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 $\bigcap _{n\in \mathbb{\mathbb{N}}}{I}_n$ is nonempty. Moreover, if lengths $\left|I_n\right|$ of the intervals satisfy $\left|I_n\right|\to 0$ as $n\to \mathrm{\infty }$, then the intersetion consists of a single point.

Proof. Let $I_n=[a_n,b_n]$ for $n\in \mathbb{\mathbb{N}}$. 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 $n\to \mathrm{\infty }$. Since $a_n\le b_n$ for all $n\in \mathbb{\mathbb{N}}$, the Comparison Theorem implies that $a\le b$.

We claim that $\bigcap _{n\in \mathbb{\mathbb{N}}}{I}_n=[a,b]$. Indeed, we have $a_n\le a$ for all $n\in \mathbb{\mathbb{N}}$ (by Comparison Theorem applied to $a_1,a_2,\dots$ and constant sequence $a_n,a_n,\dots$). Similarly, $b\le b_n$ for all $n\in \mathbb{\mathbb{N}}$. Therefore $[a,b]$ is contained in the intersection.

On the other hand, if $x<a$, then $x<a_n$ for some $n$ so that $x\in ̸I_n$. Similarly, if $x>b$ then $x>b_m$ for some $m$ so that $x\in ̸I_m$. This proves the claim.

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

quod erat demonstrandum