I am told and have seen proof of:

Theorem. Let $X$ be a metric space. If the set $M ⊂ X$ is relatively compact, then for any $\epsilon > 0$ there exists a finite $\epsilon$-net for $M$. If $X$ is complete, then the converse holds as well."

We denote by "relative-compactness" of $M$, the notion that for an arbitrary sequence $F \subset M$, we can find at least one convergent sub-sequence (convergent to some point $x$ in $X$). Of course, if $M$ is also closed, it follows that $M$ is compact.

Also we define a $\epsilon$-net for $M$ to be:

"Definition: Let $M \subset X$ ,and let $\epsilon > 0$ be some number. A set $N \subset X$ is said to be an $\epsilon$-net of the set $M$, if for any $x\in M$, there is a point $y \in N$ such that $x \in B(y,\epsilon) = \{r \in X : \rho(y,r) < \epsilon\}$."

If N is a finite set as well, then it is called a finite $\epsilon$-net.

Now there is a stated corollary (without proof) to our theorem I need your help to wrap my head around:

Corollary. The subset $M$ of a complete metric space $X$ is relatively compact if and only if for any $\epsilon$ > 0 it has a relatively compact $\epsilon$-net"

I started a proof for the backwards implication:

By our assumption, note that for any $\epsilon > 0$, there is a "$\epsilon/2$"-net of $M$ that itself is relatively compact; denote such a "$\epsilon/2$"-net by $N(\epsilon/2)$. So $N(\epsilon/2)$ itself, by relative compactness, must have a finite $\epsilon/2$-net (we stick to the very same value for "$\epsilon$", "$\epsilon/2$" as we have every right to do so), which we will denote by "$f$". Let "$F$" be the union of all open balls of radius "$\epsilon/2$" centered at the (finite) points of "$f$".

This finite $\epsilon/2$-net of $N(\epsilon/2)$ is not quite an $\epsilon/2$-net for $M$, but it must be "close" to being one: note that any point in $M$ is at most less than "$\epsilon/2$" distance away from the set $F$, as $F$ covers $N(\epsilon/2)$. So now we inflate the balls up: all up to a radius of "$\epsilon$" (they were all of radius "$\epsilon/2$" before this procedure). Call this new set of inflated balls "$D$". Now it must be true that any "$m$" in $M$ is also in $D$ and so it follows that we have found a finite $\epsilon$-net for $M$.

Does this proof make sense or it is non-sense?

Also can someone provide the forwards-implication please? I tried an idea and it was rubbish (I tried using completeness and the fact that relatively compact sets are bounded), so I'd like your help to see how it's true.

Thank you very much and have a nice day.


In this question and its answer we see proved that $M$ is relatively (sequentially) compact iff $\overline{M}$ is (sequentially) compact. As part of this answer I showed that if $M$ totally bounded (for every $\varepsilon >0$, $M$ has a finite $\varepsilon$-net) then $\overline{M}$ is totally bounded. And a closed subset of a complete metric space is complete.

So then if $X$ is complete and $M$ is totally bounded then $\overline{M}$ is complete and totally bounded and thus compact (this is your reverse implication). And this implies that $M$ is relatively compact. If $M$ is relatively compact, then $\overline{M}$ is compact, and thus totally bounded which implies $M$ is totally bounded. This does not need completeness.


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