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%\newcommand{\leon}{<_{\rm On}} %\newcommand{\qsd}{Q_{\mbox{\tiny ${\sigma}$-discrete}}} %\newcommand{\qssc}{Q_{\mbox{\tiny${\sigma}$-$M_2$}}} %\newcommand{\qsd}{Q_0} %\newcommand{\qssc}{Q_1} %\newcommand{\cont}{2^{\omega}} %\newcommand{\irr}{\<\ncal,{\varepsilon}\>} %\newcommand{\Psz}{P^{\mbox{\tiny zero}}} %\newcommand{\fg}{\} %\def\fgi#1;{\} %\newcommand{\fgh}{\} %\def\fghi#1;{\} %\newcommand{\Pmt}{P^{\mbox{\tiny $M_2$}}} %\newcommand{\ro}{\operatorname{RO}} %\newcommand{\dd}{\operatorname{d}} %\theoremstyle{definition} %\newtheorem{ocaaxiom}{Open Coloring Axiom} %\renewcommand{\theocaaxiom}{} \newcommand{\ccdot}{\dot c} \def\ddd#1;#2;{X^#1_#2} \def\ccc#1;#2;{C^#1_#2} \newcommand{\dkl}{\ddd {\kappa};{\lambda};} \newcommand{\cckl}{\ccc {\kappa};{\lambda};} \newcommand{\dm}{D_{\mu}} \newcommand{\lvk}{{\lambda}^{<{\kappa}}} \newcommand{\vkl}{{}^{<{\kappa}}{\lambda}} \newcommand{\kl}{{}^{\kappa}{\lambda}} \newcommand{\lk}{{\lambda}^{\kappa}} \def\sss#1;{#1^*} \def\eee#1;#2;{E^{#1}_{#2}} \def\emn{\eee {\mu};{\nu};} \newcommand{\pid}{\operatorname{\pi_{\delta}}} \newcommand{\pin}{\operatorname{\pi_n}} \newcommand{\px}{\operatorname{\pi_{\xi}}} \newcommand{\pmm}{\operatorname{\pi_m}} \newcommand{\swp}{\operatorname{swp}} %\newcommand{\D}{\operatorname{D}} \myheads Forcing countable networks $\dots$ %for spaces satisfying %$\protect\operatorname{R}(X^{\omega})={\omega}$ ; Forcing countable networks $\dots$ %for spaces satisfying %$\protect\operatorname{R}(X^{\omega})={\omega}$ ; % I. Juh\'asz, L. Soukup, and Z. Szentmikl\'ossy; \title{Forcing countable networks for spaces \\ satisfying $\RR(X^{\omega})={\omega}$} \thanks{The preparation of this paper was supported by the Hungarian National Foundation for Scientific Research grant no. 16391 } \author{I. Juh\'asz} \address{Mathematical Institute of the Hungarian Academy of Sciences} \email{juhasz@@math-inst.hu} \author{L. Soukup} \thanks{The second author was supported by DFG (grant Ko 490/7-1)} \address{Mathematical Institute of the Hungarian Academy of Sciences} \email{soukup@@math-inst.hu} \author{Z. Szentmikl{\'o}ssy} \address{E{\"o}tv{\"o}s Lor{\'a}nd University, Department of Analysis} \subjclass{54A25,03E35} \keywords{ net weight, weakly separated, Martin's Axiom, forcing} \begin{document} %\maketitle %\begin{center}\large\today\end{center} \begin{abstract} We show that all finite powers of a Hausdorff space $X$ do not contain uncountable weakly separated subspaces iff there is a c.c.c poset $P$ such that %$1_P\force$ {``\em in $V^P$ $X$ is a countable union of 0-dimensional subspaces of countable weight. %''.} We also show that this theorem is sharp in two different senses: (i) we can't get rid of using generic extensions, (ii) we have to consider all finite powers of $X$. \end{abstract} \maketitle %\input{sp} \section{Introduction} \label{sc:intr} We use standard topological notation and terminology throughout, cf \cite{J}. The following definitions are less well-known. \begin{definition} Given a topological space $\$ and a subspace $Y\subs X$ a function $f$ is called a {\em \nea on Y} iff $f:Y\to {\tau}$ and $y\in f(y)$ for each $y\in Y$. \end{definition} \begin{definition} A space $Y$ is {\em weakly separated} if there is a \nea $f$ on $Y$ such that $$ \forall y\ne z\in Y\ ( y\notin f(z) \lor z\notin f(y) ), $$ moreover $$ \RR(X)=\sup\{|Y|:Y\subs X\mbox{ is weakly separated}\}. $$ \end{definition} The notion of weakly separated spaces and the cardinal function $\RR$ were introduced by Tka{\v c}enko in \cite{Tk}, where the following question was also raised: does $\RR(X^{\omega})={\omega}$ (or even $\RR(X)={\omega}$) imply that $X$ has a countable network (i.e. $\nw(X)={\omega}$)? (Note that $\RR(X^{\omega})=\omega$ is equivalent to $\RR(X^n)=\omega$ for all $n\in\omega$, moreover $\nw(X)=\omega$ implies $\RR(X^\omega)=\nw(X^\omega)=\omega$.) Several consistent counterexamples to this were given, e.g. in \cite{Cie}, \cite{HJ}, \cite{JSSz} and \cite[p. 43]{Stevo2}, but no ZFC counterexample is known. (In \cite {Stevo2} it is stated that under PFA the implication is valid, but no proof is given.) The counterexamples given in \cite{JSSz} and \cite{Stevo2} from CH are also first countable. Our main result here says that, at least for $T_2$ spaces, a weaker version of Tka{\v c}enko's conjecture is valid, namely $\RR(X^{\omega})={\omega}$ implies that $\nw(X)={\omega}$ holds in a suitable c.c.c and hence cardinal preserving generic extension! In fact, in this extension $X$ becomes ${\sigma}$-second countable, i.e. $X$ is the union of countably many subspaces of countable weight. In section \ref{sc:sharp} we show that the main result is sharp in different senses. Firstly, we force, for every natural number $n$, a 0-dimensional, first countable space $X$ such that $\RR(X^n)={\omega}$, but $\nw(X)>{\omega}$ in any cardinal preserving extension of the ground model. Secondly, we construct in ZFC a 0-dimensional $T_2$ space $X$ such that $\chi(X)=\nw(X)=\omega$ but $X$ is not $\sigma$-second countable. \begin{comment} We disprove the consistency of the statement \begin{equation} \tag{$*$}\text{\em If ${\chi}(X)\nw(X)={\omega}$, then $X$ is ${\sigma}$-second countable}. \end{equation} \end{comment} It easily follows from the proof of our main result that if $MA_{\kappa}$ holds and $X$ is a Hausdorff space with $|X|+\w(X)\le {\kappa}$, then $\RR(X^{\omega})={\omega}$ if and only if $X$ is ${\sigma}$-second countable. We also prove that $MA(Cohen)$ is not enough to yield this equivalence. To do this we use a result of Shelah (the proof of which presented here with his kind permission) saying that in any generic extension by Cohen reals the ideal of the first category subspaces of a space from the ground model is generated by the subspaces of first category from the ground model. \begin{comment} If $A$ and $B$ are sets, $\Fin(A,B)$ denotes the family of functions mapping a finite subset of $A$ into $B$. A topological space is {\em ${\sigma}$-second countable} if it is the union of countably many subspaces of countable weight. \end{comment} \begin{comment} We also present the proof of a theorem of Shelah with his kind permission saying that the second categoricity of a topological space is preserved by adding Cohen reals to the ground model. \end{comment} \section{The main result}\label{sc:equiv} \begin{theorem}\label{tm:main} Given a Hausdorff topological space $X$ the following are equivalent: \begin{enumerate}\arablabel \item $\RR(X^{\omega})={\omega}$, \item there is a c.c.c poset $P$ such that $V^P\models$ ``$\nw(X)={\omega}$'', \item there is a c.c.c poset $P$ such that \newline $V^P\models$ ``{$X$ is a countable union of 0-dimensional subspaces of countable weight.}'' %\begin{multline}\notag %V^P\models\text{$X$ is a countably union of %0-dimensional subspaces of countable weight.} %\end{multline} \end{enumerate} \end{theorem} \proof Since the implications (3) $\Rightarrow$ (2) $\Rightarrow$ (1) are clear, it remains to prove only that (1) implies (3). So assume that $X$ is Hausdorff with $\RR(X^{\omega})={\omega}$, and fix a base $\bcal$ of $X$ and a well-ordering $\prec$ on $X\cup\bcal$. The space $X$ contains at most countably many isolated points by $\RR(X)={\omega}$, so it is enough to force an appropriate partition of $X'$, the set of non-isolated points of $X$. We say that a 4-tuple $\nau$ is in $P$ provided (i)--(v) below hold: \begin{enumerate}\rlabel \item $A\in\br X';<{\omega};$ and $\ucal\in \br \bcal;<{\omega};$, \item $f$ and $g$ are functions, \item $f:A\to {\omega}$, $g:\ucal\to {\omega}\times {\omega}$, \item if $g(U)=\$ and $f(x)=n$ then $x\notin(\overline{U}\setm U)$ whenever $x\in A$ and $U\in \ucal$, \item if $g(U)=g(V)=\$ and $f(x)=n$ then $x\in U$ iff $x\in V$ whenever $x \in A$ and $U,V\in \ucal$. \end{enumerate} Our idea is that $f$ will guess the partition of $X'$ into countably many pieces, $\{F_n:n<{\omega}\}$; if $g(U)=\$ then $U\cap F_n$ will be clopen in the subspace $F_n$, and $g(U)=g(V)=\$ implies $U\cap F_n=V\cap F_n$. Consequently, each $F_n$ will have a countable clopen base. For $p\in P$ we write $p=\naui p;$. If $p,q\in P$ we set $p\le q$ iff $\fp\supset \fq$ and $\gp\supset \gq$. Two conditions $p$ and $q$ from $P$ are called {\em twins} provided $|\ap|=|\aq|$, $|\up|=|\uq|$ and denoting by ${\eta}$ and by ${\rho}$ the unique $\prec$-preserving bijections between $\ap$ and $\aq$, and between $\up$ and $\uq$, respectively, we have \begin{enumerate}\arablabel \item ${\eta}\rest \ap\cap \aq=\id$, ${\rho}\rest \up\cap \uq=\id$, \item $\forall x\in \ap$ $\fp(x)=\fq({\eta}(x))$. \item $\forall U\in\up$ $\gp(U)=\gq({\rho}(U))$. \item $\forall x\in \ap$ $\forall U\in\up$ ( $x\in U$ iff ${\eta}(x)\in {\rho}(U)$, and $x\in \overline{U}$ iff ${\eta}(x)\in \overline{{\rho}(U)}$ ). \end{enumerate} \begin{lemma}\label{lm:ccc} $\pcal=\$ satisfies c.c.c. \end{lemma} \Proof{lemma \ref{lm:ccc}} Let $\ooset p;{\alpha};\subs P$. Write $p_{\alpha}=\naui {\alpha};$. Using standard $\Delta$-system and counting arguments we can assume that these conditions are pairwise twins. Let $k=|A^{\alpha}|$ and $\{a_{{\alpha},i}:i \land a_{{\alpha},i}\in U\},\\ \notag\ucal^1_{{\alpha},i}=\{U\in\ucal^{\alpha}: \exists j\ g^{\alpha}(U)=\<(f^{\alpha}(a_{{\alpha},i}),j\> \land