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Let $A \subset \mathbb R$ be countable and let $f_n: A\to \mathbb R,\,$ $n\in\mathbb N,\,$be a sequence of functions, such that there exists $M \ge 0$ with $\,\lvert\,f_n(x)\rvert\le M,\,$ for all $n\in\mathbb N$ and $x\in A.\,$ Show that there exists a subsequence $f_{n_k}$ of $f_n$ which converges pointwise, i.e., $\{f_{n_k}(x)\}$ converges for every $x\in A$.

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?

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1 Answer 1

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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 $j\in\mathbb N$.

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    $\begingroup$ Nice answer. But I don't understand why $f_{k,n}$ is subsequence of $f_{j,n}$ please can you illustrate? $\endgroup$ Commented Nov 17, 2018 at 11:46
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    $\begingroup$ @MSE_Lover Both $j$ and $k$ are indexes of members in the set $A$. Since Not only does $j$ represents the j-th member of $A$, but it also represents the j-th $f$ sequence. Now, recall k-th sequence is a convergent subsequence of the j-th sequence. As a result, the subsequence relation exists. $\endgroup$
    – Andes Lam
    Commented Jul 9, 2021 at 15:44

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