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\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 
$0<k<\omega$ there is no 
$(k,l)$ meager $T_1$ topology.
}

In what follows we will use also the following example.

\ex{conpoints}{ Let $\lambda$ be a limit ordinal and let 
$\B = \P (\lambda )
\cup \{(\lambda +1\setminus \alpha\colon \alpha < 
\lambda \}$.  
A $T_1$ topology generated by \B\/ will be denoted by 
$\tau_{ord}(\lambda )$.  It is easy to see that 
$\tau_{ord}(\lambda )$ makes $\P(\{\lambda \})$ nowhere
dense.
}

The next corollary follows immediately from 
Theorem \ref{singleton}.

\cor{Comega}{ $\tau_{ord}(\omega )$ is a 
separable metrizable topology
making 
$\P(\{\omega \})$ nowhere dense.
In particular, it is $(1,\omega)$ nowhere dense.}

In fact, $\tau_{ord}(\omega )$ is just an order 
topology on $\omega+1$.

The following examples indicate that even in the 
simplest possible
case, discussed in Theorem \ref{singleton}, there 
is no unique
$(1,\omega)$ nowhere dense topology.

\ex{Comegasq}{ The topology induced from 
$\real^{2}$ on a countable set 
\[
X=\{(1/n,1/m)\colon 0<n,m<\omega \}\cup\{(0,0)\}
\]
makes $\P(\{(0,0)\})$ 
nowhere dense. Space $X$ is not homeomorphic to 
$\tau_{ord}(\omega)$.  
However, the diagonal $D=\{(1/n,1/n)\colon 0<n<\omega\}$  
is homeomorphic to $\tau_{ord}(\omega)$.
}

\ex{ultra}{Let \F\ be a non-principal
ultrafilter on $\omega$. 
The
topology $\tau$ on $\omega +1$ generated by 
$\P(\omega)\cup\{F\cup\{\omega \}\colon F\in\F\}$ makes 
$\P(\{\omega \})$ nowhere dense.
The space $(\omega +1,\tau)$ is a $(1,\omega)$ 
nowhere dense
and does not have any
subspace 
homeomorphic to $\tau_{ord}(\omega)$.  
This is because in $\tau_{ord}(\omega )$ only 
cofinite sets containing $\omega$ are open 
unlike in any 
infinite subspace of
$(\omega +1, \tau)$ containing $\omega$.}

Corollary \ref{Comega} and Examples \ref{Comegasq} and
\ref{ultra} 
show a variety of $(1,\omega)$ nowhere dense
topologies. 
It 
is easy to show that the space from Example 
\ref{ultra} is normal
but
not metrizable, since $\omega$ has no countable basis.

\bigskip

The next theorem gives a necessary and sufficient
condition for the existence of metrizable topology which
is $(\kappa,\lambda)$ nowhere dense.

\thm{Metr}{ Let $\kappa>0$ 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
\[
U_{\varepsilon}^{\cal J}=\bigcap_{a\in{\rm dom}(\varepsilon)} %change
U^{\cal J}_{<a,\varepsilon(a)>}
\]
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}
<<f_r>_{r\in A},\D> & \leq & <<f_r^\prime>_{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
$<<f_r>_{r\in A},\D>,
<<f_r^\prime>_{r\in A^\prime},\D^\prime>\in{\Bbb P}$.

It is easy to see that ${\Bbb P}$ is $\omega$-closed. 
To see that ${\Bbb P}$ satisfies $\omega_2$-cc take a sequence
$\{<<f_r^\xi>_{r\in A^\xi},\D^\xi>\in{\Bbb P}\colon
\xi<\omega_2\}$
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, 
$<<f_r^\xi\cup f_r^\zeta>_{r\in A},\D^\xi\cup\D^\zeta>$
extends 
$<<f_r^\xi>_{r\in A^\xi},\D^\xi>$
and
$<<f_r^\zeta>_{r\in A^\zeta},\D^\zeta>$.

Now, let $G$ be a $V$-generic filter over ${\Bbb P}$ and define
mapping $e\colon\real\to 2^{\omega_2}$
by putting for $s\in \real$,
\[
e(s)=\bigcup\{g_s\colon
(\exists <<f_r>_{r\in A},\D>\in G)(s\in A\ \&\ g_s=f_s)\}.
\]
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 $\xi<\omega_1$
\[
D_s=\{<<f_r>_{r\in A},\D>\in{\Bbb P}\colon s\in A\}
\]
and
\[
D_s^\xi=
\{<<f_r>_{r\in A},\D>\in{\Bbb P}\colon s\in A\ \&\ 
\xi\in\text{dom}(f_s)\}.
\]

To see that $D_s$ is dense in ${\Bbb P}$ take 
$<<f_r>_{r\in A},\D>\in{\Bbb P}$ 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 $\bigcup_{k<m}\e^k_{n_k}\in H(\omega_2)$
for every $m<\omega$. This can be
done by the definition 
(\ref{conE}) of $\E$. Then, 
$f_s=\bigcup_{k<\omega}\e^k_{n_k}\in H_\omega(\omega_2)$
and 
$<<f_r>_{r\in A\cup\{s\}},\D>$ extends $<<f_r>_{r\in A},\D>$.

