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The book from which I am learning analysis states cantor's completeness principle as follow. ;

"Consider a nest of closed intervals I1,I2,I3...In , each being denoted as [an,bn]. As n tends to infinity .. the common intersection of all the intervals is a point which is equal to the limit to which both an and bn converges to."

I can prove the converegence of the sequences an and bn using Weierstrass principle.

Now for the two sequences we can see that for any two given positive integers m and n a(m)>b(n)...........(1)

Now what I want to know is that does the cantor's principle imply that given any two sequences ,one of which is monotonically increasing and another is monotonically decreasing then the two sequences converge to the same limit ? Obviously this is wrong, and there is a gaping hole in my understanding. Can I know where my understanding misses the right track ? Secondly, how to put this principle into use, when we just know that there exist a common limit for both the sequences but don;t know its value?

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You need the diameter of the intervals to go to $0$ for all this work. Said another way, you need $\lim_{n\rightarrow\infty} b_n - a_n = 0$. –  Jason DeVito Oct 12 '12 at 17:49
    
You should give the entire statement of the principle from your book, rather than an edited version. –  Chris Eagle Oct 12 '12 at 17:54
    
@JasonDeVito Yes, The Weierstrass principle provides a reason for their convergence and the the convergence of the diameter of the intervals to zero, does the finishing move. But a wasp sitting on my understanding prevents me to see where is it wrong in saying that cantor's completeness principle can be applied to any two sequences ,with one of them monotonically increasing and another monotonically decreasing, thereby leading them to common limit. I hope you can provide a reply to clear this doubt . –  danny gotze Oct 12 '12 at 18:01
    
@ChrisEagle Well it means the same as the statement in book. –  danny gotze Oct 12 '12 at 18:05
    
Your two sequences, say x_n monotone increasing and y_n monotone decreasing, need two other things to get convergence. You seem to mention these but explicitly we need x_j <= y_k for all j and k (this is equivalent to the intervals [x_n,y_n] being "nested") and you need lim (y_n - x_n) = 0 (this being the assumption that lengths go to zero). I can't see from your comment that you see this or not, just trying to help... –  coffeemath Oct 12 '12 at 19:26

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The statement as you’ve quoted it is simply false; here is a counterexample. For $n\in\Bbb Z^+$ let $$I_n=\left[-\frac1n,1+\frac1n\right]\;;$$ then $I_1\supsetneqq I_2\supsetneqq I_3\supsetneqq\ldots$, but $\bigcap_{n\ge 1}I_n=[0,1]$, which contains more than one point, and the sequences $$\left\langle-\frac1n:n\in\Bbb Z^+\right\rangle\quad\text{and}\quad\left\langle 1+\frac1n:n\in\Bbb Z^+\right\rangle$$ don’t converge to a common limit.

In order to make it a true statement, you must add the further hypothesis that $\lim\limits_{n\to\infty}(b_n-a_n)=0$.

Without that extra hypothesis, you can only conclude that $\bigcap_{n\ge 1}I_n\ne\varnothing$. This isn’t hard: you know that

$$a_1\le a_2\le a_3\le\ldots\le b_3\le b_2\le b_1\;,$$

so the non-decreasing sequence $\langle a_n:n\in\Bbb Z^+\rangle$ is bounded above by $b_1$, and the non-increasing sequence $\langle b_n:n\in\Bbb Z^+\rangle$ is bounded below by $a_1$. Thus, the sequences converge, say to $a$ and $b$ respectively, and it’s easy to show (1) that $a\le b$, and (2) that $\bigcap_{n\in\Bbb Z^+}I_n=[a,b]\ne\varnothing$.

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M.scott Thanks a lot for your help. But when you don't know beforehand whether the sequences converge to the same limit , how can you add this assumption. Because adding this assumption is same as assuming the convergence of both the sequences to the same limit. Adding this assumption to the counter example you have put forth will also lead to the wrong conclusion that I have made in my question. –  danny gotze Oct 14 '12 at 5:16
    
And could you please explain the second part of my question ,i.e.how to put this principle into use, when we just know that there exist a common limit for both the sequences but don;t know its value? –  danny gotze Oct 14 '12 at 5:17
    
@danny: I’m not saying that you can arbitrarily add that assumption. I’m saying that the theorem that you stated is false because it doesn’t include that assumption. In order to make it a true theorem, you must add that assumption to its hypotheses. I don’t really understand the second part of your question: what do you mean by put this principle to use? –  Brian M. Scott Oct 14 '12 at 6:13
    
I guess including this assumption doesn't imply that the two sequences are convergent.right? So , in the end I got clarified that only when we add the assumption that the lengths are shrinking to zero we can say that the Cantor's completeness principle holds good. Now , the second part of the question is what exactly is the use of this theorem? I know that I can use it to prove the Bolzanno-Weierstrass theorem.But suppose a sequence is given, so in any situation can you prove that it converges to a limit L using cantor's principle. –  danny gotze Oct 14 '12 at 8:50
    
@danny: The two sequences of endpoints, $\langle a_n:n\in\Bbb Z^+\rangle$ and $\langle b_n:n\in\Bbb Z^+\rangle$, are convergent with or without the extra assumption; the only difference is that with it, you can conclude that \lim_na_n=\lim_nb_n$, and this is the only point in the intersection of the intervals, while without it you can only conclude that $\lim_na_n\le\lim_nb_n$, and that the intersection of the intervals is non-empty (but possibly larger than one point). Both versions are used as tools in proving other results. The version with the extra hypothesis, for instance, is used ... –  Brian M. Scott Oct 14 '12 at 9:05

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