Let $X$ be a locally compact Hausdorff space such that $C_c(X),$ the space of all continuous functions with compact support is complete. Show that $X$ is compact.

My attempt:

I have shown that $C_c(X)$ is dense in $C_0(X),$ the space of all continuous functions vanishing at infinity. Since $C_c(X)$ is complete, therefore $$C_c(X)=C_0(X).$$

Now to conclude that $X$ is compact, it suffices to find a function in $C_0(X)$ which vanishes nowhere on $X.$ Is this always possible?

  • $\begingroup$ If I'm not mistaken the set of countable ordinals is a locally compact Hausdorff space such that every element of $C_0$ vanishes somewhere. $\endgroup$ Dec 6, 2017 at 0:30
  • $\begingroup$ In fact come to think of it it's seeming to me like $\omega_1$ is actually a counterexample to the original question. Are you certain it's true that if $C_c(X)$ is complete then $X$ is compact? $\endgroup$ Dec 6, 2017 at 0:33
  • $\begingroup$ In the book "A Course in Commutative Banach Algebras-Kaniuth", it is stated in example $1.1.2$ that $C_c(X)$ is complete only when $X$ is compact. $\endgroup$ Dec 6, 2017 at 0:36
  • $\begingroup$ Possible duplicate of $C_{c}(X)$ is complete. then implies that $X$ is compact. $\endgroup$ Dec 6, 2017 at 1:04

2 Answers 2


This is false. Maybe true under extra assumptions: Metrizable, $\sigma$-compact, whatever? Also I bet it's not hard to show that $C_c(X)$ complete does imply that $X$ is sequentially compact (and/or various other "countable compactness" conditions.)

Let $X=\omega_1$ (the set of countable ordinals), with the order topology. Then $X$ is locally compact but not compact. A standard argument shows that $C_c(X)=C_0(X)$, hence $C_c(X)$ is complete.

Standard argument: Say $f\in C_0(X)$. Then $\{|f|\ge1/n\}$ is compact, hence bounded: There exists $\alpha_n\in\omega_1$ such that $$|f(x)|<1/n\quad(\alpha_n<x\in\omega_1).$$There exists $\alpha\in\omega_1$ with $\alpha>\alpha_n$ for every $n$. Hence $f$ is supported on $[0,\alpha]$, so $f\in C_c(X)$.

  • $\begingroup$ What is $\omega_1$? $\endgroup$
    – user99914
    Dec 6, 2017 at 1:03
  • $\begingroup$ @JohnMa $\omega_1$ is the set of countable ordinals. Which is to say it's an uncountable well-ordered set where every element has only countably many predecessors. $\endgroup$ Dec 6, 2017 at 1:07

If $\Omega$ is $\sigma$-compact, then the space is always complete when endowed with the strict inductive limit topology of Fréchet spaces. You have to specify which topology you are considering on that space.


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