Let M be a complete and $A \subset M$ be totally bounded. $\Rightarrow$ cl(A) is compact.

Asked Jun 02 '17 at 06:24

Active Jun 02 '17 at 22:20

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Claim

Let $M$ be a complete and $A \subset M$ be totally bounded. $\Rightarrow$ cl(A) is compact.

to prove above claim, please give me any starting point.

What I know about compactness is regarding finite sub-cover for every open cover of $\operatorname{cl}(A)$.

I know M is complete and $A$ is totally bounded, which means contained in the union of finite open disks.

but with only this two facts, how could I get to the conclusion of compactness of $\operatorname{cl}(A)$?

edited Jun 02 '17 at 22:20

Davide Giraudo

asked Jun 02 '17 at 06:24

delog

Try proving that $cl(A)$ is totally bounded. Then use this: https://math.stackexchange.com/questions/109550/totally-bounded-complete-implies-compact – Prahlad Vaidyanathan Jun 02 '17 at 06:29
are you sure presupposing that is ok? – Asinomás Jun 02 '17 at 06:35

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