%%%%%%% FOR LATEX ONLY \documentstyle[12pt]{article} \def\integer{\mbox{$\bf Z$}} \def\natural{\mbox{$\bf N$}} \def\rational{\mbox{$\bf Q$}} \def\real{\mbox{$\bf R$}} \newcommand{\text}{\mbox} \newcommand{\Bbb}{\bf} \newcommand{\supp}{\mbox{\rm supp}} \newcommand{\interior}{\mbox{\rm int}} \newcommand{\closure}{\mbox{\rm cl}} \newcommand{\cf}{\mbox{\rm cf}} \newcommand{\non}{\mbox{\rm non}} \newcommand{\cel}{\mbox{\rm c}} \newcommand{\weight}{\mbox{\rm w}} %%%%%%% END FOR LATEX ONLY %% Theorems, etc. \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{problem}{Problem}[section] \newtheorem{example}{Example}[section] \newtheorem{definition}{Definition} \newtheorem{remark}{Fact}[section] \newcommand{\thm}[2]{\begin{theorem}\label{#1}#2\end{theorem}} \newcommand{\cor}[2]{\begin{corollary}\label{#1}#2\end{corollary} } \newcommand{\prop}[2]{\begin{proposition}\label{#1}#2 \end{proposition}} \newcommand{\lem}[2]{\begin{lemma}\label{#1}#2\end{lemma}} \newcommand{\pr}[2]{\begin{problem}\label{#1}#2\end{problem}} \newcommand{\ex}[2]{\begin{example}\label{#1}#2\end{example}} \newcommand{\defi}[2]{\begin{definition}\label{#1}#2 \end{definition}} \newcommand{\rem}[2]{\begin{remark}\label{#1}#2\end{remark}} %%%%%%%%%%%%%%%%%%%%% MISCELLANEOUS \renewcommand{\P}{\mbox{$\cal P$}} \newcommand{\A}{\mbox{$\cal A$}} \newcommand{\B}{\mbox{$\cal B$}} \newcommand{\C}{\mbox{$\cal C$}} \newcommand{\D}{\mbox{$\cal D$}} \newcommand{\E}{\mbox{$\cal E$}} \newcommand{\F}{\mbox{$\cal F$}} \newcommand{\J}{\mbox{$\cal J$}} \newcommand{\U}{\mbox{$\cal U$}} \newcommand{\e}{\varepsilon} % characteristic function \newcommand{\charf}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}} \title{Topologies making a given ideal nowhere dense or meager} \author{Krzysztof Ciesielski\\ \small Dept. of Mathematics, West Virginia Univ., Morgantown, WV 26506-6310 \\ \small (kcies@wvnvms.wvnet.edu)\\ Jakub Jasinski\\ \small Dept. of Mathematics, University of Scranton, Scranton, PA 18510-4666\\ \small (jsj303@jaguar.ucs.uofs.edu)} \date{} \pagestyle{myheadings} \markright{Topologies making a given ideal ... %\ \ \today } \begin{document}\maketitle %\footnotetext[\null]{1991 {\em Mathematics Subject %Classification.} %Primary 54D80. Secondary 54A35, 03E35.} \insert\footins{\vskip .7\footnotesep\footnotesize 1991 {\em Mathematics Subject Classification.} Primary 54D80; Secondary 54A35, 03E35. {\em Key words and phrases.} Nowhere dense, meager, perfectly meager and universally null sets.} \begin{abstract} Let $X$ be a set and let $\J$ be an ideal on $X$. In this paper we show how to find a topology $\tau$ on $X$ such that $\tau$-nowhere dense (or $\tau$-meager) sets are exactly the sets in $\J$. We try to find the ``best'' possible topology with such property. % %change % %%%%%%%%%% added In Section 1 we discuss the ideals $\{\emptyset\}$ and $\P(X)$. We also show that for every ideal $\J\neq\P(X)$ there is a topology $T_0$ making it nowhere dense and that this topology is $T_1$ if $\bigcup\J=X$. Section 2 concerns principal ideals $\P(S)$ for $S\subset X$. It contains characterization of cardinal pairs $(\kappa,\lambda)=(|S|,|X\setminus S|)$ for which $\P(S)$ can be made nowhere dense or meager by compact Hausdorff, metric, and complete metric topologies. Section 3 deals with the ideals containing all singletons. We prove there that it is consistent with ZFC+CH that for every $\sigma$-ideal $\J$ on $\real$ containing all singletons and such that every element of $\J$ is either null or meager, there exists a Hausdorff zero dimensional topology making $\J$ nowhere dense. Section 4 contains the discussion of the above theorem. In particular, it is noticed there that the theorem follows from CH for the ideals with the cofinality $\leq\omega_1$. % %%%%%%%%%% END added \end{abstract} \section{Preliminaries} Most notation used in this paper will follow \cite{Kunen}. If $\tau$ is a topology on $X$ then $N(\tau,X)$ (resp. $M(\tau,X)$) is the family of all $\tau$-nowhere dense (resp. $\tau$-meager) sets. We also write $N(\tau )$ or $N(X)$ if the other parameter is clear from the context. For an ideal $\J$ on a nonempty set $X$ we say that a topology $\tau$ on $X$ {\em makes \J\ meager} ({\em nowhere dense}, respectively) if $\J=M(\tau,X)$ ($\J=N(\tau,X)$, respectively). We will start with some easy remarks. \rem{tauofjay2}{ \raggedright If $\tau $ makes \J\ nowhere dense then:\\ \centering $\tau $ makes \J\ meager if and only if \J\ is a $\sigma$-ideal.} Since every ideal on a finite set is a $\sigma$-ideal we immediately conclude \rem{remFiniteIdeals1}{ \raggedright Let $S$ be a finite subset of $X$ and let $\J\subset\P(S)$ be an ideal. If $\tau$ is a topology on $X$ then:\\ \centering $\tau$ makes \J\ meager if and only if $\tau$ makes \J\ nowhere dense.} Let us begin with considering two special cases of ideals: the trivial ideal $\{\emptyset\}$ and the improper ideal $\P(X)$. For the trivial ideal the following is true. \rem{emptyideal}{ If $\tau$ is a $T_0$ topology on $X$ then the following conditions are equivalent: \begin{description} \item[(1)] $\tau$ makes $\{\emptyset\}$ nowhere dense; \item[(2)] $\tau$ makes $\{\emptyset\}$ meager; \item[(3)] $\tau$ is a discrete topology on $X$. \end{description} } Proof. Equivalence of (1) and (2) and the implication ``(3)$\Rightarrow$(1)'' are obvious. To see that (1) implies (3) first notice that the set $\closure(\{x\})$ is clopen for every $x\in X$, since otherwise a nowhere dense set $\closure(\{x\})\setminus\interior(\closure(\{x\}))$ would be nonempty. Now, if $y\in\closure(\{x\})$ then also $x\in\closure(\{y\})$ since $\closure(\{y\})$ is open and contains $y$. But $X$ is $T_0$, so $y=x$. This means that $\{x\}=\closure(\{x\})\in\tau$ for every $x\in X$. Fact \ref{emptyideal} has been proved. \medskip Notice that in the above the assumption of $X$ being $T_0$ is important since the indiscrete %change space $(X,\{\emptyset,X\})$ also makes the trivial ideal nowhere dense. In the case of improper ideal $\P(X)$ on a set $X$ the situation is a little bit more complicated, as described in the following fact. \rem{fullideal}{ Let $X$ be a nonempty set. \begin{description} \item[(1)] There is no topology on $X$ making $\P(X)$ nowhere dense. \item[(2)] If $X$ is finite then there is no topology on $X$ making $\P(X)$ meager. \item[(3)] There is neither compact $T_2$ nor complete metrizable topology on $X$ making $\P(X)$ meager. \item[(4)] If $X$ is infinite then there is a metrizable topology $\tau$ on $X$ making $\P(X)$ meager. \end{description} } Proof. (1) is obvious, since $X$ is dense in itself. (2) follows from (1), by Fact \ref{remFiniteIdeals1}. (3) follows immediately from the Baire Category Theorem. To see (4) let $Y$ be a set with the same cardinality that $X$ and identify $X$ with $Y\times\rational$, where $\rational$ stands for the set of rational numbers considered with the natural topology. If we equip $Y$ with the discrete topology (or any other metrizable topology), then the product topology on $Y\times\rational$ is metrizable, and it makes $\P(X)=\P(Y\times\rational)$ meager, since the sets $Y\times\{q\}$ are nowhere dense in $Y\times\rational$. \bigskip Facts \ref{emptyideal} and \ref{fullideal} fully describe the situation with the trivial and improper ideals. Thus, in what follows we will consider only proper and non-trivial ideals. Now, let $\J$ be an ideal on $X$. Define $\tau (\J )=\{ X\setminus A\colon A \in \J\} \cup \{\emptyset\}$. The following fact is easy to verify. (See \cite{Janko}. Compare also Lemma \ref{Ni1}.) \thm{tauofjay}{If \J\ is a proper ideal on $X$, then $\tau(\J)$ is a topology on $X$ making \J\ nowhere dense.