# If $X$ is locally compact and $\mathcal C_c (X)$ is separable, then $X$ is separable

Let

• $$X$$ be a metric space,
• $$\mathcal C_b(X)$$ the space of real-valued bounded continuous functions,
• $$\mathcal C_0(X)$$ the space of real-valued continuous functions that vanish at infinity, and
• $$\mathcal C_c(X)$$ the space of real-valued continuous functions with compact supports.

Then $$\mathcal C_b(X)$$ and $$\mathcal C_0(X)$$ are real Banach space with supremum norm $$\|\cdot\|_\infty$$. It's mentioned here that "if $$E$$ is locally compact and separable, then $$\mathcal C_0 (E)$$ is separable". Now I would like to prove a somehow reverse direction, i.e.,

Theorem: If $$X$$ is locally compact and $$\mathcal C_c (X)$$ is separable, then $$X$$ is separable.

Could you have a check on my below attempt?

Proof: Because $$\mathcal C_0(X)$$ is the closure of $$\mathcal C_c(X)$$ in $$\mathcal C_b(X)$$, it suffices to show that

If $$X$$ is locally compact and $$\mathcal C_0 (X)$$ is separable, then $$X$$ is separable.

Let $$\mathcal M(X)$$ the space of finite signed Radon measures on $$X$$. Let $$[\cdot]$$ be the total variation norm on $$\mathcal M(X)$$. We define a map $$f:X \to \mathcal M(X), x \mapsto \delta_x.$$

Notice that $$f(X)$$ is a subset of the closed unit ball of $$\mathcal M (X)$$. Let $$E := \mathcal C_0 (X)$$. By Riesz–Markov–Kakutani theorem, $$(\mathcal M(X), [\cdot])$$ is isometrically isomorphic to $$E^*$$ though a canonical map $$\Phi:\mathcal M(X) \to E^*$$. Let $$B$$ be the closed unit ball of $$E^*$$. Clearly, $$\Phi \circ f (X) \subset B$$.

By Banach-Alaoglu's theorem, $$B$$ is compact in the weak$$^*$$ topology $$\sigma(E^*, E)$$. Because $$E$$ is separable, the subspace topology $$\sigma_B(E^*, E)$$ that $$\sigma(E^*, E)$$ induces on $$B$$ is metrizable. It follows that $$\sigma_B(E^*, E)$$ is compact metrizable and thus separable.

If one proves that $$\Phi \circ f$$ is a homeomorphism from $$X$$ (together with metric topology) onto $$\Phi \circ f(X)$$ (together with its subspace topology induced by $$\sigma_B(E^*, E)$$), It would follow that $$X$$ is separable.

• Why is $\Phi\circ f$ a homeomorphism between $X$ and $\Phi(f(X))$? It is clear a continuous bijection. But I think a few words need to be said about $(\Phi\circ f)^{-1}$. Commented Nov 2, 2022 at 14:44
• @OliverDíaz You're right! I'm thinking of how $\delta_{x_n} \to \delta_x$ in weak$^*$ topology implies $x_n \to x$ in metric topology. Commented Nov 2, 2022 at 14:50
• That requires a proof. Commented Nov 2, 2022 at 15:11
• you may try to polish what I wrote in my posting regarding the continuity of $(\Phi\circ f)^{-1}$. Commented Nov 2, 2022 at 16:11
• (+1) The question is a good one. I am aware of the similar result for weak convergence in probability: namely that $\delta_{x_n}\Longrightarrow\delta_x$ iff $x_n\rightarrow x$. In the same setting, it can be shown that $\{\delta_x:x\in X\}$ is closed in $\mathcal{M}^+(X)$ with the topology $\sigma(\mathcal{M}(X),\mathcal{C}_b(X))$. Commented Nov 2, 2022 at 20:08

To show that $$\Phi\circ f: X\rightarrow\Phi(f(X))$$ is a homeomorphism it seems that is enough to show that for any sequence $$(x_m:m\in\mathbb{N})\subset X$$, $$\delta_{x_m}\stackrel{v}{\longrightarrow}\delta_x$$ iff $$x_m\xrightarrow{m\rightarrow\infty} x$$ in $$X$$.

Sufficiency is obvious. As for necessity, I suggest the OP to consider a sequence of open and relatively compact neighborhoods $$V_n$$ around $$x$$ such that $$V_{n+1}\subset\overline{V_{n+1}}\subset V_n$$, and $$\operatorname{diam}(V_n)\xrightarrow{n\rightarrow\infty}0$$.

