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

qju
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1 Answers1

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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.

Luiz Cordeiro
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  • 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?) – Ramen Dec 07 '17 at 16:05
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    @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. – Luiz Cordeiro Dec 08 '17 at 17:24
  • 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? – Ramen Dec 09 '17 at 21:07
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    @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 $K_n$. – Luiz Cordeiro Dec 10 '17 at 05:25
  • It is not clear at all why the fact that $R_{K_{n}}(C_{n})$ is separable implies that $C_{n}$ should be separable. – Sqrt Oct 12 '21 at 12:03
  • Here are two ways to see this. 1. Any isometry induces an isometric inverse defined on the range of the original isometry. In particular, it is continuous so that it maps separable sets to separable sets. Hence $C_{n}$ is the continuous image of the separable space $R_{K_{n}}(C_{n})$. 2. If $C_{n}$ was not separable then it would have an uncountable discrete subset. The isometry would preserve such a subset so that $R_{K_{n}}(C_{n})$ would not be separable. As this leads to a contradiction, it follows that $C_{n}$ must be separable. – Dean Miller Dec 01 '24 at 13:13
  • @DeanMiller As a sidenote, your second approach seems a little more advanced than I would expect for a solution to this question, as I can only see a proof for the implication "nonseparability$\Rightarrow$uncountable discrete subset" with transfinite induction on $\omega_1$. Is there any simpler way to see it without ordinals? – Luiz Cordeiro Dec 02 '24 at 04:07
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    @LuizCordeiro I was, at first, thinking this was a (reasonably) trivial well-known result and hadn't thought about proving it for a long time. The only approaches I can come up with for that implication involve either Zorn's lemma or transfinite induction. So perhaps the first approach should be preferred since it makes use of more elementary methods. – Dean Miller Dec 04 '24 at 05:28
  • On another note, I really like what you did here by defining those $C_{n}$ sets and using isometries combined with each $C(K_{n})$ being separable to obtain the countable dense subset. I modified the approach you presented to answer this related question. – Dean Miller Dec 06 '24 at 08:09