} It is easy to see that \rem{tauofjay1}{ If\/ $\bigcup \J = X$ then $\tau (\J )$ is a $T_{1}$ but not $T_{2}$ topology.} % However, in general, the elements of $X \setminus \bigcup \J$ are not separated by $\tau (\J )$. Thus, to prove the next theorem, we need to modify topology $\tau (\J )$. \thm{tauofjayT0}{ For every proper ideal \J\ on $X$ there is $T_0$ topology $\tau$ on $X$ making \J\ nowhere dense.} Proof. Extend topology $\tau(\J)$ to $\tau_0(\J)=\tau(\J)\cup\P(X\setminus\bigcup \J)$. It is easy to see that $\tau_0(\J)$ is a $T_0$ topology on $X$. It is also not difficult to see that all sets from $\J$ remain closed nowhere dense, while no new nowhere dense sets are added. \bigskip Since Theorem \ref{tauofjayT0} closes the problem of the existence of $T_0$ topological spaces making a given ideal nowhere dense (meager), for the rest of the paper we will study the spaces that are at least $T_1$. The paper is organized as follows. In Section \ref{secPrincipal} we consider the case where $\J$ is principal. This case seems to be well understood. However, two open problems are stated near the end of Section \ref{secPrincipal}. Section \ref{SecNonPrinc} is devoted to the ideals containing all singletons. Here two main results are based on additional axioms whose role, in several instances, is not entirely clear. \section{Principal Ideals}\label{secPrincipal} Recall that an ideal $\J \subseteq \P (X)$ is said to be {\em principal} if there exists a subset $S\subseteq X$ such that $\J =\P (S)$. This section is devoted to a problem of making such ideals nowhere dense or meager. Since such problem depends only of the cardinality of sets $S$ and $X\setminus S$ the following definition will be useful. For cardinal numbers $\kappa$ and $\lambda$ we say that a topological space $X$ (or its topology) is {\em $(\kappa,\lambda)$ nowhere dense (meager)} provided there exists $S\in[X]^\kappa$ such that $|X\setminus S|=\lambda$ and $N(X)=\P(S)$ ($M(X)=\P(S)$). Let us note the following obvious facts, that follow directly from Fact \ref{tauofjay2}. \rem{RemPrinObv1}{ If $k<\omega$ then topology $\tau$ is $(k,\lambda)$ nowhere dense if and only if it is $(k,\lambda)$ meager.} \rem{RemPrinObv2}{ If a topology $\tau$ is $(\kappa,\lambda)$ nowhere dense then it is $(\kappa,\lambda)$ meager.} Now, suppose that a topology $\tau $ on $X$ makes $\J =\P (S)$ meager. For any $y\in X\setminus S$ the singleton $\{ y\}$ must be dense in some nonempty open set $U$. Thus, $y\in V$ for all nonempty open subsets $V\subseteq U$. It follows that either $U=\{ y\}$ or $\tau$ is not $T_{1}$. This observation can be stated as follows. \rem{RemDiscrete}{ If a $T_1$ topological space $X$ makes $\P(S)$ meager then $\{ y\}$ is open for every $y\in X\setminus S$.} \lem{principal}{Let $\tau$ be a $T_{1}$ topology on $X$ and let $S\subset X$. Topology $\tau$ makes $\J =\P (S)$ nowhere dense iff \begin{description} \item[(1)] $\{ y\}$ is open for every $y\in X\setminus S$; and, \item[(2)] $\interior(S)=\emptyset$; (equivalently, $X\setminus S$ is dense in $X$.) \end{description} } Proof. ``$\Rightarrow$'' Assume that $N(\tau )=\J$. (1) follows from Fact \ref{RemDiscrete}. Thus, $S$ is closed. Now, if $U\subset S$ is open then it is empty since $S$ is nowhere dense, which proves (2). ``$\Leftarrow$'' Assume (1) and (2). For any $A\in N(\tau )$ we must have $A\subseteq S$ because of (1). On the other hand, if $S$ was dense in an open set $U$ then, by (1), $U\subseteq S$ and, by (2), U is empty. \bigskip Lemma \ref{principal} shows that any $T_{1}$ topology making $\J =\P (S)$ nowhere dense is a union of $\P (X\setminus S)$ and a family \F\ of sets meeting $S$. In general \F\ does not need to be closed under supersets but in case of $S$ being a singleton these topologies have a nice characterization. \thm{singleton}{Let $X$ be a set and let $s\in X$. Topology $\tau$ is a $T_{1}$ topology making $\P (\{ s\} )$ nowhere dense if and only if it has a basis of the form \begin{equation} \B =\{ \{ x\}\colon x\in X\setminus \{ s\} \} \cup \F, \label{eq:basis} \end{equation} where \F\ is some non-maximal filter on $X$ such that $\bigcap \F =\{ s \}$.} Proof. ``$\Rightarrow$'' Assume that $\tau$ is a $T_{1}$ topology on $X$ making $\P (\{ s\} )$ nowhere dense. Take $\F =\{ U\in \tau\colon s\in U\}$. Evidently $\F$ is closed under finite intersections. It is closed for taking supersets since, by Lemma \ref{principal}, $\{ x\}$ is open for every $x\in X\setminus \{ s\}$. Thus, $\F$ is a filter. It is not maximal, since $\{ s\} \not\in \F$. Moreover, $\bigcap \F = \{ s\}$, because $\tau$ is $T_{1}$. Clearly \B\/ as in the equation (\ref{eq:basis}) is a basis for $\tau$. ``$\Leftarrow$'' Assume that \B\/ is as in (\ref{eq:basis}) above. It is easy to see that $\tau =\{ \bigcup \B'\colon \B' \subseteq \B\} = \P (X \setminus S) \cup \F$ is a $T_{1}$ topology making $\P (\{ s\} )$ nowhere dense. \cor{t1ist2}{ Any $T_{1}$ topology making $\P(\{s\})$ nowhere dense is also a $T_{2}$ topology.} Notice that Corollary \ref{t1ist2} is false for the ideal $\P(S)$ even if $S$ has only two elements. For example, if we take different $a,b\in\real\setminus\omega$ and define topology $\tau$ on $X=\omega\cup\{a,b\}$ as generated by sets $\{c\}\cup\omega\setminus F$, where $c\in\{a,b\}$ and $F\in[\omega]^{<\omega}$ then $\tau$ is a topology making $\P(S)=\P(\{a,b\})$ nowhere dense which is $T_1$ but not $T_2$. In general, for an arbitrary nonempty $S\subseteq X$, $T_{1}$ topologies on $X$ making $\J=\P(S)$ nowhere dense do not have any simple characterization analogous to Theorem \ref{singleton}. However, for finite sets $S$ we have the following characterization. \thm{finite}{ If $S=\{s_{1}, s_{2}, \ldots, s_{n}\}$ is finite nonempty subset of $X$ and $\tau$ is a $T_{1}$ topology on $X$ then, $\tau$ makes $\P(S)$ nowhere dense if and only if $X=X_{1}\cup X_{2}\cup \ldots \cup X_{n}$, where \begin{description} \item[(a)] $X_{k}\cap S=\{s_{k}\}$, \item[(b)] $X_{k}\in \tau$, \item[(c)] $\tau_{k}=\tau|_{X_{k}}$ makes $\P(\{s_{k}\})$ nowhere dense, \end{description} for every $k=1, 2, \ldots, n$. } Proof. ``$\Leftarrow$'' follows easily from Theorem \ref{singleton}. ``$\Rightarrow$'' For $k\in\{1, 2, \ldots, n\}$ let $X_k=(X\setminus S)\cup\{s_k\}$. Clearly (a) and (b) are satisfied, since $\tau$ is $T_1$. The easy checking of (3) is left to the reader. \bigskip It is worth to mention that the analog of Theorem \ref{finite} does not need to be true for infinite sets $S$, because $\tau|_{S}$ does not need to be discrete. For example, consider $\omega+1$ with the order topology and take $(\omega +1)^{2}$ with the product topology. Clearly, it makes $\P([\{\omega\}\times(\omega +1)]\cup [(\omega +1)\times\{\omega\}])$ nowhere dense while $\{(\omega,\omega)\}$ is not open in $\{\omega\}\times(\omega +1)$. The following facts follow easily from Theorems \ref{singleton} and \ref{finite} and Fact \ref{tauofjay2}. \cor{finitex}{ If $X$ is finite and $S\subset X$ is nonempty then there is no $T_{1}$ topology on $X$ making $\P(S)$ either nowhere dense or meager. In particular, for any $l<\omega$ and $00$ and $\lambda$ be cardinal numbers. The following conditions are equivalent. \begin{description} \item[(i)] There exists a metrizable space $X$ which is $(\kappa,\lambda)$ nowhere dense. \item[(ii)] $\lambda^\omega\geq\kappa$ and $\lambda\geq\omega$. \end{description} } Proof. ``(i)$\Rightarrow$(ii)'' Let $\tau$ be a metrizable topology on $X$ making $\J=\P(S)$ nowhere dense with $|S|=\kappa$ and $|X\setminus S|=\lambda$. Then, by Lemma \ref{principal}, $X\setminus S$ is dense in $X$. Since $S\neq\emptyset$, as $\kappa>0$, we conclude that $\lambda$ is infinite. Moreover, for every $x\in S$ there exists a sequence $s(x)=\{d_n\}\in(X\setminus S)^\omega$ that converges to $x$. Thus, the function $s\colon S\to (X\setminus S)^\omega$ is one-to-one and so, $\kappa=|S|\leq|X\setminus S|^\omega=\lambda^\omega$. ``(ii)$\Rightarrow$(i)'' On $\lambda^\omega$ define metric $d$ by putting for different $f,g\in\lambda^\omega$ \[ d(f,g)=\frac{1}{1+\min\{n\in\omega\colon f(n)\neq g(n)\}}. \] Let $Z_0$ be the family of all $f\in\lambda^\omega$ such that $f$ is equal to zero for almost all $n<\omega$. Thus, \[ |Z_0|=|\lambda^{<\omega}|=\lambda \] and $Z_1=\lambda^\omega\setminus Z_0$ has cardinality $\lambda^\omega$. Define function $G\colon\lambda^\omega\to\real$ by \[ G(f)= \left\{ \begin{array}{cl} [1+\max\{n\in\omega\colon f(n)\neq 0\}]^{-1} & f\in Z_0\\ 0 & f\in Z_1\\ \end{array}\right. \] Now, consider graph of $G$ as a subspace of $\lambda^\omega\times\real$ with the natural product topology. Notice that $G_0=G\cap(Z_0\times\real)$ is dense in $G$ and points of $G_0$ are isolated in $G$. Thus, $G_1=G\cap(Z_1\times\real)=Z_1\times\{0\}$ is closed and nowhere dense in $G$. So, $N(G)=\P(Z_1)$. Take $G_2\in[G_1]^\kappa$ and define $X=G_0\cup G_2$ with the subspace topology of $G$. It is easy to see that $X$ has the desired properties. This finishes the proof. \bigskip Theorem \ref{Metr} shows that there are no $(2^{c}, \omega)$ and $(2^{c}, c)$ nowhere dense metrizable spaces.\footnote{ Letter $c$ stands for the cardinality of continuum, i.e., $c=2^\omega$.} On the other hand, $\beta\natural$ is a separable Hausdorff compact space which is $(2^{c}, \omega)$ nowhere dense. The existence of $(2^{c}, c)$ nowhere dense compact $T_2$ topology is stated in Corollary \ref{corAlex2}. Let us also recall that the density of a topological space $X$ is defined by \[ d(X)=\min\{|D|\colon D \text{ is dense in } X\}+\omega. \] \thm{alexandroff}{ Let $\mu$ be the density of a compact Hausdorff topological space $Y$ without isolated points. If $\mu\leq\lambda\leq|Y|$ then there exists a compact space $X$ which is $(|Y|,\lambda)$ nowhere dense.} Proof. The proof below is essentially a repetition of Alexandroff's duplicated sphere construction. Let $D$ be a dense subset of $Y$ with cardinality $\lambda$. We define set $X$ as $(D\times\{0\})\cup(Y\times\{1\})$. Topology on $X$ is defined by local basis system. Points of $D\times\{0\}$ will be considered as isolated points. For $y\in Y$ a local base of $(y,1)$ is defined as a family of all sets of the form \[ X\cap(U\times\{0,1\})\setminus\{(y,0)\} \] for every open set $U$ in $Y$ containing $y$. It is easy to see that $X$ defined that way is Hausdorff and compact. Also, the set $D\times\{0\}$ is discrete and dense in $X$, since $D$ was dense in $Y$ and $Y$ did not have isolated points. Hence, subspace $S=Y\times\{1\}$ is nowhere dense in $X$, while $X\setminus S=D\times\{0\}$ is an open discrete subspace of $X$ of cardinality $\lambda$. Thus, $X$ is $(|Y|,\lambda)$ nowhere dense. The proof of Theorem \ref{alexandroff} has been completed. \cor{corAlex1}{ Let $\omega\leq\lambda\leq\kappa$. If $\kappa\geq c$ then the following conditions are equivalent: \begin{description} \item[(i)] There exists a compact Hausdorff space which is $(\kappa,\lambda)$ nowhere dense. \item[(ii)] There exists a compactification $\gamma\lambda$ of $\lambda$, considered with the discrete topology, such that $|\gamma\lambda\setminus\lambda|=\kappa$. \item[(iii)] There exists a compact Hausdorff space $X$ with cardinality $\kappa$ and density less then or equal to $\lambda$. \end{description} } Proof. ``(i)$\Rightarrow$(ii)'' follows directly from Lemma \ref{principal}. ``(ii)$\Rightarrow$(iii)'' is obvious. ``(iii)$\Rightarrow$(i)'' follows from Theorem \ref{alexandroff} used with $X^\prime=X\times[0,1]$ considered with the product topology, since then $X^\prime$ does not have isolated points. \medskip Since the density of $\beta\natural$ is equal to $\omega$ and its cardinality is $2^c$ we conclude immediately that \cor{corAlex2}{ There exists a compact Hausdorff topological space which is $(2^c,c)$ nowhere dense.} So, let us consider a general problem for which cardinal numbers $\kappa$ and $\lambda$ there exists a compact Hausdorff topological space which is $(\kappa,\lambda)$ nowhere dense. If $\kappa=0$ then the only $T_0$ space which is $(\kappa,\lambda)$ nowhere dense is a discrete space of cardinality $\lambda$. Thus, $\lambda$ must be finite. If $\kappa>0$ then $\lambda\geq\omega$, since no finite set can have an accumulation point. So, assume that $\kappa>0$ and $\lambda\geq\omega$. Now, if $\kappa\leq\lambda$ then we have the following example. \ex{exComp}{ If $\kappa>0$ and $\lambda\geq\omega$ are cardinal numbers such that $\kappa\leq\lambda$ then there is a compact Hausdorff topological space $X$ which is $(\kappa,\lambda)$ nowhere dense.} Construction. To construct such a space, let $Y$ be a free union of $\kappa$ copies of one point compactification of a discrete space of cardinality $\lambda$. If $\kappa$ is finite, take $X=Y$. If $\kappa$ is infinite, define $X$ as a one point compactification of $Y$. \medskip Notice also that if there is $(\kappa,\lambda)$ nowhere dense compact Hausdorff space then $\kappa\leq 2^{2^\lambda}$, since $|X|\leq 2^{2^{d(X)}}$ for every Hausdorff space $X$ \cite[Theorem 3.2]{Hodel}. Thus, the problem is reduced to $\lambda\geq\omega$ and $\lambda<\kappa\leq 2^{2^\lambda}$. Notice also, that for $\kappa=2^{2^\lambda}$ and $\kappa=2^\lambda$ there is a compact Hausdorff space which is $(\kappa,\lambda)$ nowhere dense. It follows from Corollary \ref{corAlex1} used with $X=\beta\lambda$ and $X=2^\lambda$, respectively. This discussion can be summarized as follows. \pagebreak \cor{corKLCpt}{ {\rm (GCH)} If $\kappa>0$ and $\lambda$ are cardinal numbers then the following conditions are equivalent. \begin{description} \item[(i)] There exists a compact Hausdorff space which is $(\kappa,\lambda)$ nowhere dense. \item[(ii)] There exists a Hausdorff space which is $(\kappa,\lambda)$ nowhere dense. \item[(iii)] $\lambda\geq\omega$ and $\kappa\leq 2^{2^\lambda}$. \end{description} } It is unknown to the authors whether Corollary \ref{corKLCpt} remains true in ZFC. By Corollary \ref{corAlex1} this question can be reformulated as the following problem. \pr{pr1}{ Let $\lambda$ be an infinite cardinal considered with the discrete topology. Can we prove in ZFC that for every cardinal $\lambda<\kappa< 2^{2^\lambda}$ there exists a compactification $\gamma\lambda$ of $\lambda$ such that $|\gamma\lambda\setminus\lambda|=\kappa$? } Notice that the answer for Problem \ref{pr1} is positive (in ZFC) for $\lambda=\omega$ and $\kappa=\omega_1$ \cite[Corollary 6.16]{Walker}. A more general problem can be stated that way. \pr{pr2}{ Characterize completely regular spaces for which: \begin{itemize} \item[(a)] for all cardinal numbers $\kappa\leq 2^{2^{|X|}}$ there is a compactification $\gamma\lambda$ of $X$ such that $|\gamma X\setminus X|=\kappa$; or \item[(b)] there is a compactification $\gamma X$ %change of $X$ such that $|\gamma X\setminus X|=|X|$. \end{itemize} } Notice, that there are completely regular non compact spaces for which (b) in