote, respectively,
the $\sigma $-ideals all meager (i.e. of the first
category) sets and of all \leb null sets in $\real$.
We shall consider
the \h topologies $\tm$ and $\tn$ where
$T$ stands for the natural topology on
$\real$.
Let $\D_\N$ denote the density topology on $\real$. (See e.g. \cite{O} or
\cite{CLO}.) Wilczy\'nski in \cite{W1} introduced the category analogue of
the density topology which will be
denoted by the $\D_\M$. Since topology $\D_M$
is less known, let us give necessary definitions.
A number $x\in \real $ is called a {\it category density point} of a set
$A\subset\real$ with the Baire property if each increasing sequence
$\{ n_k\} $ of positive integers has a subseqence
$\{ n_{m_k}\}$ such that the
sequence of characteristic functions
$$ \chi _{[-1,1]\cap n_{m_k}(A-x)}(t)$$
(where $n_{m_k}(A-x)=\{ n_{m_k}(a-x):a\in A\}$)
tends to $\chi _{[-1,1]}(t)$ for
all points $t\in \real$ except for those belonging to a meager set.
If $[-1,1]$
is replaced by $[-1,0]$ or $[0,1]$,
we get the respective notions of one-sided category density points.
Topology $\dm$ consists of all sets $A\subset\real$ with the
Baire property, such
that each point of $A$ is a category density point of $A$. There are many
analogies between $\dm$ and $\dn$;
for details, see \cite{PWW}, \cite{W2} and
\cite{CLO}.

It is known that the only possible inclusions between the above-mentioned
topologies are the following
$T\zr \Tm \zr\Dm$ and $T\zr\Tn\zr\Dn $. (See \cite{LJW}.)

The following proposition is due to W. Wilczy\'nski (oral communication).
\begin{pr}\label{p3.1}
Every $\dm$-per\-fect set has nonempty $T$-inte\-rior.
\end{pr}

\bp
 Let $F$ be a $\dm$-perfect set. Then $F$ is nonmeager. Indeed, if
$F\in \M$ and $x\in F$ then $U=(\real \setminus F)\cup \{ x \}\in\Dm $ and
$F\cap U= \{ x \} $ which contradicts the fact that $F$ is a
$\dm$-perfect set.
Since $F$ is a nonmeager set with the Baire property,
there is an open interval
$V$ such that $F\cap V $ is comeager in $V$. We claim that $V\subset F$.
Let $x\in V$ and let $W$ be a $\dm$-neighbourhood of $x$.
Thus $x$ is a category
density point of the both sets $V$ and $W$. Consequently,
$V\cap W \notin \M$.
Since $F\cap V$ is comeager in $V$, we thus have $F\cap W \neq\emptyset $.
Hence $x$ belongs to the the $\dm$-closure of $F$ (equal to $F$).
\ep

Let $\Sigma _\M$ and $\Sigma _\N$ denote, respectively,
the $\sigma$-fields of all sets with the
Baire property and of all \leb measurable sets in $\real$.
As in Section 1, $\Sigma (T)$
stands for the field of all subsets of $\real$
with nowhere dense boundary, and
$NWD(T)$ - the ideal of nowhere dense subsets of $\real$.

\begin{tw}\label{tw3.1}
\bd
\item[(a)]
$S(Perf(\Dn ))=\Sigma _\N$, $S^0(Perf(\Dn ))=\N$. (See \cite{R}.)
\item[(b)]
$S(Perf(\Tn ))=\Sigma _\N$, $S^0(Perf(\Tn ))=\N$. (See \cite{BR}.)
\item[(c)]
$S(Perf(\Tm ))=\Sigma (T)$, $S^0(Perf(\Tm ))=NWD(T)$. (See \cite{BR}.)
\item[(d)]
$S(Perf(\Dm ))=\Sigma (T)$, $S^0(Perf(\Dm ))=NWD(T)$.
\ed
\end{tw}

\bp
(a) Let $\f$ denote the family of all
$T$-perfect sets of positive measure.
Burstin \cite{Bu} proved that $\Sigma_\N =\Sf $.
It is not hard to prove that $\f$
and $\mbox{Perf}(\Dn )$ are \mc (cf. \cite[Lemma 3.2]{R}).
Thus by our scheme we have
$S(\mbox{Perf}(\Dn ))=\Sigma_\N$.
Moreover $\Sof\subset H(\Sf)= H(\Sigma_\N) =\N$ and
$\N\subset\Sigma_\N =\Sf $.
But $\N\subset\Sf $ easily implies that $\N\subset
\Sof $. Hence $\N =\Sof = S^0(\mbox{Perf}(\Dn ))$.

(b) If $\f$ is as in (a), Lemma \ref{l3.1} easily
implies that $\f$ and $\mbox{Perf}(\Tn )$
are \mc (cf. \cite{BR}). The rest is the same as in (a).

(c) From Lemma \ref{l3.1} it follows that $\tm$-perfect
sets have nonempty $T$-interior.
Also, a nonempty $T$-open set contains a $\tm$-perfect set (a closed
nondegenerate interval). Hence $\mbox{Perf}(\Tm )$ and $T\zp $
are mutually coinitial.
Thus the assertion follows from Proposition \ref{p4}.

(d) As in (c) it suffices to prove that $\mbox{Perf}(\Dm )$
and $T\zp $ are mutually
coinitial. Firstly, by Proposition \ref{p3.1}, every $\dm$-perfect set has
nonempty $T$-interior. Secondly, let us show that a nondegenerate interval
$[a,b]$  is a $\dm$-perfect set. Indeed, $[a,b]$ is obviously $\dm$-closed.
Let $x\in [a,b]$ and $x\in U\in \Dm $.
Then $x$ is a category density point of
$U$ and $x$ is at least a one-sided category density point of $[a,b]$.
Hence $U\cap [a,b]\notin \M  $ and so, $x$ is a $\dm$-accumulation
point of $[a,b]$.
\ep

Note that statements (a) and (b),(c) of Theorem \ref{tw3.1}
can be extended to cases dealing
with spaces more general than $\Bbb R$, as it was mentioned in \cite{R}
and \cite{BR}. Topology $\dm$ can be considered in certain linear
topological spaces and statement (d) of Theorem \ref{tw3.1} then holds.

Assertions (a),(d) and (b),(c) of Theorem \ref{tw3.1} show a kind
of asymmetry between measure and category. Knowing (a),(b) we
rather expected to obtain $\Sigma_{\cal M}$ and $\cal M$
as the respective Marczewski families in (c),(d).
So the following problem appears:
\begin{prb}
Find a topology $\tau$ on $\Bbb R$ such that $\Sigma_{\cal M}$ (the
$\sigma$-field of all subsets of $\Bbb R$ with the Baire property)
is of the form $S(\mbox{Perf}(\tau ))$.
\end{prb}
Note that Brown \cite{Br1} (see also \cite{BET}) showed the equalities
$\Sigma_{\cal M}=S({\cal G})$ and ${\cal M}=S^0({\cal G})$ provided that
$\cal G$ consists of sets of the form $U\setminus F$ where $U$ is open
and $F$ is an $F_{\sigma}$ meager set. This easily implies that
$\Sigma_{\cal M}=S(\Tm )$ and ${\cal M}=S^0(\Tm )$ since $\cal G$
and $\Tm $ are mutually coinitial. Thus $\Sigma_{\cal M}$ is a topological
field.

\bigskip
\noindent
{\bf Acknowledgements.} We would like to thank the referee
for several valuable remarks.

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\end{document}
