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 $a_n \wedge b_n \in I_n$.

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    $\begingroup$ Er...so "trivially we have that $\,a_n\,,\,b_n\in I_n\,$...so?? You haven't proved anything at all yet. $\endgroup$ – DonAntonio Nov 10 '12 at 0:52
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    $\begingroup$ ‘For each $n$ there is a point in $I_n$’ is not the same as ‘there is a point that is in every $I_n$’. $\endgroup$ – Brian M. Scott Nov 10 '12 at 0:53
  • $\begingroup$ @BrianM.Scott and if I add the statement that this point is in $I_m$ for all $m \leq n$? $\endgroup$ – providence Nov 10 '12 at 0:59
  • $\begingroup$ 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 $0$. $\endgroup$ – Brian M. Scott Nov 10 '12 at 1:11
  • $\begingroup$ 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. $\endgroup$ – Brian M. Scott Nov 10 '12 at 1:25

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

  • $\begingroup$ 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? $\endgroup$ – providence Nov 10 '12 at 1:20
  • $\begingroup$ 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? $\endgroup$ – Benjamin Dickman Nov 10 '12 at 1:23
  • $\begingroup$ @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 $\forall a_i, b_j$, $\;i, j\in \mathbb{N}\;$, $a_i<b_j$? $\endgroup$ – Namaste Nov 10 '12 at 1:26
  • $\begingroup$ @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. $\endgroup$ – providence Nov 10 '12 at 1:29
  • $\begingroup$ 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... $\endgroup$ – Namaste Nov 10 '12 at 1:30

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?


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