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.
Actually, the Axiom of Choice is not required in its full strength, but the weaker Axiom of Countable Choice (CC) will suffice.