Probably dumb limit

Question

I have a sequence of continuous functions $f_n : I^k \rightarrow I^k$ converging uniformly to a continuous function $f$. Then for each $n$ I choose a point $x_n$ and since they're chosen in $I^n$ which is sequence-compact (not sure this is the right terminology) there will be a converging subsequence (which for simplicity we will call $x_n$ again) to some $x$. Will I then have that the sequence $f_n (x_n)$ converges to $f(x)$??

Answer 1

Yes, because the convergence is uniform. So, if $\varepsilon>0$, take $N\in\mathbb N$ such than $n\geqslant N$ and $y\in I^n$ implies that $\bigl\lvert f(y)-f_n(y)\bigr\rvert<\frac\varepsilon2$. Since the convergence is uniform, $f$ is continuous and so there is some $M\in\mathbb N$ such that $n\geqslant M\implies\bigl\lvert f(x)-f(x_n)\bigr\rvert<\frac\varepsilon2$. So, if $n\geqslant\max\{N,M\}$,$$\bigl\lvert f(x)-f_n(x_n)\bigr\rvert\leqslant\bigl\lvert f(x)-f(x_n)\bigr\rvert+\bigl\lvert f(x_n)-f_n(x_n)\bigr\rvert<\varepsilon.$$