Let $D$ be a metric space. If $D$ is $\sigma$-compact, does this imply that $D$ is separable? I thought I had a proof, but I think it is wrong. my proof:
Let $K_n$ the compact sets such that $K_n \nearrow D$. Then $K_n$ are sub metric spaces of $D$ so all the $K_n$ are separable. Let $\{ (x_p^n): p \in \mathbb{N} \}$ the countable dense set of $K_n$. I think now $\{ (x_p^n): p,n \in \mathbb{N} \}$ is a countable dense set of $D$. The fac, it is countable is clear. If $G$ is an open set, then there exists a $K_n$ such that $G \subset K_n$ and because every $K_n$ separable, there exist a $p \in \mathbb{N}$ such that $x_n^p \in D$. So $D$ is seperable.
The problem is that i can't prove the statement: $G$ open then there exists a $K_n$ such that $G \subset K_n$. Can someone prove this , or is this statement false ?
The statement is of course not true. Take $G$ to be the entire space, for example.
But what you can prove is that there is some $n$ such that $G\cap K_n$ is not empty, and therefore relatively open there. And then it meets the relevant dense set. (Of course, assuming $G\neq\varnothing$, which is of course the initial assumption.)