15
$\begingroup$

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

How does the following look?

Proof:

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 $\overline{E} = K$.

Q.E.D.

$\endgroup$
4
  • 1
    $\begingroup$ You don't need open covers. Sequential compactness plus a contradiction proof do the trick $\endgroup$ Feb 3, 2014 at 6:54
  • 1
    $\begingroup$ Looks fine to me. $\endgroup$
    – copper.hat
    Feb 3, 2014 at 6:57
  • $\begingroup$ In general if X is a metric space and Y is a dense subset of X then the set of open balls centered at members of Y with rational radii is a base for X. So if X is separable then X is second-countable. $\endgroup$ Feb 17, 2019 at 22:27
  • $\begingroup$ Nice proof. Thanks for sharing! $\endgroup$
    – jakab922
    Mar 26, 2023 at 14:52

1 Answer 1

0
$\begingroup$

I liked it a lot.

But to prove that $E$ is a dense subset of $K$, you can do it in the following way, for each open set $U$ of $K$, there is a $V_{n}\subseteq U$ so, you will have $x_{n}\in U\cap E$, that complete the proof.

I used an equivalence of density: $E$ is a dense subset of $K$ if and only if for each open subset $U$ of $K$ exists a $x\in U\cap E$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .