# Totally bounded space.

I'm studying the Riesz-Fréchet-Kolmogorov Theorem, and I have a problem with the first line of the proof (cfr.Brezis) who said:

"Let $F\subset L^p(\Omega)$ a relatively compact space $\Rightarrow$ $F$ is totally bounded, i.e. forall $\varepsilon>0$ there exist $f_1,...,f_n\in F$ such that $F\subset\cup_{i=1}^n B(f_i,\varepsilon)$."

I know (cfr. Brezis) that in a metric space $F$ is totally bounded if and only if for all $\varepsilon>0$ there exist a finite covering of $F$ by finite balls of radius $\varepsilon$; and
$F$ is relatively compact if and only if its closure is compact (if and only if its closure is totally bounded).

My idea to understand the line of the proof is this:

if $F$ is relative compact, so $\overline{F}$ is totally bounded $\Rightarrow$ forall $\varepsilon>0$ there exist $g_1,...,g_n\in\overline{F}$ such that $F\subseteq\overline{F}\subseteq\cup_{i=1}^nB(g_i,\varepsilon)$. Now the problem is that if $F$ is not closed, $g_i$ may be an element of $\overline{F}$\ $F\subseteq\partial F$.

By the definition of $\overline{F}=F \cup\partial F$, for all $g_i\in\partial F$, $\exists r>0$ such that $B(g_i,r)\cap F\neq\emptyset\Rightarrow \exists f_i\in F$ such that $f_i\in B(g,r)\cap F°\subset F.$
Now, the above assertion should prove that $B(g_i,\varepsilon)\subset B(f_i,\varepsilon+r)\Rightarrow F\subset\overline{F}\subset \cup_{i=1}^nB(f_i,\varepsilon+r)$ as desired.

You kind of guessed the idea, but you do not seem to have done it quite right: namely, you've considered some $r>0$, whereas you should have pointed out that such $r$ could be arbitrarily small, regardless of $\varepsilon$. I'd rather have done it directly like: consider $g_i\in \overline F$ such that $\{ B(g_i,\varepsilon/2)\}_{i=1}^k$ covers $\overline F$. For all $i$, consider $h_i\in F$ such that $d(h_i,g_i)<\frac\varepsilon 2$. Then, $B(h_i,\varepsilon)\supseteq B(g_i,\varepsilon/2)$, which implies that $\{B(h_i,\varepsilon)\}_{i=1}^k$ covers $\overline F$.