Let $S = \{x \in \mathbb{R}| x < a_n \hspace{2mm}\text{for infinitely many $a_n$}\}$. Find a subsequence converging to $\sup S$ Let $(a_n)_{n\in\mathbb N}$ be a bounded sequence. Let $S = \{x \in \mathbb{R}| x < a_n \hspace{2mm}\text{for infinitely many $a_n$}\}$. Prove there exists a subsequence converging to supremum of $S$.
The part about "infinitely many terms" is confusing me a bit. Any help would be appreciated.

Define
$$
\xi_N:=\sup_{k\ge N}a_n
$$
So $\{\xi_N\}_{N\in\Bbb N}$ is a decreasing sequence, thus it admits limit $L$, which is finite, since $\{a_n\}_n$ is bounded by hypothesis.
Prove by yourself that $\{\xi_N\}_{N\in\Bbb N}$ is (one of the possible) sequence you are searching for, and $L=\sup S$.
Usually one refers to such a limit as $\limsup_{n\to\infty}a_n$.
