Let $A:=[a_n,b_n],n\in\mathbb{N}$, be a nested sequence of closed intervals, i.e. $a_{n+1}>a_n$ and $b_{n+1}<b_n$ for all $n\in\mathbb{N}$. Show that the intersection $\cap_{n\in\mathbb{N}}A_n\neq \emptyset$ is non-empty. Moreover, if $\lim (b_n-a_n)=0$, then $\cap_{n\in\mathbb{N}}A_n=\{x_0\}$ consists of a single point. Is such a statement generally true for a nested sequence of non-closed intervals?

I have absolutely no idea how to do this one. Never worked with nested intervals before.

  • $\begingroup$ This is called the Nested Interval Theorem. $\endgroup$ – T.J. Gaffney Feb 23 '14 at 4:53
  • $\begingroup$ so it really just comes down to a monotonic bounded sequence then $\endgroup$ – terrible at math Feb 23 '14 at 4:58
  • $\begingroup$ $\;\cup A_n\;$ is the union of the $\;A_n\;$ , not their intersection, which is $\;\cap A_n\;$ ... $\endgroup$ – DonAntonio Feb 23 '14 at 5:10
  • $\begingroup$ @terribleatmath, google Nested Interval Theorem, or Cantor theorem on nested intervals, etc. It uses the basic Bolzano-Weierstrass Theorem. $\endgroup$ – DonAntonio Feb 23 '14 at 5:11
  • $\begingroup$ @DonAntonio personal.bgsu.edu/~carother/cantor/Nested.html I found that site, and ... "Clearly, both a and b are elements of , because both are an elements of the closed interval for any n. (Why?) " No, that is not clear, Could you explain that to me? $\endgroup$ – terrible at math Feb 23 '14 at 10:16

(1) Let $\{a_n: n\in \mathbb{N}\}$ be the set of all the left-hand endpoints. Then the set is non-empty and because the intervals are nested each $b_n$ is an upper bound. Let $x= \sup\{a_n: n\in \mathbb{N}\}$. Then $a_n\le x\le b_n$ for all $n \in \mathbb{N}$ (why?). Hence $x\in \bigcap_n[a_n,b_n]$.

(2) If $(a_n-b_n) \rightarrow 0$ we need to show that the intersection just contain a single point. Suppose to the contrary that there exists some other $x'$ such that $x\not=x'$. Let $\varepsilon= |x-x'|/4$. Then there exists a $N\in \mathbb{N}$ such that $|a_n-b_n|\le \varepsilon$ for all $n\ge N$. Since $x,x'\in \bigcap_n[a_n,b_n]$, then $a_n\le x\le b_n$ and $a_n\le x'\le b_n$. Thus

\begin{align}|x-x'|\le |x-b_N|+|b_N-a_N|+|a_N-x'|\\ \le 3\varepsilon=3|x-x'|/4\end{align}

a contradiction.

  • $\begingroup$ Thank you. Looking at this now it is helping a lot. It is late an I will give it more of a throrough reading tomorrow. I have never seen nested intervals before so I am trying to get a basic understanding. The hardest part for me to understand is what the intersection operation actually does here. Could you explain exactly what a nested interval is and what the intersection does here? (Of course i know what intersection does in general, i'm just confused by this) $\endgroup$ – terrible at math Feb 23 '14 at 10:19
  • $\begingroup$ @terribleatmath: A nested interval is "collection" of intervals $I_n:= [a_n,b_n]$ such that $I_{n+1}\subset I_n$. The resulting nested sequence looks like $I_0 \supset I_1 \supset ... \supset I_n \supset I_{n+1}\supset...$ (try to draw a picture may help you). The intersection give us the set of point which are in all the intervals, i.e., the set of point $x$ such that $a_n\le x \le b_n$ for all $n\in \mathbb{N}$. $\endgroup$ – Jose Antonio Feb 23 '14 at 16:48
  • $\begingroup$ Looking at this again, the problem says $(b_n - a_n) \implies 0$ not $(a_n - b_n) \implies 0$ or does it not matter? $\endgroup$ – terrible at math Feb 23 '14 at 19:57
  • 1
    $\begingroup$ By the limit laws if $(b_n-a_n )\rightarrow 0$ then $(a_n-b_n)=-(b_n-a_n)\rightarrow-1\cdot0=0$, i.e., $(a_n-b_n)\rightarrow 0$. $\endgroup$ – Jose Antonio Feb 23 '14 at 20:24

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.