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I am trying to prove:

If a sequence is monotone and bounded then it converges.

My idea is: Assume $a_n$ is monotone and not converges and then show that it is not bounded. But: my problem is that I fail to prove it is not bounded. Please can you help me?

I found a different proof in a book but I want to know if my proof can work.

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4 Answers 4

up vote 3 down vote accepted

Let $\langle x_n:n\in\Bbb N\rangle$ be a bounded, monotone sequence in $\Bbb R$; without loss of generality assume that the sequence is non-decreasing. To prove that it converges, at some point you’re going to have to use the fact that $\Bbb R$ is complete: the theorem is not true in $\Bbb Q$. The most straightforward way to use completeness is simply to let $a=\sup\{x_n:n\in\Bbb N\}$, which exists by completeness because the sequence is bounded, and show that $\langle x_n:n\in\Bbb N\rangle\to a$. I don’t see any way to carry out your idea without essentially incorporating this argument as part of it, and if you do that, you’re just needlessly complicating things.

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Without loss of generality assume that $(a_n)$ is increasing and bounded above (the other case is similar) then the set $$A=\{a_n\;|\; n\in\mathbb N\}$$ has a supremum $s=\sup A$ and we know by the characterization of this supremum: $$\forall \epsilon>0,\; \exists a_p\in A \;|\; s-\epsilon\le a_p\le s$$ but since $(a_n)$ is increasing then $$\forall \epsilon>0,\; \exists p\in\mathbb N, \forall n\ge p \;|\; s-\epsilon\le a_p\le a_n\le s$$ which means that $\displaystyle\lim_{n\to\infty}a_n=s$.

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Is that little vertical line $\;\mid\;$ supposed to mean "s.t.=such that" , or what? I've never seen that. –  DonAntonio Dec 16 '13 at 13:06
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Yes it means such that and I think it's well known. –  Sami Ben Romdhane Dec 16 '13 at 13:07
    
Never saw it...anyway, +1 . –  DonAntonio Dec 16 '13 at 13:08
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@Don: It's well known with that meaning in set builder notation, of course ... but is not a usual choice of punctuation to use together with quantifiers. –  Henning Makholm Dec 16 '13 at 13:10
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This is probably the "different proof" the OP wrote about already having seen. –  Henning Makholm Dec 16 '13 at 13:11

Note that this statement would not be true if we were working exclusively with rational numbers, because a monotone and bounded sequence of rational numbers does not necessarily have a rational limit. So you will have to use a special property of the reals here. In which way that property was formulated in your class we do not know.

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The original post asks: My idea is: Assume $a_n$ is monotone and not converges and then show that it is not bounded. But: my problem is that I fail to prove it is not bounded. Please can you help me?

If you want to prove the statement, if a sequence is monotone and bounded then it converges, the logically equivalent contrapositive would be, if a sequence is divergent then either it is not monotone or it is not bounded. So, your idea would only get you halfway there. You would also need to prove that divergent bounded sequences cannot be monotone.

To the question asked though, you are seeking a direct proof of monotone and divergent $\implies$ unbounded. This will be a bit awkward as divergent and unbounded are defined in the negative. Without loss of generality assume the sequence $(a_n)$ is divergent and increasing. A divergent sequence in the reals is not a Cauchy sequence, which means there is a small distance $\epsilon>0$ such that there is an infinite subsequence $(b_n)$ whose members are at least $\epsilon$ apart from each other. Then, since $(a_n)$ and hence $(b_n)$ are increasing, $b_n \ge b_0 + n \epsilon$ grows without bound.

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