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 $T$.


I assume that in (1) you mean that there is a subsequence $\langle f_{n_k}:k\in\Bbb Z^+\rangle$ of the original sequence such that $\langle f_{n_k}(x):k\in\Bbb Z^+\rangle$ converges; yes, this is a direct consequence of the Bolzano-Weierstrass theorem.

For (2), let $T=\{x_1,\dots,x_m\}$. Use (1) to get a subsequence of $\langle f_n:n\in\Bbb Z^+\rangle$ that converges at $x_1$. Then apply (1) to that subsequence to get a 'sub-subsequence' converging at $x_1$ and $x_2$. Repeat as needed. You shouldn't have any real trouble showing that your final subsequence converges uniformly on $T$, simply because $T$ is finite.

  • $\begingroup$ I'm not sure I completely follow. I understand that for each element in T, our subsequence converges. But why does it necessarily converge uniformly? Thank you for your help! $\endgroup$ – Mark V. May 12 '12 at 19:49
  • $\begingroup$ @MarkV.: Fix $\epsilon>0$. For each $x\in T$ let $\langle f_{n_k}(x):k\in\Bbb Z^+\rangle\to f(x)$; there is a $k_x(\epsilon)$ such that $|f_{n_k}(x)-f(x)|<\epsilon$ for $k\ge k_x(\epsilon)$. Let $k(\epsilon)=\max\{x\in T:k_x(\epsilon)\}$. What happens for $k\ge k(\epsilon)$? $\endgroup$ – Brian M. Scott May 12 '12 at 19:56
  • $\begingroup$ I added my attempt to the solution to the question. If you could look it over, I would be greatly appreciative. Thank you again for your help! $\endgroup$ – Mark V. May 12 '12 at 20:33
  • $\begingroup$ @MarkV.: Yes, that's exactly the idea that I had in mind. $\endgroup$ – Brian M. Scott May 12 '12 at 20:40

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