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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!

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2 Answers 2

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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.)

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    $\begingroup$ 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
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    $\begingroup$ @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
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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.

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