TITLE: The equivalence of Alexandroff convergence and continuity
QUESTION [2 upvotes]: Let $\{f_n\}$ be a sequence of continuous functions from a topological space $X$ to a metric space $(Y,e)$ which pointwise converges to a limit function $f$. It is known that the continuity of $f$ is equivalent to Alexandroff convergence of the sequence $\{f_n\}$ to $f$. The latter means that for every $\varepsilon > 0$ and natural $n'$ there exist a countable open cover $\{G_n\}$ of $X$ and a sequence $\{n_k\}$ of positive integers greater than $n'$ such that $e(f_{n_k}(x), f(x)) <\varepsilon$ for each $x\in G_k$. I know how to prove that Alexandroff convergence of the sequence $\{f_n\}$ implies continuity of $f$, but I ask you for help to prove the converse implication. Thanks.

REPLY [3 votes]: Let $x\in X$. Since a sequence $\{f_n(x)\}$ converges to $f(x)$, there exists a number $n(x)>n’$ such that $e(f_n(x), f(x))<\varepsilon$. Since both $f_n$ and $f$ are continuous at $x$, there exists an open neighborhood $G_x$ of $x$ such that $e(f_n(y), f(y))<\varepsilon$ for each $y\in G_x$. It remains to put $\{n_k\}=\{n(x): x\in X\}$ and $G_k=\bigcup \{G_x: n(x)=n_k\}$ for each $k$.
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