My quick question is this: I know it's true that any sequence in a compact metric space has a convergent subsequence (ie metric spaces are sequentially compact). Also, any arbitrary compact topological space is limit point compact, ie every (infinite) sequence has a limit point.(adsbygoogle = window.adsbygoogle || []).push({});

So in general, are the compact spaces that are not sequentially compact?

This is part of a larger problem: If a real-valued function on a topological space X is proper, show the image of f is closed. My idea was to chose a limit point y of f(X) and a sequence f(x_n) in f(X) converging to Y. Cover this sequence by a closed interval I. The the preimage of I is compact and contains the sequence x_n. Now, x_n has a limit point, say x.

If I could show f(x)=y, I'd be done. But i get stuck without being able to use sequential compactness.

THanks ahead of time

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# Convergent subsequences in compact spaces

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