How to prove that every extremally disconnected collectionwise Hausdorff space is discrete?

Can someone at least give me a hint as to how to construct the proof?

  • $\begingroup$ According to Wikipedia (no reference given there) this holds for first countable spaces, not general ones? $\endgroup$ Jul 15 '18 at 5:24
  • $\begingroup$ This is false as stated: $\beta \omega$ is not discrete, but it is e.d. and cwH. $\endgroup$ Jul 15 '18 at 9:42

Suppose $X$ is extremally disconnected (this implies the following property: if $U$ and $V$ are disjoint open sets then $\overline{U}$ and $\overline{V}$ are also disjoint (and open)), Hausdorff and cwH as well.

Suppose that $(x_n)$ is a sequence in $X$ of all distinct points, such that $x_n \to x \in X$, where $x \notin \{x_n: n \in \omega\}$. Then as $X$ is cwH, and the set $\{x_n: n \in \omega\}$ is discrete (this uses that $X$ is a Hausdorff space), we can find open sets $U_n$ such that $x_n \in U_n$ and such that the $U_n$ are pairwise disjoint.

But then $U = \bigcup_{n} U_{2n}$ and $V = \bigcup_{n} U_{2n+1}$ are open and disjoint and $x \in \overline{U} \cap \overline{V} \neq \emptyset$ and this contradicts $X$ being extremally disconnected.

It follows that $X$ can only have trivial convergent sequences: ones that are eventually constant, in essence, or we could extract a sequence as above from it.

So if $X$ were also first countable or a sequential space (more generally), if $X$ were non-discrete, there would be a non-sequentially closed set $A$, and we'd have $x_n$, all from $A$, converging to some $x \notin A$, and we could apply the above argument to get a contradiction. But I don't think "$X$ cwH and e.d. implies $X$ discrete" holds in full generality. A counterexample is $\beta \omega$, the Čech-Stone compactification of the countable discrete $\omega$, which is even collectionwise normal.


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