The Nested Interval Theorem in Higher Dimensions

Nested Intervals

Definition: A sequence of intervals $I_{n}$ where $n\in \mathbb {N}$ is said to be nestedif $I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq ...\supseteq I_{n}\supseteq I_{n+1}\supseteq ...$.

For example, consider the interval $I_{n}=[0,{\frac {1}{n}}]$ We note that $I_{1}=[0,1]$, $I_{2}=[0,{\frac {1}{2}}]$, $I_{3}=[0,{\frac {1}{3}}]$, … As we can see $I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq ...\supseteq I_{n}\supseteq I_{n+1}\supseteq ...$ and so the sequence of intervals $I_{n}$ is nested. The diagram above illustrates this specific nesting of intervals.

Sometimes a nested interval will have a common point. In this specific example, the common point is $0$ since $0\leq 0<{\frac {1}{n}}$ for all $n\in \mathbb {N}$. We denote the set theoretic intersection of all these intervals to be the set of common points in a nested set of intervals:

${\begin{aligned}\bigcap _{n=1}^{\infty }I_{n}=C\end{aligned}}$

Sometimes a set of nested intervals does not have a common point though. For example consider the set of intervals $I_{n}=(n,\infty )$. Clearly $I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq ...\supseteq I_{n}\supseteq I_{n+1}\supseteq ...$ since $I_{1}=(1,\infty )$, $I_{2}=(2,\infty )$, … However, there is no common point for these intervals.

Example 1

Determine the set $C$ of points to which the set of nested intervals $I_{n}=(1-{\frac {1}{n}},2+{\frac {1}{n}})$ have in common.

We first note that $I_{1}=[0,3]$, $I_{2}=[0.5,2.5]$, $I_{3}=[0.66...,2.33...]$, … We conjecture that the set of points $C=\bigcap _{n=1}^{\infty }I_{n}=[1,2]$ are contained within all the intervals. This can be informally deduced since $\lim _{n\to \infty }1-{\frac {1}{n}}=1$ and $\lim _{n\to \infty }2+{\frac {1}{n}}=2$.

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 $J=[c,d]$. Then $I\subseteq J$ if and only if $c\leq a<b\leq d$.

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 $a_{n}\leq b_{1}$.

• 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 $a_{n}\leq \xi$ for all $n\in \mathbb {N}$.

• We now want to show that $\xi \leq b_{n}$ for all $n\in \mathbb {N}$. We will do this by showing that $a_{n}\leq b_{k}$ for all $n,k\in \mathbb {N}$.

• First consider the case where $n\leq k$. We thus have that $I_{n}\supseteq I_{k}$ by the definition of nested intervals and so by lemma 1 we get that $a_{n}\leq a_{k}\leq b_{k}\leq b_{n}$. Here we see that $a_{n}\leq b_{k}$.

• 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 $a_{k}\leq a_{n}\leq b_{n}\leq b_{k}$. Once again we have that $a_{n}\leq b_{k}$

• So then $a_{n}\leq b_{k}$ for all $n,k\in \mathbb {N}$, and so $b_{k}$ is an upper bound to the set $A$ and so $\sup A=\xi \leq b_{k}$ for all $k\in \mathbb {N}$. Furthermore we have that $a_{k}\leq \xi$ 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 $a_{k}\leq b_{n}$ for all $n,k\in \mathbb {N}$.

• First consider the case where $n\leq k$. We thus have that $I_{n}\supseteq I_{k}$ by the definition of nested intervals and so by lemma 1 we get that $a_{n}\leq a_{k}\leq b_{k}\leq b_{n}$. Here we see that $a_{k}\leq b_{n}$.

• 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 $a_{k}\leq a_{n}\leq b_{n}\leq b_{k}$. Once again we have that $a_{k}\leq b_{n}$

• So then $a_{k}\leq b_{n}$ 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 $[\xi ,\eta ]=\bigcap _{n=1}^{\infty }I_{n}$. $\blacksquare$

Licensing

Content obtained and/or adapted from:

• Nested Intervals, mathonline.wikidot.com under a CC BY-SA license
• The Nested Intervals Theorem, mathonline.wikidot.com under a CC BY-SA license