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If $X$ is a complete metric space. I want to show that:

$A \subset X$ is totally bounded $\Rightarrow \overline{A}$ is totally bounded.

Definition of totally bounded: a set $A$ is totally bounded if, for all $\varepsilon > 0$, there is a finite $F\subset A$ such that $A \subset \bigcup_\limits{x \in F} B(x, \varepsilon) $

I don't think completeness is needed in order to show this, but it's part of a bigger problem I want to proof.

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    $\begingroup$ $A \subset \bigcup_{x\in F}B(x,\varepsilon/2) \implies \overline{A}\subset \bigcup_{x\in F}\overline{B(x,\varepsilon/2)} \subset \bigcup_{x\in F}B(x,\varepsilon)$. $\endgroup$ – Vinícius Novelli Oct 27 '17 at 20:51
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For completeness, let me post an answer to conclude the question.

First of all, we want to prove that for any $\varepsilon>0,$ there exists a finite set $F\subseteq \overline{A}$ such that $$\overline{A}\subseteq\bigcup_{x\in F}B(x,\varepsilon).$$ Since $A$ is totally bounded, there exists a finite set $F\subseteq A\subseteq \overline{A}$ such that $$A\subseteq\bigcup_{x\in F} B(x,\varepsilon/2).$$ Observe that $$\overline{A}\subseteq\bigcup_{x\in F}\overline{B(x,\varepsilon/2)}\subseteq \bigcup_{x\in F}B(x,\varepsilon).$$ Therefore, $\overline{A}$ is totally bounded.

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