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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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    @coffeemath I have mentioned the first condition.I guess we add an extra assumption to it,i.e,, the length of the intervals tend to zero. 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 ?2012-10-14

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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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    @danny: You’re very welcome.2012-10-24