Let $X$ be a locally compact metric space which is also $\sigma$-compact. Let $C_{c}(X)$ be the continuous functions on $f$ from $X$ to $\mathbb{R}$ with compact support. Is $C_{c}(X)$ separable?
My work so far:
If $X$ is a compact metric space, then by Urysohn's Lemma and Stone-Weierstrass, the continuous functions $C(X)$ on $X$ are separable and hence the result follows as $C_{c}(X) = C(X)$.
Suppose $X = \mathbb{R}$. Write $\mathbb{R} = \bigcup_{N = 1}^{\infty}[-N, N]$. Let $f \in C_{c}(\mathbb{R})$. Then $f$ is supported on a compact set $K \subset [-N, N]$ for some $N$. Thus $C_{c}(\mathbb{R}) =\bigcup_{N = 1}^{\infty}C([-N, N])$. Each $C([-N, N])$ has a countable dense subset $\{\psi_{N, n}\}_{n = 1}^{\infty}$ and so $\bigcup_{N, n = 1}^{\infty}\{\psi_{N, n}\}$ is a countable dense subset of $C_{c}(\mathbb{R})$.
In the general case, $X = \bigcup_{i = 1}^{\infty}X_{i}$ where each $X_{i}$ is compact and $X_{1} \subset X_{2} \subset \cdots$. Let $f \in C_{c}(X)$. Then $f$ is supported on a compact set $K = \bigcup_{i = 1}^{\infty}K \cap X_{i}$. Is it still true that $C_{c}(X) = \bigcup_{i = 1}^{\infty}C(X_{i})$?
Best Answer
Your idea is good, and you must only take a little more care for it to work. As you said, if $K$ is a compact metric space, then $C(K)$ is separable.
Now, suppose that $X$ is is a locally compact, $\sigma$-compact metric space. You can find a sequence $\left\{K_n\right\}_{n\in\mathbb{N}}$ of compact subsets of $X$ satisfying:
$K_n\subseteq\text{int}K_{n+1}$ for every $n$;
$X=\bigcup_{n\in\mathbb{N}}K_n$.
For every $n$, let $C_n=\left\{f\in C_c(X):\text{supp}f\subseteq K_n\right\}$. Notice that $C_c(X)=\bigcup_{n\in\mathbb{N}}C_n$, so it is sufficient to show that each $C_n$ is separable.
Fixed $n$, consider the function $R_{K_n}:C_n\rightarrow C(K_n)$, $f\mapsto f|_{K_n}$. This is a linear isometry (not necessarily surjective), so $R_{K_n}(C_n)$ is a subspace of the separable space $C(K_n)$, so it is also separable, hence $C_n$ is also separable.