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Let $X$ be a topological space. Then we can define $$C_0(X):=\{f \in C(X)\mid \forall \epsilon > 0: \exists K \subseteq X \mathrm{\ compact}: \forall x \notin K: |f(x)| < \epsilon\}$$

If $X$ is locally compact, I have also seen the following definition, if $X$ is locally compact: $$C_0'(X) = \{f \in C(X)\mid \forall \epsilon > 0: \{x \in X: |f(x) | \geq \epsilon\} \mathrm{\ compact}\}$$

What is the relation between these two definitions of $C_0?$ Clearly $C_0'(X) \subseteq C_0(X)$. Do we have equality? Why do we require that $X$ is locally compact in the second definition?

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  • $\begingroup$ They are equivalent no matter how twisted the space $X$ is. The locally compactness is added probably to show that after a (one point) compactoification, the space remains locally compact. but that is irrelevant your problem. $\endgroup$ – Oliver Diaz Jul 6 at 17:23
  • $\begingroup$ If $X$ is locally compact, then $C_0(X)$ is the set of functions in $C(X_a)$ that vanish at the point at infinity in the Aleksandrov compactification $X_a$ of $X$. $\endgroup$ – egreg Jul 6 at 17:31
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    $\begingroup$ The definitions are fine (and equivalent) even if $X$ is not locally compact, but the space $C_0(X)$ may not be very interesting in this case. For example when $X$ is an infinite-dimensional Banach space, $C_0(X)$ contains nothing but the zero function. The assumption of local compactness is probably added because that is the only case the author is interested in studying, and that assumption may be used for later results. $\endgroup$ – Nate Eldredge Jul 6 at 17:41
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Yes, the two definitions are equivalent, even without assuming local compactness. (A usual terminology to describe this behavior is to say that such a function vanishes at infinity.)

To see that $C_0'(X)\subseteq C_0(X)$, take $f\in C_0'(X)$. Then, $K_{\varepsilon}\equiv\{x\in X\,|\,|f(x)|\geq\varepsilon\}$ is compact for every $\varepsilon>0$. Use this $K_{\varepsilon}$ to verify that $f\in C_0(X)$.

Conversely, suppose that $f\in C_0(X)$ and fix an arbitrary $\varepsilon>0$. There must exist some compact $K\subseteq X$ such that $x\in X\setminus K$ implies $|f(x)|<\varepsilon$. Therefore, the set $E\equiv\{x\in X\,|\,|f(x)|\geq\varepsilon\}$ is a subset of $K$. But $E$ is closed due to the continuity of $f$ and a closed subset of a compact set is compact in any topological space. The conclusion is that $E$ is compact, which implies that $f\in C_0'(X)$.

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  • $\begingroup$ Obviously... Thanks! Any idea why my book defines $C_0(X)$ for a locally compact space if everything works without that assumption? $\endgroup$ – user745578 Jul 6 at 17:24
  • $\begingroup$ @user745578 It might have been an incorrect hunch on the author’s part that one of the definitions is more restrictive than the other and additional topological assumptions would be needed to establish equivalence. Which textbook is it? $\endgroup$ – triple_sec Jul 6 at 17:30
  • $\begingroup$ C*-algebras and operator theory, but maybe the author assumes $X$ is locally compact because later he shows that every abelian C*-algebra is of the form $C_0(X)$ with $X$ locally compact $\endgroup$ – user745578 Jul 6 at 17:32

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