# Probably dumb limit

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 $$f(x)$$??

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 $$n\geqslant\max\{N,M\}$$,$$\bigl\lvert f(x)-f_n(x_n)\bigr\rvert\leqslant\bigl\lvert f(x)-f(x_n)\bigr\rvert+\bigl\lvert f(x_n)-f_n(x_n)\bigr\rvert<\varepsilon.$$