# 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}

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.

• Nearly the same question: math.stackexchange.com/questions/3631729/… (3) clearly rules out (1), so it remains to show (2) is false and (4) is false. May 14 '20 at 23:12
• Might as well make it $0$ for $x \le \frac{1}{m+10}$, it won't change anything.
– Sam
May 15 '20 at 16:27

(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$$.