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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asked Dec 7, 2010 at 18:39
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$\begingroup$ The theorem needs to assume $U$ is bounded. As it is phrased now the theorem is not well posed. $\endgroup$– Carl MummertDec 7, 2010 at 18:45
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$\begingroup$ Have you tried using the definition of sup? btw, is this homework? $\endgroup$– AryabhataDec 7, 2010 at 18:47
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$\begingroup$ Thanks for your comment, could you please point out, where to find it. I have searched in some analysis books, but couldnt find it! $\endgroup$– Johannes GererDec 7, 2010 at 18:48
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$\begingroup$ Its not homework, I just need it all the time and wanted a reference to be sure. $\endgroup$– Johannes GererDec 7, 2010 at 18:49
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2$\begingroup$ I don't think this is suitable for the tag set-theory, but at this moment I'm uncertain about other tags. $\endgroup$– Asaf Karagila ♦Dec 7, 2010 at 21:18
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4 Answers
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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.
answered Dec 7, 2010 at 18:48
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3$\begingroup$ +1 for pointing out that you need the Axiom of Choice. It would never have occurred to me! $\endgroup$– TonyKDec 7, 2010 at 18:59
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4$\begingroup$ Is it possible without the axiom of choice? $\endgroup$ Dec 7, 2010 at 19:18
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6$\begingroup$ The statement is equivalent to Form 94F in the Howard-Rubin database, 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. $\endgroup$ Dec 7, 2010 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 $$y_n\xrightarrow{n\to\infty}\lim_{n\to\infty}x_n=s\tag4\;.$$
answered Jun 14, 2017 at 20:26
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$\begingroup$ 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 $s$. $\endgroup$– guest1Feb 11, 2022 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 $U$.
answered Dec 7, 2010 at 18:49
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$\begingroup$ Ok, so you need the Axiom of Choice. With that it is easy, you are right! $\endgroup$ Dec 7, 2010 at 18:58
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$\begingroup$ 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. $\endgroup$ Jun 14, 2017 at 19:57
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2$\begingroup$ @0xbadf00d Should be $\cap$, not $\subset$. Thanks for finding a 7 year old typo... $\endgroup$ Jun 14, 2017 at 20:03
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$\begingroup$ @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. $\endgroup$– KunNov 16, 2017 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.
