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For Euclidean spaces, we have that a compact subspace has to be closed (and bounded.) But how about an arbitrary metric space? Or how about an arbitrary topology space?

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A compact subset of a Hausdorff space is closed (exercise), and any metric space is Hausdorff. In general this need not be the case. The simplest counterexample is the $2$-point space with the indiscrete topology.

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    @Voldemort: No, a subset that contains all of its limit points is necessarily closed in any space. What Qiaochu is pointing out is that in a non-Hausdorff space you can have a compact set that doesn’t contain all of its limits points and therefore isn’t closed.2012-10-13
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It's also true that any compact subset of a metric space is bounded, since $\{B_n( 0) : n \in \mathbb{N} \}$ is an open cover.

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    That I already knew. Actually, in metric space, a subset is compact if and only if it is totally-bounded and complete.2012-10-13