the above fails. For example, it is easy to see that the only compactification of $\omega_1$, considered with the order topology, is its one point compactification $\omega_1+1$. (This is the case, since any continuous function from %change a compactification of $\omega_1$ into $[0,1]$ is eventually constant on $\omega_1$.) Some results of this section concerning $(\kappa,\lambda)$ nowhere dense topological spaces are summarized in the next theorem. \pagebreak \thm{table1}{ In the following table we examine the existence of $(\kappa,\lambda)$ nowhere dense topological spaces in the following classes: all topological spa\-ces, $T_0$ spaces, $T_1$ spaces, $T_2$ spaces, compact spaces C, metrizable spaces M, complete metrizable spaces M$^c$ and separable spaces S. Table 1 lists the best possible classes in which particular $(\kappa,\lambda)$ nowhere dense spaces exist. \bigskip %\begin{table}%[h] \centering\begin{tabular}{||l|c|c|c|c|c||} \hline \hspace{.6in} $\kappa$ & $\kappa=0$ & $0<\kappa<\omega$ & $\kappa=\omega$ & $\kappa=c$ &$\kappa=2^{c}$\\ \hspace{.2in} $\lambda$ & & & & & \\ \hline $\lambda=0$ & -- & -- & -- & -- & -- \\ \hline $0<\lambda<\omega$ &M$^c$C & $T_{0}$ &$T_{0}$ &$T_{0}$& $T_{0}$\\ \hline $\lambda=\omega$ &M$^c$S & CM & CM & CM & C\,$T_{2}$S\\ \hline $\lambda=c$ & M$^c$ & M$^c$ or C\,$T_{2}$ & M$^c$ or C\,$T_{2}$ & M$^c$ or C\,$T_{2}$ &C\,$T_{2}$\\ \hline $\lambda=2^{c}$ & M$^c$ & M$^c$ or C\,$T_{2}$ & M$^c$ or C\,$T_{2}$ & M$^c$ or C\,$T_{2}$ &M$^c$ or C\,$T_{2}$\\ \hline \end{tabular} %\nopagebreak \bigskip \centering {\rm Table 1:} $(\kappa,\lambda)$ nowhere dense spaces. %\end{table} } Proof. Row of $\lambda=0$ follows from Fact \ref{fullideal}(1). Column of $\kappa=0$ is discussed in Fact \ref{emptyideal}. Examples from row $0<\lambda<\omega$ and $\kappa>0$ are as described in Theorem \ref{tauofjayT0}. They cannot be $T_1$ by Lemma \ref{principal}, since no finite subset of $T_1$ space can have an accumulation point. Evidently, by Lemma \ref{principal}, all the spaces from row $\lambda=\omega$ are separable (compact metric spaces are separable), while these from rows with $\lambda>\omega$ cannot be separable. For $\lambda=\omega$ and $0<\kappa\leq\omega$ the spaces from Example \ref{exComp} are also metrizable, since they are countable. For $\lambda=\omega$ and $\kappa=c$ notice that \[ X=([0,1]\times\{0\})\cup \{(\frac{m}{n},\frac{1}{n})\in[0,1]^2\colon m \text{ and } n \text{ are relatively prime}\} \] consider with the subspace topology of the plane is a compact metric $(c,\omega)$ nowhere dense space. Since, by Theorem \ref{Metr}, there is no metrizable $(2^c,\omega)$ nowhere dense space, $\beta\natural$ is the best possible $(2^c,\omega)$ nowhere dense space. Similarly, by Theorem \ref{Metr}, there is no metrizable $(2^c,c)$ nowhere dense space, so the example from Corollary \ref{corAlex2} is the best possible. In the remaining seven cases the existence of compact $T_2$ spaces from Example \ref{exComp}. There is no compact metrizable example in these cases, since such an example would be separable. Moreover, in all this cases $\kappa\leq\lambda$. Thus, the complete metrizable spaces can be constructed as a free sum of $\kappa$ many spaces $\omega+1$ and of discrete space of cardinality $\lambda$. This finishes the proof of Theorem \ref{table1}. \bigskip For the similar theorem on $(\kappa,\lambda)$ meager spaces we need also the following fact. \rem{CompactMeager}{ If $X$ is a compact $T_2$ or complete metrizable space then $X$ is $(\kappa,\lambda)$ nowhere dense if and only if $X$ is $(\kappa,\lambda)$ meager.} Proof. Implication ``$\Rightarrow$'' follows from Fact \ref{RemPrinObv2}. For the other direction let $X$ be a compact $T_2$ which is $(\kappa,\lambda)$ meager and let $S\subset X$ be such that $M(X)=\P(S)$. By Fact \ref{RemDiscrete} $\{y\}$ is open for every $y\in X\setminus S$. So, $S$ is closed in $X$. It is also meager, by our assumption. Hence, by Baire Category Theorem, $\interior(S)=\emptyset$. So, $S$ is nowhere dense and $\P(S)=N(X)$. \bigskip The results on $(\kappa,\lambda)$ meager topological spaces similar to these of Theorem \ref{table1} are summarized in the next theorem. \thm{table2}{ In the following table we examine the existence of $(\kappa,\lambda)$ meager topological spaces in the following classes: all topological spaces, $T_0$ spaces, $T_1$ spaces, $T_2$ spaces, compact spaces C, metrizable spaces M, complete metrizable spaces M$^c$ and separable spaces S. Table 2 lists the best possible classes in which particular $(\kappa,\lambda)$ meager spaces exist. \bigskip %\begin{table}%[h] \centering\begin{tabular}{||l|c|c|c|c|c||} \hline \hspace{.6in} $\kappa$ & $\kappa=0$ & $0<\kappa<\omega$ & $\kappa=\omega$ & $\kappa=c$ &$\kappa=2^{c}$\\ \hspace{.2in} $\lambda$ & & & & & \\ \hline $\lambda=0$ & -- & -- & MS & MS & M \\ \hline $0<\lambda<\omega$ &M$^c$C & $T_{0}$ & MS & MS & M \\ \hline $\lambda=\omega$ &M$^c$S & CM & CM & CM & M or C\,$T_{2}$S\\ \hline $\lambda=c$ & M$^c$ & M$^c$ or C\,$T_{2}$ & M$^c$ or C\,$T_{2}$ & M$^c$ or C\,$T_{2}$ &M or C\,$T_{2}$\\ \hline $\lambda=2^{c}$ & M$^c$ & M$^c$ or C\,$T_{2}$ & M$^c$ or C\,$T_{2}$ & M$^c$ or C\,$T_{2}$ &M$^c$ or C\,$T_{2}$\\ \hline \end{tabular} \bigskip \centering {\rm Table 2:} $(\kappa,\lambda)$ meager spaces. } Proof. The two first columns of Table 2 are identical to those of Table 1 by Fact \ref{remFiniteIdeals1}. For the remaining cases first notice that, by Fact \ref{CompactMeager}, the compact $T_2$ entries and complete metrizable entries in Table 2 must be the same as that these in Table 1. The entries for $\lambda=0$ and $\kappa\in\{\omega,2^c\}$ are justified by Fact \ref{fullideal} (4). For $\lambda=0$ and $\kappa=c$ take $\real\times\rational$ with the subspace topology of the plane. The non compact metrizable $(\kappa,\lambda)$ meager spaces for $\lambda>0$ and $\kappa\geq\omega$ can be obtain by taking a free union of $(\kappa,0)$ meager metrizable example and a discrete space of cardinality $\lambda$. These spaces will be separable for $\lambda<\omega$ and $\kappa\in\{\omega,c\}$. They cannot be separable for $\kappa=2^c$ since separable metric spaces have cardinality less then or equal to $c$. They cannot be separable for $\lambda>\omega$ by Fact \ref{RemDiscrete}. This finishes the proof of Theorem \ref{table2}. \section{Ideals containing all singletons}\label{SecNonPrinc} Assume that $X$ is an infinite set and $\J=[X]^{<\omega}$. By Fact \ref{tauofjay1} the family $\tau(\J)=\{X\setminus N\colon N\in\J\}\cup\{\emptyset\}$ is a $T_{1}$ topology with $N(\tau,X)=\J$. Taking free union of such spaces we obtain the following. \ex{jayfin}{For any $k<\omega$ there exists a $T_{1}$ space X with k disjoint open sets and $N(\tau, X)=[X]^{<\omega}$. } This stands in contrast with the next lemma. \lem{disopen1}{Let $(\tau,X)$ be a $T_{1}$ space with $[X]^{<\omega}\subseteq N(\tau,X)$. If there exists an infinite family $\{B_{n}\colon n<\omega\}$ of nonempty disjoint open sets, then there is an infinite set $Y\in N(\tau,X)$. } Proof. Let $Y=\{y_{n}\colon n<\omega\}$ be a selector from $\{B_{n}\colon n<\omega\}$. If $V\in\tau$ is nonempty, then $V\cap B_{n}\cap Y\subseteq\{y_{n}\}$ for each $n<\omega$. Since $\{y_{n}\}\in N(\tau,X)$, $Y$ cannot be dense in $V$. \lem{disopen2}{If $X$ is an infinite Hausdorff space, then there exists an infinite family $\{U_{n}\colon n<\omega\}$ of nonempty pairwise disjoint open sets. } Proof. A folklore diagonal argument is left to the reader. \thm{jayfinHausdorff}{There is no Hausdorff topology on $X$ making the ideal $[X]^{<\omega}$ nowhere dense. } Proof. For finite $X$ it is obvious. So, suppose that $\tau$ is a Hausdorff topology on an infinite set $X$ such $[X]^{<\omega}\subseteq N(\tau,X)$. By Lemma \ref{disopen2} there exists an infinite family of open nonempty disjoint sets in $X$ and, by Lemma \ref{disopen1}, there exists an infinite nowhere dense subset of $X$. \bigskip Notice also that $M(X)$ is always a $\sigma$--ideal. So we cannot have $M(X)=[X]^{<\omega}$. Hence, \rem{finmeager}{There is no topology making $[X]^{<\omega}$ meager.