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.
Best 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.