$C_c(X)$ is complete, then $X$ is compact

Solution 1:

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 $f\in C_c(X)$.

Solution 2:

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.