} By Fact \ref{tauofjay1} for any uncountable set $X$ there exists a (not second countable) $T_{1}$ topology on $X$ making $[X]^{\leq\omega}$ nowhere dense. Theorem \ref{lusin} proved below shows that the existence of such second countable topologies is independent of ZFC axioms. (This also follows from Kunen's theorem quoted below as Fact \ref{kun}.) Recall that an uncountable separable metric space is called a {\em Lusin space } if $M(X)=[X]^{\leq\omega}$. It is known that the Continuum Hypothesis CH implies the existence of Lusin spaces while under the Martin's Axiom MA such spaces do not exist. (See \cite[p.205]{Miller} or \cite{Kunen2}.) Thus, the existence of Lusin spaces is independent of ZFC axioms. \thm{lusin}{If there exists an uncountable, second countable $T_{1}$ space $X$ such that $M(\tau,X)\subseteq[X]^{\leq\omega}$ then there exists a Lusin space. } Proof. Let $S$ be a union of boundaries from all sets of some countable open basis for $X$. Then, $X\setminus S$ is an uncountable $T_{1}$ zero dimensional space. By Urysohn Metrization Theorem \cite[p.256]{Croom} $X\setminus S$ is metrizable, hence it is a Lusin space. \rem{kun}{{\rm (Kunen \cite{Kunen2})} %change MA$+\neg$CH implies that there is no uncountable Hausdorff space $X$ with $M(X)=[X]^{\leq\omega}$. } A topology $\tau$ on $X$ is said to be {\em compatible with an ideal $\J$} on $X$ if for any $Y\subseteq X$ the following condition holds: if for every $x\in Y$ there exists an open neighborhood $x\in U$ such that $U\cap Y\in \J$, then $Y\in\J$. The following facts are due to Nij\aa stad. (See \cite{Janko} and \cite{Nijastad}.) \lem{Ni1}{If a topology $\tau$ is compatible with an ideal \J\ on $X$, then $\tau^{\cal J}=\{U\setminus N\colon U\in\tau$ and $N\in\J\}$ is a topology on $X$. } Notice that any second countable topology on $X$ is compatible with any $\sigma$-ideal on $X$. Thus, the family $\tau^{\cal J}$ in the next lemma is indeed a topology. \lem{Ni2}{ Let $(X,\tau)$ be a second countable topological space and let \J\ be a $\sigma$-ideal on $X$ such that $\J\cap\tau=\{\emptyset\}$. If $\tau^{\cal J}$ is a topology as in Lemma \ref{Ni1} then $N\in N(\tau^{\cal J},X)$ if and only if $N=M\cup I$ for some $M\in N(\tau,X)$ and $I\in \J$. } Proof. ``$\Rightarrow$'' Assume that $N\in N(\tau^{\cal J}, X)$. Let $\{U_{n}\colon n<\omega\}$ be a basis for $\tau$. For each $n<\omega$ there exist a nonempty open $V_{n}\subseteq U_{n}$ and $I_{n}\in\J$ such that $N\cap (V_{n}\setminus I_{n})=\emptyset$. Take $M=N\setminus \bigcup\{I_{n}\colon n<\omega\}$ and $I=N\cap\bigcup\{I_{n}\colon n<\omega\}$. We have $M\in N(\tau,X)$ and $I\in \J$. Clearly $N=M\cup I$. ``$\Leftarrow$'' Now let $N=M\cup I$ where $M\in N(\tau, X)$ and $I\in\J$. For any nonempty open set $U\in \tau$ there exists a nonempty open set $V\subseteq U$ which is disjoint with $M$. Clearly, if $I'\in \J$ then $V'=V\setminus(I\cup I')\subseteq U\setminus I'$, $V'\in\tau^{\cal J}$, and $V'\cap N=\emptyset$. Also $V'\neq \emptyset$, since $\J\cap \tau=\{\emptyset\}$. Thus, $N\in N(\tau^{\cal J},X)$. \lem{T2top1}{{\rm (CH)} Let $A$ be a set with $|A|=c$. If \J\ is a $\sigma$-ideal on $A$ containing all singletons, then there exists a Hausdorff topology $\tau^\prime$ making \J\ meager. } Proof. Let $\tau_0$ be a topology on $A$ such that $(A,\tau_0)$ is a Lusin space. Then, $M(\tau_0,A)=[A]^{\leq\omega}$. Let $D\subset A$ be a countable dense subset and let $\J^\prime=\J|_{A\setminus D}=\{J\cap(A\setminus D)\colon J\in\J\}$. Clearly $\J^\prime$ is a $\sigma$-ideal on $A$ such that $\J^\prime\cap \tau_0=\{\emptyset\}$. Hence, by Lemma \ref{Ni1}, $\tau^\prime=\tau_0^{{\cal J}^\prime}$ is a topology on $A$. Moreover, by Lemma \ref{Ni2}, $M(\tau^\prime,A)=\{M\cup I\colon M\in M(\tau_0,A)$ and $I\in \J^\prime\}$. It follows that \begin{eqnarray*} M(\tau^\prime, A) & = & \{M\cup I\colon M\in M(\tau_0,A) \text{ and } I\in\J^\prime\}\\ & = & \{D'\cup I\colon D'\in[A]^{\leq\omega} \text{ and } I\in\J^\prime\}\\ & = & \J. \end{eqnarray*} Clearly $\tau^\prime$ is a Hausdorff, since $\tau_0$ was. \thm{T2top}{{\rm (CH)} For any $\sigma$-ideal \J\ on a set $X$ of cardinality continuum there exists a Hausdorff topology $\tau$ making \J\ meager. } Proof. If $A=\bigcup\J$ is uncountable let $\tau^\prime$ be a topology on $A$ as in Lemma \ref{T2top1}. Define $\tau$ as a topology on $X$ generated by $\tau^\prime\cup\P(X\setminus A)$. It is easy to see that $\tau$ has desired properties. If $\bigcup\J$ is at most countable then a metrizable topology $\tau$ exists by Theorem \ref{table2}. \bigskip Notice that the topology $\tau$ from Theorem \ref{T2top} does not need to be regular. To see this take $\J=[A]^{\leq\omega}$. Then, $\tau=\tau^\prime=\tau_0^{{\cal J}^\prime}$, as in Lemma \ref{T2top1}. Let $a\in A$ be any condensation point for $\tau_0$. There is a sequence $X=\{x_{n}\colon n<\omega\}\subset A\setminus (\{a\}\cup D)$ of $\tau_0$-condensation points such that $\lim_{n \rightarrow \infty} x_{n}=a$, where $D$ is a dense countable set from the proof of Lemma \ref{T2top1}. Now take any $\tau_0$-open neighborhood $U$ of $a$ and a set $C=(A\setminus U)\cup X$. The set $C$ is closed in $(A,\tau^\prime)$ but it cannot be separated from $a$ since if $C\subset W\in \tau^\prime$ and $a\in V=V_0\setminus A\in \tau^\prime$, where $V_0\in\tau_0$ and $I\in[A\setminus D]^{\leq\omega}$, then there exists $x_n\in V_0$ and $U\in \tau^\prime$ such that $x_{n}\in U\subset T$ and $U\cap V\neq\emptyset$. The following theorem implies that for a large number of natural $\sigma$-ideals $\J$ on $\real$ it is consistent with ZFC that there exists a zero dimensional {\em regular} topological space $X$ making $\J$ nowhere dense. \thm{thForcing}{ There exists a model of ZFC+GCH in which the following holds. There exists a one-to-one mapping $e\colon\real\to 2^{\omega_2}$ such that $e[\real]$ is a dense subspace of $2^{\omega_2}$ having the following properties: \begin{description} \item[(i)] every nowhere dense subset of $e[\real]$ is at most countable; \item[(ii)] for every $\alpha<\omega_2$ and every $r\in\real$ there exists an ordinal $\xi<\omega_1$ such that $e(r)(\alpha+\xi)=0$; \item[(iii)] $e[A]$ has an empty interior in $e[\real]$ %change for every meager and every null subset $A$ of $\real$. \end{description} } Before we prove Theorem \ref{thForcing} we will show the following corollary. \cor{corMeagering}{ There exists a model of ZFC+GCH in which the following holds. For every $\sigma$-ideal $\J$ of $\real$ containing all singletons and such that every element of $\J$ is either meager or null, there exists a Hausdorff zero dimensional topology $\tau_{\cal J}$ on $\real$ making $\J$ nowhere dense. In particular, there exist Hausdorff zero dimensional topologies on $\real$ making nowhere dense the following ideals: perfectly meager sets, universally measure zero sets, strong measure zero sets, null sets, etc. } Proof. Let $\tau$ be