Proofs Regarding Nested Intervals

Recall from the The Nested Intervals Theorem page the definition of sequence of nested intervals $I_n = [a_n, b_n]$ where $n \in \mathbb{N}$ is such that $I_1 \supseteq I_2 \supseteq I_3 \supseteq ... \supseteq I_n \supseteq I_{n+1} \supseteq ...$. Also recall that if $A = \{ a_n : n \in \mathbb{N} \}$ and $B = \{ b_n : n \in \mathbb{N} \}$ where $\xi = \sup A$ and $\eta = \inf B$, and that $I_n = [a_n, b_n]$ is a sequence of closed bounded nested intervals then $[\xi, \eta] = \bigcap_{n=1}^{\infty} I_n$.

We will now look at some proofs involving nested intervals.

Example 1

Using the Archimedean Property, show that if $I_n = [0, \frac{1}{n}]$ then $\bigcap_{n=1}^{\infty} I_n = \{ 0 \}$.

It should be rather clear that $0 \in I_n = [0, \frac{1}{n} ]$ for all $n \in \mathbb{N}$ since $0 ≤ 0 ≤ \frac{1}{n}$ for all $n \in \mathbb{N}$. It should also be clear that any number less than 0 cannot be in the intersection of these intervals since if $x < 0$ then $x \not \in I_1 = [0, 1]$. Suppose that $x > 0$ such that $x \in I_n$ for all $n \in \mathbb{N}$. By one of the Archimedean Property Corollaries, since $x > 0$ then there exists an $n_x \in \mathbb{N}$ such that $0 < \frac{1}{n_x} < x$, so then $x \not \in [0, \frac{1}{n_x}]$ which is a contradiction to our assumption that $x \in I_n$ for all $n \in \mathbb{N}$, so we have that the only number in all of these intervals is $0$.

Example 2

Prove that if $I_n = [n, \infty)$ then $\bigcap_{n=1}^{\infty} I_n = \emptyset$.

We first note that $I_1 = [1, \infty)$, $I_2 = [2, \infty)$, … is a sequence of nested intervals.

Assume that $x \in \bigcap_{n=1}^{\infty} I_n$, in other words, this set is nonempty. Then for all $n \in \mathbb{N}$ we have that $n ≤ x < \infty$, which implies that $x$ is an upper bound to the set of natural numbers. We know by one of the Archimedian properties that for every positive number $x \in \mathbb{R}$ there exists a natural number $n_x \in \mathbb{N}$ such that $n_x - 1 ≤ x ≤ n_x$. This implies that $x \not \in I_{n_x} = [n_x, \infty)$ though, which is a contradiction to the assumption that $x \in \bigcap_{n=1}^{\infty} I_n$. Therefore $\bigcap_{n=1}^{\infty} I_n = \emptyset$.

Example 3

Prove that if $I_n = [2, 3 + \frac{1}{n} ]$ then $\bigcap_{n=1}^{\infty} I_n = [2, 3]$.

We note that $I_1 = [2, 4]$, $I_2 = [2, 3.5]$, … is a sequence of closed bounded nested intervals. Let $A := \{ 2 \}$ and let $B := \{ 3 + \frac{1}{n} : n \in \mathbb{N} \}$, and let $\xi = \sup A$ and $\eta = \inf B$. By the nested intervals theorem we have that $[\xi, \eta] = \bigcap_{n=1}^{\infty} I_n$. To prove that $[2, 3] = \bigcap_{n=1}^{\infty}$ we need to show that $\xi = 2$ and that $\eta = 3$.

First we will show that $\sup A = \xi = 2$. The set $A = \{ 2 \}$ contains only one element and so $\sup A = \xi = 2$

We will now show that $\inf B = \eta = 3$. We must first show that $3$ is a lower bound to $B$. We note that $0 < \frac{1}{n}$ for all $n \in \mathbb{N}$. Adding $3$ to both sides of this inequality and we get that $3 < 3 + \frac{1}{n}$ for all $n \in \mathbb{N}$. Therefore $3$ is a lower bound to the set $B$. We now want to show that $3$ is the greatest lower bound. Suppose that $3 < w$. Then there exists a $b \in B$ such that $b < w$. Our first case is when $4 < w$. Then choose $b = 3 + \frac{1}{n}$ for any $n \in \mathbb{N}$ and so $b < w$. Our second case is when $3 < w ≤ 4$. We want to show that $b = 3 + \frac{1}{n}$ for some $n \in \mathbb{N}$ exists such that $b < w$. We thus want to show that $3 < b = 3 + \frac{1}{n} < w ≤ 4$ which is equivalent to showing that $0 < \frac{1}{n} < w - 3 ≤ 1$. We know that since $w - 3 > 0$ (since $3 < w ≤ 4$) then such a natural number $n \in \mathbb{N}$ exists by one of the Archimedian properties and so such a $b \in B$ exists such that $b < w$. Therefore $\inf B = \eta = 3$.

By the nested intervals theorem $[2, 3] = \bigcap_{n=1}^{\infty} I_n$.

Example 4

Prove that if $I_n = \left (1, 1 + \frac{1}{n} \right)$ then $\bigcap_{n=1}^{\infty} I_n = \emptyset$.

We note that this sequence of intervals is nested since $I_1 = (1, 2)$, $I_2 = \left (1, \frac{1}{2} \right)$, …

Now assume that $x \in \bigcap_{n=1}^{\infty} I_n$, in other words, that the intersection is nonempty. Then it follows that $1 < x < 1 + \frac{1}{n}$ for all $n \in \mathbb{N}$. We note that this is equivalent to $0 < x - 1 < \frac{1}{n}$ for all $n \in \mathbb{N}$. We also note that $x - 1 > 0$ and so by one of the Archimedean properties that there exists a natural number $n_x \in \mathbb{N}$ such that $0 < \frac{1}{n_x} < x - 1$ which implies that $1 < 1 + \frac{1}{n_x} < x$, which is a contradiction since then $x \not \in \left (1, \frac{1}{n_x} \right)$. So our assumption that $x \in \bigcap_{n=1}^{\infty} I_n$ was wrong and so $\bigcap_{n=1}^{\infty} I_n = \emptyset$