# Does $C_0(X)$ determine the topology for a locally compact space $X$?

Given a locally compact Hausdorff space, does $$C_0(X)$$, the continuous functions vanishing at infinity, determine the topology of $$X$$?

For example, for a net $$\{x_{\alpha}\}\subset X$$ if I have $$f(x_\alpha)\to f(x)$$ for all $$f\in C_0(X)$$, does it follow that $$x_{\alpha} \to x$$ ?

I cannot find any reference to the Banach-Stone theorem that proves this. I would really appreciate some feedback.

Thanks!

Yes. If $$X$$ is a locally compact Hausdorff space, then $$C_0(X)$$ is an Abelian C*-algebra. Moreover, the Gelfand spectrum of $$C_0(X)$$ is homeomorphic to $$X$$.
In another word, if $$C_0(X)$$ and $$C_0(Y)$$ are C*-isomorphic, then $$X$$ and $$Y$$ are homeomorphic.
(In fact, every Abelian C*-algebra is of the form $$C_0(X)$$. This is known as the Gelfand Theorem.)
• Every abelian $C^*$ algebra is actually of the form $C(K)$, for some compact space $K$, which is stronger than $C_0(X)$ with a locally compact $X$. This theorem is known as Gelfand-Naimark theorem. – uniquesolution Jun 6 '19 at 16:41
• @uniquesolution For abelian C*-algebra $A$, if $A$ is unital, then $A=C(K)$ for some compact Hausdorff space $K$. If $A$ is non-unital, we only have $A=C_0(X)$ for some locally compact Hausdorff space. – Danny Pak-Keung Chan Jun 6 '19 at 23:57
Let $$X^\ast$$ be the one-point compactification of $$X$$, with compactifying point $$\infty$$, then if $$C$$ is closed in $$X$$ and $$x \in X\setminus C$$, note that $$C^\ast:=C \cup \{\infty\}$$ is compact in $$X^\ast$$ and as $$X^\ast$$ is normal we can find a continuous $$f: X^\ast \to \mathbb{R}$$ such that $$f(x)=1$$ and $$f[C^\ast]=\{0\}$$. And then $$g=f\restriction X \in C_0(X)$$ and $$g(x) \notin \overline{g[C]}$$ and so the set $$C_0(X)$$ separates points and closed sets. By well-known general topology facts means that it determines the topology on $$X$$ (the topology on $$X$$ is the unique smallest topology that makes all functions in $$C_0(X)$$ continuous) and what you state about nets is a consequence of that fact.