Let $X$ be a metric space. Prove if $X$ is compact, then $X$ is separable.
- X separable $\iff X$ contains a countable dense subset.
- $E \subset X $ dense in $X \iff \overline{E} = X$.
- $X$ compact $\iff$ every open cover of $X$ admits a finite subcover.
one might also use that $X$ compact in a metric space implies closed and bounded.
Proof
We want to show $\exists E \subset X. X = \overline{E} = E \cup E'$ where $E'$ denotes the set of limit points. If $X$ is compact then a subset $E$ of $X$ would be compact, so that subset is closed and bounded. Since $E$ is closed, $E \supseteq E'$. Now we need to show that $E$ is a countable dense subset... But I don't know where to go from here. I have the following:
Hint: cover $X$ with neighborhoods of radius $\frac{1}{n}$ - so for every positive integer $n$, there are finitely many neighborhoods of radius $\frac{1}{n}$ whose union covers $X$. So maybe it would be better to work with the open cover definition of compact.