There exist infinitely many subsequences of $(f_{m})_{m \geq 1}$ which converge at every point of $E$ Let $E=\{ \frac{1}{n} | n \in \mathbb{N}\}$. For each $m \in \mathbb{N}$ define $f_{m} : E \to \mathbb{R} $ by
$$
f_{m}(x) = 
\begin{cases}
\cos{(m x)} & \text{if }\,x \geq \frac{1}{m}\\
0 & \text{if }\,\frac{1}{m+10}<x<\frac{1}{m}\\ 
x&\text{if } x \le \frac{1}{m+10}\\
\end{cases}
$$ 
Then which of the following statements is true?
$(1)$ No subsequence of $(f_{m})_{m \geq 1}$ converges at every point of $E.$
$(2)$Every subsequence of $(f_{m})_{m \geq 1}$ converges at every point of $E.$
$(3)$There exist infinitely many subsequences of $(f_{m})_{m \geq 1}$ which converge at every point of $E.$
$(4)$There exist a subsequence of $(f_{m})_{m \geq 1}$ which converges to $0$ at every point of $E.$
Here is what I tried :
Let $\frac{1}{k} \in E$ .Then $\forall n \geq k$, we have $f_{n}(\frac{1}{k})= \cos{(\frac{n}{k})}$. I don't understand how to approach further. Any help would be appreciated. Thanks in advance.
 A: (3) is true (so (1) is false): 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 $\{f^n_n\}$.
(2) is false: $\{f_m(1)\}_{m\ge 1} = \{\cos(m)\}_{m\ge 1}$ diverges. This can be proved by assuming it converges and using identities for $\cos(2n)$ and $\cos(3n)$ to get a contradiction (since both $\cos(2n)$ and $\cos(3n)$ would also converge to the same limit).
(4) is false: Suppose such subsequence exists, say $\{f_{m_k}\}_{k\ge 1}$. Then, using $\cos(m_k) = \cos(2\cdot\frac{m_k}{2}) = 2\cos^2(m_k\cdot\frac{1}{2}) - 1$ we get $f_{m_k}(1) = 2f^2_{m_k}(\frac{1}{2})-1$, which implies, by taking the limit, that $0 = -1$. 
