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Let $M$ be a Hausdorff space and $U$ be its open subset. If a subset $V$ of $U$ has compact closure 'in $U$', then why is the closure in $U$ closed in $M$? I think I have to use the Hausdorff condition of $M$, but cannot get the desired result...Could anyone please help me?

Keith
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1 Answers1

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Recall that a compact subspace of an Hausdorff space is closed (proof). Compactness is an intrinsic property of a topological space, hence it is independent of the ambient space (proof). Thus the compact closure of $V$ in $U$ is also a compact subspace of the Hausdorff space $M$, hence it's closed in $M$.

Fabio Lucchini
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  • $\overline{V}$ is closed in $U$ implies there exists a subset $A$ of $M$, which is closed in $M$ such that $A\cap U=\overline{V}$. You cant directly infer that $\overline{V}$ is compact in $U$ implies $\overline{V}$ is also compact in M. For example $(0,1)$ is an open subset of $\Bbb{R}$, and $(0,1/2]$ is closed in $(0,1)$, with the topology induced by $\Bbb{R}$, but $(0,1/2]$ is not closed in $\Bbb{R}$ – QED Jan 09 '18 at 18:23
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    Compactness is an intrinsic property of a topological space: it doesn't dependes on the outer spaces. – Fabio Lucchini Jan 09 '18 at 18:26
  • I think that is what is to be proved here: that compactness is an intrinsic property of a topological space. Otherwise it was a trivial problem. – QED Jan 09 '18 at 18:29
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    But this is clear from the definition of compactness... – Fabio Lucchini Jan 09 '18 at 18:36
  • Note that in your example $(0,1/2] $ is not compact... – Fabio Lucchini Jan 09 '18 at 18:37
  • I know it is not compact. I just gave you amn example that closed in a subset doesn't not imply closed in the entire space. You need to say in the proof whatever you find is 'clear' from the definition of compactness. – QED Jan 09 '18 at 18:39