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It is well known that in an arbitrary metric space a sequentially compact subset is closed and bounded, under which hypothesis does the converse hold? Or in other words, are there spaces, other than $\mathbb{R}^n$ (or any finite dimensional Banach space) with this property?

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    $\begingroup$ Sequential compactness = Compactness = Complete and totally bounded. So we are looking for a space where closed and bounded implies complete and totally bounded. Thus, the space should at least be complete, so you should look for complete spaces where bounded and totally bounded are same thing. $\endgroup$ – Vishal Gupta Feb 27 '17 at 16:24
  • $\begingroup$ Since you know that in metric spaces, Compact $\iff$ Sequentially compact. So I refer you here. $\endgroup$ – Error 404 Feb 27 '17 at 17:26
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Every finite dimensional normed vector space. Since each of them is isometrical isomporphic to $\mathbb{R}^N$. Conversely every infinite dimensional n.v.s. has NOT this property since the closed unit ball is closed bounded but not compact

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  • $\begingroup$ I assumed La Flaca meant "that are not homeomorphic to $\mathbb R^n$" when they said "apart from". But maybe they didn't. $\endgroup$ – Mark S. Feb 27 '17 at 16:35
  • $\begingroup$ @MarkS. Yes, you're right, that's what I meant. But it's my fault, I didn't specified it. I will edit $\endgroup$ – la flaca Feb 27 '17 at 16:58

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