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 countablethen 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 $G \subset X$ beanyset 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.