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I am trying to show that if $A$ and $B$ are compact in a Hausdorff space, then their intersection is compact.

The way I did it is as follows. $A$ and $B$ are compact in a Hausdorff space, so they are closed. As a consequence, their intersection (denote it by $C$) is closed. But $C$ is closed in $A$, which is compact, so $C$ is compact.

I am pretty sure this is correct, but so far compactness seems a bit tricky, so I want to ensure I get everything right.

Thank you!

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Looks good! ${}$ – Ayman Hourieh Feb 22 '13 at 23:23
A little detail about "$C$ is closed in $A$": $A$ and $B$ are closed in the Hausdorff space (denoted $H$), so $C = A \cap B$ is closed in $H$. In the subspace $A$, $C \cap A = C$ is closed in $A$. – hengxin Feb 11 '14 at 5:35

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up vote 3 down vote accepted

Yes, that's correct. Your proof relies on Hausdorffness, and it's worth noting that this is necessary: here you will find an example of a non-Hausdorff space with two compact subsets $A$, $B$ having a non compact intersection. Neither $A$ nor $B$ is closed.

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To be more precise, the proof relies on the fact that compact sets are closed. There are non-Hausdorff spaces in which all compact sets are closed, so the proof would work fine there. – Austin Mohr May 12 at 17:05

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