a topology on $\real$ generated by mapping $e\colon\real\to 2^{\omega_2}$ from Theorem~\ref{thForcing}. Then, a base of $(\real,\tau)$ induced by a standard product basis of $2^{\omega_2}$ is given by sets \begin{eqnarray*} U_{\varepsilon} & = & \{r\in\real\colon\e\subset e(r)\}\\ & = & \{r\in\real\colon(\forall a\in\text{dom}(\e))(e(r)(a)=\e(a))\} \end{eqnarray*} for all $\e\in H(\omega_2)$, where $H(A)$ is defined as a set of all functions from finite subsets of $A$ into $2$. Notice that sets $U_{\varepsilon}$ are nonempty, since $e[\real]$ is dense in $2^{\omega_2}$. Let $\J$ be a $\sigma$-ideal as stated in the assumptions. We will define $\tau_{\cal J}$ by modifying $\tau$. Let $\beta<\omega_2$ be such that the sets $\{U_{\varepsilon}\colon \e\in H(\beta)\}$ separate points of $\real$. Such $\beta$ can be found, since $|\real|=\omega_1$. Let \[ \Lambda=\{\alpha<\omega_2\colon\text{ cofinality of $\alpha$ is } \omega_1\} \] and let $\{J_\alpha\colon \alpha\in\Lambda\}$ be an enumeration of $\J$ such that \begin{equation}\label{conBeta} J_\alpha=\emptyset\ \ \ \text{for all}\ \ \ \alpha<\beta. \end{equation} For $\alpha\in\Lambda$, $\xi<\omega_1$ and $i<2$ we put \[ U^{\cal J}_{<\alpha+\xi,i>}= \left\{ \begin{array}{cl} U_{\{<\alpha+\xi,0>\}}\setminus J_\alpha & i=0\\ U_{\{<\alpha+\xi,1>\}}\cup J_\alpha & i=1\\ \end{array}\right. \] and define a base of $\tau_{\cal J}$ as a family of all sets \[ U_{\varepsilon}^{\cal J}=\bigcap_{a\in{\rm dom}(\varepsilon)} %change U^{\cal J}_{} \] for $\e\in H(\omega_2)$. Notice that \begin{equation}\label{conIdeal} U^{\cal J}_{\varepsilon}\triangle U_{\varepsilon}\in\J\ \ \ \text{for every}\ \ \ \e\in H(\omega_2). \end{equation} Clearly, $(\real,\tau_{\cal J})$ is zero dimensional. It is Hausdorff, since, by (\ref{conBeta}), $U^{\cal J}_{\e}=U_{\e}$ for all $\e\in H(\beta)$ and $\{U_{\e}\colon \e\in H(\beta)\}$ separates points. Thus, it is enough to show that the nowhere dense sets of $(\real,\tau_{\cal J})$ are precisely the sets from $\J$. It is easy to see that sets from $\J$ are closed, since if $J\in\J$ and $\alpha\in\Lambda$ is such that $J=J_\alpha$ then, by (ii) of Theorem~\ref{thForcing}, \[ \real\setminus\bigcup_{\xi<\omega_1} U^{\cal J}_{\{<\alpha+\xi,0>\}}= \real\setminus\left( \bigcup_{\xi<\omega_1} U_{\{<\alpha+\xi,0>\}}\setminus J_\alpha \right) =\real\setminus(\real\setminus J_\alpha) =J. \] Also, the sets from $\J$ are nowhere dense, since, by (iii) of Theorem~\ref{thForcing}, $U_{\e}\not\in\J$, while, by (\ref{conBeta}), $U^{\cal J}_{\e}\triangle U_{\e}\in\J$ for every $\e\in H(\omega_2)$. So, we have proved that every set from $\J$ is nowhere dense in $(\real,\tau_{\cal J})$. To finish the proof choose a closed nowhere dense set $F$ in $(\real,\tau_{\cal J})$. We will show that $F\in\J$. Let $\{\e_s\in H(\omega_2)\colon s\in S\}$ be such that $F=\real\setminus\bigcup_{s\in S}U^{\cal J}_{\e_s}$. Since $F$ is nowhere dense, for every $\e\in H(\omega_2)$ there is $s\in S$ such that $U^{\cal J}_{\e}\cap U^{\cal J}_{\e_s}\neq\emptyset$. This implies that for every $\e\in H(\omega_2)$ there exists $s\in S$ such that \[ U_{\e}\cap U_{\e_s}\neq\emptyset. \] By a simple closure operation we can find a countable infinite set $T\subset \omega_2$ such that for every $\e\in H(T)$ there is $s\in S$ such that \[ U_{\e}\cap U_{\e_s}\neq\emptyset\ \ \ \&\ \ \ \e_s\in H(T). \] Let $S_0=\{s\in S\colon \e_s\in H(T)\}$. Thus, $S_0$ is at most countable and for every $\e\in H(\omega_2)$ there is $s\in S_0$ such that \[ U_{\e}\cap U_{\e_s}\neq\emptyset, \] i.e., the set $F_0^\prime=\real\setminus\bigcup_{s\in S_0}U_{\e_s}$ is nowhere dense in $(\real,\tau)$. So, by (i) of Theorem~\ref{thForcing}, $F_0^\prime$ is at most countable and, in particular, $F_0^\prime\in\J$. But if $F_0=\real\setminus\bigcup_{s\in S_0}U^{\cal J}_{\e_s}$ then, by (\ref{conIdeal}), $F_0\triangle F_0^\prime\in\J$. Hence, $F_0\in\J$. But $F\subset F_0$. Thus, $F\in\J$. This finishes the proof of the Corollary \ref{corMeagering}. \bigskip {\bf Proof of Theorem \ref{thForcing}.} Let $V$ be a model of ZFC+GCH. We will find a generic extension of $V$ in which the theorem holds. The forcing we will use is $\omega$-closed and satisfies $\omega_2$-cc. Thus, it preserves cardinal numbers and the real numbers from $V$ and from its extension are the same. We will denote this set of real numbers by $\real$. Let \begin{equation}\label{conE} \E=\{\{\e_n\}_{n<\omega}\in[H(\omega_2)]^{\leq\omega}\colon (\forall \e\in H(\omega_2))(\exists n<\omega)(\e\cup\e_n\in H(\omega_2))\} \end{equation} and let \[ {\Bbb S}=\{\{f_r\in H_\omega(\omega_2)\colon r\in A\}\colon A\in[\real]^{\leq\omega}\}, \] where $H_\omega(X)$ stands for the functions from at most countable subsets of $X$ into $2$. The forcing notion we will use is defined as ${\Bbb P}={\Bbb S}\times[\E]^{\leq\omega}$. The partial order on ${\Bbb P}$ is defined by \begin{eqnarray} <_{r\in A},\D> & \leq & <_{r\in A^\prime},\D^\prime> \nonumber \\ & \Longleftrightarrow & \D\supset\D^\prime\ \ \&\ \ A\supset A^\prime\ \ \&\ \ (\forall r\in A^\prime)(f_r\supset f_r^\prime) \nonumber \\ & \& & (\forall r\in A\setminus A^\prime)(\forall \{\e_n\}\in \D^\prime)(\exists n<\omega)(\e_n\subset f_r) \label{conFor} \end{eqnarray} for all $<_{r\in A},\D>, <_{r\in A^\prime},\D^\prime>\in{\Bbb P}$. It is easy to see that ${\Bbb P}$ is $\omega$-closed. To see that ${\Bbb P}$ satisfies $\omega_2$-cc take a sequence $\{<_{r\in A^\xi},\D^\xi>\in{\Bbb P}\colon \xi<\omega_2\}$ and notice that we can choose a subset $I\in[\omega_2]^{\omega_2}$ such that $A^\xi=A^\zeta$ for all $\xi,\zeta\in I$. Now, using CH and a standard $\Delta$-system argument, we can find $\xi,\zeta\in I$ such that $f_r^\xi\cup f_r^\zeta\in H_\omega(\omega_2)$ for every $r\in A=A^\xi=A^\zeta$. Clearly, $<_{r\in A},\D^\xi\cup\D^\zeta>$ extends $<_{r\in A^\xi},\D^\xi>$ and $<_{r\in A^\zeta},\D^\zeta>$. Now, let $G$ be a $V$-generic filter over ${\Bbb P}$ and define mapping $e\colon\real\to 2^{\omega_2}$ by putting for $s\in \real$, \[ e(s)=\bigcup\{g_s\colon (\exists <_{r\in A},\D>\in G)(s\in A\ \&\ g_s=f_s)\}. \] In order to show that this definition is correct it has to be argued that the following sets are dense in ${\Bbb P}$ for every $s\in \real$ and $\xi<\omega_1$ \[ D_s=\{<_{r\in A},\D>\in{\Bbb P}\colon s\in A\} \] and \[ D_s^\xi= \{<_{r\in A},\D>\in{\Bbb P}\colon s\in A\ \&\ \xi\in\text{dom}(f_s)\}. \] To see that $D_s$ is dense in ${\Bbb P}$ take $<_{r\in A},\D>\in{\Bbb P}$ and assume that $s\not\in A$. Let $\D=\{\{\e_n^k\}\colon k<\omega\}$ and define by induction on $m<\omega$ a sequence $\{n_m\}_{m<\omega}$ such that $\bigcup_{k_{r\in A\cup\{s\}},\D>$ extends $<_{r\in A},\D>$. To see that $D_s^\xi$ is dense in ${\Bbb P}$ take $<_{r\in A},\D>\in{\Bbb P}$. By the density of $D_s$ we can assume that $s\in A$. If $\xi\not\in\text{dom}(f_s)$, extend $f_s$ onto $\xi$ arbitrarily and notice that such obtained condition extends $<_{r\in A},\D>$. Thus, we proved that indeed $e\colon\real\to 2^{\omega_2}$. To see that $e$ is one-to-one it is enough to see that the set \begin{eqnarray*} D_{s,t}= \{<_{r\in A},\D> & \in{\Bbb P}\colon & s,t\in A \\ & \&\ & (\exists \xi\in\text{dom}(f_s) \cap\text{dom}(f_t))(f_s(\xi)\neq f_t(\xi))\} \end{eqnarray*} is dense in ${\Bbb P}$ for every distinct $s,t\in\real$. Similarly, the density of $e[\real]$ in $2^{\omega_2}$ follows from the density of \[ D_{\e}= \{<_{r\in A},\D>\in{\Bbb P}\colon (\exists r\in A)(\e\subset f_r)\} \] for every $\e\in H(\omega_2)$. The density of both