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.
 A: 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 $\limsup_{n\to\infty}a_n$.
A: Hint: Let $\alpha = \sup S$. There are either (i) infinitely many $a_n$ with $a_n\geq \alpha$ or (ii) infinitely many $a_n$ with $a_n<\alpha$ (both could be true).
Suppose (i) is true. Then for all $k$, we find that $\alpha+\frac1k\not\in S$ so there are only finitely many $a_n$ with $a_n>\alpha+\frac1k$. Considering (i), this means there are infinitely many $a_n$ satisfying $\alpha\leq a_n\leq \alpha+\frac1k$. Construct a subsequence $a_{n_k}$ with $a_{n_k}\to \alpha.$
If (ii) is true, a similar approach is applied to $\alpha-\frac1k$. Since $\alpha-\frac1k\in S$, there are infinitely many $a_n$ with $a_n>\alpha-\frac1k$. If $(\alpha-\frac1k,\alpha)$ contains an infinite number of $a_n$ terms, then the argument is similar to case (i). Otherwise, it must be true that an infinite number of $a_n$ satisfy $a_n\geq \alpha$, in which case we are in case (i).
