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I think the this is not that difficult question, but I'm in a trouble. I was just reading Rudin's "Real and Complex Analysis", page 70 and was trying to understand $$C_c(X)=C_0(X)\quad(X:\text{compact})$$ where $$C_c(X)=\text{the collection of continuous functions which have compact support}$$ $$C_0(X)=\text{the collection of continuous functions which vanish at infinity}$$ and where we are dealing with complex functions.

The inclusion $\subset$ seems obvious even though $X$ is not compact. The reverse inclusion is exactly the title of this question ; Continuous complex functions which vanish at infinity have compact support.

But, my exact question is that how $f$ has a compact support, if $X$ is compact? The set $\{0\}$ is a closed subset of complex plane, and it's inverse image $f^{-1}(\{0\})$ is also closed since $f$ is continuous. Then supp($f$), which is the complement of $f^{-1}(\{0\})$, is open. If supp($f$) were closed, it should be compact, being a closed subset of compact set $X$. But supp($f$) is open. What was wrong in my argument?

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When $X$ is a topological space, the support of $f$ most commonly means the closure of $X\setminus f^{-1}(\{0\})$, not $X\setminus f^{-1}(\{0\})$ itself. Sometimes, but rarely, this notion is referred to as closed support.

You may be confusing the terminology with the notion of support for when $X$ is a set with no topological structure; then the support of $f$ is indeed defined to be $X\setminus f^{-1}(\{0\})$.

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  • $\begingroup$ Sorry, I missed the term "closure". Rudin defined the support of a complex function on a topological space $X$ is the closure of the set $\{x:f(x)\neq0\}$ like you said. $\endgroup$
    – shyzealot
    Commented Sep 18, 2019 at 3:51
  • $\begingroup$ But then, support of $f$ is always closed, and it is a closed subset of a compact set X. So supp($f$) is compact and we are done. Is it right? I think I didn't use the fact that $f$ vanishes at infinity. $\endgroup$
    – shyzealot
    Commented Sep 18, 2019 at 3:55
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    $\begingroup$ @shyzealot Yes, that's correct. In a compact space the notion of compact support is a bit silly - every function has compact support. $\endgroup$ Commented Sep 18, 2019 at 4:01
  • $\begingroup$ I see, thanks a lot $\endgroup$
    – shyzealot
    Commented Sep 18, 2019 at 5:36

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