$C_{c}(X)$ is complete. then implies that $X$ is compact.

Question

Let $X$ is locally compact Hausdorff space .If $C_{c}(X)$ is complet,then $X$ is compact (this is to be proved). I know that $C_{c}(X)$ is dense in $C_{0}(X)$. As $C_{c}(X)$ is complete implies that $C_{c}(X)=C_{0}(X)$. and I know only urysohn's lemma and Tiez's extension theorem on locally compact Hausdorff spaces and other basic things.

Answer 1

This isn't true. For instance, if $X=\omega_1$ is the first uncountable ordinal, then $C_c(X)=C_0(X)$ is complete (since every continuous map $X\to \mathbb{R}$ is eventually constant), but $X$ is not compact.