\documentclass[12pt]{amsart}
\usepackage{verbatim}
\usepackage{enumerate}
\input{l2esldef.new}


\def\irtc{\ccal}
\def\irtn{\ncal}
\def\irtt{\tcal}
\def\irtf{\fcal}
\def\irth{\hcal}
\def\irtk{\kcal}

%\def\oda{\overset\cdot\to A}
%\def\odb{\overset\cdot\to B}
%\def\odx{\overset\cdot\to X}
\newcommand{\oda}{\dot A}
\newcommand{\odb}{\dot B}
\newcommand{\odx}{\dot X}
\newcommand{\tght}{\operatorname{t}}
\newcommand{\wght}{\operatorname{w}}
\newcommand{\sprd}{\operatorname{s}}
\newcommand{\hd}{\operatorname{hd}}
\newcommand{\hght}{\operatorname{h}}
%\topmatter
\title{ What is left of CH after you add Cohen reals?}
\author {I. Juh\'asz} 
\author{L. Soukup} 
\author{Z. Szentmikl\'ossy}

%\date{ October 18, 1996 }
\thanks{Research supported by OTKA grant no. 16391}


\begin{document}
\begin{abstract}The principle CH* concerning elementary submodels is formulated and is shown to be valid in any generic extension obtained by adding any number of Cohen reals to a ground model satisfying CH. 

CH* has interesting topological consequences, e.g.: 
\begin{enumerate}
\begin{enumerate}[(i)]
\item Every initially $\omega_1$ compact, countably tight $T_3$ space is compact.
\item Let $X$ be a countably tight compact $T_2$ space; then
\begin{enumerate}[(a)]
\item if $S$ is $G_\delta$-dense in $X$ then every point of $X$ is the
limit  \newline of a converging $\omega_1$-sequence from $S$;\
\item if $Y\subset X$ with $\sprd(Y)\le\omega_1$ then $\hght(Y)\le\omega_1$
\item $X$ contains no complete binary tree of closed sets of height $\omega_2$.
\end{enumerate}
\item If $X$ is a compact $T_2$ space with small diagonal then 
$X$ is metrizable.
\end{enumerate}
\end{enumerate}

\end{abstract}

\maketitle

\section{The Principle CH*}

\smallskip

Our aim here is to formulate a principle that may be considered as a weakening of CH because it will be shown to hold in any generic extension of a ground model satisfying CH by adding any number of Cohen reals. This principle, that will be called CH*, will concern elementary submodels but will include no mention of forcing.

Let us start with the following trivial observation: 
If CH holds then for every uncountable regular cardinal $\lambda$ 
there are "many" {\em countably closed} elementary submodels of size
$\omega_1$ of 
$H(\lambda)$. More precisely, they form a stationary (hence cofinal) 
subset of $[H(\lambda)]^{\omega_1}$. In fact, the existence of such 
countably closed elementary submodels is equivalent to CH. 

What happens to these elementary submodels if Cohen reals are added? 
We are going to denote by $\irtc _\kappa$ the natural notion of
forcing that 
adds $\kappa$ many Cohen reals, i.e. $\irtc_\kappa=\irtf n(\kappa,
2)$. 
The following is well known (cf. e.g. \cite{DTW} or \cite{Sh}).


\begin{fact} \label{1.1}
Assume that our ground model $V$ satisfies CH, $\kappa\in M\prec
H(\lambda)$, 
$[M]^\omega\subset M$ and $G$ is $\irtc_\kappa$-generic over $V$. 
Then, in $V[G]$, 
$$
N=M[G]\prec H(\lambda)^{V[G]},
$$
where
$$
M[G]=\{val_G(\tau)\colon\tau\in M\}.
$$
\end{fact}




Of course, if $\kappa>\omega_1$ then $|M|=\omega_1$ implies $|N|=|M|<\kappa$ hence $N$ can no longer be countably closed. However, it will turn out that $N$ has some properties that are weaker than countable closedness but still strong enough to yield interesting consequences. In order to formulate them we first need a definition.

\begin{definition}\label{1.2} 
Let $N$ be any set and $A\subset N$. We say that $B$ is an $N$-cover of $A$ if the following three conditions are satisfied.
\begin{enumerate}[(a)]
\item $A\subset B\subset N$;
\item $N\cap[B]^\omega$ is cofinal in $[B]^\omega$;
\item for every $S\in N\cap [B]^\omega$ we have $|S\cap A|=\omega$.
\end{enumerate}
\end{definition}


Clearly, if $[N]^\omega\subset N$ then every set $A\subset N$ is its own 
$N$-cover. Hence it is clear that both (i) and (ii) in the following theorem are weakenings of the property of countable completeness.

\begin{theorem}\label{1.3} If $N=M[G]$ is as in \ref{1.1} then
\begin{enumerate}[(i)]
\item $N\cap [N]^\omega$ is cofinal in $[N]^\omega$;
\item every set $X\in [N\cap On]^{\omega_1}$ has an uncountable subset $A\in [X]^{\omega_1}$ that has an $N$-cover.
\end{enumerate}
\end{theorem}


\begin{proof} (i) is well-known and follows easily from the fact that in a
CCC extension every countable set of ordinals is contained in a countable ground model set, moreover from the countable closedness of $M$.

