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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!

Martin Sleziak
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Johannes Gerer
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4 Answers4

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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.

Carl Mummert
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    +1 for pointing out that you need the Axiom of Choice. It would never have occurred to me! – TonyK Dec 07 '10 at 18:59
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    Is it possible without the axiom of choice? – Johannes Gerer Dec 07 '10 at 19:18
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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. – Nate Eldredge Dec 07 '10 at 20:04
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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 $n\in\mathbb N$ $\Rightarrow$ $s-1/n<s$ 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 $$y_n\xrightarrow{n\to\infty}\lim_{n\to\infty}x_n=s\tag4\;.$$

0xbadf00d
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  • Would equation (2) still hold if we were to exchange the "less" by a "less or equal", i.e., $s - \frac{1}{n} \leq x_n \leq s$? I guess it should because "less" implies "less or equal" right? So my actual question is, why do we specifically choose "less", i.e., < ? I think it is clear that it can happen that $s - \frac{1}{n} = x_n$ if $x_n$ is not (yet) equal to $s$. – guest1 Feb 11 '22 at 14:05
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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 $U$.

bzc
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  • Ok, so you need the Axiom of Choice. With that it is easy, you are right! – Johannes Gerer Dec 07 '10 at 18:58
  • Why do you think that $\left(\frac{M+a_{i-1}}2,M\right)\subseteq U$? Just take any finite $U$ and you see that this cannot be assumed. – 0xbadf00d Jun 14 '17 at 19:57
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    @0xbadf00d Should be $\cap$, not $\subset$. Thanks for finding a 7 year old typo... – bzc Jun 14 '17 at 20:03
  • @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. – Kun Nov 16 '17 at 19:57
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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.

daOnlyBG
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George Chailos
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