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I'm struggling to prove the following fact:

Suppose that $X$ is locally compact metric space. Let us denote with $C_0(X)$ the space of functions vanishing at infinity (i.e., $\forall f \in C_0(X)$ $\forall \varepsilon > 0$ $\exists \, E\subset X$ s.t. $E$ is compact and $|f(x)|<\varepsilon$ for $x \in X\setminus E$). Then $C_0(X)$ is separable.

I've proven that $C_0(X)$ equipped with a supremum norm is a Banach space, and that $C_c(X)$ (functions with compact support) are dense in $C_0(X)$, so my guess would be to somehow use those facts to prove that $C_0(X)$ is separable. However, I can't exactly see how. I've seen the cases for compact spaces or using the assumption of $\sigma$-compactness. Any help is highly appreciated.

Wham Bang Shang-a-Lang
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    Can you use the Stone-Weierstrass theorem and construct a countable subalgebra that separates points? – Matthew Leingang Sep 06 '19 at 14:12
  • @MatthewLeingang wouldn't you need it to contain some non-zero constant function? But it wouldn't vanish at infinity – GSofer Sep 06 '19 at 14:18
  • @GSofer My question wasn't a hint; it was a guess. ;^) – Matthew Leingang Sep 06 '19 at 14:52
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    Are you missing a separability assumption? As written I think this is false. For example, let $X$ be an uncountable set with the discrete metric. Then for each $x \in X$, define $f_x(x) = 1$ and $f_x(y) = 0$ for $y \neq x$. Then ${f_x: x \in X} \subseteq C_0(X)$ and $|f_x - f_y|_\infty = 1$ for $y \neq x$ which means that $C_0(X)$ has an uncountable discrete subset and so isn't separable. With the added assumption of separability of $X$ this is true since you can e.g. embed $C_0(X)$ into $C(\tilde{X})$ where $\tilde{X}$ is the one-point compactification of $X$. – Rhys Steele Sep 06 '19 at 15:26
  • https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Locally_compact_version does not require any constant functions. So you can try to modify the C(X) proof, but that requires that $X$ is separable. Let $D\subset X$ be dense and countable. Set $d_x(y):=1/(1+d(x,y))\ (x\in D)$. Let $A$ be the subalgebra of $C_0(X)$ generated by the functions $d_x\in C_0$,* i.e., the set of linear combinations of their products. Let $A'$ be the subset with rational coefficients. Then $A'$ is countable and dense in $A$. But $A$ separates points and vanishes nowhere. So we are done if * is true. Is it? – user3810316 Jun 10 '20 at 20:34
  • So my above comment answers @MatthewLeingang that the non-zero constant function is not needed. Additionally, it provides a possible sketch of a proof to an answer, where the part "*" was left open. Indeed, not every $d$ yields that "$d_x\in C_0$", which requires that all (D-centric) closed balls are compact. It seems not true for all locally compact spaces, just for "proper spaces": https://en.wikipedia.org/wiki/Metric_space#Locally_compact_and_proper_spaces So the sketched proof only seems to work for them. – user3810316 Jun 17 '20 at 12:10

2 Answers2

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As Rhys Steele mentions, this is not true unless you assume $X$ to be second countable (or, equivalently for metric spaces, separable). Rhys gives a counterexample showing the theorem can fail without this assumption, but more is true: it always fails without this assumption.

Proposition. Let $X$ be a locally compact Hausdorff space. If $C_0(X)$ is separable then $X$ is second countable.

Proof. Let $\{f_n\}$ be a countable dense subset of $C_0(X)$, and for each $n$ let $U_n = \{x \in X: f_n(x) > 1/2\}$, which is an open subset of $X$. I claim that $\{U_n\}$ is a countable base for the topology of $X$. For let $x \in X$ and let $V$ be an open neighborhood of $x$. Then by Urysohn's lemma for locally compact Hausdorff spaces, there exists a function $f$ compactly supported inside $V$ with $f(x) = 1$. In particular $f \in C_c(X) \subset C_0(X)$, so by density, we can find some $f_n$ with $\|f-f_n\|_\infty < 1/2$. Then we have $f_n(x) > 1/2$ so $x \in U_n$. Moreover, if $y \in U_n$ then $f_n(y) > 1/2$ and so $f(y) > 0$, which implies $y \in V$. Therefore $U_n \subset V$. This proves that $\{U_n\}$ is a base.

Now, supposing that $X$ is second countable, you can proceed in a similar way to the compact case, applying the locally compact version of Stone-Weierstrass. Using the second countability and local compactness, you should be able to construct a countable family $f_n$ of compactly supported functions which separates points and vanishes nowhere. Then consider the algebra $\mathcal{A}_0$ generated over $\mathbb{Q}$ by the $f_n$; i.e. all functions consisting of finite rational linear combinations of finite products of the $f_n$. Show that $\mathcal{A}_0$ is countable, and that the closure of $\mathcal{A}_0$ is a closed algebra over $\mathbb{R}$. Stone-Weierstrass then implies that the closure of $\mathcal{A}_0$ equals $C_0(X)$, so $C_0(X)$ is separable.

Nate Eldredge
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In the case of $X = \mathbb{R}$. Consider the (countable) set of function $$G = \{(P I_n) * \eta_m: P \in \mathbb{Q}(x), m, n \in \mathbb{N}\} \subset C_0(X),$$ where $I_n (x) = \mathbf{1}_{[-n, n]}(x)$ and $\eta_m = \frac{1}{m} \eta(\frac{x}{m})$, $\eta$ is a mollifier and $*$ means convolution. For arbitray $f \in C_0(X)$, assuming it is supported on $[-N, N]$, by Weiterstrass theorem we can find a sequence of $P_n \in \mathbb{Q}(x)$ such that $P_n(x) I_N (x)$ approximates $f(x)$ uniformly. Using the property of mollifier, we conlcude $G$ is dense in $C_0(X)$. Therefore, $C_0(X)$ is separable.

lye012
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