Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that every compact metric space $K$ has a countable base, and that $K$ is therefore separable.

How does the following look?


For each $n \in \mathbb{N}$, make an open cover of $K$ by neighborhoods of radius $\frac{1}{n}$, and we have a finite subcover by compactness, i.e.

$$K \subset \bigcup_{x \in K} N_{\frac{1}{n}}(x) \ \Rightarrow \ \exists x_1, ..., x_N \in K \text{ such that } K \subset \bigcup_{i=1}^{N} N_{\frac{1}{n}} (x_i)$$

Doing this for every $n \in \mathbb{N}$, we get a countable union of finite collections of sets, so the collection of these sets, call it $S$, is countable.

We claim that $S$ is a countable base for $K$, which is defined as a countable collection of open sets such that for any $x \in K$ and any open set $G$ with $x \in G$, there is some $V \in S$ such that $x \in V \subset G$.

Let $x \in K$ and let $G$ be any open set with $x \in G$. Then since $G$ is open, there is some $r > 0$ such that $N_r(x) \subset G$. Choose $n \in \mathbb{N}$ such that $\frac{1}{n} < \frac{r}{2}$, so that the maximal distance between points in a neighborhood of radius $\frac{1}{n}$ is $r$. Then there must be some $i$ such that $x \in N_{\frac{1}{n}}(x_i) \subset N_r(x)$ because any neighborhood of radius $\frac{1}{n}$ containing $x$ cannot contain points a distance more than $r$ away. This shows that $S$ is a countable base.

The second part of the question asks us to show that $K$ is separable. Let $\{V_n\}$ be our countable base for $K$. For each $n \in \mathbb{N}$, choose $x_n \in V_n$, and let $E = \{ x_n \ | \ n \in \mathbb{n} \}$. We claim that $E$ is a countable dense set, which would show that $K$ is separable.

First, note that $E$ is clearly countable. To show that it's dense, we need to show that $\overline{E} = K$. This is equivalent to showing that $(\overline{E})^c = \emptyset$. Now $(\overline{E})^c$ is an open set because it's the complement of a closed set, $\overline{E}$. If $(\overline{E})^c$ is nonempty, then there is some $x \in (\overline{E})^c$, which is open, so since $\{V_n\}$ is a base, there is some $n$ such that $x \in V_n \subset (\overline{E})^c$, which implies that $x_n \in (\overline{E})^c$, a contradiction, because

$x_n \in E \implies x_n \in \overline{E} \implies x_n \notin (\overline{E})^c$.

Therefore, $(\overline{E})^c = \emptyset$, so that $\overline{E} = K$.


share|cite|improve this question
You don't need open covers. Sequential compactness plus a contradiction proof do the trick – LeGrandDODOM Feb 3 '14 at 6:54
Looks fine to me. – copper.hat Feb 3 '14 at 6:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.