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Let $X$ be a compact Hausdorff space and let $Y$ be a locally compact subspace. Does it follow that $Y$ is closed in $X$?

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    $\begingroup$ From Wikipedia: "All open or closed subsets of a locally compact Hausdorff space are locally compact in the subspace topology." Pick a non-closed, open example. $\endgroup$ – Theo Bendit Apr 30 at 11:31
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    $\begingroup$ Note that a closed subset of a compact space would again be compact. The general fact: if $X$ is locally compact Hausdorff then $Y \subseteq X$ is locally compact in the subspace topology iff $Y$ is locally closed i.e. it can be written $Y=O \cap C$ where $C$ is closed and $O$ is open in $X$. $\endgroup$ – Henno Brandsma Apr 30 at 12:34
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No. Take $X=[0,1]$ and $Y=(0,1)$ with the standard topology.

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