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.
Best Answer
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.