Let ${a_n}$ be a bounded sequence of real numbers, and let P be the set of limit points. Prove that $\limsup a_n = \sup P$.
I have proved that there must be a subsequence that converges to limsup$a_n$. But I am stuck connecting this to the maximum limit point!
Best Answer
Let $b_n:=\sup\{a_k:k\ge n\}$. By definition $\lim_{n\to \infty}b_n=\limsup_{n\to \infty}a_n$. Note that since $(b_n)$ converges, every subsequence converges to the same limit. Now let $(a_{n_k})$ be a convergent subsequence. Then $a_{n_k}\le b_{n_k}$ for all $k\in \mathbb{N}$. It follows that $\lim_{k\to \infty}a_{n_k}\le \lim_{k\to \infty}b_{n_k}=\limsup_{n\to \infty}a_n$.