To see that $D_s^\xi$ is dense in ${\Bbb P}$ take 
$<<f_r>_{r\in A},\D>\in{\Bbb P}$. 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 
$<<f_r>_{r\in A},\D>$.

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}=
\{<<f_r>_{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
\[
D_{\e}=
\{<<f_r>_{r\in A},\D>\in{\Bbb P}\colon
(\exists r\in A)(\e\subset f_r)\}
\]
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
\[
D_{\e}^J=
\{<<f_r>_{r\in A},\D>\in{\Bbb P}\colon
(\exists r\in A\setminus J)(\e\subset f_r)\}
\]
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
\[
E_s^\alpha=\{<<f_r>_{r\in A},\D>\in{\Bbb P}\colon s\in A\
\&\ (\exists \xi<\omega_1)(<\alpha+\xi,0>\in f_s)\}
\]
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 
\[
E_D=\{<<f_r>_{r\in A},\D>\in{\Bbb P}\colon D\in\D\}
\]
is dense in ${\Bbb P}$ we can find 
$<<f_r^\prime>_{r\in A^\prime},\D^\prime>\in G$
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 $<<f_r>_{r\in A},\D>\in G$ 
extending $<<f_r^\prime>_{r\in A^\prime},\D^\prime>$
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 $[\real]^{<c}$ if measure
$\mu$ is continuum additive. 
Thus, the next fact tells us, 
in particular, that for a continuum additive measure $\mu$ in 
Example \ref{realmeasurable}
the ideal $\J$ cannot be made meager by a metrizable
topology.

\rem{nometric}{Let $|X|= c$ and let 
$\J$ be an $\omega_1$-saturated
$\sigma$-ideal on $X$ such that $[X]^{<c}\subset\J$. If a Hausdorff
topology $\tau$ on $X$ is making $\J$
meager then $\weight(X,\tau)\leq c$. 

In particular, there is no metrizable topology making
$\J$ meager.
}

Proof. Let $\tau$ be a Hausdorff topology making $\J$
meager and, by way of contradiction, assume that 
$\weight(X,\tau)\leq c$. 
Condition (\ref{conXXX}) from Fact \ref{sigmamaximal}
shows that the set of all points which have
meager open neighborhoods is meager.  Therefore, without loss of
generality, we may assume that $\tau\cap\J =\{\emptyset\}$. 
Hence, $\cel(X)\leq\omega$, since $\J$ is $\omega_1$-saturated.
So, by
Lemma \ref{cofinality}, we obtain $\cf(M(\tau,X)) \leq c$. 
Let
$\{M_{\beta}\colon \beta < c \}$ be  a cofinal family in 
$M(\tau,X)$.  For $\alpha < \omega_{1}$ and $\beta < c$, select
\[
x_{\alpha}^{\beta}\in X\setminus [M_\beta
\cup
\{x_{\delta}^{\gamma}\colon\gamma=\beta\text{ and }\delta < \alpha
\text{ or }\gamma < \beta \}].
\]
Clearly, the sets $X_{\alpha} = \{ x_{\alpha}^{\beta}\colon \beta < c \}
\not\in \J$ are pairwise disjoint subsets of $\real$. 
But this contradicts $\omega_1$-saturation of $\J$.

\bigskip

We would like to conclude this discussion 
with few remarks on universally null and perfectly meager
sets.  
Recall that a set $X\subseteq \real$ is {\em perfectly
meager} ($X\in PM$) if $P\cap  X$ is meager in $P$ for every
perfect set 
$P\subseteq \real$.  A set $Y\subseteq \real$ is {\em
universally
null} ($Y\in UN$) if $\mu(Y)=0$ for every continuous Borel 
probability measure
on $\real$. For more information on $PM$ and $UN$ see
\cite{Miller}
or \cite{Brown}.  In general these two ideals have many
similar properties. 
In particular, it is an open problem whether $UN \not= PM$ is
provable in ZFC \cite{Corazza}. In this situation it would be
particularly desirable to know more about topologies making $UN$
or $PM$ meager.

For an ideal $\J$ on a set $X$ we define
\[
\non(\J)=\min\{|Y|\colon Y\subseteq X
        \text{ and }Y\not\in\J\}.
\]


\lem{Grzegorek}{$\cf(UN)>\non(UN)$ 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}

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\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. Croom,
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\end{document}

