I happened accross this excercise:
Let $a_n$ be a bounded, divergent sequence. Proove that there are at least two subsequences converging to different limits.
As a first strategy I thought of doing this:
Let $b_n \rightarrow L$ be a convergent subsequence (B-W theorem). Choose $\epsilon$ such that $\forall N, \exists n>N$ so that $|a_n - L|>\epsilon$. Using this post, build a subsequence as follows:
$n_1 = \min(n>1 / |a_n-L|>\epsilon)$
$n_k = \min(n>n_{k-1}/ |a_n-L|>\epsilon)$
The set $a_{n_k}, k>=1$ is a subsequence of $a_n$, therefore bounded, therefore contains a convergent subsequence, which cannot converge to $L$ (first question: is this ok?)
Moreover I wondered if $sup$ and $inf$ of the sequence can be used, and what are their properties. Is the following reasoning correct?
Let $S_k = {a_n, n>=k}$, since all $S_k$ are bounded sequences, $sup(S_k)$ exists. Furthermore, since $S_1\supset S_2 \supset S_3 ...$ then $\sup(S_1)\geq \sup(S_2)\geq \sup(S_3) ...$. Therefore, the sequence $\sup(S_1), \sup(S_2)...$ is monotone and bounded (therefore convergent).
Third question, is it true that $\displaystyle L = \lim_{n\rightarrow \infty}\inf(S_n) = \lim_{n\rightarrow \infty}\sup(S_n) \Leftrightarrow a_n\rightarrow L$?
Thanks