The Nested Intervals Theorem

# The Nested Intervals Theorem

We have just looked at what exactly a Nested Interval is, and we are about to look at a critically important theorem in Real Analysis. Before we look at the Nested Intervals Theorem let's first look at the following important lemma that will be used to prove the Nested Intervals Theorem.

 Lemma 1: Let $a ≤ b$ and let $c ≤ d$ and let $I = [a, b]$ and $J = [c, d]$ be intervals (which are nonempty). Then $I \subseteq J$ if and only if $c ≤ a ≤ b ≤ d$.

Now let's look at the Nested Intervals theorem.

 Theorem 1: If the interval $I_n = [a_n, b_n]$ for $n \in \mathbb{N}$ is a sequence of closed bounded nested intervals then there exists a real number $\xi = \sup \{ a_n : n \in \mathbb{N} \}$ such that $\xi \in \bigcap_{n=1}^{\infty} I_n$.
• Proof of Theorem: We note that by the definition of nested intervals that $I_n \subseteq I_1$ for all $n \in \mathbb{N}$ so then $a_n ≤ b_1$.
• Now consider the nonempty set $A = \{ a_n : n \in \mathbb{N} \}$ that is bounded above by $b_1$. Thus this set has a supremum in the real numbers and denote it $\sup A = \xi$ so that $a_n ≤ \xi$ for all $n \in \mathbb{N}$.
• We now want to show that $\xi ≤ b_n$ for all $n \in \mathbb{N}$. We will do this by showing that $a_n ≤ b_k$ for all $n, k \in \mathbb{N}$.
• First consider the case where $n ≤ k$. We thus have that $I_n \supseteq I_k$ by the definition of nested intervals and so by lemma 1 we get that $a_n ≤ a_k ≤ b_k ≤ b_n$. Here we see that $a_n ≤ b_k$.
• Now consider the case where $n > k$. We thus have that $I_k \supseteq I_n$ by the definition of nested intervals and so by lemma 1 once again we have that $a_k ≤ a_n ≤ b_n ≤ b_k$. Once again we have that $a_n ≤ b_k$
• So then $a_n ≤ b_k$ for all $n, k \in \mathbb{N}$, and so $b_k$ is an upper bound to the set $A$ and so $\sup A = \xi ≤ b_k$ for all $k \in \mathbb{N}$. Furthermore we have that $a_k ≤ \xi$ for all $k \in \mathbb{N}$, and so $\xi \in I_k$ for every $k \in \mathbb{N}$ and thus $\xi \in \bigcap_{n=1}^{\infty} I_n$ and so the set theoretic union is nonempty. $\blacksquare$
 Theorem 2: If the interval $I_n = [a_n, b_n]$ for $n \in \mathbb{N}$ is a sequence of closed bounded nested intervals then there exists a real number $\eta = \inf \{ b_n : n \in \mathbb{N} \}$ such that $\eta \in \bigcap_{n=1}^{\infty} I_n$.
• Proof: We note that by the definition of nested intervals that $I_n \subseteq I_1$ for all $n \in \mathbb{N}$ so then $a_1 \leq b_n$.
• Now consider the nonempty set $B = \{ b_n : n \in \mathbb{N} \}$ that is bounded below by $a_1$. Thus this set has an infimum in the real numbers and denote it $\inf B = \eta$ so that $\eta \leq b_n$ for all $n \in \mathbb{N}$.
• We now want to show that $a_n \leq \eta$ for all $n \in \mathbb{N}$. We will do this by showing that $a_k ≤ b_n$ for all $n, k \in \mathbb{N}$.
• First consider the case where $n ≤ k$. We thus have that $I_n \supseteq I_k$ by the definition of nested intervals and so by lemma 1 we get that $a_n ≤ a_k ≤ b_k ≤ b_n$. Here we see that $a_k ≤ b_n$.
• Now consider the case where $n > k$. We thus have that $I_k \supseteq I_n$ by the definition of nested intervals and so by lemma 1 once again we have that $a_k ≤ a_n ≤ b_n ≤ b_k$. Once again we have that $a_k ≤ b_n$
• So then $a_k ≤ b_n$ for all $n, k \in \mathbb{N}$, and so $a_k$ is an upper bound to the set $A$ and so $a_k \leq \eta = \inf B$ for all $k \in \mathbb{N}$. Furthermore we have that $\eta \leq b_k$ for all $k \in \mathbb{N}$, and so $\eta \in I_k$ for every $k \in \mathbb{N}$ and thus $\eta \in \bigcap_{n=1}^{\infty} I_n$. $\blacksquare$
 Theorem 3: Let $A := \{ a_n : n \in \mathbb{N} \}$ and $B := \{ b_n : n \in \mathbb{N} \}$. If $\sup A = \xi$ and $\inf B = \eta$ then if the interval $I_n = [a_n, b_n]$ is a sequence of closed bounded nested intervals then $[\xi, \eta] = \bigcap_{n=1}^{\infty} I_n$.
• Proof: We first prove that $[\xi, \eta] \subseteq \bigcap_{n=1}^{\infty} I_n$. Now let $x \in [\xi, \eta] = \{ x \in \mathbb{R} : \xi \leq x \leq \eta$. Therefore $\xi \leq x \leq \eta$. But we know that $a_n \leq \xi$ for all $n \in \mathbb{N}$ and we know that $\eta \leq b_n$ for all $n \in \mathbb{N}$ and so $a_n \leq \xi \leq x \leq \eta \leq b_n$ for all $n \in \mathbb{N}$. Therefore $x \in [a_n, b_n]$ for all $n \in \mathbb{N}$ or in other words $x \in \bigcap_{n=1}^{\infty} I_n$. Therefore $[\xi, \eta] \subseteq \bigcap_{n=1}^{\infty} I_n$.
• We will now prove that $\bigcap_{n=1}^{\infty} I_n \subseteq [\xi, \eta]$. Let $x \in \bigcap_{n=1}^{\infty} I_n$. Then $a_n \leq x \leq b_n$ for all $n \in \mathbb{N}$. We also know that $a_n \leq \xi \leq \eta \leq b_n$ for all $n \in \mathbb{N}$. Suppose that $x$ is such that $a_n \leq x \leq \xi$. Since $\xi$ is the supremum of the set $A$ then there exists an element $a_x \in \{ a_n : n \in \mathbb{N} \}$ such that $x < a_x$ and so $x \not \in [a_x, b_x]$ and therefore $x \not \in \bigcap_{n=1}^{\infty} I_n$, a contradiction. Now suppose that $x$ is such that $\eta \leq x \leq b_n$ for all $n \in \mathbb{N}$. Since $\eta$ is the infimum of the set $B$ then there exists an element $b_x \in \{ b_n : n \in \mathbb{N} \}$ such that $b_x < x$ and so $x \not \in [a_x, b_x]$ and so $x \not \in \bigcap_{n=1}^{\infty} I_n$, once again, a contradiction. We must therefore have that $a_n \leq \xi \leq x \leq \eta \leq b_n$ and so $x \in [\xi, \eta]$ and so $\bigcap_{n=1}^{\infty} I_n \subseteq [\xi, \eta]$.
• Since $[\xi, \eta] \subseteq \bigcap_{n=1}^{\infty} I_n$ and $\bigcap_{n=1}^{\infty} I_n \subseteq [\xi, \eta]$ we have that $[\xi, \eta] = \bigcap_{n=1}^{\infty} I_n$. $\blacksquare$