# Nested sequence of closed intervals

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.

• This is called the Nested Interval Theorem. – T.J. Gaffney Feb 23 '14 at 4:53
• so it really just comes down to a monotonic bounded sequence then – terrible at math Feb 23 '14 at 4:58
• $\;\cup A_n\;$ is the union of the $\;A_n\;$ , not their intersection, which is $\;\cap A_n\;$ ... – DonAntonio Feb 23 '14 at 5:10
• @terribleatmath, google Nested Interval Theorem, or Cantor theorem on nested intervals, etc. It uses the basic Bolzano-Weierstrass Theorem. – DonAntonio Feb 23 '14 at 5:11
• @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? – terrible at math Feb 23 '14 at 10:16

## 1 Answer

(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.

• 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) – terrible at math Feb 23 '14 at 10:19
• @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 $n\in \mathbb{N}$. – Jose Antonio Feb 23 '14 at 16:48
• 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? – terrible at math Feb 23 '14 at 19:57
• 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., $(a_n-b_n)\rightarrow 0$. – Jose Antonio Feb 23 '14 at 20:24