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We say that $K \subset X$ is relatively compact, if for every sequence $\{x_n\}_{n \in \mathbb{N}} \subset K$ has a Cauchy sub-sequence.

We say that $N_\varepsilon$ is an $\varepsilon\text{-net}$ of $A \subset X$ if for every $a \in A$ there is $y \in N_\varepsilon$ such that $d(a, y) < \varepsilon$.

Now, let $X$ be a metric space. I want to show that if $K \subset X$ is relatively compact, then for every $\varepsilon > 0$ there is a finite $\varepsilon\text{-net}$ for $K$.

My problem here is that $X$ is not necessarily complete, and that means that $\overline K$ is not necessarily compact.

$\overline K$ not being compact makes it harder for me to build the finite $\varepsilon\text{-net}$ for $K$.

I thought maybe covering $K$ with open balls of radius $\varepsilon$ but I am not sure what to do from there, or how the fact that each sequence in $K$ has a Cauchy sub-sequence helps here.

Help would be appreciated.

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HINT: Suppose that $\epsilon>0$, and there is no finite $\epsilon$-net in $K$; then you can recursively choose an infinite sequence whose points are all at least $\epsilon$ away from one another.

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  • $\begingroup$ Thank you! This hint was very helpful! $\endgroup$
    – Gabi G
    Dec 26, 2020 at 21:59
  • $\begingroup$ @GabiG: You’re welcome. $\endgroup$ Dec 26, 2020 at 22:23

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