To show (ii) it will suffice to prove that if $p\models\odx\in
[M\cap On]^{\omega_1}$ then some extension $q$ of $p$ forces that some uncountable subset of $\odx$ has an $N$-cover. To this end we first define distinct ordinals $\alpha_\xi\in M\cap On$ and conditions $p_\xi\in
\irtc_\kappa$ for $\xi\in\omega_1$ such that $p_\xi\le p$ and $p_\xi\models\alpha_\xi\in\odx$. We may assume that $\{p_\xi\colon\xi\in\omega_1\}$ is a $\Delta$-system with root $q$; of course then $q\le p$.

Let us set for each $\xi\in\omega_1$
$$
r_\xi=p_\xi\cap M,
$$
and define the $\irtc_\kappa$-names $\oda$ and $\odb$ as follows:
$$
\oda=
\{\langle p_\xi,\alpha_\xi\rangle\colon
\xi\in\omega_1\},\quad \odb=
\{\langle r_\xi,\alpha_\xi\rangle\colon\xi\in\omega_1\}.
$$
Clearly, $q\force\oda\in [\odx]^{\omega_1}\,\, \& \,\,
\oda\subset\odb\subset N$, hence in view of (i) it will suffice to show
that $q$ forces part c) of \ref{1.2}. 
This, on the other hand is an easy consequence of the fact 
that every extension of $q$ can only be incompatible with 
only finitely many of the conditions $p_\xi$.
\end{proof}

Let us now introduce the following piece of notation.

\begin{definition}\label{1.4} For any regular cardinal 
$\lambda$ we denote by $\irtn_\lambda$ the collection of 
those elementary submodels $N\prec H(\lambda)$ that satisfy 
$|N|=\omega_1\subset N$ as well as (i) and (ii) of \ref{1.3}.
\end{definition}


Finally, we are ready to formulate our principle CH*.

\begin{definition}\label{1.5} CH* is the statement that for every 
large enough regular cardinal $\lambda$ the collection 
$\irtn_\lambda$ is cofinal in $[H(\lambda)]^{\omega_1}$.
\end{definition}


Clearly, CH* is a consequence of CH, moreover from \ref{1.1} and 
\ref{1.3} we immediately get the following result.

\begin{theorem}\label{1.6} If $V\models CH$ then for any cardinal $\kappa$ we have $V^{\irtc _\kappa}\models$ CH*.
\end{theorem}


In fact, what can be shown is that for any regular $\lambda>\kappa$, in $V^{\irtc_\kappa}$, the collection $\irtn_\lambda$ is even stationary in $[H(\lambda)]^{\omega_1}$.

\bigskip

\section{The Main Lemma}

\smallskip

It will turn out that all the results that we deduce from CH* 
make use of a 
single lemma concerning initially $\omega_1$-compact and 
countably tight $T_3$ spaces belonging to an elementary submodel
$N\in\irtn_\lambda$. 
Therefore we call this our Main Lemma. 
Before formulating and proving it, however, we have 
to give several auxiliary results.

\begin{lemma}\label{2.1} Let $N$ be an elementary submodel of some 
$H(\lambda)$ with $\omega_1\subset N$ and $X$ be a 
topological space such that $X\in N$. 
Assume that $A,B$ are subsets of $X$ with 
$\overline{A}\cap\overline{B}=\emptyset$ and $\tilde A$, 
$\tilde B$ are $N$-covers of $A$ and $B$, respectively. 
Then $|\tilde A\cap \tilde B\cap X|\ge\omega$ implies that 
there is an $\omega_2$-dyadic system of closed sets in $X$.
\end{lemma}


\begin{proof} By our assumption there is a set 
$T\subset\tilde A\cap\tilde B\cap X$ with $|T|=\omega$. 
In view of \ref{1.2} b)  there are sets $S_A\in N\cap
[\tilde A]^{\omega}$ and $S_B\in N\cap [\tilde B]^\omega$ with 
$T\subset S_A\cap S_B$. Then, clearly, $S=S_A\cap S_B\cap X$ 
is also in $N$ and $|S|=\omega$ as $T\subset S$.

The $\omega_2$-dyadic system will be of the form $\{\overline{S_\alpha^i}
\colon\langle\alpha, i\rangle\in\omega_2\times 2\}$ with 
$S_\alpha^i\subset S$ for all $\langle\alpha,i\rangle$. 
It can be defined by a straightforward transfinite recursion using 
the following claim.

