Let $X$ be a metric space. Prove if $X$ is compact, then $X$ is separable.

  1. X separable $\iff X$ contains a countable dense subset.
  2. $E \subset X $ dense in $X \iff \overline{E} = X$.
  3. $X$ 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.


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.

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    $\begingroup$ The collection of the centers of these balls of radius 1/n, n varying among all positive integers, is a countable and dense set. WARNING: definition 1 is only true in $\mathbb{R}^n$, false in general. Definition 2 is OK. $\endgroup$ – Pietro Majer Oct 15 '14 at 1:11
  • $\begingroup$ Ooh yes, you're right, thanks. $\endgroup$ – user181728 Oct 15 '14 at 1:13
  • $\begingroup$ Related: Prove that every compact metric space is separable (Although it seems that in that question the OP asks mainly about verification of their own proof.) $\endgroup$ – Martin Sleziak Dec 1 '17 at 20:28

One approach is to prove that if $X$ is a compact metric space then $X$ is totally bounded. This means that for every $\varepsilon > 0$ there is a finite number, say $n(\varepsilon)$, of points, call them $x_1,\dots,x_{n(\varepsilon)}$, such that the balls $B_\varepsilon(x_1),\dots,B_\varepsilon(x_{n(\varepsilon)})$ cover $X$. This is actually quite simple to prove, because if $X$ is a compact metric space, then given $\varepsilon > 0$, the cover $\{ B_{\varepsilon}(x) : x \in X \}$ has a finite subcover of the desired form.

From there, cover $X$ with finitely many balls of radius $1$; extract the center of each. Now every point is within $1$ of a point in your (finite) set. Cover $X$ with finitely many balls of radius $1/2$; extract the center from each. Now every point is within $1/2$ of a point in your (still finite) set. Repeat for each $1/m$ for $m \in \mathbb{N}$ and take the countable union. A countable union of finite sets is countable, so you have your countable dense subset.


Hint: For $\delta$, we see that $\{ B_{\delta}(x) : x \in X \}$ is an open cover of $X$. There is a finite subcover of this cover since $X$ is compact, so there is a finite set $E_\delta$ such that $\{B_\delta(x) : x \in E_\delta\}$ is an open cover of $X$. In particular, this means that for any $y \in X$, there exists a $x \in E_\delta$ such that $d(x,y) < \delta$.

Find a countable union of such $E_\delta$ to construct a dense subset of $X$.

  • $\begingroup$ Okay, I understand that. How do you prove something is dense? By showing that the closure of the countable collection of finite subcovers of $X$ equals $X?$ $\endgroup$ – user181728 Oct 15 '14 at 1:14
  • $\begingroup$ If $A\subset X$, then $A$ is said to be dense in $X$, if $\overline{A}=X$. $\endgroup$ – Sujaan Kunalan Oct 15 '14 at 1:16
  • $\begingroup$ The same definition as what Sujaan said but in different words. $A$ is dense in $X$ iff for every open set $O$ in $X$ we have that $A\cap O \not= \emptyset$. $\endgroup$ – user171177 Oct 15 '14 at 1:17
  • $\begingroup$ Wait, we're dealing with a countable collection of finite subcovers - these subcovers already equal $X$, correct? So the union of all these finite subcover would equal $X$ and so would the closure? $\endgroup$ – user181728 Oct 15 '14 at 1:18
  • $\begingroup$ No, $E_\delta$ is not the subcover. It is the centers of the balls which compose the subcover. The subcover itself covers $X$, so it's useless as it has uncountably many points. $\endgroup$ – Bruce Zheng Oct 15 '14 at 1:18

What I think is an easier approach is to prove the stronger property that $X$ has a countable basis (using this hint about the covering $X$ with all open balls with a radius $1/n$ and then the existence of a countably dense subset of $X$ is a consequence of $X$ being second-countable (having a countable basis).

  • $\begingroup$ I haven't learned about countable basis or second-countable, yet. $\endgroup$ – user181728 Oct 15 '14 at 1:24
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    $\begingroup$ Ok, then never mind. It is just a slightly stronger property that implies being separable (and in this case the proof is much more straightforward like you have already constructed a countable basis for your space so you could be done) $\endgroup$ – user171177 Oct 15 '14 at 1:27

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