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sequentially compact

topology, sequence, convergence
sequential compactness
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Mathematics Subject Classification

40A05 no label found54D30 no label found


Are you sure that
"Sequential compactness is equivalent to compactness when $X$ is a metric space."?

I can prove this claim only when $X$ is metric and separable.


Munkres proves this as part of Theorem 28.2 in Topology, 2nd ed. (pp. 179-180).

In particular, Munkres proves that if $X$ is sequentially compact metric, then $X$ is totally bounded, that is, given a positive $\epsilon$, there is a finite covering of $X$ by $\epsilon$-balls. Hence $X$ is Lindel\"{o}f. Because $X$ is also metric, this implies that $X$ is separable. So the assumption that $X$ is separable is not necessarily to show that sequential compactness implies compactness.

The proof that compactness implies sequential compactness does not use separability.

I try to prove that "Sequentially compactness is equivalent to compactness when X is metric space", if you interested in this problem, can you give me some help?

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