Nested_Interval_Property

For each $n \in \mathbb{N}$, assume we are given a closed interval $I_{n} = \left. \left\lbrack a_{n},b_{n} \right\rbrack \right. = \{ x \in \mathbb{R}:a_{n} \leq x \leq b_{n}\}$. Assume also that each $I_{n}$ contains $I_{n + 1}$. Then, the resulting nested sequence of closed intervals

$$ I_{1} \supseteq I_{2} \supseteq I_{3} \supseteq I_{4} \supseteq \cdots $$

has a non-empty interinterion. That is to say

$$ {\bigcap(n = 1)}^{\infty}I_{n} \neq \varnothing $$

Proof

In order to show that ${\bigcap(n = 1)}^{\infty}I_{n}$ is not empty, we are going to use the Axiom_of_Completeness (AoC) to produce a single real number $x$ satisfying $x \in I_{n}$ for every $n \in \mathbb{N}$. Now, the AoC is a statement about bounded sets, and the one we want to consider is the set

$$ A = \{ a_{n}:n \in \mathbb{N}\} $$

of left-hand endpoints of the intervals.

Because the intervals are nested, we see that every $b_{n}$ serves as an upper_bound for $A$. Thus, we are justified in setting

$$ x = \sup A $$

Now, consider a particular $I_{n} = \left. \left\lbrack a_{n},b_{n} \right\rbrack \right.$. Because $x$ is an upper_bound for $A$, we have $a_{n} \leq x$. The fact that each $b_{n}$ is an upper_bound for $A$ and that $x$ is the least upper bound implies $x \leq b_{n}$.

All together then, we have $a_{n} \leq x \leq b_{n}$, which means $x \in I_{n}$ for every choice of $n \in \mathbb{N}$. Hence, $x \in {\bigcap(n = 1)}^{\infty}I_{n}$, and the interinterion is not empty.