Prove that there exist infinitely many subsequences of $(f_m)_{m≥1}$ which converge at every point of $E$. Let $E=\{1/n | n\in \mathbb {N}\}$. For each $m\in \mathbb {N}$ define $f_m: E\to \mathbb{R}$ by
$$f_m(x)= \begin{cases} 
      \cos (mx) & x\geq 1/m \\
      0 & 1/(m+10)\leq x < 1/m \\
      x & x < 1/(m+10) 
   \end{cases}
$$
then prove that there exist infinitely many subsequences of $(f_m)_{m\geq 1}$ which converge at every point of $E$.
For fix $1/k, k\in \mathbb{N}$ I can say that by Bolzano Weierstrass theorem there exists convergent subsequence. However, I don't know how to prove, in general, for every point of $E$. Any hint would be sufficient. Thanks.
 A: In general, if $E = \{a_1, a_2, ... \}$ is a countable subset of $\mathbb R$ and $f_n: E \to \mathbb R$ is a sequence of functions such that for each $x \in E$ the sequence $\{f_n(x)\}$ is bounded, then there is a subsequence $\{f_{n_k}\}$ that converges on $E$. Applying the same result to an arbitrary subsequence of the original sequence proves that there are infinitely many convergent subsequences.
Sketch of proof: $\ $ Since $\{f_n(a_1)\}$ is bounded, there is a set $N_1 \subset \mathbb N$ such that $\{f_n(a_1)\}_{n \in N_1}$ converges. Let $\{f^1_n\}$ be the corresponding subsequence of $\{f_n\}$. Likewise, there is a set $N_2 \subset N_1$ such that $\{f_n(a_2)\}_{n \in N_2}$ converges, and a corresponding subsequence $\{f^2_n\}$ of $\{f^1_n\}$. Notice that, by construction, $\{f^2_n(a_1)\}$ also converges. Continuing this way we get subsequences $\{f^k_n\}_{k \ge 1}$ of the original sequence $\{f_n\}$ such that, for each $k$, $\{f^{k+1}_n\}$ is a subsequence of $\{f^k_n\}$ and $\{f^k_n(a_i)\}$ converges for $i \in \{1, 2, ... k\}$. Now take the subsequence $\{f^n_n\}$.
