# Nested interval theorem

pyfgcr
The Nested interval theorem: If An = [an, bn] is a sequence of closed intervals such that An+1 $\subseteq$ An for all n $\in$ N, then $_{n \in n}\bigcap$A = ∅.
I think of the case where a1=a2=...=an and b1=b2=...=bn for all n, hence every set A(n+1) will be the "subset" of A(n) and the intersection is the original closed interval. So I think the theorem in my textbook have some problem. Any correction for this ?

Staff Emeritus
Homework Helper
The Nested interval theorem: If An = [an, bn] is a sequence of closed intervals such that An+1 $\subseteq$ An for all n $\in$ N, then $_{n \in n}\bigcap$A = ∅.
I think of the case where a1=a2=...=an and b1=b2=...=bn for all n, hence every set A(n+1) will be the "subset" of A(n) and the intersection is the original closed interval. So I think the theorem in my textbook have some problem. Any correction for this ?

It should be:
If $A_n=[a_n,b_n]$ is a sequence of closed intervals such that $A_{n+1}\subseteq A_n$ for all $n\in\mathbb{N}$, then $\bigcap_{n\in \mathbb{N}}A_n \neq \emptyset$.

DonAntonio
It should be:
If $A_n=[a_n,b_n]$ is a sequence of closed intervals such that $A_{n+1}\subseteq A_n$ for all $n\in\mathbb{N}$, then $\bigcap_{n\in \mathbb{N}}A_n \neq \emptyset$.

...and not only that: it must be also that $\,b_n-a_n\xrightarrow[n\to\infty]{} 0\,$ , as $\,A_n:=[n,\infty)\,$ would contradict.

DonAntonio