Why is the condition that the intervals be closed necessary? Could someone give me an example of a sequence of nonempty, bounded, nested intervals whose intersection is empty? I can't think of one, so why does the theorem require it?

Here is the theorem:

If $I_1 \supset I_2 \supset I_3 \supset \dots$ is a sequence of nested, closed, bounded, nonempty intervals, then $\bigcap_{n=1}^{\infty}I_n$ is nonempty. If, in addition, the length of $I_n$ approaches zero, then $\bigcap_{n=1}^{\infty}I_n$ consists of a single point.

My apologies if this has been asked before but I couldn't find it.

  • 12
    $\begingroup$ The moral of this story is the familiar fact that strict inequalities might not hold in the limit: you can have, for example, a sequence with all its terms less than 2 but with limit equal to 2. This is the nature of the supremum. $\endgroup$ – preferred_anon Jul 22 '15 at 23:18
  • 2
    $\begingroup$ A counter example as shown by Michael is always nice, but it doesn't always give the the moral behind the reason that it doesn't hold. It's important to note Daniel's comment that, as a general principle, if you have a strict inequality, then in the limit the equality may no longer be strict but now weak. Daniel's comment is the answer you should remember! :) (Of course, Michael's is very helpful for this specific situation, and could be adapted to others!) $\endgroup$ – Sam T Jul 23 '15 at 8:03

Consider the family $A_n = (0, \frac{1}{n})$. We have $A_{n+1} \subset A_n$ for every $n \in \mathbb{N}$, and the length of $A_n$ approaches zero as $n$ approaches infinity, but $\bigcap_{n=1}^{\infty} A_n$ is empty. To see this, note that any element of the intersection would be greater than zero, yet less than $\frac{1}{n}$ for every $n \in \mathbb{N}$; by the Archimedian property of the real numbers, there is no such element.

  • 1
    $\begingroup$ Ah, you beat me! $\endgroup$ – preferred_anon Jul 22 '15 at 23:16
  • 1
    $\begingroup$ every $A_n = (0, \frac{1}{n})$ is of positive length, then is it possible for the intersection of all nonempty interval $A_n$ becomes empty? $\endgroup$ – iMath Nov 1 '16 at 5:05
  • $\begingroup$ @iMath: I don't understand your question. My answer shows that the intersection of the $A_n$ is empty. $\endgroup$ – Michael Albanese Nov 1 '16 at 19:44
  • $\begingroup$ while I think the intersection of all nonempty interval An couldn't be empty $\endgroup$ – iMath Nov 2 '16 at 8:24
  • $\begingroup$ @iMath: Why do you think that? $\endgroup$ – Michael Albanese Nov 2 '16 at 12:09

One way to look at it is to see why the proof of the NIT fails if you try open sets - the NIT proof shows that the infinite intersection becomes the singleton closed interval $[a,a] = \{a\}$. If you try it with open sets, you'll arrive at $(a,a) = \{\}$.

  • $\begingroup$ I was trying to make a coy reference to the proof of the nested interval theorem, which shows that the infinite intersection is $[a,a] = \{a\}$, whereas if you tried it with open sets you'd get to $(a,a) = \{\}$. I guess that's not clear, I'll edit my answer. $\endgroup$ – anonymous Aug 14 '16 at 19:52
  • 4
    $\begingroup$ This is still misleading though--if the intervals are open, that doesn't mean that the intersection becomes $(a,a)$ instead of $[a,a]$. In fact, the intersection might still be $[a,a]$, or might be $[a,a)$ or $(a,a]$, depending on whether $a$ actually is an endpoint of any of the intervals you're intersecting. $\endgroup$ – Eric Wofsey Aug 14 '16 at 19:56
  • 2
    $\begingroup$ You're absolutely right, for example consider $I_n = (-1/n, 1/n)$ with intersection $[0,0]$. Should I delete this answer or leave it as a record of this wrong line of thinking? $\endgroup$ – anonymous Aug 14 '16 at 20:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.