Let {$f_n$}$^\infty_{n=1}$ be a sequence of functions defined on a set $S$. Suppose that there is a positive number $M$ such that |$f_n(x)$| $\leq M$ for every $n$ and for every $x \in S$.
1) If we choose any point $x \in S$ then show that there is a subsequence of functions {$f_{n_k}$}$^\infty_{k=1}$ such that it converges as $k \to \infty$.
Is this just a direct consequence of Bolzano-Weierstass? Or is there a little more to it? It seems to me that this is a direct consequence. Thoughts?
2) I'd like help showing that if $T$ is a finite subset of $S$ then there is a subsequence of functions which converges uniformly on $T$. I'm having trouble starting this proof. Any helpful hints?
Attempt: Let $T$ be a finite subset of $S$. Then, $T$ = {$x_1, x_2, \dots , x_m$}. Use (1) choosing $x_1$. Then, there is a sub-sequence of {$f_n$}$^\infty_{n=1}$ that converges at $x_1$. Let's say this sub-sequence converges to $g$.
Apply (1) again on this newly created sub-sequence. However, this time choose $x_2$. Then we get a "sub-sub-sequence" that converges to $g$ at both $x_1$ and $x_2$. Repeat this process for each element of $T$.
Then, we obtain a sub-sequence of {$f_n$}$^\infty_{n=1}$ that converges to $g$ on each element of $T$, call this sub-sequence {$g_n$}$^\infty_{n=1}$.
Now, let $\epsilon > 0$.
For each $x_i \in T$, we know {$g_n$} converges to $g$.
Thus, there is an $N_i > 0$ such that if $n > N_i$ then |$g_n - g$| $< \epsilon$.
Let $N = \max${$N_1, N_2, \dots, N_m$}. Then if $n > N$ and $x \in T$ then it holds that |$g_n - g$| $< \epsilon$.
Therefore, {$g_n$} is uniformly convergent on $T$.