Second countable space $\implies [x_n\to a, f(x_n)\to L \implies \lim\limits_{x\to a}f(x)=L]$

Question

Let $f:S\to Y$ where $S\subset X$. I wanted to know if $\lim\limits_{x\to a}f(x) = L$ (where $a$ is a limit point of $A$) was equivalent to $\lim\limits_{n\to \infty}f(x_n) = L$ for all sequences $x_n\to a$. I think that the first implies the second. Indeed I have to show that if I fix an open $V\subset Y$ arbitrary then $\exists N$ such that $f(x_n)\in V, \forall n>N$. Now form the definition of $\lim\limits_{x\to a}f(x) = L$, there exists an open $U$ such that $a\in U\subset S$ and $f(U)\subset V$. But then since $x_n\to a$ and $U$ contains $a$, there exists a natural $N$ such that $n>N\implies x_n\in U$. So for $n>N f(x_n)\in f(U) \subset V$ and we are done.

Now for the converse, I read from the internet that the space needs to be second countable, so with a countable basis. I don't really understand how the second countability plays a role here.

Answer 1

Actually, we only need $S$ to be first countable. Suppose $S$ is first countable, and let $(U_n)_{n=1}^\infty$ be a local basis of the point $a$. We would like it to be monotone, so let's define $W_n=\cap_{k=1}^n U_k$ for every $n$.

Now we can show that the second condition implies the first. Indeed, suppose $\lim_{x\to a} f(x)\ne L$. Then there must be some neighborhood $y\in V\subseteq Y$ such that for every neighborhood $U$ of $a$, the image $f(U)$ is not contained in $V$. In particular, for every $n\in\mathbb{N}$ there is a point $x_n\in W_n$ such that $f(x_n)\notin V$. Note that we have $x_n\to a$. Indeed, let $U$ be a neighborhood of $a$. Then there is some $n$ such that $U_{n_0}\subseteq U$. And then for every $n>n_0$:

$x_n\in W_n=\cap_{k=1}^n U_k\subseteq U_{n_0}\subseteq U$

So indeed we eventually have $x_n\in U$. However, $f(x_n)$ obviously can't converge to $L$, so we have a contradiction.

This direction is not true without first countability. For example, define the co-countable topology on $X=\mathbb{R}$. A sequence converges in this topology if and only if it is eventually constant. So in that case $x_n\to a$ always implies $f(x_n)\to f(a)$, for any function $f$. If that would imply the first condition of your statement then it would mean every function $f:X\to Y$ is continuous. But this is not true, for example take $Y=X$ with the discrete topology. The identity map $f:X\to Y$ is not continuous.