these types of sets can be proved similarly as that of $D_s^\xi$ and $D_s$, respectively. We will leave it as an exercise. To argue for (iii) let $\e\in H(\omega_2)$ and let $J$ be either meager or null. We will show that $U_\e\not\subset J$. We can assume that $J$ is a Borel set, extending it, if necessary. Thus, $J$ is already in $V$ and the set \[ D_{\e}^J= \{<_{r\in A},\D>\in{\Bbb P}\colon (\exists r\in A\setminus J)(\e\subset f_r)\} \] belongs to $V$. The density of this set is proved similarly as that of $D_{\e}$. This easily implies (iii). To prove (ii) it is enough to notice that for every $\alpha<\omega_2$ and $s\in\real$ the set \[ E_s^\alpha=\{<_{r\in A},\D>\in{\Bbb P}\colon s\in A\ \&\ (\exists \xi<\omega_1)(<\alpha+\xi,0>\in f_s)\} \] is dense in ${\Bbb P}$. The proof of this fact is identical to the proof of the density of $D_s^\xi$. To finish the proof it is enough to show (i). So, let $F$ be a closed nowhere dense subset of $e[\real]$. Choose a family $\{\e_s\in H(\omega_2)\colon s\in S\}$ such that $F=e[\real]\setminus\bigcup_{s\in S}[\e_s]$, where $[\e]=\{f\in 2^{\omega_2}\colon \e\subset f\}$. Since $F$ is nowhere dense in $e[\real]$, for every $\e\in H(\omega_2)$ there is $s\in S$ such that $e[\real]\cap[\e]\cap[\e_s]\neq\emptyset$. Similarly as in Corollary \ref{corMeagering} we can find a countable subset $S_0$ of $S$ such that for every $\e\in H(\omega_2)$ there is $s\in S_0$ such that $e[\real]\cap[\e]\cap[\e_s]\neq\emptyset$. Notice that $F_0=e[\real]\setminus\bigcup_{s\in S_0}[\e_s]$ contains $F$ and that $D=\{\e_s\colon s\in S_0\}\in\E$. We will show that $F_0$ is at most countable. Since the set \[ E_D=\{<_{r\in A},\D>\in{\Bbb P}\colon D\in\D\} \] is dense in ${\Bbb P}$ we can find $<_{r\in A^\prime},\D^\prime>\in G$ such that $D\in\D^\prime$. We will show that $F_0\subset A^\prime$. Let $t\in \real\setminus A^\prime$. It is enough to show that $e[t]\in \bigcup_{s\in S_0}[\e_s]$. Choose $<_{r\in A},\D>\in G$ extending $<_{r\in A^\prime},\D^\prime>$ and such that $t\in A$. Then, by (\ref{conFor}), there exists $s\in S_0$ such that $\e_s\subset f_t\subset e(t)$. Thus, $e(t)\in [\e_s]$. This finishes the proof of Theorem \ref{thForcing}. \section{ Remarks on Theorem 3.9.} First notice that if we assume only the Continuum Hypothesis then the following weaker version of Theorem \ref{thForcing} can be proved. \thm{thForcWeak}{{\rm (CH)} There exists a one-to-one mapping $e\colon\real\to 2^{\omega_1\times\omega_1}$ such that $e[\real]$ is a dense subspace of $2^{\omega_1\times\omega_1}$ and: \begin{description} \item[(i)] every nowhere dense subset of $e[\real]$ is at most countable; \item[(ii)] for every $\alpha<\omega_1$ and every distinct $r,s\in\real$ %change there exist $\xi<\eta<\omega_1$ %change such that $e(r)(<\alpha,\xi>)=0$ and $e(s)(<\alpha,\eta>)=1$; %change \item[(iii)] $e[A]$ has an empty interior in $e[\real]$ %change for every $A\subset\real$ which is either meager or null. \end{description} } Sketch of proof. Change the definition of forcing ${\Bbb P}$ in Theorem \ref{thForcing} by replacing $\omega_2$ with $\omega_1\times\omega_1$. Let $\F$ be the family of all dense subsets of ${\Bbb P}$ considered in the proof of Theorem \ref{thForcing}, i.e., sets $D_s$, $D_s^\xi$, $D_{s,t}$, $D_{\varepsilon}^J$, $E^\alpha_s$ and $E_D$, where indexes are chosen as in the proof. (Evidently in the definition of $E^\alpha_s$ we replace $\alpha+\xi$ with $<\alpha,\xi>$.) Then $\F$ has cardinality $\omega_1$ and we can easily construct, by transfinite induction of length $\omega_1$, a filter $G$ in ${\Bbb P}$ intersecting every set in $\F$. But this is all we need to conclude (i)-(iii). \bigskip Although Theorem \ref{thForcWeak} is very similar to Theorem \ref{thForcing} we cannot, in general, deduce from it the conclusion of Corollary \ref{corMeagering}. This is the case, since to this order we would need to modify the topology of $e[\real]\subset 2^{\omega_1\times\omega_1}$ by $|\J|$-many sets and we have only $\omega_1$ coordinates to make this adjustment. However, if there exists $\{J_\alpha\colon\alpha<\omega_1\}\subset\J$ cofinal in $\J$% \footnote{ Under CH it is not the case for the ideals of perfectly meager sets and universally null sets. This follows from Lemma \ref{Grzegorek}.} (i.e., such that for every $J\in\J$ there is $\alpha<\omega_1$ with $J\subset J_\alpha$), then similarly as in Corollary \ref{corMeagering} we can find a zero dimensional Hausdorff topology on $\real$ making $\J$ meager. In particular, we can conclude the following. \cor{corCHmeager}{ {\rm (CH)} If $\J$ is a $\sigma$-ideal on $\real$ containing all singletons and having cofinality $\omega_1$ (i.e., with cofinal subfamily of cardinality $\omega_1$) then there is a zero dimensional Hausdorff topology on $\real$ making $\J$ nowhere dense. In particular, there exist a zero dimensional Hausdorff topologies making null sets and (ordinary) meager sets nowhere dense. } In the recent years two refinements of the natural topology on $\real$ have been intensively studied: the density topology and the ${\cal I}$-density topology. (For summary of topological properties of these topologies see \cite{CLO:Book}.) They make, respectively, null sets and ordinary meager sets nowhere dense. However, both of these topologies are connected. Moreover, the ${\cal I}$-density topology is Hausdorff but not regular. The density topology is completely regular but not normal. In this context the following questions seems to be interesting. \pr{probNullSet}{ Can we find in ZFC a zero dimensional Hausdorff topology on $\real$ making meager sets nowhere dense?} % It is easy to see that under Martin's Axiom such a topology exists. \pr{probNormal}{ Can topologies from Corollaries \ref{corMeagering} or \ref{corCHmeager} be normal? compact? metrizable? In particular, can we have such topologies for the ideals of meager sets or null sets? } Notice that neither of Theorem \ref{thForcing}, Corollary \ref{corMeagering} and Corollary \ref{corCHmeager} can be proved in full generality in ZFC. More precisely, Corollaries \ref{corMeagering} and \ref{corCHmeager} fail under MA+$\neg$CH for the ideal $[\real]^{\leq\omega}$ by Fact \ref{kun}. The Corollaries also fail for the ideal $SMZ$ of strong measure zero sets under the Borel Conjecture, i.e., when $SMZ=[\real]^{\leq\omega}$. This is the case since every Lusin set has strong measure zero, i.e., Borel Conjecture implies that there are no Lusin sets. In what follows we will present few additional examples showing farther limitations on possible generalizations of Theorem \ref{thForcing} and Corollaries \ref{corMeagering} or \ref{corCHmeager}. The first fact shows that the assumptions on the ideal $\J$ in Corollary \ref{corMeagering} cannot be completely disregarded if we like to make $\J$ nowhere dense. \rem{maximal}{If $X$ has at least two elements and $\J$ is a maximal ideal on $X$, then there is no Hausdorff topology making $\J$ nowhere dense. } Proof. Suppose that $\tau$ is a Hausdorff topology on $X$ making $\J$ nowhere dense. By the maximality of $\J$ out of every two disjoint open sets one must belong to $\J$. So, $X$ cannot have two disjoint non-empty open sets. Thus, as a Hausdorff space, $X$ must be a singleton. \bigskip If we like to make $\J$ only meager in Corollary \ref{corMeagering} the situation is not so clear. The next fact shows that we cannot make measurable ideal meager by a Hausdorff topology. \rem{sigmamaximal}{Assume that there exists a measurable cardinal $\kappa$, let $|X|\geq\kappa$ and let $\J$ be a maximal $\sigma$-ideal on $X$ containing all singletons. Then there is no Hausdorff topology making $\J$ meager. } Proof. By way of contradiction assume that there is a Hausdorff topology $\tau$ on $X$ making $\J$ meager. Hausdorff property implies that all points, with possible one exception, have open meager neighborhoods. Let $\U$ be a maximal family of pairwise disjoint open meager sets. Then, $\real\setminus\bigcup\U$ is closed and nowhere dense. To get a contradiction it is enough to show that $\bigcup\U$ is meager in $X$. But this follows immediately from the following fact. \begin{equation}\label{conXXX} \text{If } \U \text{ is a family of disjoint open and meager sets then } \bigcup\U \text{ is meager.