Nested Intervals Property

(See MATH 409 Lecture 7#Nested Intervals Property→)

Theorem. If $\left\{I_{n}\right\}$ is a nested sequence of nonempty closed bounded intervals, then the intersection $\bigcap _{n\in \mathbb {N} }I_{n}$ is nonempty. Moreover, if lengths $\left|I_{n}\right|$ of the intervals satisfy $\left|I_{n}\right|\to 0$ as $n\to \infty$, then the intersetion consists of a single point.

Proof. Let $I_{n}=\left[a_{n},b_{n}\right]$ for $n\in \mathbb {N}$. Since the sequence $\left\{I_{n}\right\}$ is nested, it follows that the sequence $\left\{a_{n}\right\}$ is increasing while the sequence $\left\{b_{n}\right\}$ is decreasing. Besides, both sequences are bonded (since both are contained in the bounded interval $I_{1}$). Hence both are convergent: $a_{n}\to a$ and $b_{n}\to b$ as $n\to \infty$. Since $a_{n}\leq b_{n}$ for all $n\in \mathbb {N}$, the Comparison Theorem implies that $a\leq b$.

We claim that $\bigcap _{n\in \mathbb {N} }I_{n}=\left[a,b\right]$. Indeed, we have $a_{n}\leq a$ for all $n\in \mathbb {N}$ (by Comparison Theorem applied to $a_{1},a_{2},\ldots$ and constant sequence $a_{n},a_{n},\ldots$). Similarly, $b\leq b_{n}$ for all $n\in \mathbb {N}$. Therefore $\left[a,b\right]$ is contained in the intersection.

On the other hand, if $x<a$, then $x<a_{n}$ for some $n$ so that $x\not \in I_{n}$. Similarly, if $x>b$ then $x>b_{m}$ for some $m$ so that $x\not \in I_{m}$. This proves the claim.

Clearly, the length of $\left[a,b\right]$ cannot exceed $\left|I_{n}\right|$ for any $n\in \mathbb {N}$. Therefore $\left|I_{n}\right|\to 0$ as $n\to \infty$ implies that $\left[a,b\right]$ is a degenerate interval: $a=b$.

quod erat demonstrandum