# nested interval theorem

###### Proposition 1.

If

 $[a_{1},\,b_{1}]\;\supseteq\;[a_{2},\,b_{2}]\;\supseteq\;[a_{3},\,b_{3}]\;\supseteq\ldots$
 $\displaystyle\bigcap_{n=1}^{\infty}[a_{n},\,b_{n}]\;\neq\;\varnothing.$

If also  $\displaystyle\lim_{n\to\infty}(b_{n}\!-\!a_{n})=0$,  then the infinite intersection 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 $n\leq m$:

1. 1.

first of all, we have the inequality  $a_{n}\leq a_{m}$ for $n\leq m$, which means that the sequence $a_{1},a_{2},\ldots,a_{n},\ldots$ is nondecreasing;

2. 2.

in addition, we also have two inequalities: $a_{m}\leq b_{n}$ and $a_{n}\leq b_{m}$. In either case, we have that $a_{i}\leq b_{j}$ for all $i,j$. This means that the sequence $a_{1},a_{2},\ldots,a_{n},\ldots$ is bounded from above by all $b_{i}$, where $i=1,2,\ldots$.

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

Now, as the supremum of $(a_{i})$, $a\leq b_{i}$ for all $i$. But because $b$ is the infimum of $(b_{i})$, $a\leq b$. Therefore, the interval $[a,b]$ is non-empty (containing at least one of $a,b$). Since $a_{i}\leq a\leq b\leq b_{i}$, 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 $c, then there is some $a_{i}$, such that $c (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 $\displaystyle\lim_{n\to\infty}(b_{n}-a_{n})=0$, 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 $a=b$. 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 property for 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.

Title nested interval theorem NestedIntervalTheorem 2013-03-22 17:27:12 2013-03-22 17:27:12 pahio (2872) pahio (2872) 11 pahio (2872) Theorem msc 54C30 msc 26-00