Dtyjlui

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.

I liked it a lot.

But to proof 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 $x_{n}in Ucap E$, 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 $xin Ucap E$.