The Arzela-Ascoli Theorem says that

If $X$ is a compact metric space and $F$ a subset of $C(X)$, then $F$ is compact if and only if $F$ is closed, uniformly bounded, and equicontinuous.

My textbook gives the following corollary:

If $X$ is a compact metric space, and $\langle f_n\rangle_{n\in \mathbb{N}}$ a uniformly bounded, equicontinuous sequence in $C(X)$, then some subsequence converges uniformly on $X$.

Can someone explain how this corollary follows? Obviously the set of all $f_n$ is itself uniformly bounded and equicontinuous, but why must it be closed?

  • 1
    $\begingroup$ Use \langle and \rangle for $\langle$ and $\rangle$, not < and >. The latter are relation symbols; they look different, and produce different spacing. $\endgroup$ – Harald Hanche-Olsen Dec 10 '17 at 20:23

The closure of the set of all $f_n$ is uniformly bounded and equicontinuous, hence compact by Arzela–Ascoli.

  • $\begingroup$ Why does the subsequence converge uniformly? $\endgroup$ – IntegrateThis Jun 17 '18 at 1:14
  • $\begingroup$ @IntegrateThis That is pretty much by definition. You have a sequence contained in a compact subset of the metric space $C(X)$; then it has a subsequence convergent in that metric; but convergence in the standard metric of $C(X)$ is nothing but uniform convergence. $\endgroup$ – Harald Hanche-Olsen Jun 17 '18 at 5:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.