nested interval theorem

Proposition 1.

If

$$[a_1,b_1]\supseteq [a_2,b_2]\supseteq [a_3,b_3]\supseteq \mathrm{\dots }$$

is a sequence of nested closed intervals, then

${\displaystyle \bigcap _{n=1}^{\mathrm{\infty }}}[a_n,b_n]\ne \mathrm{\varnothing }.$

If also  $\underset{n\to \mathrm{\infty }}{lim}(b_n-a_n)=0$,  then the infiniteintersection consists of a unique real number.

Proof.

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

1. first of all, we have the inequality $a_n\le a_m$ for $n\le m$, which means that the sequence $a_1,a_2,\mathrm{\dots },a_n,\mathrm{\dots }$ is nondecreasing;

2. 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 $i,j$. This means that the sequence $a_1,a_2,\mathrm{\dots },a_n,\mathrm{\dots }$ is bounded from above by all $b_i$, where $i=1,2,\mathrm{\dots }$.

Therefore, the limit of the sequence $(a_i)$ exists, and is just the supremum, say $a$ (see proof here (http://planetmath.org/NondecreasingSequenceWithUpperBound)). Similarly the sequence $(b_i)$ is nonincreasing and bounded from below by all $a_i$, where $i=1,2,\mathrm{\dots }$, and hence has an infimum $b$.

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

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

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

Remark.  This result is called the nested interval theorem.It is a restatement of the finite intersection propertyfor the compact set  $[a_1,b_1]$.  The result may also be proven by elementary methods:namely, any number lying in between the supremum of all the $a_n$ and the infimum of all the $b_n$will be in all the nested intervals.