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.



2 Answers 2


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.)

  • 1
    $\begingroup$ You can search textbooks about "operator algebra", or keywords like "C*-algebra, von Neumann algebra" $\endgroup$ Jun 6, 2019 at 1:14
  • $\begingroup$ 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. $\endgroup$ Jun 6, 2019 at 16:41
  • 2
    $\begingroup$ @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. $\endgroup$ Jun 6, 2019 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.