% This is lecture no. 2 of MH3100/: Real Analysis I.

\chapter{Nested Interval Property}%
\label{cha:nested_interval_property}

\section{Nested Interval Property}%
\label{sec:nested_interval_property}

Let us first examine an example of a nested interval.
\begin{exmp}
	Bisection Method.
	Given $f\vcentcolon[a,b]\mapsto\mathbb{R}$ with $f(a)< 0 <f(b)$, the bisection method can be used to find some $c\in[a,b]$ satisfying $f(c) = 0$.
	\begin{mdframed}
		\underline{The Bisection Method:}
		\begin{enumerate}
			\item Set the interval $I_1\vcentcolon = [a_1,b_1]\vcentcolon =[a,b]$.
			\item For $k = 1,2,3,\dotsc$, set $m_k = \frac{1}{2}(a_k +b_k)$ and set the interval
				\[
				I_{k+1}\vcentcolon = [a_{k+1}, b_{k+1}] \vcentcolon =
				\begin{cases}
					[m_k,b_k] & \text{if } f(m_k) < 0,\\
					[a_k,m_k] & \text{if } f(m_k) \geqslant 0.\\
				\end{cases}
				\]
		\end{enumerate}
	\end{mdframed}
	This gives us a nested sequence of closed intervals
	\[
	I_1 \supseteq I_2 \supseteq I_3 \supseteq I_4 \supseteq \dotsb
	.\]
\end{exmp}

\begin{thrm}
	Nested Interval Property.
	If $I_n = [a_n, b_n]$ is a closed and bounded interval for each $n\in\mathbb{N}$, for which $I_n\supseteq I_{n+1}$ for each $n\in\mathbb{N}$, then the resulting \textbf{nested sequence of closed and bounded intervals}:
	\[
	I_1 \supseteq I_2 \supseteq I_3 \supseteq I_4 \supseteq \dotsb
	.\]
	has a \textbf{nonempty intersection}, that is, there is some $c\in\mathbb{R}$ with $c\in I_n$, $\forall n\in\mathbb{N}$.
\end{thrm}

Note that the theorem is not true in general if the intervals are not closed:
\[
	(0,1] \supset(0,\frac{1}{2}] \supset(0,\frac{1}{3}] \supset(0,\frac{1}{4}] \supset\dotsb
.\]
The theorem is also not true in general if the intervals are not bounded:
\[
[1,\infty)\supset[2,\infty)\supset[3,\infty)\supset[4,\infty)\supset\dotsb
.\]
The Nested Interval Property does not hold for rational numbers.
For example, let $a_n$ denote the decimal expansion of $\sqrt{2}$ up to $n$ decimal places, and let $b_n$ denote $a_n + 10^{-n}$:
\[
\begin{aligned}
	a_1 & = 1.4,\\
	a_2 & = 1.41,\\
	a_3 & = 1.414,\\
	a_4 & = 1.4142,\\
	a_5 & = 1.41421,\\
	    & \nvdots
\end{aligned}
\qquad
\begin{aligned}
	b_1 & = 1.5,\\
	b_2 & = 1.42,\\
	b_3 & = 1.415,\\
	b_4 & = 1.4143,\\
	b_5 & = 1.41422,\\
	    & \nvdots
\end{aligned}
\]
Then the nested sequence of closed bounded intervals in $\mathbb{Q}$ 
\[
	I_1 \supseteq I_2 \supseteq I_3 \supseteq I_4 \supseteq \dotsb
,\]
where $I_n\vcentcolon = \left\{q\in\mathbb{Q}\vcentcolon a_n\leqslant q\leqslant b_n\right\}$ for each $n\in\mathbb{N}$, has an empty intersection since $\sqrt{2}\notin\mathbb{Q}$.