Then define functions $$f_n\in\mathcal{C}_{00}(X)$$ with $$0\leq f_n\leq 1$$ such that $$f_{n+1}=1$$ on $$\overline{V_{n+1}}$$ and $$f_n=0$$ on $$X\setminus V_n$$. Then If $$\delta_{x_m}\stackrel{v}{\rightarrow}\delta_x$$, for any $$n$$, $$f_n(x_m)\xrightarrow{m\rightarrow\infty} f_n(x)=1$$. This means that for all $$m$$ large enough, the $$x_m$$ are close to $$x$$, i.e. $$x_m\xrightarrow{m\rightarrow\infty}x$$.

• Luckily, I have just posted a different approach as an answer. Could you please have a check on it? Commented Nov 2, 2022 at 15:54

As mentioned by @OliverDíaz, the fact that $$f^{-1}$$ is a homeomorphism (in weak$$^*$$ topology of $$\mathcal M(X)$$) from $$f(X)$$ onto $$X$$ is not clear. I have added a proof below.

WLOG, we assume $$d \le 1$$. For $$x \in X$$ and $$r>0$$, let

• $$B_r (x)$$ be the open ball centered at $$x$$ with radius $$r$$.
• $$\overline B_r (x)$$ the closed ball centered at $$x$$ with radius $$r$$.
• $$\overline{B_r (x)}$$ the closure of $$B_r (x)$$.

Notice that $$\overline{B_r (x)} \subset \overline B_r (x)$$ but not necessarily that $$\overline{B_r (x)} = \overline B_r (x)$$. Assume $$a, x_n \in X$$ such that $$\delta_{x_n} \to \delta_a$$ in weak$$^*$$ topology, i.e., $$\forall f \in \mathcal C_0(X) : \int_X f \mathrm d \delta_{x_n} \to \int_X f \mathrm d \delta_{a} \quad \text{as} \quad n \to \infty.$$

Because $$X$$ is locally compact, there is a sequence $$(r_m) \subset \mathbb R_{>0}$$ such that $$r_m \searrow 0$$ and $$\overline B_{r_m} (a)$$ is compact. Clearly, $$\overline{B_{r_m} (a)}$$ is compact. Let $$C_m := X \setminus B_{r_m} (a)$$, and $$f_m (x) := d(x, C_m) \quad \forall x \in X.$$

Then $$f_m \in \mathcal C_b(X)$$. Because $$C_m$$ is closed, $$f_m (x) \neq 0 \iff d(x, C_m)>0 \iff x \notin C_m \iff x \in B_{r_m} (a).$$

So $$\operatorname{supp} (f_m) = \overline{B_{r_m} (a)}.$$

Hence $$f_m \in \mathcal C_c (X)$$ and thus $$f_m(x_n) \xrightarrow{n \to \infty} f_m(a) \quad \forall m \in \mathbb N.$$

This implies $$\lim_{n \to \infty} d(x_n, C_m) = d(a, C_m) \quad \forall m \in \mathbb N.$$

As such, $$\lim_{n \to \infty} d(x_n, a) \le \lim_{n \to \infty} d(x_n, C_m) + d(a, C_m) = 2 d(a, C_m) \le 2r_m \quad \forall m \in \mathbb N.$$

The proof is completed by taking the limit $$m \to \infty$$.

• Using $g(y)=\min(1,d(x,y))$ works if you were considering $C_b(X)$ instead of $C_0(X)$ as your space of test functions. $g$ is not in $C_0(X)$. Commented Nov 2, 2022 at 15:56
• @OliverDíaz I may be wrong, but I proved that $A_\varepsilon := \{x \in X : |g(x)| \ge \varepsilon \}$ is compact for all $\varepsilon >0$. This is the definition of $\mathcal C_0(X)$. Commented Nov 2, 2022 at 15:58
• Consider $X=\mathbb{R}$. $g(x)=\min(1,|x|)$ does not vanish at infinity. Commented Nov 2, 2022 at 16:01
• @OliverDíaz You are right! I made a stupid mistake. Commented Nov 2, 2022 at 16:07
• The use of the distance function in my argument is in the fact that $\operatorname{diam}(V_n)\xrightarrow{n\rightarrow\infty}0$ by choice. Using relatively compact balls around $x$ was a good way to fix your previous argument. Commented Nov 2, 2022 at 20:50