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Suppose that $E$ is a countable set. For every $n \geq 1$, consider a function $f_n: E \to \mathbb{R}$ in such a way that the sequence $(f_n)$ is pointwise bounded. Show that there is a subsequence $(f_{n_k})$ with the property that $f_{n_k}(x)$ converges for all $x \in E$.

I know that if $\mathcal{F} = \{f_n\}$ is equicontinuous and uniformly bounded, then we can find a subsequence $(f_{n_k})$ that converges uniformly. I also know that total boundedness implies equicontinuity and uniform boundedness.

I would love a hint.


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Have you considered using Bolzano-Weierstrass theorem? Every bounded sequence has a convergent subsequence? – Gautam Shenoy Feb 15 '13 at 9:16

Mimic the proof of the Ascoli-Arzela theorem. Let $E=\{e_n\}_{n=1}^\infty$.

The sequence $\{f_n(e_1)\}$ is bounded. There is a convergent subsequence that I denote $\{f_{n^1_k}(e_1)\}$.

The sequence $\{f_{n^1_k}(e_2)\}$ is bounded. It has a convergente subsequence that I will denote $\{f_{n^2_k}(e_2)\}$. Observe that $\{f_{n^2_k}(x)\}$ converges if $x\in\{e_1,e_2\}$.

Iterate this argument. In the $m$-th step we will have constructed a sequence $\{f_{n^m_k}\}$ such that $\{f_{n^m_k}(x)\}$ onverges if $x\in\{e_1,\dots,e_m\}$.

The diagonal sequence $\{f_{n_k^k}\}$ converges for all $x\in E$.

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