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Let $H$ be a complete metric space, we equip $H$ with the induced metric norm. Let $S$ be a set of $H$.

Assume that for any $\epsilon>0$, we can find compact set $K_\epsilon$, such that $S$ is contained in a $\epsilon$ neighborhood of $K_\epsilon$.

Q Can we say $\bar S$, i.e. the closure of $S$, is compact?

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    $\begingroup$ What notion of "dimension" are you using? What do you mean by "the induced metric norm"? $\endgroup$ – Eric Wofsey Sep 25 '18 at 4:11
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Answer assuming completeness: the hypothesis is very confusing but the result is true in any complete metric space. Since compact sets are totally bounded, the hypothesis tells you that $S$ is totally bounded. Hence its closure is compact.

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