Suppose a subspace of $X$ is compact iff closed. Then is $X$ compact hausdorff?

Asked Sep 02 '19 at 13:05

Active Sep 02 '19 at 15:09

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Let $X$ be a topological space.

So if $X$ is compact hausdorff, then a subspace is closed if and only if it is compact. Does the converse hold?

If a subspace of a topological space $X$ is compact if and only if it is closed, then is $X$ hausdorff? (clearly it must be compact).

Note: this is not a duplicate of "If every compact set is closed, then is the space Hausdorff?", because $X$ must be compact here too.

edited Sep 02 '19 at 15:09

asked Sep 02 '19 at 13:05

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I think the interval $[-1,1]$ with two origins could be a counterexample – Alessandro Codenotti Sep 02 '19 at 13:24
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@AlessandroCodenotti Is it? Removing one of the origins yields a compact subset that isn't closed. – Andrew Dudzik Sep 02 '19 at 14:22
@slade you're right – Alessandro Codenotti Sep 02 '19 at 14:24
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One quick observation: $X$ must be $T_1$, as any singleton is compact and therefore closed. – Andrew Dudzik Sep 02 '19 at 14:34
@Slade what about a counterexample that is $T_0$ but not $T_1$? i.e. can we assume that $X$ must be $T_0$, and would that be enough? – Sep 02 '19 at 14:40
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I showed that $X$ is $T_1$ under your assumptions, so if we're looking for counterexamples we have to look for spaces that are $T_1$ but not Hausdorff. – Andrew Dudzik Sep 02 '19 at 14:44
@EricWofsey oh, I see. Sorry. – Sep 02 '19 at 14:49
This answer seems to give a counterexample. – Arnaud D. Sep 02 '19 at 14:57
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@EricWofsey Right, my bad. But the accepted answer to this question gives again the counterexample mentioned above, and this seems to be related. – Arnaud D. Sep 02 '19 at 15:15

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