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Let $\{s_n\}$ be a sequence of reals. Let $E=\{x\in \overline{\mathbb{R}}|s_{n_k}→x\}$

($\overline{\mathbb{R}}$ denotes extended real number)

Definition of upper limit of $\{s_n\}$ is $\sup E$.

I know that if $E$ is nonempty, it is well-defined.

If $\{s_n\}$ is bounded, then $E$ is nonempty.

However, how do i show that $E$ is nonempty when $\{s_n\}$ is not bounded?

If $\{s_n\}$ is not bounded, $\forall M>0$, there exists $N\in \mathbb{N}$ such that $|s_N|>M$.

It's obvious that at least one of 'set of $s_n$ such that $s_n > M$' and 'set of $s_n$ such that $s_n < M$' must be infinite, but how do i show this?

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Do you want this result in $ZF$ or $ZFC$? Just kidding! – William Aug 17 '12 at 21:13
@William: actually, I thought of exactly the same question. :-) – tomasz Aug 17 '12 at 21:14
Maybe I misunderstand something, but is $s_n$ is unbounded, then isn't it obvious that it has a subsequence convergent to either $+\infty$ or $-\infty$? – tomasz Aug 17 '12 at 21:17
@tomasz It's obvious, but i had some trouble with showing it explicitly.. – Katlus Aug 17 '12 at 21:23
I think the easiest proof is to note that the extended real line is compact. – MartianInvader Aug 17 '12 at 21:24
up vote 1 down vote accepted

Suppose that for each $M>0$ there is an $n\in\Bbb N$ such that $s_n>M$. (The argument is similar if the sequence is unbounded below.) For each $k\in\Bbb Z^+$ let $n_k$ be the smallest natural number such that $s_{n_k}>k$. Now let $S=\{s_{n_k}:k\in\Bbb Z^+\}$. For every $M>0$, $s_{n_{\lceil M\rceil}}>\lceil M\rceil\ge M$, so $\sup S=+\infty$. (Here $\lceil x\rceil$ is the ceiling of $x$, i.e., the smallest integer $m$ such that $x\le m$.)

With just a little modification of the idea you can find a subsequence converging to $+\infty$: just let $n_{k+1}$ be the smallest natural number greater than $n_k$ such that $s_{n_{k+1}}>k+1$. Then $\langle s_{n_k}:k\in\Bbb Z^+\rangle$ is a subsequence of the original sequence that increases without bound.

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Either $s_n$ is bounded or it is not. If it is unbounded, then there must be a subsequence such that either $s_{n_k} \to +\infty$ or $s_{n_k} \to -\infty$. Hence $\pm \infty \in E$.

If the sequence is bounded, then for some $K \in \mathbb{R}$, $\{ s_n \} \subset [-K, K]$ which is compact, hence it has a convergent subsequence, with $s_{n_k} \to x$, with $|x| \leq K$. Hence $x \in E$.

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