Let $\left(a_n\right)_{n = 1}^\infty $ be bounded sequence. Define the sequence $b_n=\sup \{ a_k:k \geqslant n\}$ for $n \geqslant 1$. Prove that $(b_n)$ converges.
Proof.Define$$A_k=\{a_k,a_{k + 1},a_{k + 2},\dots\} $$ for $k \in \{ 1,2,3,\dots\} $$$b_n = \sup (A_n).$$Then, for $n \in \{ 1,2,3,...\}$,$${A_{n + 1}} \subseteq {A_n}$$hence$$b_{n + 1}\leqslant b_n,$$
i.e. the sequence $\left(b_n\right)_{n=1}^\infty$ is decreasing. Further, we know $(a_n)_{n = 1}^\infty $ is bounded, so there exists a $B \geqslant 0$ such that $|{a_n}| \leqslant B$ for $n \in\mathbb{N}$, implying$ -B \leqslant {a_n}$and so we can conclude $$-B \leqslant a_n\leqslant b_n\leqslant b_1.$$So we see that sequence $\left(b_n\right)$ is bounded. As $(b_n)$ is bounded and decreasing, by the monotone convergence theorem, $\left(b_n\right)$ converges.
Is my proof correct?