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Is the closure of a compact subspace of a topological space always compact?

I want to say no, but i can't think of/find any counterexamples. I think its probably true with added conditions, like Hausdorff property? Because the closure is always closed and closed sets in a compact hausdorff space are compact.

Any counterexamples for a non-Hausdorff space?

Antonio Maria Di Mauro
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bmmcutet12
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  • Are you asking "Is the closure of a compact subspace of a topological space always compact"? – Lee Mosher Aug 19 '17 at 21:25
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    This is true in the Hausdorff setting, since there compact sets are closed, and closed sets are their own closure: https://math.stackexchange.com/questions/83355/how-to-prove-that-a-compact-set-in-a-hausdorff-topological-space-is-closed – Kaj Hansen Aug 19 '17 at 21:48
  • Also see this. –  Aug 20 '17 at 04:15

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Consider the space $X = \mathbb{Z}$ equipped with the topology where the non-empty open sets are precisely those containing $0$.

As a finite set $\{ 0 \} $ is compact. Further we have that $\overline{\{0\}} = X$ and $X$ is not compact since $\bigcup_{x \in X} \{0,x\}$ is an open cover with no finite subcover.

It is however always the case that the closure of a compact set in a Hausdorff space is compact since in Hausdorff spaces compact sets are closed. (not because of the similar fact that you give that closed sets in compact spaces are compact, even if those spaces aren't Hausdorff)

Rhys Steele
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