Let $E = \left \{\frac {1} {n}\ \big |\ n \in \Bbb N \right \}.$ For each $m \in \Bbb N$ define $f_m : E \longrightarrow \Bbb R$ by $$ f_m(x) = \left\{ \begin{array}{ll} \cos (mx) & \quad \text {if}\ \ x \geq \frac {1} {m} \\ 0 & \quad \text {if}\ \ \frac {1} {m+10} \lt x \lt \frac {1} {m} \\ x & \quad \text {if}\ \ x \leq \frac {1} {m+10} \end{array} \right. $$ 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 exists infinitely many subsequences of $(f_m)_{m \geq 1}$ which converge at every point of $E.$
$(4)$ There exists a subsequence of $(f_m)_{m \geq 1}$ which converges to $0$ at every point of $E.$
My attempt $:$ Let us fix $x = \frac {1} {n} \in E.$ Let us take any subsequence $\left (f_{m_k} \right )_{k \geq 1}$ of the given sequence $\left (f_m \right )_{m \geq 1}.$ Then since $(m_k)_{k \geq 1}$ is a strictly increasing sequence of natural numbers it follows that there exists $N \in \Bbb N$ such that for all $k \geq N$ we have $m_k \geq n.$ But that means $x=\frac {1} {n} \geq \frac {1} {m_k},$ for all $k \geq N.$ Hence we have $f_{m_k} \left (\frac {1} {n} \right ) = \cos \left (\frac {m_k} {n} \right),$ for all $k \geq N.$ But then $\lim\limits_{k \to \infty} f_{m_k} \left (\frac {1} {n} \right )$ doesn't exist since $\lim\limits_{x \to \infty} \cos x$ doesn't exist. So there doesn't exist any subsequence of the given sequence $(f_m)_{m \geq 1}$ which converges at any point of $E.$ But the answer key to this question suggests that the correct answer is option $(3).$ I wonder how can it be! Am I missing something? Any help or suggestion in this regard will be appreciated.
Thanks for your time.