# Totally bounded set implies its closure is totally bounded

Let $$X$$ be a metric space. I want to show that:

If a subset $$A \subset X$$ is totally bounded, then its closure $$\overline{A}$$ is totally bounded.

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

This is part of a bigger problem I want to prove.

• $A \subset \bigcup_{x\in F}B(x,\varepsilon/2) \implies \overline{A}\subset \bigcup_{x\in F}\overline{B(x,\varepsilon/2)} \subset \bigcup_{x\in F}B(x,\varepsilon)$. Commented Oct 27, 2017 at 20:51
• This is also the $\Longrightarrow$ direction of math.stackexchange.com/questions/2958346/… . Commented Sep 19, 2019 at 16:32

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.