\renewcommand{\theClaim}{}
\begin{Claim} If $\irtt\subset[S]^\omega$ with $|\irtt|\le\omega_1$ 
then there are sets $S_0,S_1\subset S$ with $\overline{S_0}
\cap\overline{S_1}=\emptyset$ such that $|S_0\cap T|=|S_1\cap T|=\omega$ 
for each $T\in\irtt$.
\end{Claim}


Assume, indirectly, that there is a collection $\irtt$ contradicting the claim. Since both $X$ and $S$ belong to $N$, by elementarity, we may assume that $\irtt\in N$ as well. But then $\omega_1\subset N$ and $|\irtt|\le\omega_1$ imply that $\irtt\subset N$. So for each $T\in\irtt$ we have $T\in N\cap 
[\tilde A]^\omega$ and $T\in N\cap [\tilde B]^\omega$, 
hence part c) of \ref{1.2} implies $|T\cap A|=|T\cap B|=\omega$. But then $S_0=S\cap A$ and $S_1=S\cap B$ clearly satisfy $\overline{S_0}\cap\overline{S_1}=\emptyset$ and $|S_0\cap T|=|S_1\cap T|=\omega$ for each $T\in\irtt$, contradicting the choice of $\irtt$. This, in turn, completes the proof of the claim and thus of lemma \ref{2.1}.
\end{proof}

From \ref{2.1} we easily deduce the following lemma.

\begin{lemma}\label{2.2} Assume $N\in \irtn_\lambda$ for some regular $\lambda$, moreover $X$ is a regular space with $X\in N$ and not containing an 
$\omega_2$-dyadic system of closed sets. We also assume that the underlying set of $X$ is an ordinal.
Then, if $A\in [N\cap X]^{\omega_1}$ and $\tilde A\subset X$ is an $N$-cover of $A$, every complete accumulation point of $\tilde A$ in $X$ belongs to $\overline A$, the closure of $A$.
\end{lemma}



\begin{proof} Let $x$ be a complete accumulation point of $\tilde A$
and assume that $x\not\in\overline A$. Then, by the regularity of $X$,
we have an open set $U$ with $x\in U$ and 
$\overline U\cap\overline A=\emptyset$.

We also have $|U\cap\tilde A|=\omega_1$, hence applying (ii) of \ref{1.3} 
there is a set $B\in[U\cap\tilde A]^{\omega_1}$ that has an $N$-cover 
$\tilde B$. But then we also have $\overline A\cap\overline
B=\emptyset$, moreover $B\subset\tilde A\cap\tilde B\cap X$, 
hence $|\tilde A\cap\tilde B\cap X|=\omega_1$. 
Consequently Lemma \ref{2.1} applies and yields an $\omega_2$-dyadic system 
of closed sets in $X$, which is impossible.
\end{proof}

\begin{remark} It can be shown that in \ref{2.2} every complete accumulation point of $\tilde A$ is actually a complete accumulation point of $A$, however we will not use this stronger result.
\end{remark}

The following purely topological lemma is folklore, but we supply its proof because perhaps it has not been published in this generality.

\begin{lemma}\label{2.3} Assume that $X$ is a topological space with
$\tght(X)\le F(X)=\omega$ and $\irtf$ is an $\omega_1$-directed family of 
closed sets in $X$. 
Then for any $A\subset X$ with $\overline A\in\irtf^+$ (i.e.  
$\overline A\cap F\ne\emptyset$ for all $F\in\irtf$) there is a countable set $S\subset A$ with $\overline S\in\irtf^+$.
\end{lemma}


\begin{proof} Assume, indirectly, that $\overline A\in\irtf^+$ but for every countable $S\subset A$ we have $\overline S\not\in\irtf^+$. Then by transfinite recursion on $\alpha\in\omega_1$ we define points $p_\alpha\in\overline A$, sets $S_\alpha\in[A]^{\le\omega}$ and closed sets $F_\alpha\in\irtf$ as follows.

If $\alpha\in\omega_1$ and $p_\beta,S_\beta,F_\beta$ 
have been defined for all $\beta\in\alpha$ then 
$\overline{\cup\{S_\beta\colon\beta\in\alpha\}}\not\in\irtf^+$, 
hence we can pick $F_\alpha\in\irtf^+$ such that 
$F_\alpha\cap\overline{\cup\{S_\beta\colon\beta\in\alpha\}}=\emptyset$
and $F_\alpha\subset\cap\{F_\beta\colon\beta\in\alpha\}$
by the $\omega_1$-directedness of $\irtf$.

Let $p_\alpha\in F_\alpha\cap\overline A$ and, using $\tght(X)\le\omega$, choose $S_\alpha\in[A]^{\le\omega}$ such that $p_\alpha\in\overline{S_\alpha}$. This completes the recursion.

