Let $F$ be a subset of a metric space $(X, d)$ such that $\overline{F}$ is compact. Show that $F$ is totally bounded. Let $F$ be a subset of a metric space $(X, d)$ such that $\overline{F}$ is compact. Show that $F$ is totally bounded.

how can I able to solve this.I have no idea.thanks for your help
 A: $\newcommand{\cl}{\operatorname{cl}}$Fix $\epsilon>0$. Since $\cl F$ is compact, it is also totally bounded, and there is a finite $A\subseteq\cl F$ such that $\cl F\subseteq\bigcup_{x\in A}B\left(x,\frac{\epsilon}2\right)$. Let $x\in A$; if $x\in F$, set $x'=x$. Otherwise, $x\in(\cl F)\setminus F$, so there is a point $x'\in F\cap B\left(x,\frac{\epsilon}2\right)$. Show that $F\subseteq\bigcup_{x\in A}B(x',\epsilon)$, and conclude, since $\epsilon>0$ was arbitrary, that $F$ is totally bounded.
Added: The intuition that you should have is something like this. For each $\epsilon>0$ you want to find a finite $A\subseteq F$ such that $F\subseteq\bigcup_{x\in A}B(x,\epsilon)$. You know that $\cl F$ is totally bounded, and $F\subseteq\cl F$, so you might start with a finite $A\subseteq\cl F$ such that $\cl F\subseteq\bigcup_{x\in A}B(x,\epsilon)$. If it happens that $A\subseteq F$, you’re done: this set $A$ works fine. The problem is that some points of $A$ may be in $(\cl F)\setminus F$. Of course those points are limit points of $F$, so they’re ‘very close to’ $F$: we can find points of $F$ as close to them as we want. But no matter how close a point $y$ is to $x$, we can’t guarantee that $B(y,\epsilon)$ will cover every part of $F$ that $B(x,\epsilon)$ covered. There’s a standard trick to handle this problem: we’ll choose $A$ so that smaller balls centred at points of $A$ cover $\cl F$. Specifically, I chose $A$ so that the $\frac{\epsilon}2$-balls centred at points of $A$ cover $\cl F$. Now I can move the centre of a ball a little bit, say from $x$ to $x'$, and while $B\left(x',\frac{\epsilon}2\right)$ won’t necessarily cover the same part of $F$ that $B\left(x,\frac{\epsilon}2\right)$ did, the larger ball $B(x',\epsilon)$ will.
