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This is a question from my real analysis textbook -

Let $(x_n)$ be a bounded sequence and for each $n \in \mathbb{N}$ let $s_n:=\sup\{x_k:k\geq n\}$ and $S:=\inf \{s_n\}$. Show that there exists a subsequence of $(x_n)$ that converges to $S$.

So, my first thought was that since $s_{n+1}\geq s_n$, it is a nonincreasing sequence and since $(x_n)$ is bounded, from monotone convergence theorem the sequence $s_n$ must converge to its infinum which is $S$. But the question asks to prove that there exists a subsequence of $x_n$ that converges to $S$. I do not have any idea how to do that.

PhysicsWizardUd
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2 Answers2

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Take $n_0=1$ and for $k\geq 1$, define $$n_{k+1}:=\inf\left\{n>n_{k}\mid s_{n_k}-\frac{1}{n_k}\leq x_{n}\leq s_{n_k} \right\}.$$ Then $(x_{n_k})$ is a subsequence of $(x_k)$ that converges to $S$.

Surb
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Take $m\in\Bbb N$. Since $S+\frac1m>S=\inf_{n\in\Bbb N}s_n$, there is some $n\in\Bbb N$ such that$$S\leqslant s_n<S+\frac1m.\tag1$$And, since $s_n=\sup_{k\geqslant n}x_k$, you can deduce from $(1)$ that there is some $n_m\in\{n,n+1,n+2,\cdots\}$ such that $S\leqslant x_{n_m}<S+\frac1m$. So, $\lim_{m\to\infty}x_{n_m}=S$.

José Carlos Santos
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