Now, it is obvious that $\{p_\alpha\colon\alpha\in\omega_1\}$ is a free sequence in $X$, contradicting $F(X)=\omega$.
\end{proof}

We are now ready to prove our main lemma.

\begin{lemma}[{Main Lemma}]\label{2.4} Let $X$ be an initially $\omega_1$-compact $T_3$ space with $\tght(X)\le\omega$ and $\irtf$ be an $\omega_1$-directed family of closed sets in $X$. We also assume that the underlying set of $X$ is an ordinal. Assume that $N\in\irtn_\lambda$ and $X,\irtf\in N$, moreover set
$$
\irtf_N=\{\overline{F\cap N}\colon F\in\irtf\cap N\}.
$$
Then
$$
\cap\irtf_N\subset\cap \irtf.
$$
\end{lemma}



\begin{proof} It is well-known that the assumptions on $X$ 
imply $F(X)=\omega$. Also, there is no $\omega_1$-dyadic system of 
closed sets in $X$. Indeed, otherwise we could find a continuous map
$f$  of some closed subspace $Y$ of $X$ onto $2^{\omega_1}$, 
as e.g. in \cite{J}, p. 70. Since $f$ preserves initial 
$\omega_1$-compactness and 
$\wght(2^{\omega_1})=\omega_1$ 
it would follow that $f$ is also a closed map. 
However, as is well-known, closed maps preserve countable tightness, 
hence we are led to the absurd conclusion that $\tght(2^{\omega_1})=\omega$!

Now let $x\in\cap\irtf _N$; to show that $x\in\cap\irtf$ by the regularity of $X$ it clearly suffices to show that for any open neighbourhood $U$ of $x$ we have $\overline U\in\irtf^+$. To this end let us fix a smaller neighbourhood $V$ of $x$ with $\overline V\subset U$.

Using \ref{1.3} (i) and the $\omega_1$-directedness of $\irtf$ it is easy to see that $\irtf\cap N$ is also $\omega_1$-directed. Consequently one can easily define a sequence $\{F_\alpha\colon\alpha\in\omega_1\}\subset\irtf\cap N$ such that $\alpha<\beta$ implies $F_\alpha\supset F_\beta$ and for every $F\in\irtf\cap N$ there is $\alpha\in\omega_1$ with $F_\alpha\subset F$ (note that $|\irtf\cap N|\le|N|=\omega_1$).

Now let us pick for each $\alpha\in\omega_1$ a point 
$x_\alpha\in V\cap F_\alpha\cap N$, this is possible because 
$x\in\overline{F_\alpha\cap N}$ and $x\in V$. Note that 
then for every set $a\in[\omega_1]^{\omega_1}$ we have 
$\{x_\alpha:{\alpha}\in a\}\in\irtf_N^+$. Thus if there is a 
point $y$ with $|\{\alpha\in\omega_1\colon x_\alpha=y\}|=\omega_1$, 
then $\{y\}\in\irtf_N^+$, hence using $y\in N$ also 
$\{y\}\in\irtf^+$. But then $y\in V\subset U$ implies 
$\overline U\in\irtf^+$ as well.

If, on the other hand, for all $y$ we have 
$|\{\alpha\in\omega_1\colon x_\alpha=y\}|\le\omega$ then 
$Y=\{x_\alpha\colon\alpha\in\omega_1\}\in[N\cap X]^{\omega_1}$ 
hence using (ii) of \ref{1.3} there is a set $A\in[Y]^{\omega_1}$ that 
has an $N$-cover $\tilde A$. Since $X\in N$ we can clearly assume that 
$\tilde A\subset X$, because $\tilde A\cap X$ is also an $N$-cover of $A$.

Now all the assumptions of Lemma \ref{2.2} are satisfied, hence every complete accumulation point of $\tilde A$ is contained in $\overline A\subset
\overline V\subset U$. Consequently, as $X$ is initially $\omega_1$-compact, we must have $|\tilde A\setminus U|\le\omega$. Let us fix a set 
$T\in N\cap[\tilde A]^\omega$ such that $\tilde A\setminus U\subset T$.

Then $\tilde A\setminus T\in\irtf_N^+$, hence applying Lemma \ref{2.3}
and property b) of the $N$-covers we can find a set 
$S\in N\cap[\tilde A\setminus T]^\omega$ such that 
$\overline S\in\irtf_N^+$ as well. But then $S\in N$, 
hence $\overline S\in N$ imply that $\overline S\in\irtf^+$, 
consequently from $S\subset U$ we also get $\overline U\in\irtf^+$, 
and thus the proof is completed.
\end{proof}

\bigskip

\section{Applications of CH*}

\smallskip

In this section we are going to present several interesting consequences of CH*. Since CH* implies CH, it is natural that most of these consequences, though not all, have been known to follow from CH. Many of them were also proven to hold in generic extensions obtained by adding $\omega_2$ Cohen reals to a ground model satisfying CH, especially in \cite{D} that gave the main impetus to our present work.

