$\mathcal{M}$ metric and locally compact. It is $\sigma-$compact if and only if it is separable.

Question

I have been able to prove one direction.

$(\Rightarrow)$ If $\mathcal{M}$ is separable there is $ D=\{x_i\}_{i\in \mathbb{N}} $ a countable dense set. It is a straightforward computation to show that $B=\{B_{x_i}^{q}| x_i \in D \: q\in \mathbb{Q}\}$ is a countable basis.

If we have a countable basis the space must be Lindelof (every cover admits a countable subcover). For every $m\in \mathcal{M}$ we create $B_m^r$ an open ball with a compact closure $\overline{B_m^r}$ (this is possible because of local compacity). Clearly (because of the Lindelof property) $\mathcal{M}=\cup_{i=1}^\infty B_{m_i}^{r_i}$.

Let $C_1=\overline{B_{m_1}^{r_1}}$, this means $C_1\subset\cup_{i=1}^\infty B_{m_i}^{r_i}$, but because $C_1$ is compact, we actually have $C_1=\cup_{i=1}^nB_{m_i}^{r_i}$ where $n\geq 2$.

$$C_1\subset\cup_{i=1}^nB_{m_i}^{r_i}\subset \overline{\cup_{i=1}^nB_{m_i}^{r_i}}\subset \cup_{i=1}^n \overline{B_{m_i}^{r_i}}$$

Because $\overline{\cup_{i=1}^nB_{m_i}^{r_i}}$ is a closed set inside a compact one, it must also be compact and we define $C_2=\overline{\cup_{i=1}^nB_{m_i}^{r_i}}$. We proceed inductively to create compact sets such that $\mathcal{M}=\cup_{n\in \mathbb{N}}C_n$ and $C_n\subset \text{int}C_{n+1}$.

$(\Leftarrow)$ I know it suffices to prove that there is a countable basis (take an element inside each element of the basis and voila, we have a dense countable set) but I have no idea how to construct such a basis when I only have this $C_i$ sequence and a metrizable locally compact space.

Answer 1

Let $M=\bigcup_{i=1}^\infty C_i$ where each $C_i$ is a compact subspace of M. Consider the cover $U_\delta(C_i)=\{B(m,\delta)\}_{m\in C_i}$ of each $C_i$. Since $C_i$ is compact there is a finite subcover $U_\delta(C_i)$, consider $P_\delta(C_i)=\{m_1,...,m_n\}$ the centers of the balls of the finite subcover (It depends on i and $\delta$ but i dont write it in order to keep the notation simple). When we consider \begin{equation} D=\bigcup_{i=1}^\infty\bigcup_{n=1}^\infty P_{1/n}(C_i) \end{equation} It will be a countable dense subset of $M$. I leave this so you can prove it.

Answer 2

$M=\bigcup_{n\in\Bbb N}C_n$ where each $C_n$ is a compact subspace of $M$ hence separable.

Let $D_n$ be a dense countable subset of $C_n$, then $\bigcup_{n\in\Bbb N}D_n$ is a dense countable subset of $M.$