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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?

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

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  • $\begingroup$ 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

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