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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!

  • 0
    The theorem needs to assume $U$ is bounded. As it is phrased now the theorem is not well posed.2010-12-07
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    Have you tried using the definition of sup? btw, is this homework?2010-12-07
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    Thanks for your comment, could you please point out, where to find it. I have searched in some analysis books, but couldnt find it!2010-12-07
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    Its not homework, I just need it all the time and wanted a reference to be sure.2010-12-07
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    Why does U have to be bounded? I its not bounded $sup U=\infty$ and then the theorem would postulate the existence of a divergent sequence in U.2010-12-07
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    @user4514 The near-universal convention in elementary real analysis is that unbounded sets do not have a supremum, not that the supremum is $\infty$.2010-12-07
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    @Carl: Really? Even the books includes this case explicitly!2010-12-07
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    Is it possible without the axiom of choice?2010-12-07
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    @Carl Mummert: That's not my experience. Many books are happy to work over the extended reals.2010-12-07
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    I don't think this is suitable for the tag set-theory, but at this moment I'm uncertain about other tags.2010-12-07
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    @user4514: You can usually choose the first rational which is close "enough" and the nominator or denominator is the smallest, and thus avoid using the axiom of choice.2010-12-07
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    @Asaf: the set $U$ may not contain any rationals, though.2010-12-08
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    @Carl: True. I thought it was some open set, my bad.2010-12-08

4 Answers 4

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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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    +1 for pointing out that you need the Axiom of Choice. It would never have occurred to me!2010-12-07
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    Is it possible without the axiom of choice?2010-12-07
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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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    Ok, so you need the Axiom of Choice. With that it is easy, you are right!2010-12-07
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    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.2017-06-14
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    @0xbadf00d Should be $\cap$, not $\subset$. Thanks for finding a 7 year old typo...2017-06-14
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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/nsqueeze 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.