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Proposition: Show that every singleton in a hereditarily compact Hausdorff space is clopen

Recall, a space is hereditarily compact if every subset is compact.

My attempt:

Let $x \in X$, then $\{x\}$ as a singleton is compact since every single subset is compact. It is compact in a Hausdorff space, so it is closed. Then given $x\neq y \in X$, $X \backslash \bigcup_{x \in X\backslash \{y\}} \{x\} = \{y\}$

The next step is that I wish to show $\{y\}$ is open. But this is only true if $X$ is finite $\implies \bigcup_{x \in X\backslash \{y\}} \{x\}$ is closed (finite union of closed is closed).

But to prove that $X$ is finite, I must show that all singletons are open. By contrapositive, $\{\{x\}|x\in X\}$ forms an open cover without finite subcover hence $X$ is not compact iff it is not finite.

How to resolve this chicken-egg problem?!

Thanks!

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Recall that Hausdorff spaces are $T_1$, but also that compact subsets of Hausdorff spaces are closed.

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  • $\begingroup$ Oh...so just consider $X\backslash (X \backslash \{y\}) = \{y\}$ $\endgroup$ – Olórin Aug 10 '16 at 15:25
  • $\begingroup$ Yes. You already had the ingredients for a quickie proof. You just missed your exit, and had to take a longer route. $\endgroup$ – Asaf Karagila Aug 10 '16 at 15:26

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