# On the cardinality of perfect spaces with the countable chain condition

QUESTION: Does every regular perfect space with the countable chain condition have cardinality bounded above by the continuum? Is this at least true for perfectly normal ccc spaces?

Recall that a space is perfect if every closed set is a $G_\delta$ set (that is, a countable intersection of open sets). We say that a space has the countable chain condition if every family of pairwise disjoint non-empty open sets is countable. Perfectly normal just means perfect and normal.

If one strengthens the ccc to the condition every discrete set is countable then the answer is yes. This follows from Hajnal and Juhasz's result that every space where singletons are $G_\delta$ and discrete sets are countable has cardinality at most continuum. Regularity is not needed for this to be true ($T_1$ is enough).

(It's easy to prove that in a perfect space where closed discrete sets are countable, it's even true that every discrete set is countable. So from the above statement it follows that every Lindelof $T_1$ perfect space has cardinality at most the continuum, something that was proved by Alexandroff and Urysohn in their Memoire, at least for compact Hausdorff spaces, if I'm not mistaken).

On the other hand:

There are Hausdorff ccc perfect spaces of arbitrarily large cardinality.

For example, let $\kappa$ be any cardinal and $D_n=\{x \in 2^\kappa: |x^{-1}(1)|=n \}$. Set $X=\bigcup_{n<\omega} D_n$. Let $\tau$ be the refinement of the usual topology on $X$ obtained by making every $D_n$ closed discrete. In other words, a basic open set has the form $U \setminus \bigcup_{n\in F} D_n$ where $U$ is open in the topology on $X$ inherited from $2^\kappa$ and $F \subset \omega$ is finite. This space is ccc: indeed, let $\{U_\alpha: \alpha<\aleph_1\}$ be a pairwise disjoint family of open sets of cardinality $\aleph_1$. By the pigeonhole principle we can assume that for some fixed finite set $F$ we have $U_\alpha=V_\alpha \setminus \bigcup_{n \in F} D_n$, where $V_\alpha$ is open in the usual topology on $X$, for every $\alpha<\aleph_1$. So for all $\alpha, \beta <\aleph_1$ such that $\alpha \neq \beta$ we have that $V_\alpha \cap V_\beta \subset \bigcup_{n \in F} D_n$ which implies that $V_\alpha \cap V_\beta$ is empty, as $\bigcup_{n\in F} D_n$ is nowhere dense. But this is a contradiction since $X$ with the topology inherited from $2^\kappa$ is dense in the ccc space $2^\kappa$ and thus it is also ccc.

The space $(X, \tau)$ is more than perfect. As a matter of fact, let $G \subset X$ be any set and set $G_n=D_n \setminus G \cap D_n$. Note that $G_n$ is closed. Then $G=\bigcap_{n<\omega} (X \setminus G_n)$, which proves that $G$ is a $G_\delta$ set.

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I don't know if this helps, but a related result is that MA + nonCH implies that a locally compact first countable ccc space is separable, and thus has cardinality at most the continuum. I think it is due to I. Juhasz. Ref if needed: I. Juhasz: Cardinal functions in Topology. Number 34 in Mathematical Centre Tract. Mathematisch Centrum, 1971. –  Mathieu Baillif May 13 '13 at 22:53
Also, I just saw a paper by Larson and Tall ("Locally compact perfectly normal spaces may all be paracompact") in which their Theorem 2 states that is is consistent with ZFC that every first countable hereditarily normal countable chain condition space is hereditarily separable. –  Mathieu Baillif May 13 '13 at 23:23
Thank you, Mathieu, these are two very nice results. In fact, every first countable ccc space has cardinality at most continuum in ZFC, by an old result of Hajnal and Juhasz. –  Santi Spadaro May 14 '13 at 0:03
Thanks for this result, Santi. I was not aware of that. –  Mathieu Baillif May 14 '13 at 18:44
Any T_3 counterexample must have large pi-character, just as your T_2 counterexample does. If X is T_3, has all points G_delta, and has more than c points, then X has more than c RO sets. By the corollary to 2.37 in Juhasz' book, if X is T_3, ccc, and has more than c RO sets, then X has an open set of points all with pi-character at least c^+. I speculate that if X is also T_{3.5}, then you might get some traction by working with a compactification of X, as most of the interesting consequences of large pi-character are for compact spaces. –  David Milovich Jun 11 '13 at 2:00

The reason why my Hausdorff non-regular counterexample above is perfect is that it is a countable union of closed discrete sets. Now, Uspenskij constructed $\sigma$-closed discrete ccc regular spaces of arbitrarily large cardinality. So that answers my question in the negative.