# Does $\sigma$ -compact imply separable?

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 ?

• Why do you need to prove that statement? To show your set $\{ (x_{p}^{n} : p, \, n \in \Bbb N \}$ is dense, you just need to show its closure equals $D$. But this set is the union of dense sets. Is it true that the closure of an arbitrary union equals the union of the closures? If so, then the closure of that sequence equals the union of closures where you view that set as the union of dense subsets of $K_{n}$. Then the closure of each subset is actually $K_{n}$, so the union of the closures would be the union of $K_{n}$'s, which would equal $D$. – layman Apr 21 '15 at 13:39
• So in my opinion, you should try to prove (assuming it's true) that $\overline{\bigcup \limits_{\alpha} A_{\alpha}} = \bigcup \limits_{\alpha} \overline{A_{\alpha}}$. – layman Apr 21 '15 at 13:40
• Your statement that the closure of an arbitrary union equals the union of the closures, does not hold. Check: math.stackexchange.com/questions/195311/… – Koen Apr 21 '15 at 13:45
• Ok, my approach was wrong, but I don't understand where you are going with your approach. What's your reasoning for trying to find an open set $G$? – layman Apr 21 '15 at 13:49
• a set $\{ x_n | n \in \mathbb{N} \}$ is dense in $D$ if for every open set in $D$ there is a $x_n$ such that $x_n \in G$ . So I want to prove that for an arbitrary open set $G$, there is an $x_n$ such that $x_n \in G$ . – Koen Apr 21 '15 at 13:54

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.)
• So , if $G$ is open in $D$ then there exists an $n \in \mathbb{N}$ such that $G \cap K_n$ is not empty and so $G \cap K_n$ is open in $K_n$, so there exists a $x_p^n \in G \cap K_n \subset G$ ? – Koen Apr 21 '15 at 14:05
• Yes, that is the idea. Now show that there is such $n$. – Asaf Karagila Apr 21 '15 at 14:06
• a proof by contradiction: If there is no $n$ such that $K_n \cap G = \emptyset$ and $K_n \nearrow D$, then $\bigcup K_n \cap G = \emptyset$. This is a contradiction because $G \subset D$. ? – Koen Apr 21 '15 at 14:13
• Yes. And why does $G\cap K_n$ must be open in $K_n$? – Asaf Karagila Apr 21 '15 at 14:15
• $K_n$ is a subspace of $D$, so the topology on K_n is $T_{K_n} = \{ B \subset K_n : \exists A \mbox{ open }: B = A \cap K_n \}$ and this is metrizable too . – Koen Apr 21 '15 at 14:25