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.
Best Answer
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 $x\in U\cap E$.