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Theorem 7.11: Suppose $f_n \to f$ uniformly on a set $E$ in a metric space. Let $x$ be a limit point of $E$, and suppose that $$\lim_{t\to x} f_n(t) = A_n \qquad (n \in N).$$ Then $\{A_n\}$ converges, and $$\lim_{t\to x} f(t) = \lim_{n\to \infty} A_n.$$

Sketch of the proof: After showing that $\{A_n\}$ converges, let $A = \lim_{n\to \infty} A_n$. Then $$|f(t)-A| \le |f(t) - f_n(t)| + |f_n(t) - A_n| + |A_n - A| \qquad(1).$$ We first choose $n$ such that $$|f(t) - f_n(t)| \le \frac {\epsilon}{3}$$ for all $t\in E$, and such that $$|A_n - A| \le \frac {\epsilon}{3}.$$ Then, for this $n$, we choose a neighbourhood $V$ of $x$ such that $$|f_n(t) - A_n| \le \frac {\epsilon}{3}$$ if $t \in V \cap E$, $t \neq x$. Substituting the above inequalities, we see that $|f(t)-A| \le \epsilon$ , provided that $t \in V \cap E$, $t \neq x$. Q.E.D

I understand the above proof except for one perhaps trivial part that has been bothering me for some time. He concludes the proof by saying that $|f(t)-A| \le \epsilon$ if $t \in V \cap E$, $t \neq x$ , but I do not see how he concludes that because we have just shown that if $n$ is sufficiently large then there exists a suitable neighbourhood for the corresponding $n$ which we could use to conclude that $|f(t)-A| \le \epsilon$. However eliminating the part that involves the condition that $n$ needs to be sufficiently large and just considering that $t \in V \cap E$, $t \neq x$ will not necessarily enable us to conclude that $|f(t)-A| \le \epsilon$ as for that to happen the inequality formed in $(1)$ needs to be made small enough and so if this is so then we cannot immediately conclude that the limit is $A$. Do please correct me if I am wrong otherwise inform me of what I am missing out, Thank you.

HAT
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    I think you are missing that the neighborhood $V$ depends on $n$. And that's why in the proof, he begins by choosing $n$ (which becomes fixed in the rest of the proof) and after that he can choose the $V$. – Kroki Apr 18 '18 at 19:20
  • @Youem So if $t$ is a member of the neighbourhood $V \cap E$ then that would also imply that the other three inequalities that are formed after $(1)$ would also hold? – HAT Apr 18 '18 at 19:27
  • In the first two inequalities only the inequality $|f(t) - f_n(t)|\le \frac\epsilon 3$ depends on $t$. But having uniform convergence this inequality holds for every $t$ in $E$ so for $t \in V\cap E$. – Kroki Apr 18 '18 at 19:32
  • @Youem I agree with you, but if this is so then when he considers $t \in V \cap E$ he is also implicitly considering $n$ to also be sufficiently large at the same time , yes? – HAT Apr 18 '18 at 19:37
  • Yes but at this step of the proof $n$ is fixed already. – Kroki Apr 18 '18 at 19:38
  • I think I have understood it now, thank you. – HAT Apr 18 '18 at 19:43

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