PlanetMath: nested interval theorem

Proof. There are two consequences to nesting of intervals: $[a_m,\,b_m]\subseteq[a_n,\,b_n]$ for $n\le m$ :

first of all, we have the inequality $a_n\le a_m$ for $n\le m$ , which means that the sequence $a_1, a_2, \ldots, a_n, \ldots$ is nondecreasing;
in addition, we also have two inequalities: $a_m\le b_n$ and $a_n\le b_m$ . In either case, we have that $a_i\le b_j$ for all . This means that the sequence $a_1, a_2, \ldots, a_n, \ldots$ is bounded from above by all , where $i=1,2,\ldots$ .

Therefore, the limit of the sequence

exists, and is just the supremum, say

(see proof here). Similarly the sequence

is nonincreasing and bounded from below by all

, where $i=1,2,\ldots$ , and hence has an infimum

Now, as the supremum of , $a\le b_i$ for all . But because is the infimum of , $a\le b$ . Therefore, the interval is non-empty (containing at least one of ). Since $a_i\le a\le b\le b_i$ , every interval contains the interval , so their intersection also contains , hence is non-empty.

If is a point outside of , say , then there is some , such that (by the definition of the supremum ), and hence $c\notin [a_i,b_i]$ . This shows that the intersection actually coincides with .

Now, since $\displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$ , we have that $b-a=\displaystyle\lim_{n\to\infty}b_n - \displaystyle\lim_{n\to\infty} a_n = \displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$ . So . This means that the intersection of the nested intervals contains a single point . $\qedsymbol$

Remark. This result is called the nested interval theorem. It is a restatement of the finite intersection property for the compact set

. The result may also be proven by elementary methods: namely, any number lying in between the supremum of all the

and the infimum of all the

will be in all the nested intervals.