The first result of this section was proved under CH, independently, by 
van Dowen and Dow, see e.g. \cite{D}.

\begin{theorem}\label{3.1} (CH*)\quad Every initially $\omega_1$-compact $T_3$ space $X$ of countable tightness is compact.
\end{theorem}


\begin{proof} We may of course assume that the underlying set of $X$ is an ordinal. Now let $\irtf$ be any maximal filter of closed sets in $X$, we shall show that $\cap\irtf\ne\emptyset$.

To this end we apply CH* and choose a large enough regular $\lambda$
and an elementary submodel $N\in\irtn_\lambda$ with both $X,\irtf\in
N$. Now, since $X$ is initially $\omega_1$-compact and $\irtf$ is
maximal, $\irtf$ is clearly even $\omega_2$-directed. Also, if $F$ is
a closed set with $F\in\irtf^+$ then actually $F\in\irtf$. So using
Lemma \ref{2.3} for $\irtf$ we get that for every 
$F\irtf$ there is an $S\in[F]^{\le\omega}$ such that $\overline 
S\in\irtf$ as well. Reflecting this statement down to $N$ 
we may now conclude that for every $F\in\irtf\cap N$ 
there is a set $S\in N\cap [F]^{\le\omega}$ such that 
$\overline S\in\irtf\cap N$ as well. But from $S\in N$ 
we have $S\subset N$ as well, so 
$\overline S=\overline{\overline S\cap N}\in\irtf_N$. 
Consequently we clearly have $\cap (\irtf\cap N)=\cap\irtf_N$, 
and here the left hand side is non-empty because 
$|\irtf\cap N|\le\omega_1$ and $X$ is initially 
$\omega_1$-compact. Thus we have by our Main Lemma \ref{2.4} that 
$\cap\irtf\supset\cap\irtf_n\ne\emptyset$ and 
this completes the proof that $X$ is compact.
\end{proof}


Our next result seems to be new even under CH. It concerns compact
$T_2$ spaces of countable tightness (in s spaces), but of
course, in view of \ref{3.1}, it could be formulated for 
initially $\omega_1$-compact $T_3$ spaces instead of compact $T_2$ ones.

\begin{theorem}\label{3.2} (CH*)\quad Let $X$ be a CCT space and $S\subset X$ be $G_\delta$-dense in $X$, i.e. $S\cap H\ne\emptyset$ for every non-empty $G_\delta$ set $H\subset X$. Then $X=\ell_{\omega_1}(S)$, i.e. every point $x$ of $X$ is the limit of a converging $\omega_1$-sequence of points of $S$.
\end{theorem}


\begin{proof} If $x\in S$ this is trivial, so assume $x\in X\setminus S$. (Of course, we may assume again that $X\subset On$.) Let $\irtf$ be the family of all those closed $G_\delta$ sets in $X$ that contain $x$.

Using CH* choose a regular $\lambda$ and an elementary submodel $N\in\irtn_\lambda$ such that $X,\irtf,x,S\in N$. Since $\irtf$ is clearly $\omega_1$-directed, the Main Lemma \ref{2.4} can be applied, hence we have
$$
\emptyset\ne\cap\irtf_N\subset\cap\irtf=\{x\}.
$$ 
Consequently, for each $F\in\irtf\cap N$ we have $x\in\overline{F\cap N}$. In fact, we claim that $x\in\overline{F\cap N\cap S}$ is also true. Indeed, first of all we have $x\in\overline{F\cap S}$ since otherwise we had an open set $G$ with $x\in G$ and $G\cap F\cap S=\emptyset$ which is impossible as $G\cap F$ is a non-empty $G_\delta$ set. So by countable tightness of $X$ there is a countable set $T\subset F\cap S$ with $x\in\overline T$ and by elementarity and $x,S,F\in N$ we may assume that $T\in N$ as well. But then $T\subset N$ is also true, hence indeed
$$
x\in\overline T\subset\overline{F\cap N\cap S}.
$$

Now, similarly as in the proof of \ref{2.4} we can define a decreasing 
$\omega_1$-sequence  $\{F_\alpha\colon\alpha\in\omega_1\}\subset\irtf\cap N$ such that for every $F\in\irtf\cap N$ we have $F_\alpha\subset F$ for some $\alpha\in\omega_1$. But then we clearly have
$$
x\in\cap\{\overline{F_\alpha\cap N\cap S}\colon\alpha\in\omega_1\}
\subset\cap\irtf_N=\{x\},
$$
hence if $x_\alpha \in F_\alpha\cap N\cap S$ for each $\alpha\in\omega_1$ then the $\omega_1$-sequence $\{x_\alpha\colon\alpha\in\omega_1\}\subset S$ must converge to $x$, and the proof is completed.
\end{proof}

