5

Theorem 4.43 is Folland's Real Analysis is

Arzelà-Ascoli Theorem I. Let $X$ be a compact Hausdorff space. If $\mathscr{F}$ is an equicontinuous, pointwise bounded subset of $C(X)$, then $\mathscr{F}$ is totally bounded in the uniform metric, and the closure of $\mathscr{F}$ in $C(X)$ is compact.

I could not understand where the assumption that $X$ is Hausdorff is used in the proof. Is this assumption necessary?

Jyotirmoy Bhattacharya
  • 5,518
  • I'd guess the assumption is not neccessary, but you don't gain much by dropping it since, if $X$ a noncompact Hausdorff space, then $C(X) \cong C(X')$ where $X'$ is compact Hausdorff by Gelfand duality. Moreover, I think $X'$ is some reasonable quotient of $X$ and the isomorphism $C(X) \cong C(X')$ is induced by this quotient mapping. – Mike F Feb 22 '15 at 07:00

0 Answers0