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> 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?
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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*.



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> 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).
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(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:



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> There are Hausdorff ccc perfect spaces of arbitrarily large cardinality.
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> 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.
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> 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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On the cardinality of perfect spaces with the countable chain condition
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. 


See this paper: <http://dml.cz/dmlcz/106296>