The following two results were established in \cite{D} in 
$V^{\ccal_{\omega_2}}$ for a ground model $V$ satisfying CH.

\begin{corollary}\label{3.3} (CH*)\quad If $X$ is CCT and $p\in X$ with $\chi(p,X)>\omega$ then $p\in\ell_{\omega_1}(X\setminus\{p\})$, i.e. $p$ is a nontrivial $\omega_1$-limit in $X$.
\end{corollary}


\begin{proof} This is immediate from \ref{3.2} because 
$\chi(p,X)>\omega$ is clearly equivalent to $X\setminus\{p\}$ 
being $G_\delta$-dense in $X$.
\end{proof}

\begin{corollary}\label{3.4} (CH*)\quad If $X$ is CCT and the 
subspace $Y\subset X$ is pseudocompact, then $\overline Y=\ell_{\omega_1}(Y)$.
\end{corollary}


\begin{proof}Indeed, this follows from \ref{3.2} because if $Y$ is pseudocompact then it is $G_\delta$-dense in $\overline Y$.
\end{proof}

The following result was know problem under CH and was proven, under CH, in \cite{JSz}. As it turns out almost the same proof now yields the same conclusion from CH*.

\begin{theorem}\label{3.5} (CH*)\quad  If a compact $T_2$ space $X$ has small diagonal then $X$ is metrizable.
\end{theorem}


\begin{proof} If $\tght(X)>\omega$ held then, 
by the main result of \cite{JSz}, 
$X$ would contain a nontrivial convergent $\omega_1$-sequence 
which contradicts having small diagonal.

So $X$ is CCT and consequently so is $X_\Delta$ obtained by 
collapsing the diagonal $\Delta$ to a single point in $X^2$. 
Indeed, this is so because the CCT property is preserved 
under products and continuous images. But then we must have 
$\chi(\Delta,X_\Delta)=\omega$, since otherwise, by \ref{3.3}, 
$X_\Delta\setminus\{\Delta\}$ would contain an 
$\omega_1$-sequence converging to $\Delta$, hence, 
equivalently, $X^2\setminus\Delta$ would contain such a sequence 
converging to the diagonal $\Delta$, contradicting that 
$X$ has small diagonal. 
However, we clearly have $\chi(\Delta,X^2)=\chi(\Delta,X_\Delta)$, 
hence $X$ has a $G_\delta$ diagonal and so is metrizable.
\end{proof}

The following consequence of CH* is again easily deduced from our main
lemma. It will, however, play an essential role below.

\begin{theorem}\label{3.6} (CH*)\quad Assume that $X$ is CCT and 
$\irtf$ is an $\omega_1$-directed family of closed sets in $X$ 
such that $d(F)\le\omega_1$ for each $F\in\irtf$. 
Then there is a subfamily $\irtf '\in[\irtf]^{\le\omega_1}$ 
such that $\cap\irtf '=\cap\irtf$.
\end{theorem}


\begin{proof} We may assume again that the underlying set of $X$ is an ordinal and from CH* we can find some regular $\lambda$ and $N\in\irtn_\lambda$ with both $X,\irtf\in N$.

Now, for each $F\in\irtf\cap N$ by elementarity there is a dense subset $S$ of $F$ with $|S|\le\omega_1$ and $S\in N$. From $\omega_1\subset N$ then we also have $S\subset N$, hence $\overline{F\cap N}=\overline S=F$. In other
words, we have $\irtf\cap N=\irtf_N$, so by the main lemma $\irtf '=\irtf\cap N$ is as required.
\end{proof}

An immediate consequence of \ref{3.6} is that in a CCT space $X$ there is no strictly decreasing $\omega_2$-sequence of closed sets of density $\le\omega_1$. This fact is used in the next result.

\begin{theorem}\label{3.7} (CH*)\quad If $X$ is CCT and $Y\subset X$ satisfies $\sprd(Y)\le\omega_1$ then $\hght(Y)\le\omega_1$ as well.
\end{theorem}


\begin{proof} Applying \cite[2.26]{J} just as in the proof 
of  \cite[3.13]{J}, we have 
$\operatorname{z}(Y)=\hd(Y)\le \sprd(Y)\cdot \tght(X)^+\le\omega_1$ 
by our assumptions. 
Assume, indirectly, that $\hght(Y)>\omega_1$, 
then there is a strictly decreasing $\omega_2$-sequence 
$\{H_\alpha\colon\alpha\in\omega_2\}$ of closed subsets of $Y$. 
But then the sequence $\{\overline{H_\alpha}\colon\alpha\in\omega_2\}$
is also strictly decreasing, moreover we have 
$d(\overline{H_\alpha})\le d(H_\alpha)\le\omega_1$, 
and this is impossible by our above remark.
\end{proof}

Our last consequence of CH* concerns complete binary trees of 
closed sets in a CCT space $X$. The proof will use \ref{3.6} 
and some ZFC results
that seem to be new and interesting in themselves.

