If $X$ is first-countable and for $x_n\to x$ we have that $f(x_n)\to f(x)$, then $f$ is continuous.

Question

Let $X$ and $Y$ be topological spaces, $f:X\to Y$ a function.If $X$ is first countable, $x\in X$, then $f$ is continuous in $x$ if and only if for every sequence in $X$ with $x_n\to x$ we have that $f(x_n)\to f(x)$.

Proof ($\leftarrow $ side).

Fix a local basis $\{b_n:n\in\mathbb N\}$ of $X$ at $x$ such that $B_n\subset B_m$ for $n> m$. Suppose $f$ is not continuous, then there exists an open set B in Y with $f(x)\in Y$ such that for all open set A of X with $x\in A$ we have $f(A)\neq B$.

Particularly, $\forall n\in\mathbb N,$ must exists a point $x_n\in B_n$ such that $f(x_n)\not\in B.$

We have $x_n\to x,$ but $f(x_n)\not\to f(x).$ Hence $f$ is continuous.

I was reading this proof and I got stuck in the bold statement, I was trying to understand it but got confused with the $B_n$ and B and the existence of the point $x_n\in B_n$ ? ? why ?

And also why to put this $B_n\subset B_m$ for $n> m$ condition on the local basis ?

Please help me to understand it.

You might be interested in this post: https://math.stackexchange.com/questions/808910/how-to-convert-from-net-to-sequence-in-a-first-countable-space Look at the second answer, this solves your problem. — Mathematician 42, Feb 15 '18 at 07:50
@TheoreticalEconomist: Yes, that's what I said (though I editted the comment several times after mixing up the (1) and (i) in the other post). — Mathematician 42, Feb 15 '18 at 08:01

score 1 · Accepted Answer · answered Feb 15 '18 at 08:06

1

So for any open $A$ around $x$, we have that $f(A)\neq B$. In particular, for any $n$ we have $f(B_n)\neq B$. Hence for any $n$, we have a $x_n\in B_n$ such that $f(x_n)\notin B$. Since the $B_n$'s form a local basis of $x$, the sequence $x_n$ must converge to $x$. For the last statement, you really have to use the definition of a local base.

answered Feb 15 '18 at 08:06

Mathematician 42

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ok thank you @Mathematician 42 :) – Feb 15 '18 at 08:20
Why is it true that $f(B_n) \neq B \rightarrow \exists x_n \in B_n$ such that $f(x_n) \notin B$? I mean, what if $f(B_n) \subset B$??? – Jun 06 '19 at 19:43
@MathematicalMushroom : You are right, we were a bit sloppy when writing down the proof. We suppose that $f$ is not continuous at $x$. By definition this implies that there exists a neighbourhood $B$ of $f(x)$ in $Y$ such that for all neighbourhoods $A$ of $x$ in $X$, we have that $f(A)\not\subseteq B$. Not just $f(A)\neq B$ as we wrote down! – Mathematician 42 Jun 18 '19 at 07:49

If $X$ is first-countable and for $x_n\to x$ we have that $f(x_n)\to f(x)$, then $f$ is continuous.

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