# $f_n(x_n)\to f(x)$ implies $f$ continuous – a question about the proof

My question refers to If $f_n(x_n) \to f(x)$ whenever $x_n \to x$, show that $f$ is continuous. (the first answer)

It’s written there that the hypotheses imply $(f_n)$ converges to $f$ pointwise. From this, we can choose a subsequence $(f_{n_{k}})$ of $(f_n)$ such that

$\left|f_{n_k} (x_k) -f(x_k)\right|<\epsilon$ for every $k$

Could you explain to me why is that. I’ve already posted a comment there but I’m not sure if anyone will reply, because it was last active 3 weeks ago.

I would really appreciate all your help.

Thank you.

#### Solutions Collecting From Web of "$f_n(x_n)\to f(x)$ implies $f$ continuous – a question about the proof"

$f_n$ converges pointwise because you can think of $\{x\}_{n\in\Bbb N}$ as a constant sequence that converges to $x$. It follows from the assumptions that:
$$\lim_{n\to\infty}f_n(x) = f(x)$$

Now for the second part. Fix $\epsilon > 0$ and $k \in \Bbb N$. Since $f_n(x_k)$ converges to $f(x_k)$ as $n \to \infty$, we can find $n_k \in \Bbb N$ so that:
$$\left|f_{n_k}(x_k) – f(x_k)\right| < \epsilon$$

Pick such $n_k$ for every $k \in \Bbb N$ and continue the proof in that question.