\begin{proof}
	Nested Interval Property.
	Let $A$ denote the set $\left\{a_n\vcentcolon n\in\mathbb{N}\right\}$.
	For each $n\in\mathbb{N}$ 
	\begin{align*}
		\forall m\in\mathbb{N}\quad (I_n\supseteq I_m) \text{ or } (I_m\supseteq I_n) & \implies \forall a_m\in A,\quad a_m\leqslant b_n\\
													& \iff b_n\ \text{is an upper bound for } A.
	\end{align*}
	Therefore the Axiom of Completeness asserts that $A$ has a supremum $c$.
	The definition of $c=\sup A$ implies that for each $n\in\mathbb{N}$ 
	\[
	a_n\in A\implies a_n\leqslant c
	,\]
	and
	\[
	b_n\ \text{is an upper bound for A}\ \implies c\leqslant b_n
	,\]
	hence $c\in I_n$.
\end{proof}

\section{Archimedean Property}%
\label{sec:archimedean_property}

The following examples highlight the meaning of the Archimedean Property.

\begin{exmp}
	Convergence to 0.
	The sequence $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},\dotsc$ converges to 0---a fact that can be shown via the definition of convergence:
	\[
	\forall\varepsilon > 0,\ \exists N\in\mathbb{N}\backepsilon\forall n\in\mathbb{N}\quad n>N\implies \frac{1}{n}<\varepsilon
	.\]
\end{exmp}

\begin{exmp}
	Divergence to $\infty$.
	The sequence $1,2,3,4,\dotsc$ diverges to $\infty$---a fact that can be shown via the definition of divergence to $\infty$:
	\[
	\forall M>0,\ \exists N\in\mathbb{N}\backepsilon\forall n\in\mathbb{N}\quad n>N\implies n>M
	.\]
\end{exmp}

\begin{thrm}
	Archimedean Property.
	Given any number $a\in\mathbb{R}$, there is some $n\in\mathbb{N}$ satisfying $n>a$.
	Equivalently, given any number $b>0$, there is some $n\in\mathbb{N}$ satisfying $\frac{1}{n}<b$.
\end{thrm}
\begin{proof}
	By contradiction.
	Suppose $a\in\mathbb{R}$ is an upper bound for $\mathbb{N}$.
	Then the Axiom of Completeness asserts that $\mathbb{N}$ has a supremum $s$.
	The definition of $s=\sup\mathbb{N}$ implies that $s-1$ is not an upper bound for $\mathbb{N}$.
	Therefore, there is some $n\in\mathbb{N}$ for which $n>s-1$.
	This means $s<n+1\in\mathbb{N}$, which contradicts the definition of $s=\sup\mathbb{N}$.
\end{proof}
\begin{thrm}
	Density of $\mathbb{Q}$ in $\mathbb{R}$.
	Given any two numbers $a,b\in\mathbb{R}$ for which $a<b$, there is some $q\in\mathbb{Q}$ satisfying $a<q<b$.
	That is, $\mathbb{Q}$ is \textbf{dense} in $\mathbb{R}$.
\end{thrm}
\begin{proof}
	For the case where $a>1$, the Archimedean Property asserts that there is some $n\in\mathbb{N}$ satisfying
	\[
	\frac{1}{n}<b-a
	.\]
	The Archimedean Property asserts that there is some $m\in\mathbb{N}$ satisfying
	\[
	m>na
	.\]
	Let $m$ denote the least natural nubmer satisfying the above inequality.
	Then $m-1\leqslant na$, which implies
	\[
	m\leqslant na + 1<na +n(b-a) = nb
	.\]
	Hence $q=\frac{m}{n}\in\mathbb{Q}$ satisfies $a<q<b$.
	For the case where $a\leqslant 1$, the Archimedean Property asserts that there is some $n\in\mathbb{N}$ satisfying
	\[
	1-a<n
	.\]
	Since $1<a+n<b+n$, the first case concludes that there is some $q^{\prime}\in\mathbb{Q}$ satisfying $a+n<q^{\prime}<b+n$.
	Then there is some $q=q^{\prime}-n\in\mathbb{Q}$ satisfying $a<q<b$.
\end{proof}
