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.

  • 15
    $\begingroup$ 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. $\endgroup$ Jul 22, 2015 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 OT
    Jul 23, 2015 at 8:03
  • 1
    $\begingroup$ On the other hand, if the intervals are of the form $(a_n, b_n)$ with $a_n < a_{n+1} < b_{n+1} < b_n$, then the intersection is always non-empty. Proof: Let $a$ be the limit of the $a_n$, and $b$ the limit of the $b_n$. We have that $[a,b] \subset (a_n, b_n)$ for all $n$, and $[a,b] \not = \emptyset$ as $a \leq b$. $\endgroup$
    – Areawoman
    Feb 2, 2020 at 0:32

2 Answers 2


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! $\endgroup$ Jul 22, 2015 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, 2016 at 5:05
  • $\begingroup$ @iMath: I don't understand your question. My answer shows that the intersection of the $A_n$ is empty. $\endgroup$ Nov 1, 2016 at 19:44
  • $\begingroup$ while I think the intersection of all nonempty interval An couldn't be empty $\endgroup$
    – iMath
    Nov 2, 2016 at 8:24
  • 1
    $\begingroup$ @Linda Thanks, you are right ! the intersection of a finite number of nested nonempty intervals is nonempty , while the intersection of an infinite number of nested nonempty intervals might be empty, like the above shown via Archimedian property of the real numbers, I couldn't believe the latter case could be true just by intuition long ago, now I've realized I was wrong , it is infinity that caused the unbelievable fact. $\endgroup$
    – iMath
    Oct 28, 2017 at 15:18

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, 2016 at 19:52
  • 6
    $\begingroup$ 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. $\endgroup$ Aug 14, 2016 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, 2016 at 20:10
  • $\begingroup$ @EricWofsey "The intersection might be be $[a,a)$ or $(a,a]$." What does $[a,a)$ and $(a,a]$ mean? I think they are a contradiction in terms. $ a \le x < a $ or $ a < x \le a $ are contradictory statements. It's a nonsense statement like $ 2 < x < 1$. No such real number exists. $\endgroup$
    – john
    Jul 1, 2022 at 4:29
  • $\begingroup$ @john: That just means that $[a,a)$ and $(a,a]$ are the empty set. $\endgroup$ Jul 1, 2022 at 4:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .