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.

Q: Consider a sequence of closed intervals $I_1 = [a_1, b_1], I_2 = [a_2, b_2], \dots$. Suppose that $\forall n \in \mathbb{N} \left(a_n \leq a_{n + 1} \wedge b_{n + 1} \leq b_n \right)$. Prove that there is a point $x$ in every $I_n$.

Proof: This is equivalent to proving that $\forall n$ we have some $x$ such that $x \in I_n$. Trivially we have $a_n \wedge b_n \in I_n$.

share|improve this question
    
Er...so "trivially we have that $\,a_n\,,\,b_n\in I_n\,$...so?? You haven't proved anything at all yet. –  DonAntonio Nov 10 '12 at 0:52
2  
‘For each $n$ there is a point in $I_n$’ is not the same as ‘there is a point that is in every $I_n$’. –  Brian M. Scott Nov 10 '12 at 0:53
    
@BrianM.Scott and if I add the statement that this point is in $I_m$ for all $m \leq n$? –  providence Nov 10 '12 at 0:59
    
That doesn’t help: it need not be in $I_{n+1}$. What if $I_n=\left[-\frac1n,\frac1n\right]$? None of the endpoints is in all of the intervals; the only point in their intersection is $0$. –  Brian M. Scott Nov 10 '12 at 1:11
    
Yes, of course $a_n\in I_m$ if $m\le n$; that’s an immediate consequence of the nesting. But it doesn’t get really get you any closer to showing that the intersection of the intervals is non-empty. Look at the example that I just gave: not one of the endpoints is in the intersection. You’re looking in the wrong place. Hold on a bit, and I’ll write up a bit of hint. –  Brian M. Scott Nov 10 '12 at 1:25
add comment

2 Answers

up vote 2 down vote accepted

The Nested Interval Theorem is sometimes known as the Nested Interval Property.

Why could it be either a 'theorem' or a 'property'?

Answer: Because it is equivalent to several other properties that come up when you are trying to show the real numbers are "complete."

If you are going to consider it a theorem (as opposed to a property) then you should identify which other properties you have at your disposal in order to prove it.

Edit: Knowing you have the Least Upper Bound Property, here is a proof (of my own) that implies the Principle of Nested Closed Intervals: (replace $\mathbb{F}$ with $\mathbb{R}$)

enter image description here

share|improve this answer
    
I am not sure what properties I have at my disposal. Suppose I have none other than the standard properties of the reals (the field axioms essentially). How should I proceed? –  providence Nov 10 '12 at 1:20
    
The issue is that any axiomatic development of the reals will assume, at some point, something equivalent to the Nested Interval Property (or something that can prove it as a result). When you say "the standard properties" I'm not quite sure what you are referring to. Maybe you should list all the properties in your question? –  Benjamin Dickman Nov 10 '12 at 1:23
    
@providence Did you click on the link B.D. provided? Have you encountered the least upper bound property? Also, did you leave out the premise that $\forall a_i, b_j$, $\;i, j\in \mathbb{N}\;$, $a_i<b_j$? –  amWhy Nov 10 '12 at 1:26
    
@amWhy Yes, I have clicked. I have also encountered the l.u.b property. No I did not leave out that premise. That need not be true. –  providence Nov 10 '12 at 1:29
    
Are you sure? We've trying to prove the Nested Interval Theorem, after all. To prove the Theorem, we must assume that $\forall a_i, b_j\;$ $i, j \in \mathbb{N},\; $a_i < b_j$ or as in Brian's answer, we must assume nesting... –  amWhy Nov 10 '12 at 1:30
show 3 more comments

HINT: From the hypothesis that the intervals are nested you know that

$$a_1\le a_2\le a_3\le\ldots~\ldots\le b_3\le b_2\le b_1\;.$$

Thus, you have two monotonic sequences, $\langle a_k:k\in\Bbb Z^+\rangle$ and $\langle b_k:k\in\Bbb Z^+\rangle$. Moreover, both are bounded: $a_1$ and $b_1$ are lower and upper bounds for both sequences. What fundamental fact about bounded monotonic sequences do you know?

share|improve this answer
add comment

Your Answer

 
discard

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.