a_{{\alpha},i}\notin U\},\\ \intertext{and finally}\notag V_{{\alpha},i}=\bigcap\{U:U\in\ucal^0_{{\alpha},i}\}\cap \bigcap\{X\setm \overline{U}:U\in\ucal^1_{{\alpha},i}\}. \end{gather} By (iv) we have $a_{{\alpha},i}\notin \overline{U}$ for $U\in \ucal^1_{{\alpha},i}$, so $a_{{\alpha},i}\in V_{{\alpha},i}$. Since $\RR(X^k)={\omega}$ there are ${\alpha}<{\beta}<\oo$ such that \begin{equation} \tag{\dag} \text{$a_{{\alpha},i}\in V_{{\beta},i}$ and $a_{{\beta},i}\in V_{{\alpha},i}$ for each $i$ and $a_{{\alpha},i}\in\overline{V}$. We have to show that $a_{{\alpha},i}\in V$. Since $a_{{\alpha},i}\in %\overline{V}\cap V_{{\beta},i}$ by $(\dag)$, we have $V\notin \ucal^1_{{\beta},i}$. But $f^{\alpha}(a_{\alpha,i})=f^{\beta}(a_{\beta,i})$ since $p_\alpha$ and $p_{\beta}$ are twins, hence $g^{\beta}(V)=\$ implies $V\in\ucal^0_{{\beta},i}\cup \ucal^1_{{\beta},i}$, so $V\in \ucal^0_{{\beta},i}$. %So taking $U={\rho}^{-1}(V)$ we have %$U\in \ucal^0_{{\alpha},i}$, i.e., $a_{{\alpha},i}\in U$. %By the claim this Hence $a_{{\alpha},i}\in V_{\beta,i}\subs V$ by $(\dag)$, which was to be proved. %Taking $U={\rho}^{-1}(V)$ we have $a_{{\alpha},i}\notin U$ by the claim. %Thus $U\in \ucal^1_{{\alpha},i}$ and so %$V\in \ucal^1_{{\beta},i}$. %But this contradicts to fact that %$a_{{\alpha},i}$ is in this intersection. So (iv) is proved. Finally we check (v). Assume that $a_{{\alpha},i}\in A^{\alpha}$ and $U,V\in\ucal^{\alpha}\cup\ucal^{\beta}$ are such that $g(U)=g(V)=\$. Define the function $\rstr:\ucal\to \ucal^{\alpha}$ by the formula $\rstr=\id\rest{\ucal^{\alpha}}\cup ({\rho})^{-1}$ Then $a_{{\alpha},i}\in U$ iff $a_{{\alpha},i}\in \rstr(U)$ and $a_{{\alpha},i}\in V$ iff $a_{{\alpha},i}\in \rstr(V)$ by the previous claim. But $g(\rstr(U))=g(U)=g(V)=g(\rstr(V))=\$, so $a_{{\alpha},i}\in\rstr(U)$ iff $a_{{\alpha},i}\in\rstr(V)$ for $p_{\alpha}$ satisfies (v). Thus $a_{{\alpha},i}\in U$ iff $a_{{\alpha},i}\in\rstr(U)$ iff $a_{{\alpha},i}\in\rstr(V)$ iff $a_{{\alpha},i}\in V$, which proves (v). \Eproof Now let $\gcal$ be a $P$-generic filter and let $F=\bigcup\{f^p:p\in\gcal\}$ and $G=\bigcup\{g^p:p\in\gcal\}$. For $n<{\omega}$ let $F_n=F^{-1}\{n\}$. \begin{lemma} $\dom(F)=X'$ and $\dom(G)=\bcal$. \end{lemma} \proof Let $p=\nau\in P$, $x\in X'\setm A$ and $U\in \bcal\setm \ucal$. Let $A^*=A\cup\{x\}$, $\ucal^*=\ucal\cup \{U\}$ and $n=\max\ran(f)+1$. Let $\dom(f^*)=A^*$, $f^*\supset f$ and $f^*(x)=n$, $\dom(g^*)=\ucal^*$, $g^*\supset g$ and $g^*(U)=\$. %$f^*=f\cup\{\\}$, and define the function %$g^*=g\cup\\>$. Then it is easy to check $p^*=\naui *;\in P$ and obviously $p^*\le p$. So the lemma holds because a generic filter intersects every dense set. \eproof For $m\in {\omega}$ let $$ \bcal_m=\{U\cap F_m:U\in\bcal\text{ and } G(U)=\ \text{for some } i\in {\omega} \}. $$ \begin{lemma} $\bcal_m$ is a countable, clopen base of the subspace $F_m$ of $X'$. \end{lemma} \proof If $U\in\bcal$, $G(U)=\$ then $U\cap F_m$ is clopen in $F_m$ by (iv). If $U,V\in\bcal$, $G(U)=G(V)=\$ then $U\cap F_m=V\cap F_m$ by (v). So $\bcal_m$ is countable. Finally we show that it is a base of $F_m$. So fix $x\in F_m$ and $V\in\bcal$ with $x\in V$. Let $p=\nau\in P$ such that $f(x)=m$. Since $X$ is Hausdorff and $x$ is non-isolated in $X$, we can choose $U\in\bcal\setm \ucal $ such that $x\in U\subs V$ and $\overline{U}\cap A=\{x\}$. Let $\ucal^*=\ucal\cup \{U\}$. Define the function $g^*:\ucal^*\to {\omega}\times {\omega}$ such that $g^*\supset g$ and $g^*(U)=\$ where $k=\min\{l:\ran g\subs l\times l\}$. Then $p^*=\$ is an extension of $p$ in $\pcal$ and $p^*\force x\in U\cap F_m\subs V\cap F_m\land U\cap F_m\in\bcal_m$. Consequently, if $p\force$ ``$x\in F_m\cap V$'' then we also have $p\force$ ``$\exists U\in\bcal_m (x\in U\subs F_m\cap V)$'', which completes the proof. \eproof Thus theorem \ref{tm:main} is proved. \eproof It is easy to check that the above proof needs the genericity of $\gcal$ over $|X|+|\bcal|$ many dense sets only, and this immediately yields the following result. \begin{corollary}\label{cor:ma} If $MA_{\kappa}$ holds then for a Hausdorff space $X$ with $|X|+\w(X)\le {\kappa}$ the following are equivalent: \begin{enumerate}\arablabel \item $\RR(X^{\omega})={\omega}$, \item $\nw(X)={\omega}$, \item $X$ is the union of countably many 0-dimensional subspaces of countable weigh, \item $X$ is the union of countably many separable metrizable subspaces. \end{enumerate} \end{corollary} \section{Sharpness of the main result}\label{sc:sharp} Our aim in this section is to examine how sharp the above main result is. The co-finite topology on any uncountable set $X$ clearly satisfies $\RR(X^\omega)=\omega$, while, in any extension, $\nw(X)=|X|$. This show that in the proof of $(1)\to (2)$ of \ref{tm:main} the Hausdorffness of $X$ cannot be replaced by $T_1$. The next result in this section implies that, at least in ZFC, the exponent ${\omega}$ in proving $(1)\to (2)$ in theorem \ref{tm:main} can not be lowered. \begin{theorem} \label{tm:rn} For each uncountable cardinal ${\kappa}$ and natural number $m$ there is a c.c.c poset $\pcal$ of cardinality ${\kappa}$ such that in $V^{\pcal}$ there is a 0-dimensional first countable topological space $X=\<{\kappa},{\tau}\>$ such that $\RR(X^m)={\omega}$ but $\RR(X^{m+1})={\kappa}$, hence $\nw(X)=\kappa$ in any cardinal preserving extension. \end{theorem} In \cite[theorem 3.5]{JSSz} we constructed a c.c.c poset $\<\Pko,\le\>$ which adds to the ground model a 0-dimensional, first countable topology $\tau$ on $\kappa$ such that $\RR(X^{\omega})={\omega}$ and $\w(X)={\kappa}$ for $X=\<\kappa,\tau\>$. The conditions in $\Pko$ are finite approximations of the space $X$ and the property $\RR(X^{\omega})={\omega}$ is guaranteed by some $\Delta$-system and amalgamation arguments. Here we will use a subset $P$ of $\Pko$ with the inherited order. To ensure $\RR(X^{m+1})={\kappa}$ we thin out $\Pko$ in the following way. We fix a family $\dcal=\{d_{\alpha}:{\alpha}<{\kappa}\}$ of pairwise disjoint elements of ${\kappa}^{m+1}$ with the intention to make $\dcal$ discrete in $X^{m+1}$. A condition $p\in \Pko$ is put into $P$ if and only if every neighbourhood given by $p$ witnesses that $\dcal$ is discrete. The main step of the proof is to show that $P$ is large enough to allow the $\Delta$-system and amalgamation arguments to work in showing $\RR(X^m)={\omega}$. \Proof{theorem \ref{tm:rn}} First we recall some definitions and lemmas from the proof of \cite[theorem 3.5]{JSSz}. %To construct a first countable, 0-dimensional %space $Y$ with ${\omega}={\chi}(Y)=\RR(Y^{\omega})<\nw(Y)$ %we defined the posets $\<\Pkoo,\le\>$ and $\<\Pko,\le\>$. A quadruple $\anfg$ is said to be in $\Pkoo$ provided $(a)$--$(b)$ below hold: \begin{enumerate}\alabel \item $A\in\br {\kappa};<{\omega};$, $n\in {\omega}$, $f$ and $g$ are functions, \item $f:A\times A\times n\to 2$, $g:A\times n\times A\times n\to 3$, \end{enumerate} For $p\in \Pkoo$ we write $p=\anfgi p;$. If $p, q\in \Pkoo$ we set $p\le q$ iff $f^p\supseteq f^q$ and $g^p\supseteq g^q$. If $p\in \Pkoo$, ${\alpha}\in \ap$, $i<\np$ we defined $ U({\alpha},i)=U^p({\alpha},i)=\{{\beta}\in \ap:\fp({\beta},{\alpha},i)=1\} $. A quadruple $\anfg\in \Pkoo$ is put in $\Pko$ iff (i)--(ii) below are also satisfied: \begin{enumerate}\rlabel %\item $\forall {\alpha}\in A$ $\forall i$. Here we will apply a subset $P$ of $\Pko$. To define it let $\{d_{\alpha}:{\alpha}<{\kappa}\}$ be a family of pairwise disjoint elements of $\br {\kappa};m+1;$ such that ${\kappa}\setm \bigcup\{d_{\alpha}:{\alpha}<{\kappa}\}$ is still infinite. Write $d_{\alpha}=\{d_{{\alpha},i}:i\le m\}$. \begin{definition}\label{df:pcal} A condition $p=\anfg\in \Pko$ is in $P$ iff it satisfies (1) and (2) below: \begin{enumerate}\arablabel \item $d_{\alpha}\subs A$ or $d_{\alpha}\cap A=\empt$ for each ${\alpha}<{\kappa}.$ \item if ${\alpha}<{\beta}<{\kappa}$, $d_{\alpha}\cup d_{\beta}\subs A$, then there is an $i=i_{{\alpha},{\beta}}\le m$ such that $$ d_{{\alpha},i}\notin U^p(d_{{\beta},i},0)\text{ and } d_{{\beta},i}\notin U^p(d_{{\alpha},i},0). $$ \end{enumerate} Let $\pcal=\$. \end{definition} We define $X$ as expected. Let $\gcal$ be a $\pcal$-generic filter and let $F=\bigcup\{f^p:p\in\gcal\}$. For each ${\alpha}<{\kappa}$ and $n\in{\omega}$ let $V({\alpha},i)=\{{\beta}<{\kappa}:F({\beta},{\alpha},i)=1\}$. Put $\bcal_{\alpha}=\{V({\alpha},i):i<{\kappa}\}$ and $\bcal=\bigcup\{\bcal_{\alpha}:{\alpha}<{\kappa}\}$. We choose $\bcal$ as the base of $X=\<{\kappa},{\tau}\>$. By standard density arguments we can see that $X$ is first countable and 0-dimensional. It is easy to see that $\RR(X^{m+1})={\kappa}$, in fact $\operatorname{s}(X^{m+1})={\kappa}$. Indeed, by \ref{df:pcal}(2), $\{d_{\alpha}:{\alpha}<{\kappa}\}$ is discrete in $X^{m+1}$, as witnessed by the open neighborhoods $V(d_{{\alpha},0},0)\times V(d_{{\alpha},1},0)\dots \times V(d_{{\alpha},m},0)$. Finally we need to show that $\pcal$ satisfies c.c.c and $V^{\pcal}\models\RR(X^m)={\omega}$. Clearly, both of these statement follow from the next lemma. \begin{lemma}\label{lm:ccc+rm} If $\{p_{\gamma}:{\gamma}<\oo\}\subs \pcal$, %$1_{\pcal}\force$ %``$\{\ccdot_{\gamma}:{\gamma}<{\omega}_1\}\subs{\kappa}^m$'', $\{c_{\gamma}:{\gamma}<{\omega}_1\}\subs{\kappa}^m$ and $j_0,\dots, j_{m-1}$ are natural numbers, then there are ordinals $\alpha<\beta<\oo$ and a condition $p\in P$ such that \begin{enumerate} \item[(+)] $p\le p_{\alpha},p_{\beta}$ and $ p\force c_{\alpha}\in \prod_{im=|{\varepsilon}^{-1}\{1\}|$. So, by the definition of ${\varepsilon}$-amalgamation, $d_{{\xi},i}\notin U^p(d_{{\zeta},i},0)$ and $d_{{\zeta},i}\notin U^p(d_{{\xi},i},0)$. Thus $p\in P$. Finally we show that %\begin{equation} %\tag{++} $p\force$ ``$c_{{\alpha}}\in \prod_{i$ is a topological space, $\D(X)$ denotes the discrete topology on $X$. If $A$ and $B$ are sets, let $\Fin(A,B)$ be the family of functions mapping a finite subset of $A$ into $B$. \begin{lemma}\label{lm:somewhere-dense} If $Z\subs X^{\omega}$ and $Z$ is somewhere dense in the space $\D(X)^{\omega}$, then $\w(Z)=\w(X)$. \end{lemma} \proof Fix a natural number $n\in {\omega}$ and a function $f:n\to X$ such that $Z$ is dense in the basic open set $U_f=\{g\in X^{\omega}:f\subs g\}$ of $\D(X)^{\omega}$. This means that \begin{equation} \tag{\dag}\forall f'\in\Fin({\omega}\setm n,X) \exists g\in Z\ f\cup f'\subs g. \end{equation} From now on we forget about the $\D(X)^{\omega}$ topology, we will use only $(\dag)$. Without loss of generality we can assume that $Z\subs U_f$. Let $\zcal$ be a base of $Z$ in the subspace topology of $X^{\omega}$. Let $\pmm:X^\omega\to X$ be the projection to the $m^{\rm th}$ factor, i.e. $\pmm(g)=g(m)$. %For $U\subs X^{\omega}$ and $m<{\omega}$ let %$\pmm(U)=\{g(m):g\in U\}$. Set $\xcal=\{\pin(U):U\in\zcal\}$. Since $\w(Z)\le \w(X)$, it is enough to show that $\xcal$ is a base of $X$. \renewcommand{\theclaim}{} \begin{claim} %The elements of $\xcal$ are If $U$ is open in $Z$ then $\pin(U)$ is open in $X$. \end{claim} \Proof{the claim } Let $x\in\pin(U)$. We need to show that $\pin(U)$ contains a neighbourhood of $x$. Pick $g\in U$ with $x=g(n)$. Then, by the definition of the product topology on $X^{\omega}$, there is a function ${\sigma}$ which maps a finite subset of ${\omega}\setm n$ into the family of non-empty open subsets of $X$ such that \begin{equation}\tag{$\star$} g\in Z\cap \bigcap_{m\in\dom({\sigma})}\pmm^{-1}{\sigma}(m)\subs U. \end{equation} We can assume that $n\in\dom({\sigma})$. Let $g'=g\rest(\dom {\sigma}\setm \{n\})$. By $(\dag)$, for each $x'\in {\sigma}(n)$ there is $h_{x'}\in Z$ such that %$f\subs h_{x'}$, %$h_{x'}({\delta})=x'$ and $h_{x'}({\xi})=g({\xi})$ %for each ${\xi}\in\dom({\sigma})\setm \{{\delta}\}$. \begin{equation} \tag{$\star\star$} f\cup\{\\}\cup f'\subs h_{x'}. \end{equation} Now $(\star)$ implies $h_{x'}\in U$ and so $x'\in\pin(U)$. Thus $x\in {\sigma}(n)\subs \pin(U)$, which was to be proved. \Eproof To show that $\xcal$ is a base let $x\in V\subs X$, $V$ open. By $(\dag)$ we can find a point $g\in Z$ with $f\subs g$ and $g(n)=x$. The family $\zcal$ is a base of $Z$ in the subspace topology of $X^{\omega}$, so there is $U\in \zcal$ such that $g\in U\subs Z\cap \pin^{-1}V$. Thus $x\in \pin(U)\subs V$ and $\pin(U)$ is open by the previous claim. Thus $\xcal$ is a base of $X$, and so $\w(X)\le|\xcal|\le|\zcal|$. Since $\w(Z)\le {\omega}\w(X)=\w(X)$, we are done. \eproof After this preparation we can give the ZFC example promised above. \begin{theorem}\label{tm:need-forcing} There is 0-dimensional Hausdorff space $Y$ such that ${\chi}(Y)\nw(Y)={\omega}$, but $Y$ is not ${\sigma}$-second countable. \end{theorem} \proof By \cite[theorem 3.1]{JSSz} there is a 0-dimensional Hausdorff space $X$ of size $\cont$ such that ${\chi}(X)\nw(X)={\omega}$, but $\w(X)=\cont$. We show that $Y=X^{\omega}$ is as required. Clearly ${\chi}(Y)=\chi(X)={\omega}$ and $\nw(Y)=\nw(X)={\omega}$. Assume that $Y=\bigcup_{k<{\omega}}Z_k$. The Baire category theorem implies that some $Z_k$ is somewhere dense in $\D(X)^{\omega}$. Then $\w(Z_k)=\w(X)=\cont$ by lemma \ref{lm:somewhere-dense}. \eproof %\begin{theorem} %If $\w(X)>{\omega}$, then $X^{\omega}$ can not be ${\sigma}$ second countable. %\end{theorem} By corollary \ref{cor:ma}, if Martin's Axiom holds, then every Hausdorff space $X$ of size and weight $<2^{\omega}$ is ${\sigma}$-second countable if and only if $\nw(X)={\omega}$. %is the union of countable %many subspaces of countable weight by corollary \ref{cor:ma}. The next theorem shows that $MA(Cohen)$ is not enough to get this equivalence. Note that for a first countable space $X$ we have $\w(X)\le |X|$. \begin{theorem}\label{tm:consistent} If ZFC is consistent, then so is ZFC + MA(Cohen) +``{\em there is a first countable 0-dimensional Hausdorff space $Y$ such that $\nw(Y)={\omega}$ and $|Y|<2^{\omega}$ but $Y$ is not ${\sigma}$-second countable }''. \end{theorem} The proof is based on theorem \ref{lm:somewhere-dense} above and theorem \ref{tm:second-preserve} below. \newcommand{\bdot}{{\dot B}} \begin{comment} \begin{definition} A topological space $X$ is of {\em second category} iff it is not the union of countably many nowhere dense subspaces. \end{definition} \end{comment} \begin{definition} Given a topological space $Z$, let $\ical(Z)$ be the $\sigma$-ideal generated by the nowhere dense subsets of $Z$. The elements of $\ical(Z)$ are called {\em first category in $Z$}. \end{definition} The next result is due to Saharon Shelah \cite{Sh} and it is included here with his kind permission. \begin{comment} \begin{theorem}\label{tm:second-preserve} The second categoricity is preserved by adding arbitrarily many Cohen reals. \end{theorem} \end{comment} \begin{theorem}\label{tm:second-preserve} If $Z$ is a topological space and the forcing notion $P=\fn(\kappa,2,\omega)$ adds $\kappa$ Cohen reals to the ground model, then the ideal $\ical^{V^P}(Z)$ is generated by $\ical^V(Z)$, that is, for each $T\in \ical^{V^P}(Z)$ there is $T'\in\ical^V(Z)$ with $T\subs T'$. \end{theorem} \proof %Let $\dot{T}$ be a $P$-name of an element of %$\ical^{V^P}(Z)$. Since % %$\ical^V(Z)$ is $\sigma$-complete and $\ical^{V^P}(Z)$ is $\sigma$-generated by the nowhere dense subsets of $Z$ in $V^P$, we can assume that $T$ is nowhere dense. Let $\dot{T}$ be a $P$-name of $T$ such that $1_P\force$``{\em $\dot{T}$ is nowhere dense}''. For each $m\in {\omega}$ define, in $V$, the subset $B_m$ of $Z$ as follows: \begin{equation} \tag{$*$}B_m=\{x\in Z:\exists p\in P\ |p|=m\land p\force x\in\dot{T}\}. \end{equation} Clearly $1_P\force \dot{T}\subs\bigcup\limits_{m\in {\omega}}B_m $, so it is enough to show that every $B_m$ is nowhere dense. Assume on the contrary that there are an open set $U\subs Z$ and $m\in {\omega}$ such that $B_m$ is dense in $U$. Now, by finite induction, we can define open sets $U\supset U_0\supset U_1\dots\supset U_m$ and conditions $q_0,\dots,q_m$ with pairwise disjoint domains such that for each $j\le m$ and for each $f\in P$ if $\dom(f)=\bigcup\limits_{i\cont$ Cohen reals to the ground model. In the generic extension clearly ${\chi}(Y)=\nw(Y)={\omega}$ will remain valid. Since $Z\notin\ical^V(Z)$ by the Baire Category theorem, we have $Z\notin \ical^{V^P}(Z)$ as well by theorem \ref{tm:second-preserve}. Therefore, if $Y=\bigcup\{Y_k:k<\omega\}$ holds in $V^P$, then $Y_k\notin \ical^{V^P}(Z)$ for some $k\in\omega$, i.e.. $Y_k$ is somewhere dense in $Z$. Thus there is a natural number $n\in {\omega}$ and a function $f:n\to X$ such that \begin{equation} \tag{\dag}\forall f'\in\Fin({\omega}\setm n,X) \ \exists g\in Y_k\ f\cup f'\subs g. \end{equation} But $(\dag)$ implies that $Y_k$ is also dense in the basic open set $(V_f)^{V^P}=\{g\in X^{\omega}\cap V^P:f\subs g\}$ of $(\D(X)^\omega)^{V^P}$. Thus, applying lemma \ref{lm:somewhere-dense} in $V^P$ we have $\w(Y_k)=\w(X)>\omega$. Thus, in $V^P$, $\operatorname{MA}(Cohen)$ holds, and still the first countable, 0-dimensional $T_2$ space $Y$ is not $\sigma$-second countable, though $\w(Y)=(\cont)^V<\kappa=(\cont)^{V^P}$. \Eproof \section{Examples for higher cardinals} In this section we generalize the constructions of \cite[theorem 3.1]{JSSz} and \ref{tm:need-forcing} for cardinals greater than ${\omega}$. \begin{theorem} \label{tm:gen-con} Let ${\kappa}$ and ${\lambda}$ be cardinals, $\cf({\kappa})={\kappa}$. Then there is a 0-dimensional Hausdorff space $\dkl$ such that ${\chi}(\dkl)={\kappa}$, $\nw(\dkl)=\lvk$ and $\w(\dkl)=\lk$. \end{theorem} \proof For each $f\in \fn({\kappa},{\lambda},{\kappa})$ put $U_f=\{g\in \kl:f\subs g\}$. Write $U(g,{\alpha})=U_{g\rest {\alpha}}$ for $g\in \kl$ and ${\alpha}<{\kappa}$. For $g\ne h\in \kl$ define ${\Delta}(g,h)=\min\{{\alpha}:g({\alpha})\ne h({\alpha})\}$. Consider the topological space $\cckl=\<\kl,{\tau}\>$ that has as a base $\{U_f:f\in \vkl\}$. Let \begin{gather}\notag Y=\{g\in \kl:\exists {\alpha}<{\kappa}\ ( \ g({\beta})=0\text{ iff }{\ \beta} \ge {\alpha}\ )\}\\ \intertext{and}\notag Z=\{g\in \kl:0\notin\ran(g)\}. \end{gather} Clearly $Y$ and $Z$ are disjoint, $|Y|=\lvk$, $|Z|=\lk$, $Z$ is closed and nowhere dense in $\cckl$. Let $X=Y\cup Z$. Our required space will be $\dkl=\$, where ${\rho}$ refines the topology ${\tau}_X$. To define ${\rho}$ put $$X_g=\bigcup\{U_f:\exists {\alpha}<{\kappa}\ f=g\rest {\alpha}\cup\{\<{\alpha},0\>\}\}$$ for $g\in Z$ and $X_g=\empt$ for $g\in Y$. For $g\in X$ and ${\alpha}<{\kappa}$ let $V(g,{\alpha})=(U(g,{\alpha})\setm X_g)\cap X$. Let the neighbourhood base of $g\in X$ in ${\rho}$ be $$ \bcal_g=\{V(g,{\alpha}):{\alpha}<{\kappa}\}. $$ First we note that $\bcal=\bigcup\{\bcal_x:x\in X\}$ is a base of a topology because \begin{equation} \tag{\dag}\forall {\alpha}<{\kappa}\ \forall h\in V(g,{\alpha})\setm \{g\} \ U(h,{\Delta}(g,h)+1)\cap X\subs V(g,{\alpha}). \end{equation} Since $V(g,{\alpha})\cap Y=U(g,{\alpha})\cap Y$ and $V(h,{\alpha})\cap Z=U(h,{\alpha})\cap Z$ for each $g\in Y$, $h\in Z$ and ${\alpha}<{\kappa}$ we have that $\=\$ and $\=\$. Thus $$ \nw(\)=\nw(\)+\nw(\)\le \w(\)+\w(\)\le \w(\<\cckl\>)= \lvk. $$ Obviously ${\chi}(\)={\kappa}$. Finally we show that $\w(\)=\lk$. This will follow if we show that $\bcal$ is an irreducible base for $X$, by lemma \cite[2.6]{JSSz}. We claim that $\{\bcal_x:x\in X\}$ is an irreducible decomposition of the base $\bcal$ (see definition \cite[2.3]{JSSz}). Since $Y$ is discrete in ${\rho}$ it is enough to show that if $g\ne h$ are from $Z$ with $g\in V(h,{\alpha})$ for some ${\alpha}<{\kappa}$ then $V(h,{\alpha})\not\subs V(g,0)$. %Assume on the contrary that this situation occurs. %Let ${\beta}=\min\{{\beta}':g({\beta}')\ne h({\beta}')\}$. Let ${\delta}={\Delta}(g,h)$. Then $U(g,{\delta}+1)\subs V(h,{\alpha})$ by $(\dag)$. Consider the element $y$ of $Y$ defined by the formulas $y\rest {\delta}+1=g\rest {\delta}+1$ and $y({\delta}+1)=0$. Then $y\in X_g$ and so $y\notin V(g,0)$. On the other hand $y\in U(g,{\delta}+1)$, so $V(h,{\alpha})\not\subs V(g,0)$. \eproof \begin{comment} To simplify the formulation of the next result we introduce a new cardinal function. \begin{definition} Given a topological space $X$ define its ``small weight partition number'' $\swp(X)$ as the minimal cardinal ${\rho}$ such that $X$ has a partition into ${\rho}$-many pieces, $X=\bigcup\limits_{{\alpha}<{\rho}}X_{\alpha}$ such that $\w(X_{\alpha})\le {\rho}$ for each ${\alpha}<{\rho}$. \end{definition} Let us remark that $\nw(X)\le \swp(X)\le \w(X)$. \end{comment} \begin{lemma}\label{lm:bigsomewhere-dense} If $X$ is any topological space and $Z\subs X^{\kappa}$, ${\alpha}<{\kappa}$, $f:{\alpha}\to X$ are such that \begin{equation} \tag{\ddag}\forall f'\in\Fin({\kappa}\setm {\alpha},X) \ \exists g\in Z\ f\cup f'\subs g, \end{equation} then $\w(Z)\ge\w(X)$. \end{lemma} The proof is similar to that of lemma \ref{lm:somewhere-dense}, so we omit it. \begin{comment} \begin{theorem} \label{tm:con-swp} Let $X$ be a topological space and ${\kappa}$ be a regular cardinal. If $ %{\chi}(X)\nw(X)\le {\kappa}<\w(X)$, then product space $Y=X^{\kappa}$ has no partition of ${\kappa}$-many subspaces of weight $\le {\kappa}$. \end{theorem} \proof %The statement ${\chi}(Y)\nw(Y)={\kappa}$ is clear. Since we are to use two different topology on $Y$ denote by ${\tau}$ the product topology $X^{\kappa}$. Fix a partition $Y=\bigcup\limits_{{\alpha}<{\kappa}}Y_{\alpha}$. We show that $\w(Y_{\alpha})=\w(X)$ for some ${\alpha}$. We can assume that the underlying set of the space $X$ is some cardinal ${\lambda}$. Then the points of $Y$ are just the elements of $\kl$. So $\{Y_{\alpha}:{\alpha}<{\kappa}\}$ can be considered as a partition of the topological space $\cckl$ as well. Since ${\kappa}$ is regular we can apply an appropriate version of the Baire category theorem to show that there is an ${\alpha}<{\kappa}$ and a basic open set $U_f$ of $\cckl$ such that $Y_{\alpha}$ is dense in $U_f$. So the next lemma completes the proof of the theorem: \begin{lemma}\label{lm:dense} If $Z\subset Y$ is dense in some basic open set $U_f$ in the $\cckl$-topology, then $\w(\)=\w(X)$. \end{lemma} \Proof{lemma} Without loss of generality we can assume that $Z\subs U_f$. Let $\zcal$ be a base of $\$. Write ${\delta}=\dom(f)<{\kappa}$. For $U\subs Y$ and ${\xi}<{\kappa}$ let $\px(U)=\{g({\xi}):g\in U\}$. Take $\xcal=\{\pid(U):U\in\zcal\}$. It is enough to show that $\xcal$ is a base of $X$. \renewcommand{\theclaim}{} \begin{claim} The elements of $\xcal$ are open in $X$. \end{claim} \Proof{the claim } Let $U\in\zcal$, $x\in\pid(U)$. Pick $g\in U$ with $x=g({\delta})$. Then, by the definition of product topology, there is a function ${\sigma}$ which maps a finite subset of ${\kappa}\setm {\delta}$ into the family of non-empty open subsets of $X$ such that \begin{equation}\tag{$*$} g\in Z\cap \bigcap_{{\xi}\in\dom({\sigma})}\px^{-1}{\sigma}({\xi})\subs U. \end{equation} We can assume that ${\delta}\in\dom({\sigma})$. Since $Z$ is dense in $U_f$ in the $\cckl$-topology for each $x'\in {\sigma}({\delta})$ there is $h_{x'}\in Z$ such that $f\subs h_{x'}$, $h_{x'}({\delta})=x'$ and $h_{x'}({\xi})=g({\xi})$ for each ${\xi}\in\dom({\sigma})\setm \{{\delta}\}$. By $(*)$ we have $h\in U$ and so $x'\in\pid(U)$. But $x'$ was arbitrary, so ${\sigma}({\delta})\subs \pid(U)$, that is, $\pid(U)$ is open. \Eproof To show that $\xcal$ is a base let $x\in V\subs X$, $V$ open. Since $Z$ is dense in $U_f$ in the $\cckl$-topology we can find a point $g\in Z$ with $f\subs g$ and $g({\delta})=x$. The family $\zcal$ is a base of $Z$ in the topology ${\tau}$ so there is $U\in \zcal$ such that $g\in U\subs Z\cap \pid^{-1}V$. Thus $x\in \pid(U)\subs V$. Thus $\xcal$ is a base of $X$, and so $\w(X)\le|\xcal|\le|\zcal|$ which proves the lemma. \Eproof The theorem is proved. \eproof \end{comment} Let us recall that given a cardinal ${\mu}$ the {\em Singular Cardinal Hypothesis $($SCH$)$} is said to hold below ${\mu}$ provided ${\nu}^{\cf({\nu})}=2^{\cf({\nu})}{\nu}^+$ for each singular cardinal ${\nu}\le {\mu}$. By \cite[lemma 1.8.1]{Jech}, if $\mu$ is regular and SCH holds below $\mu$, then $\log(\mu^+)=\min\{\nu:2^\nu\ge\mu^+\}$ is also regular. It is well-known that the failure of SCH requires the consistency of large cardinals, therefore the assumption of our next lemma is quite reasonable. Also note that $SCH$ trivially holds below $\aleph_{\omega}$. \begin{comment} Assume now ${\rho}$ is singular and a contradiction will be derived. If $2^{<{\rho}}$ is regular then $2^{<{\rho}}=2^{\nu}$ for some ${\nu}<{\rho}$. But in this case ${\mu}<2^{\rho}=(2^{<{\rho}})^{\cf({\rho})}= 2^{\max({\nu},\cf({\rho}))}\le {\mu}$, which is impossible. So $2^{<{\rho}}$ is also singular. Hence $\cf(2^{<{\rho}})=\cf({\rho})<{\mu}$ and $2^{<{\rho}}<{\mu}$. Since SCH holds below ${\mu}$ we have $2^{\rho}=(2^{<{\rho}})^{\cf({\rho})}=(2^{<{\rho}})^+$. %Since $2^{<{\rho}}\le{\mu}$ but they are different, because %$2^{<{\rho}}$ is singular and ${\mu}$ is regular, we have %$2^{<{\rho}}<{\mu}$. Thus $2^{<{\rho}}<{\mu}<2^{\rho}=(2^{<{\rho}})^+$, which is absurd. \end{comment} \begin{theorem} \label{tm:sch} If ${\mu}$ and $\log({\mu^+})$ are both regular cardinals then there is a 0-dimensional $T_2$ space $Y$ such that ${\chi}(Y)\nw(Y)\le{\mu}$, but $Y$ is not the union of ${\mu}$ subspaces of weight ${\mu}$. \end{theorem} \proof Let ${\rho}=\log(\mu^+)$. \begin{comment} Then ${\rho}\le 2^{<{\rho}}\le {\mu}<2^{\rho}$. If ${\rho}$ is regular, then \end{comment} Applying theorem \ref{tm:gen-con} for ${\kappa}={\rho}$ and ${\lambda}=2$ we get a space $X$ with ${\chi}(X)\nw(X)= 2^{<{\rho}}\le {\mu}<\w(X)$. Let $Y=X^{\rho}$ and consider a partition $Y=\bigcup_{{\alpha}<{\rho}} Y_{\alpha}$. Since $\kappa={\rho}$ is regular, applying the technique of the standard proof of the Baire Category theorem we can see that the space $C^\rho_{|X|}$ is not the union of $\rho$ nowhere dense subspaces. %We can apply an appropriate version %of the Baire category theorem to show that Therefore there are ordinals ${\alpha},{\xi}<{\rho}$ and a function $f:{\alpha}\to X$ such that \begin{equation} \tag{\ddag'}\forall f'\in\fn({\rho}\setm {\alpha},X,\rho) \ \exists g\in Y_{\xi}\ f\cup f'\subs g. \end{equation} But (\ddag') clearly implies (\ddag). So, by lemma \ref{lm:bigsomewhere-dense}, $\w(Y_{\xi})\ge \w(X)>{\mu}$. Thus $Y$ satisfies the requirements. The theorem is proved. \eproof \begin{problem} $($ZFC$)$ If $\mu$ is a cardinal, is there a space $X$ such that $\chi(X)\nw(X)=\mu$ (or just $\nw(X)={\mu}$), but $X$ is not the union of $<2^\mu$ many subspaces of weight $<2^\mu$ $($or just $\w(X)=2^\mu)$? \end{problem} \begin{remark} The simplest case of Problem 1 left open by theorem \ref{tm:gen-con} is that $2^{\omega}={\omega}_2$, $2^\oo={\omega}_3$ and ${\mu}=\oo$. Indeed, if $\dkl$ is the space constructed in theorem \ref{tm:gen-con} and $\nw(\dkl)=\lvk\le\oo$, then ${\lambda}\le\oo$ and ${\kappa}\le{\omega}$. So $\w(\dkl)=\lk\le \oo^{\omega}={\omega}_2<2^{\oo}$. \end{remark} \begin{thebibliography}{9} %lm\bibitem{Buk} L. Bukowsky \bibitem{Cie} K. Ciesielski, {\em On the netweigth of subspaces}, Fund. Math., {\bf 117} (1983), no 1, 37--46. \bibitem{HJ} A. Hajnal, I. Juh{\'a}sz, {\em Weakly separated subspaces and networks}, Logic {}'78, Studies in Logic, {\bf 97}, 235--245. \bibitem{Jech} T. Jech, {\em Set Theory}, Academic Press, New York, 1978. \bibitem{J} I. Juh{\'a}sz, {\em Cardinal Functions - Ten Years Later}, Math. Center Tracts 123, Amsterdam, 1980. \bibitem{JSSz} I. Juh{\'a}sz, L. Soukup, Z. Szentmikl{\'o}ssy, {\em What makes a space have large weight?}, Topology and its Applications, {\bf 57} (1994) 271-285. \bibitem{Sh} S. Shelah, private communication \bibitem{Tk} M. G. Tka{\v c}enko, {\em Chains and cardinals}, Dokl. Akad. Nauk. SSSR 239 (l978) no 3, 546-549 %\bibitem{Stevo} private communication (????) \bibitem{Stevo2} S. Todor{\v c}evi{\v c}, {\em Partition Problems in Topology}, Contemporary Mathematics, vol 84, Providence, 1989 \end{thebibliography} \end{document}