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Theorem Every sequence {$s_n$} has a monotonic subsequence whose limit is equal to $\limsup s_n$. I think to show that there exist a monotonic subsequence is kind of straight forward but I could show that there exist such subsequences whose limit is $\limsup s_n$.

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    You probably mean real sequence (=sequence of real numbers), but perhaps it would be better to mention it explicitly in the question.2012-07-16

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Let $L=\limsup_n s_n$. (Note that $L$ can be $\infty$.) Probably the simplest approach is to prove first that $\langle s_n:n\in\Bbb N\rangle$ has a subsequence $\langle s_{n_k}:k\in\Bbb N\rangle$ converging to $L$, and then show that $\langle s_{n_k}:k\in\Bbb N\rangle$ has a monotonic subsequence.

By definition $L=\lim_{n\to\infty}\sup_{k\ge n}s_k\;,$ so for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $\left|L-\sup_{k\ge n}s_k\right|<\frac{\epsilon}2$ for all $n\ge n_\epsilon$. By the definition of supremum there is a $k_\epsilon\ge n_\epsilon$ such that $0\le\left(\sup_{k\ge n}s_k\right)-s_{k_\epsilon}<\frac{\epsilon}2\;,$ and therefore

$|L-s_{k_\epsilon}|\le\left|L-\sup_{k\ge n}s_k\right|+\left|\left(\sup_{k\ge n}s_k\right)-s_{k_\epsilon}\right|<\epsilon\;.$

For $i\in\Bbb Z^+$ let $n_i=k_{1/i}$, so that $|L-s_{n_i}|<\frac1i$; clearly the sequence $\langle s_{n_i}:i\in\Bbb Z^+\rangle$ converges to $L$. However, it may not be a subsequence of $\langle s_n:n\in\Bbb N\rangle$, because the indices $n_1,n_2,n_3,\dots$ may not be increasing. To finish the argument, you must do two things.

  1. Show that $\langle n_i:i\in\Bbb Z^+\rangle$ has an increasing subsequence $\langle n_{i_j}:j\in\Bbb Z^+\rangle$; $\langle s_{n_{i_j}}:j\in\Bbb Z^+\rangle$ is then a subsequence of $\langle s_n:n\in\Bbb N\rangle$ converging to $L$.

  2. Show that $\langle s_{n_{i_j}}:j\in\Bbb Z^+\rangle$ has a monotonic subsequence. That subsequence will necessarily have the same limit, $L$, as $\langle s_{n_{i_j}}:j\in\Bbb Z^+\rangle$ itself.

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    @did: The OP seemed less concerned about that part. I can always provide more detail if asked.2012-07-17
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Let $S=\limsup\limits_{n\to\infty} s_n$.

  • First we construct a subsequence $s_{n_k}$ such that $\lim\limits_{k\to\infty} s_{n_k}=S$ and $k\le l \Rightarrow |S-s_{n_k}| \ge |S-s_{n_l}|.$

In the other words, we construct a subsequence which converges to $S$ and has the property, that the every term of the subsequence is closer so $S$ than the previous one (or in the same distance from $S$).

To prove this we only use the fact that for every $\varepsilon>0$ there are infinitely many $n$'s such that $|S-s_n|<\varepsilon.$ (Maybe it might be simpler to treat the case that there are infinitely many $n$'s such that $s_n=S$ separately.)

  • If we have sequence with the above property, then one of the sets $\{k\in\mathbb N; s_{n_k}\ge S\}$ and $\{k\in\mathbb N; s_{n_k}\le S\}$ is infinite. We take the one, which is infinite, and we obtain a monotone subsequence converging to $S$.
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I shall prove a similar property of the infimum

Let $\left(a_n\right)$ be a bounded sequence of real numbers and $a=\limsup{a_n}$, $\epsilon>0$

The set $\left\{n\in \mathbb{N}|a-\epsilon is infinite while the set $\left\{n\in \mathbb{N}|a+\epsilon is finite (why?) and thus the sets $\left\{n\in \mathbb{N}|a-\epsilon and $\left\{n\in \mathbb{N}|a_n\le a+\epsilon\right\}$ are both infinite subsets of $\mathbb{N}$

Therefore, the set \begin{equation}\left\{n\in \mathbb{N}|a-\epsilon$X$ is infinite and $A,B\subseteq X$ are infinite and $B^c$ is finite then $A\bigcap B$ is infinite. Hint: $A=A\bigcap (B\bigcup B^c)$)

For $\epsilon=1,\frac{1}{2},...,\frac{1}{n}$ we can construct by (1) a subsequence $\left({{a}_{k_n}}\right)$ such as that $a-\dfrac{1}{n} Then, \begin{equation}\forall n\in \mathbb{N}\ \ \left|{{a}_{k_n}}-a\right|<\dfrac{1}{n}\end{equation}and thus $a_{k_n}\to a$. Next note that every sequence has a monotone subsequence, the proof can be found here: http://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem#Proof

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    Nameless: Your comment is odd on two counts. First, the previous version of your post could only have one effect on the OP, which was to believe the statement (1) was true because of the wrong reason I gave. This is misleading (on purpose ? I hope not!). Second, to use the result that every sequence has$a$monotone subsequence (something which you do not mention in your post...) seems to simply preempt the whole question.2012-07-16