# Continuous complex functions which vanish at infinity have compact support

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?

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 $$X\setminus f^{-1}(\{0\})$$.
• 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. Commented Sep 18, 2019 at 3:51
• 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. Commented Sep 18, 2019 at 3:55