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
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?
Best Answer
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.