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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.

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    $\begingroup$ 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. $\endgroup$
    – aschepler
    May 14 '20 at 23:12
  • $\begingroup$ Might as well make it $0$ for $x \le \frac{1}{m+10}$, it won't change anything. $\endgroup$
    – Sam
    May 15 '20 at 16:27
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(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 $0 = -1$.

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  • $\begingroup$ What about (2) and (4)? $\endgroup$ May 15 '20 at 8:35
  • $\begingroup$ @mathisfun Sorry, forgot about those. Added both to the answer. $\endgroup$
    – Sam
    May 15 '20 at 16:41

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