A complete binary tree of closed sets of height $\kappa$ in a space $X$ (or 
simply a Cantor $\kappa$-tree in $X$) is a collection 
$\{F_t\colon t\in 2^{<\kappa}\}$ where each $F_t$ is a nonempty 
closed set in $X$ such that $t\supset s$ implies $F_t\subset F_s$ and $F_{\overset\frown{t0}}
\cap F_{\overset\frown{t1}}=\emptyset$ for each $t\in 2^{<\kappa}$.

We denote by $\irtt r_\kappa(X)$ the family of all Cantor $\kappa$-trees in $X$. Our aim now is to show that if $\kappa$ is regular and $X$ is a CCT space in which there is a Cantor $\kappa$-tree, then there is also one whose members are all separable. First, however, we prove the following lemma.

\begin{lemma}\label{3.8} Assume $cf(\kappa)>\omega$, $X$ is a CCT space and 
$T=\{F_t\colon t\in 2^{<\kappa}\}\in\irtt r_\kappa(X)$. Then there is a countable set $S\subset X$ and an index $t\in 2^{<\kappa}$ such that for each $s\in 2^{<\kappa}$ with $s\supset t$ we have $\overline S\cap F_s\ne\emptyset$.
\end{lemma}


\begin{proof} Assume, indirectly, that there are no such $S$ and $t$. We may then define by transfinite recursion on $\alpha\in\omega_1$ points $p_\alpha\in X$ and indices $t_\alpha\in 2^{<\kappa}$ with $p_\alpha\in F_{t_\alpha}$ and $t_{\alpha_1}\subset t_{\alpha_2}$ for $\alpha_1<\alpha_2$ as follows.

Suppose that $\alpha\in\omega_1$ and $\{p_\beta,t_\beta\colon
\beta\in\alpha\}$ have already been chosen with these properties. 
From $cf(\kappa)>\omega$ it follows that 
$t=\cup\{t_\beta:{\beta}\in\alpha\}
\in 2^{<\kappa}$, hence by the indirect assumption there is an 
extension $t_\alpha\supset t$ such that
$$
\overline{\{p_\beta\colon\beta\in\alpha\}}\cap F_{t_\alpha}
=\emptyset.
$$
We then pick $p_\alpha\in F_{t_\alpha}$ arbitrarily.

Clearly, this construction goes through for all $\alpha\in\omega_1$, 
and it is also obvious that the sequence 
$\{p_\alpha\colon\alpha\in\omega_1\}$ obtained in this way is free. 
But this contradicts that $X$ is CCT.
\end{proof}

Note that this proof only used $F(X)=\omega$. In the following result, however, the compactness of $X$ is used essentially.

\begin{theorem}\label{3.9} If $\kappa$ is an uncountable regular
cardinal and $X$ is a CCT space with 
$\irtt r_\kappa (X)\ne\emptyset$ then there is a Cantor 
$\kappa$-tree $\widetilde{T}\in\irtt r_\kappa (X)$ such that every 
element of $\widetilde{T}$ is separable.
\end{theorem}


\begin{proof} Let $T=\{F_t\colon t\in 2^{<\kappa}\}$ be a Cantor $\kappa$-tree in $X$. We denote by $\irth$ the collection of all pairs $\langle H,t\rangle$ where $H$ is a closed set in $X$, $t\in 2^{<\kappa}$ moreover for any $s\in 2^{<\kappa}$ with $s\supset t$ we have $F_s\cap H\ne\emptyset$. We also consider the following partial order $\prec$ on $\irth$:
$$
\langle H,t\rangle\prec\langle H',t'\rangle\quad
\text{if}\quad H\subset H'\quad\text{and}\quad t\supset t'.
$$

Note that for any $t\in 2^{<\kappa}$ we clearly have 
$\langle F_t,t\rangle\in\irth$.

Let us now list several properties of this partial ordering $\langle\irth ,
\prec\rangle$.

1) For every $\langle H,t\rangle\in\irth$ there are $\langle H_0,t_0\rangle$, 
$\langle H_1,t_1\rangle\in\irth$ such that $H_0\cup H_1\subset H$, 
$H_0\cap H_1=\emptyset$ and $t\subset t_0$, $t\subset t_1$.

Indeed, we may simply put $t_i=\overset\frown{ti}$ and $H_i=H\cap F_{t_i}$ 
for $i\in 2$.

2) If $\nu\in\kappa$ and 
$\{\langle H_\alpha,t_\alpha\rangle\colon\alpha\in\nu\}$ is a decreasing sequence in $\irth$ (i.e. $\alpha\in\beta\in\nu$ imply $\langle H_\beta,t_\beta\rangle\prec\langle H_\alpha,t_\alpha\rangle$) then there is some $\langle H,t\rangle\in\irth$ with $\langle H,t\rangle\prec
\langle H_\alpha,t_\alpha\rangle$ for all $\alpha\in\nu$. In other words, the poset $\langle\irth,\prec\rangle$ is $\kappa$-closed.

