Arzela-Ascoli Anthony Knapp Proof STATEMENT: (Arzela Ascoli Theorem) If $\left\{f_n\right\}$ is an equicontinuous family of scalar-valued functions defined on a compact Hausdorff space $X$ and if $\left\{f_n\right\}$ has the property that $\left\{f_n(x)\right\}$ is bounded for each $x$, then $\left\{f_n\right\}$ has a uniformly convergent subsequence.
Proof: We may assume that there are infinitely many distinct functions $f_n$, since otherwise the assertion is trivial. Let $|f_n(x)|\leq c_x$ for all $n$, and form the product space $C=\prod_{x\in X}\left\{z\in \mathbb{C}\mid |z|\leq c_x\right\} $. The space $C$ is compact by Tychonoff theorem, and we are now assuming that there are infinitely many members of the sequence $\left\{f_n\right\}$ in the space. Let $S$ be the image of the sequences as a subset of $C$. If $S$ were to contain all its limit points, then each $f_n$ would have an open neighborhood in $C$ disjoint from the rest of $S$; these open sets and $S^c$ would form an open cover of $C$ with no finite sub cover, in contradiction to compactness of $C$. Thus $S$ has a limit point $f$ not in $S$. By Lemma 10.47 and the remarks before it, the family $S\cup \left\{f\right\}$ is equicontinuous.
Lemma 10.47: Let $\mathcal{F}=\left\{f_\alpha\right\}$ is equicontinuous at $x$ in $X$, then the closure $\mathcal{F}^{cl}$ of $\mathcal{F}$ in the product topology on $\mathbb{C}^X$ is equicontinuous at $x$.
QUESTION: How does Knapp conclude that each $f_n$ would have an open neighborhood in $C$ disjoint from $S$. 
 A: The assertion

If $S$ were to contain all its limit points, then each $f_n$ would have an open neighborhood in $C$ disjoint from the rest of $S$

is incorrect. If for example $(f_n)$ is a sequence of distinct functions uniformly converging to $f_0$ (or $f_1$ if for you $0\notin \mathbb{N}$), then $S$ is a closed subset of $C$, hence contains all its limit points, and every neighbourhood of $f_0$ intersects $S\setminus \{f_0\}$.
The mistake is easily fixable, however. The goal here is to show the existence of a limit point of $S$. So if $S$ contains a limit point of $S$, we are done. And if $S$ had no limit points - that is, $S$ contains all its limit points and no $f_n$ is a limit point of $S$ - then the assertion that each $f_n$ has an open neighbourhood disjoint from $S\setminus \{f_n\}$ holds. And in that case, the part of the proof shows by contradiction that $S$ must have a limit point contra the assumption.
However, it is a fundamental property of compact spaces that every infinite subset of a compact space has a limit point, so I don't see the point of that particular argument.
A: This error is corrected in Knapp’s errata, which can be found here. (The uncorrected proof can be seen here.)
The phrase “contain all its limit points” should read “have no limit point in $C$.”
