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As an example of Arzela-Ascoli theorem in intro. to topology course, the following case was exhibited:
Let $A\subset C[a,b]$ be a set of differentiable functions such that: $\forall f\in A,\ \vert f \vert \leq M_1,\ \vert f' \vert \leq M_2$ thus A is bounded and equicontinuous, thus every sequence $\{f_n\}$ has a converging subsequence.

My question is: The limit function isn't necessarily in $A$, right?
A isn't necessarily a closed set, and the more detaild chain of consequences is: bounded & equicontinuous $\rightarrow$ absolutely bounded $\rightarrow\ \bar A$ is closed and absolutely bounded $\rightarrow\ \bar A$ is compact $\rightarrow$ every sequence $\{f_n\}\subset \bar A$ has a converging sub-sequence.
So a sequence in $A$ might converge to $C[a,b]\supset \bar A \ni f \notin A$, am I right?

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    $\begingroup$ Right, take $f_n(x) \equiv 1/n$ for example. $\endgroup$ – zhw. Mar 14 '17 at 20:48
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    $\begingroup$ Another way to think about this is that Arzelà–Ascoli only tells you $A$ is precompact. So if it's not closed (and hence not compact), the sequence need not converge to a limit in $A$. $\endgroup$ – Fimpellizieri Mar 14 '17 at 20:52
  • $\begingroup$ Great, thank you both $\endgroup$ – galra Mar 14 '17 at 21:47
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As pointed out in the comments by zhw and Fimpellizieri:

We only know that $A$ is precompact for the metric induced by the uniform norm, but it may not be closed. For example, if $A=\left\{ x\mapsto 1/n,n\in\mathbb N, n\geqslant 1\right\}$, the limit of $\left(1/n\right)$ does not belong to $A$.

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