Indeed, this is immediate from the regularity of $\kappa$ and the compactness of $X$; we may just set $H=\cap\{H_\alpha\colon\alpha\in\nu\}$ and 
$t=\cup\{t_\alpha\colon\alpha\in\nu\}$.

3) Finally, for any $\langle H,t\rangle\in\irth$ there is a $\langle K,s\rangle\in\irth$ with $\langle K,s\rangle\prec\langle H,t\rangle$ such that $K$ is separable.

Indeed, by our definition, $\{F_r\cap H\colon r\supset t\}$ 
is clearly a Cantor $\kappa$-tree in $H$. But then we may 
apply Lemma \ref{3.8} which yields a countable set $S\subset H$ 
and a sequence $s\supset t$ such that $F_r\cap
\overline S\ne\emptyset$ for each $r\supset s$, hence $\langle
\overline S,s\rangle$ is as required.

Of course, 3) just says that $\irtk=\{\langle K,s\rangle\in\irth\colon d(K)=\omega\}$ is dense in the poset $\irth$.

Now, it is standard to construct using properties 1)-3) 
a full binary subtree 
$\{\langle K_t,s_t\rangle\colon t\in 2^{<\kappa}\}$ 
of $\irtk$ such that for each $t\in 2^{<\kappa}$ we have 
$K_{\overset\frown{t0}}\cap K_{\overset\frown{t1}}=\emptyset$. 
Consequently, the first "co-ordinates" $\{K_t\colon t\in 2^{<\kappa}\}$ 
of this tree yield us a Cantor $\kappa$-tree in $X$ consisting of separable 
sets, as was required.
\end{proof}

Since the weight of a separable $T_3$ space is at most 
$c$, it may not contain a well-ordered decreasing 
sequence of closed sets of length $c^+$. Consequently 
we immediately have the following ZFC result from \ref{3.9}.

\begin{corollary}\label{3.10} If $X$ is a CCT space then 
$\irtt r_{c^+}(X)=\emptyset$.
\end{corollary}


It should be noted that, as is shown e.g. by the 
Alexandrov one-point compactifications of large discrete spaces, 
there is no upper bound for the lengths of decreasing 
sequences of closed sets in CCT spaces.

Of course, under CH we have $c^+=\omega_2$ in \ref{3.10}. 
Combining \ref{3.6} with \ref{3.9} we can see that $\omega_2$ 
works as an upper bound under $CH^*$ as well.

\begin{corollary}\label{3.11} (CH*)\quad For every CCT space $X$ we have 
$\irtt r_{\omega_2}(X)=\emptyset$.
\end{corollary}


The standard proof of the \v Cech--Pospi\v sil theorem (see
e.g. \cite[3.16]{J},) yields the following statement: 
If $X$ is a compact $T_2$ space such that $\chi(p,X)\ge\kappa$ holds 
for each $p\in X$ then $\irtt r_\kappa(X)\ne\emptyset$. 
Hence \ref{3.11} implies that, under CH*, 
in every CCT  space there is a point of character at most $\omega_1$. 
It was conjectured in \cite{J2} that this consequence may be provable in
ZFC. As a matter of fact, it may turn out that even \ref{3.11} 
is provable in ZFC!

Also, the above conjecture was shown to hold in $V^{\irtc_{\omega_1}}$ for 
{\it any \/} ground model $V$. We don't know if the same can be proved about 
\ref{3.11}.

We conclude this paper with a result that says that under PFA even
$\omega_1$ is a (strict) upper bound for the height of a Cantor tree
in a CCT space. In view of our above remarks this constitutes a
strengthening of $A$. Dow's result saying that, under PFA, every CCT
space has a point of countable character. The proof is quite similar
to the proof of this result of Dow  given in \cite{J3} for .

\begin{theorem} (PFA)\label{3.12}\quad If $X$ is a CCT space then 
$\irtt r_{\omega_1}(X)=\emptyset$.
\end{theorem}


\begin{proof} Assume, indirectly, that $T\in \irtt
r_{\omega_1}(X)$. Then, if $P$ is the standard countably closed notion
of forcing that collapses $|X|$ to $\omega_1$, $X$ is no longer
compact in $V^P$. Indeed, there will be many branches of $T$ with
empty intersection. On the other hand, as was shown in the proof of
3.3 in \cite{J3}, $X$ remains both countably compact and countably tight in
$V^P$. Hence, by Balogh's lemma (see 3.1 in \cite{J3}), there is a proper
poset $Q$ that adds a free sequence of length $\omega_1$ to $X$. But
then PFA applied to $P\ast Q$ implies that $X$ contains such a
sequence already in $V$,  a contradiction.
\end{proof}


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