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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!

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    @Carl: True. I thought it was some open set, my bad.2010-12-08

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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.

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    The statement is equivalent to Form 94F in the Howard-Rubin database, http://consequences.emich.edu/CONSEQ.HTM. Equivalently, every denumerable family of nonempty subsets of $\mathbb{R}$ has a choice function; they call this $C(\aleph_0, \infty, \mathbb{R})$. (In the paper copy it appears as Form 73 but was later shown to be equivalent to 94.) It is not provable in ZF (the database lists several models where it is false) so some sort of choice is indeed necessary.2010-12-07
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Because it follows directly from the definition of supremum. You can create the sequence $\{a_n\}$ easily: Let $M = \sup U $. Assume that $M \notin U$ (otherwise just let $a_i = M \, \forall i \in \mathbb{N}$). Fix $a_0 \in U$. Choose $a_i \in (\frac{M + a_{i-1}}{2}, M) \cap U$ for $i = 1, 2, \dots$ Such $a_i$ must exist since $M$ is the supremum. Otherwise $\frac{M + a_{i-1}}{2} < M$ would be a lower upper bound for $U$.

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    @BrandonCarter each element in the sequence depends on the last element for the sake of non-decreasing sequence. But logically this is a bit awkward for me to understand because the existence of element $i$ in the sequence is quantified by the element $i-1$, which is not known at the time before applying Axiom of Choice.2017-11-16
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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 and hence $s-\frac1n\not\in M\;,\tag1$ i.e. $\exists x_n\in U$ with $s-\frac1n 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 $y_n\xrightarrow{n\to\infty}\lim_{n\to\infty}x_n=s\tag4\;.$

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