} \end{equation} To argue for (\ref{conXXX}) let $N^n_U$ be nowhere dense subsets of $X$ such that $U=\bigcup_{n< \omega}N^n_U$ for every $U\in\U$. Then every set $N_n=\bigcup_{U\in{\cal U}}N^n_U$ is nowhere dense since sets from $\U$ separate sets $\{N^n_U\}_{U\in{\cal U}}$. So, $\bigcup\U=\bigcup_{n< \omega}N_n$ is meager. Fact \ref{sigmamaximal} has been proved. \bigskip Clearly in Fact \ref{sigmamaximal} the cardinality of $X$ is greater that continuum. It could be expected that the same result could be obtained if you replace $0$-$1$ universal measure with a finite universal measure, i.e., a measurable cardinal with a real measurable cardinal. However, this is not the case, as our next example shows. \ex{realmeasurable}{If $\mu$ is a universal $\sigma$-additive measure on $\real$ which extends Lebesgue measure and $\J = \{ X\colon \mu (X) = 0 \}$, then there is a Hausdorff topology on $\real$ making $\J$ nowhere dense.} Proof. Let $\tau$ be the density topology on $\real$, i.e., $\tau$ is the family of all measurable subsets of $\real$ such that every point of $A$ is its density point. (See \cite[p. 90]{Oxtoby}.) Then $\tau^{\cal J} = \{U \setminus I\colon U\in\tau\ \&\ I\in\J\}$ is a Hausdorff topology making $\J$ nowhere dense. To see this notice that $\tau\cap\J =\{\emptyset\}$ so, by ``$\Leftarrow$'' part of Lemma \ref{Ni2} (the assumption of Lemma \ref{Ni2} that $X$ is second countable was not used in the proof of part ``$\Leftarrow$''), we have $\J\subset N(\tau,\real)$. On the other hand let $Y\in N(\tau,\real)$ and let ${\cal V}=\{V\in\tau\colon Y\cap V\in\J\} =\{V\in\tau\colon(\exists I\in\J)(Y\cap V\setminus I=\emptyset)\}$. It is easy to see that $\bigcup{\cal V}$ is $\tau $-dense in $\real$, since $\real\setminus\bigcup{\cal V}\subset Y$. Let ${\cal V}^\prime$ be some maximal pairwise disjoint subfamily of $\cal V$. Then, ${\cal V}^\prime$ is countable and $\bigcup{\cal V}^\prime$ is also $\tau $-dense in $\real$. Now, we have $Y\subset\bigcup_{V\in{\cal V}^\prime}(V\cap Y) \cup (\real \setminus \bigcup \cal{V'})$. But $\real\setminus\bigcup{\cal V}^\prime\in N(\tau,\real)$ so it is Lebesgue null. It follows that $\mu(Y) = 0$. \bigskip It is worth to notice that the topology $\tau^{\cal J}$ from Example \ref{realmeasurable} is not normal. To see it recall that the density topology $\tau$ on $\real$ is not normal \cite[p. 90]{Oxtoby}. So, let $C_{1}$ and $C_{2}$ be two disjoint $\tau $-closed subsets of $\real$ which can not be separated by $\tau$. Suppose that $C_{1}$ and $C_{2}$ can be separated by $\tau^{\cal J}$. Then $C_{1}$ and $C_{2}$ are contained in disjoint sets $E_{1} \setminus I_{1}$ and $E_{2} \setminus I_{2}$, respectively, where $E_{1}$ and $E_{2}$ are in $\tau$ and $\mu (I_{1}) = \mu (I_{2}) = 0$. We may assume that $E_{1} \cap C_{2} = E_{2} \cap C_{1} = \emptyset$. Since $E_{1} \cap E_{2} \subseteq I_{1} \cup I_{2}$ and $\mu $ is an extension of Lebesgue measure $E_{1} \cap E_{2}$ is Lebesgue null. It follows that $E_{1} \setminus E_{2}, E_{1} \setminus E_{2} \in \tau $ separate $C_{1}$ and $C_{2}$, which is a contradiction. Recall that the {\em weight} of a topological space is defined by \[ \weight(X) =\min\{|{\cal B}|\colon \B\ \text{ is an open basis for }\ \tau\}+\omega. \] The {\em celularity} of $X$ is defined as \[ \cel(X)=\sup\{|{\cal C}|\colon{\cal C}\ \text{ is a pairwise disjoint family of open sets}\}+\omega. \] If $\J$ is an ideal on a set $X$ then $\C \subset \J$ is {\em cofinal} in $\J$ if for each $I \in \J$ there exists a set $C \in \C$ such that $I \subset C$. We define {\em cofinality} of $\J$ by \[ \cf(\J) =\min\{ |{\cal C}|\colon {\cal C}\subset\J\text{ is cofinal in }\J \}. \] \lem{cofinality}{Let $\tau$ be a topology on a set $X$. Then $\cf(M(\tau,X))\leq\weight(X)^{{\rm c}(X)}$. } Proof. Let $\B$ be a basis for $\tau$ with $|\B|\leq \weight(X)$. For every $A \in N(\tau , X)$ there exists a maximal pairwise disjoint family $\C_{A} \subset \B$ such that $A \cap \bigcup \C_{A}= \emptyset$. Clearly, $\bigcup\C_{A}$ is dense in $X$. It follows that \[ \cal{I}= \{\real\setminus\bigcup\cal{C'}\colon \cal{C'}\subset\B\text{ is a family of pairwise disjoint sets}\} \] is cofinal in $N(\tau , X)$. Hence, $\cf(N(\tau,X))\leq|{\cal I}|\leq\weight(X)^{{\rm c}(X)}$. So \[ \cf(M(\tau,X))\leq[\cf(N(\tau,X))]^\omega \leq(\weight(X)^{{\rm c}(X)})^{\omega } =\weight(X)^{{\rm c}(X)}. \] \bigskip Recall that an ideal $\J$ on $X$ is {\em $\omega_1$-saturated} if every family $\F\subset\P(X)\setminus\J$ of pairwise disjoint sets is at most countable. In particular, the ideal $\J$ from Example \ref{realmeasurable} is $\omega_1$-saturated. It also contains $[\real]^{\non(UN)$ and $\cf(PM)>\non(PM)$.} Proof. Following an idea of Rec\l aw \cite{Reclaw} we take two disjoint compact perfect subsets $C$ and $D$ of linearly independent perfect set $Z\subseteq\real$. (See \cite[Theorem 19, p.206]{Morgan}.) The mapping $h\colon C\times D \to C+D, h(c,d)=c+d$, is a homeomorphism. To prove $\cf(UN) > \non(UN)$ choose a universally null set $X\subseteq D$ such that $|X|=\non(UN)$. We can find such a choice by a theorem of Grzegorek \cite{Grzegorek}. Notice that any selector $Y$ from $\{C+x\colon x\in X\} $ is also universally null because $g=\pi_{2}\circ h^{-1}$ ($\pi_{2}$ is a projection onto the second coordinate) is an injective continuous mapping from $Y$ onto $X$. Now, take any family $\C=\{C_{x}\colon x\in X\}$ of UN sets such that $|\C|=|X|=\non(UN)$. We will show that $\C$ does not cover UN, i.e., that $\cf(UN)>|\C|=\non(UN)$. To this end for every $x\in X$ select $y_x\in (C+x)\setminus C_{x}$. Then, $Y=\{y_{x}\colon x\in X\}\in UN$ but $Y\not\subseteq C_{x}$ for any $x\in X$. Hence \C\/ is not cofinal in $UN$. The proof of $\cf(PM)>\non(PM)$ is similar. We just need to take $X$ to be Grzegorek's set from Theorem 1 of \cite{Grzeg}. \bigskip Since under Martin's Axiom $\non(UN)=\non(PM)=c$, Lemma \ref{cofinality} yields the following. \thm{UNPM}{{\rm (MA)} There is no second countable topology on $\real$ making $UN$ or $PM$ meager. } The following problem seems to be interesting. \pr{UNandPM}{How good topologies making $UN$ or $PM$ meager (or nowhere dense) can be found in ZFC? } \begin{thebibliography}{22} \bibitem{Brown} J. B. Brown and G. V. Cox, {\sl Classical theory of totally imperfect spaces}, Real Analysis Exchange 7 (1981-82), 185--232. \bibitem{CLO:Book} K. Ciesielski, L. Larson and K. Ostaszewski, {\sl ${\cal I}$-density continuous functions}, Memoirs of Amer. Math. Soc. vol. 107 no. 515, 1994. %change \bibitem{Corazza} Paul Corazza, {\sl The generalized Borel conjecture and strongly proper orders}, Trans. Amer. Math. Soc. 316 (1989), 115--140. \bibitem{Croom} Fred H. 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