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From: José Carlos Santos on 7 Oct 2008 02:27
On 07-10-2008 6:43, TheGist wrote:

>>>> Let (a_n,b_n) be a sequence of nested open intervals. Assume
>>>> that intersection of all (a_n, b_n) is empty.
>>>> Let A=Sup a_n=Inf b_n
>>>> Show that there exists an n_0 such that for all n > n_0 either a_n=A or
>>>> b_n=A.
>>>> Proof:
>>>> If r,s >= n_0 the x_r, x_s both belong to
>>>> (a_n0, b_n0) . therefore |x_r - x_s| <= b_n0 - a_n0
>>>> since lim(b_n - a_n)=0 . Furthermore, since b_n - a_n =0
>>>> it follows that x_r is Cauchy and thus converges so x_r -> x.
>>>> We know that A=Sup a_n=Inf b_n thus the subsequence x_r in (a_n0, b_n0)
>>>> converges to the single value A for all n > n_0 and thus so do a_n and
>>>> b_n . That is, either a_n=A or b_n=A. Q.E.D
>>>> Problems? Am I close or is it all wrong?
>>> I am afraid I am very stuck on this one.
>>> I need to start at square one again.
>>> Can anyone provide me with a hint to get started in the
>>> right direction?- Hide quoted text -
>>>
>>> - Show quoted text -
>>
>> No problem. You know that a_1 <= a_2 <= ... and b_1 >- b_2 >= ... .
>> You also know (previous post) that sup a_n = sup b_n. Now if one
>> either of these sequences becomes stationary, say a_56 is the final
>> a, then there is nothing in the interesection because the b's march
>> all the way down to a_56, sweeping past every number in their path on
>> the way to their goal of a_56. On the other hand, if neither the a's
>> nor the b's becomes stationary, can this happen, or must there be an
>> intersection?
>
> So, let me try and formalize this...
> Would you advise a proof by contradiction?
>
> Suppose no such n_0 exists. Then neither a_n and b_n reach a limit.
> Then for a_n increasing and b_n decreasing eventually the intersection
> of all (a_n, b_n) is not null

What do you mean by "eventually"? The intersection of all intervals
either is empty (not "null") or it is not.

> which is a contraditcion.

Yes, but why is it not empty?

Best regards,

Jose Carlos Santos
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