Let $A \subseteq \mathbb R$ be countable and let $f_n: A\to \mathbb R$ be a sequence of functions such that there exists $M \ge 0$ with $\,\lvert\,f_n(x)\rvert\le M$ for all $n$. I am trying to show that there exists a subsequence $f_{n_k}$ of $f_n$ that converges pointwise.
Here is what I have so far: If $A = \{a_1, a_2, \dots \}$ then $f_n(a_1)$ is a bounded sequence hence by Bolzano Weierstrass theorem contains a convergent subsequence $f_{n_{k_1}}$. By the same argument $f_{n_{k_1}}(a_2)$ contains a convergent subsequence $f_{n_{k_2}}$.
Next I want to define $$f_{n_k} (x) = \lim_{j \to \infty} f_{n_{k_j}}(x),$$ the pointwise limit. Then $f_{n_k}(a_j)$ converges for every $a_j \in A$ (it's clear by how it was defined).
Am I done now or am I missing something? Is there anything left to show?
Best Answer
We can establish the existence of such a sequence using a standard diagonal argument (due to Cantor).
Let $A=\{a_n:n\in\mathbb N\}$. Using Bolzano-Weierstrass for the bounded sequence $\{f_n(a_1)\}_{n\in\mathbb N}$ we can find a convergent subsequence which we denote as $\{f_{1,n}(a_1)\}_{n\in\mathbb N}$. Next, as $\{f_{1,n}(a_2)\}_{n\in\mathbb N}$ is bounded, it also contains a convergent subsequence which we denote as $\{f_{2,n}(a_2)\}_{n\in\mathbb N}$.
In this way we construct recursively the following convergent sequences: \begin{align} f_{1,1}(a_1),&f_{1,2}(a_1),\ldots f_{1,n}(a_1),\ldots,\\ f_{2,1}(a_2),&f_{2,2}(a_2),\ldots f_{2,n}(a_2,\ldots,\\ \vdots&\\ f_{n,1}(a_n),&f_{n,2}(a_n),\ldots f_{n,n}(a_n),\ldots,\\ \vdots& \end{align} with $\{f_{k,n}\}_{n\in\mathbb N}$ a subsequence of all the sequences $\{f_{j,n}\}_{n\in\mathbb N}$, for $j<k$, and say that $\lim_{n\to\infty}f_{j,n}(a_j)=f(a_j)$.
The sequence $\{f_{n,n}\}_{n\in\mathbb N}$ is finally a subsequence of all the above, and hence $\{f_{n,n}(a_j)\}_{n\in\mathbb N}$ converges for all $j\in\mathbb N$, and in particularly $$ \lim_{n\to\infty}f_{n,n}(a_j)=f(a_j), $$ for all $j\in\mathbb N$.