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This is not a duplicate. Please read carefully.

First, I know every sequence has a monotone subsequence. For an answer.

Second, I know we can construct a subsequence converging to limsup. For an answer.

The question is: can we construct a monotone subsequence converging to limsup? The reason I asked this is that from the following exercises

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My answer to 27 is the following:

For $\epsilon = 1$, we have $a_n< M + 1$ for all but finitely many $n$. Pick out these finite $n$'s, and denote the largest number of these $n$'s as $N_1$, and $M - 1 < a_n$ for infinitely many $n$, choose $n_1$ which satisfies $n_1>N_1$ (we can because Archimedean property).

For $\epsilon = 1/2$, we have $a_n< M + 1/2$ for all but finitely many $n$. Pick out these finite $n$'s, and denote the largest number of these $n$'s as $N_2$, and $M - 1/2 < a_n$ for infinitely many $n$, choose $n_2$ which satisfies $n_2>max\{N_1,n_1\}$ (we can because Archimedean property). Continuing we get $\{a_{n_k}\}$ as the subsequence. QED

The author seems to encourage me to imitate the argument in 27 to prove 28.

So I guess he implies that there is a monotone subsequence converging to limsup. But how?

Hamilton
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  • If as you say $a_n<M+1$ for all but finitely many $n$ then there is no largest such $n.$ Therefore you cannot as you do denote "the largest of these $n$" as $N_1.$ [namely because there is no largest.] – coffeemath Sep 30 '21 at 03:42
  • @coffeemath, thank you, this mistake has been corrected now. – Hamilton Sep 30 '21 at 03:46
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    When you say 'Pick out these finite $n$'s', you assume that there is at least $1$ for which $a_n \geq M + 1$. It could be the case that no such $n$ exists. – Matt E. Sep 30 '21 at 04:11
  • I see no such implication in Ex. 28. The goal is only to find a monotone subsequence, and the author suggests that the proof is similar (which you already know is true). – Michał Miśkiewicz Sep 30 '21 at 07:24
  • You've shown you are capable of writing in math jax. So please stop posting images of questions. Links and images die .... Since you know very well how to format, I take it you post images when you feel too lazy to put the effort into formatting? – amWhy Oct 21 '21 at 16:53
  • @Not just lazy, because my time is limited sometimes. However, I know now it is a very irresponsible way of doing things to this community. It won't happen in my future posts. Thank you very much for your advice. – Hamilton Oct 21 '21 at 16:57

1 Answers1

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By combining your two stated facts, you can answer the question. Let $(a_n)_{n=1}^\infty$ be a sequence. Then some subsequence, say $(a_{n_j})_{j=1}^\infty$, converges to $\lim\sup a_n$ by your second statement. For ease of notation, set $b_j = a_{n_j}$. Then by your first statement, we know that the sequence $(b_j)_{j=1}^\infty$ has a monotone subsequence, say $(b_{j_k})_{k=1}^\infty$. The monotone subsequence $(b_{j_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, so we're done.

Matt E.
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