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 $C_{c}(X) = \bigcup_{i = 1}^{\infty}C(X_{i})$?


1 Answer 1


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:

  1. $K_n\subseteq\text{int}K_{n+1}$ for every $n$;

  2. $X=\bigcup_{n\in\mathbb{N}}K_n$.

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.

  • $\begingroup$ How do we need property 2. of the sequence? Do the compact sets need to be nested in each other? I know that in locally compact metric spaces the existence of a sequence satisfying 1. & 2. is equivalent to the $\sigma$-compactness of the space so why just not use any sequence of compact sets whose union gives the whole space? (Is it ok to dig questions out like that?) $\endgroup$
    – Ramen
    Commented Dec 7, 2017 at 16:05
  • 1
    $\begingroup$ @Ramen I don't see a problem "digging out" questions. Property 1. Is necessary because we want $C_c(X)=\bigcup C_n$ (in fact this condition is equivalent to the set of compacts satisfying the condition that for all $n,m$, there is some $N$ with $K_n\cup K_m\subseteq int K_N$). If we drop condition 1. then we can use Stone-Weierstrass on $\bigcup C_n$ and obtain the same result. $\endgroup$ Commented Dec 8, 2017 at 17:24
  • $\begingroup$ Of course I meant property 1. (instead of 2.) in the first phrase of my comment. I think you still have answered my question as if that was the case, no? So the point is the compact sets have to be nested in each other because we need to place the support of the $C_c$-function in a single compact set? $\endgroup$
    – Ramen
    Commented Dec 9, 2017 at 21:07
  • 1
    $\begingroup$ @Ramen Yes, that's funny, I answered your question as if it was for property 1 and didn't even notice. And your last comment is correct, we want to place the support of a $C_c$-function in a single one of the $K_n$. $\endgroup$ Commented Dec 10, 2017 at 5:25
  • $\begingroup$ It is not clear at all why the fact that $R_{K_{n}}(C_{n})$ is separable implies that $C_{n}$ should be separable. $\endgroup$
    – Sqrt
    Commented Oct 12, 2021 at 12:03

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