# Does a compact subspace have to be closed in an arbitrary metric space?

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.
Since the question is sated for metric spaces, why not use the following elementary fact. Take $u$ to be a limit point and $u_n$ a sequence in the compact subset converging to $u$. Since $u_n$ has a subsequence converging in the compact subset, $u$ must be in the compact subset. – William Oct 13 '12 at 2:37
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.