TITLE: Sequence converging to the supremum
QUESTION [7 upvotes]: Please help me, I can't find the theorem anywhere that states something like:

For a bounded set $U\subset\mathbb R$ there exists a non-descreasing sequence $(a_n)_{n\in \mathbb N}$ with $a_n \in U$ with $\lim_{n\to\infty}a_n=\sup U$.

Thank you!

REPLY [8 votes]: Let $$M:=\left\{a\in\mathbb R:a\ge x\text{ for all }x\in U\right\}\;.$$ By definition, $$s:=\sup U=\min M\;.$$ Let $n\in\mathbb N$ $\Rightarrow$ $s-1/n<s$ and hence $$s-\frac1n\not\in M\;,\tag1$$ i.e. $\exists x_n\in U$ with $$s-\frac1n<x_n\le s\tag2\;.$$ The squeeze theorem yields $$x_n\xrightarrow{n\to\infty}s\tag3\;.$$ Now, let $$y_n:=\max(x_1,\ldots,x_n)\;\;\;\text{for }n\in\mathbb N\;.$$ Note that $(y_n)_{n\in\mathbb N}\subseteq U$ is nondecreasing with $$y_n\xrightarrow{n\to\infty}\lim_{n\to\infty}x_n=s\tag4\;.$$