Proving two numbers from two sets derived from a bounded sequence are equal

Question

Let $(x_i)$ be a bounded sequence. Let $s_i=\sup\{x_j: j\ge i\}$ and $S=\inf\{s_i\}$. Let $L$ be the set of all accumulation points of $(x_i)$. Prove $S=\sup L$.

So I know and have shown that $S=\inf\{s_i\}=\lim s_n$. And I have also shown that there is a subsequence of $(x_i)$ that converges to $S$. I have gotten the problem to be:

$\sup(\lim x_{n_{k}})=\lim(\sup\{x_j: j\ge i\})$, where $x_{n_{k}}$ is a subsequence of $(x_i)$ where its limit is equal to $S$. I don't know what to do from here though.

Note, limit here always means as $n$ goes to infinity.

Answer 1

Sketch. We use the standard trick(especially when sups and infs are involved) of proving '=' by proving both '≤' and '≥'. Two claims imply the result:

1. For any $i>0$ and any $\ell ∈ L, s_i \geq \ell$.
2. There is an $n_k$ such that $x_{n_k} → \inf s_i = S$.

From 1, for any $i$ and $\ell$, \begin{align} s_i &\geq \ell\\ \implies \quad \ s_i &\geq \sup_{\ell ∈ L} \ell = \sup L \\ \implies \inf_{i>0}s_i &\geq \sup L\end{align}

On the other hand, 2 implies that $S∈ L$, immediately implying $\inf s_i≤\sup L$.