Question: Please help me, I can't find the theorem anywhere that states something like:

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> 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 $\lim\_{n\to\infty}a\_n=\sup U$.
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Thank you!

Answer: 
You can prove the theorem (under the assumption that $\sup U$ exists) directly from the definition of supremum. For each $n$ there is some point in $U$ within $1/n$ of the supremum. Use the axiom of choice to choose a sequence $(a\_n)$ so that for each $n$, $a\_n$ is within $1/n$ of the supremum. Then prove that this sequence converges to the supremum.
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*](https://en.wikipedia.org/wiki/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 $$y\_n\xrightarrow{n\to\infty}\lim\_{n\to\infty}x\_n=s\tag4\;.$$