Consequence of separability in a locally compact metric space

Question

The reason why I didn't find an answer to my problem in the first place, was that I wasn't familiar with the concept "second countable" and therefore didn't realize that the condition of my problem implied second countability. No problem if this reason is insufficient for reopening. Here is my original question :

I am trying to prove that in a locally compact metric space $E$, separability implies that $E$ is a countable union of compact sets.

I know that (in a locally compact space) every compact set (and not just every point) has a compact neighborhood. Let $\{x_1, x_2, \dots \}$ be an enumeration of a countable and dense subset of $E$. I know that the set of open balls around $x_1, x_2, \dots$ with positive rational radii constitutes a basis of the topology of $E$, and I wonder how using this fact might complete my proof.

My idea is the following : let $C_1$ be a compact neighborhood of $x_1$ and $O_1=\dot{C}_1$ (interior of $C_1$). We have $C_1=\bar{O}_1$. Then, let $C_2$ be a compact neighborhood of the compact set containing the union of $C_1$ with a compact neighborhood of $x_2$ and let $O_2=\dot{C}_2$. We have $C_2=\bar{O}_2$. By continuing in this way, I'm afraid that $\bigcup_{n \in \mathbb{Z}_{+}} C_n$ might not be equal to $E$. The reason why I introduce $O_1, O_2, \dots$ is that I might need to replace them somehow by the open balls that form a basis of the topology and use the fact that their union is $E$.

If I use the open balls which constitute a basis of the topology or the corresponding closed balls, to have the union "grow more rapidly", I'm afraid that these closed balls might not be compact and I don't know how to argue that they have a compact neighborhood.

Answer 1

If $E$ is a metric space then separabaility$\iff$second countable space. Let $\mathcal{B}=\{B_{n}: n \in \mathbb{N}\}$ a countable base. Since for every $x\in E$ by hypothesis $\exists U$ such that U is open and $x\in U_x\subset K_x$ where $K_x$ is compact. Since every metric space is T2, $K_x$ is even closed, and that Is what you need in this situation. Since $\mathcal B$ is a base for every $x \in E$ $\exists B_n \in \mathcal B$depending on x.Let $B_{n_x}=B_n$ then $B_{n_x}\subset U_x\subset K_x$. Now because $K_x$ is closed you have that $\overline B_{n_x}\subset \overline K_x\subset K_x$ so $\overline B_{n_x}$ is compact since is closed in a compact. Consider now $\Phi = \bigcup_{x \in E}\overline B_{n_x}$ it is indexed on E but $\Phi\subset \overline {B}=\{\overline B_{n}: n \in \mathbb{N}\}$ which is countable. $\square$