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

functional-analysismeasure-theorymetric-spacesweak-topology

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.

Best Answer

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. $x_m\xrightarrow{m\rightarrow\infty}x$.

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