Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
The theorem needs to assume $U$ is bounded. As it is phrased now the theorem is not well posed. –  Carl Mummert Dec 7 '10 at 18:45
Have you tried using the definition of sup? btw, is this homework? –  Aryabhata Dec 7 '10 at 18:47
Thanks for your comment, could you please point out, where to find it. I have searched in some analysis books, but couldnt find it! –  JSG Dec 7 '10 at 18:48
Its not homework, I just need it all the time and wanted a reference to be sure. –  JSG Dec 7 '10 at 18:49
I don't think this is suitable for the tag set-theory, but at this moment I'm uncertain about other tags. –  Asaf Karagila Dec 7 '10 at 21:18

3 Answers 3

up vote 8 down vote accepted

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.

share|improve this answer
+1 for pointing out that you need the Axiom of Choice. It would never have occurred to me! –  TonyK Dec 7 '10 at 18:59
Is it possible without the axiom of choice? –  JSG Dec 7 '10 at 19:18
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. –  Nate Eldredge Dec 7 '10 at 20:04

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 $a_i \in (\frac{M + a_{i-1}}{2}, M) \subset U$ 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 $U$.

share|improve this answer
Ok, so you need the Axiom of Choice. With that it is easy, you are right! –  JSG Dec 7 '10 at 18:58

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.