\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} } \newcommand{\hm}{\hspace*{0.25cm} } \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 $0\le a0$ 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. 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