% This is lecture no. 6 of MH3100/: Real Analysis I.

\chapter{Cauchy Criterion}%
\label{cha:cauchy_criterion}

\section{Cauchy Sequences}%
\label{sec:cauchy_sequences}

\begin{definition}
	Cauchy Sequence.
	A sequence $\left\{a_n\right\}$ is said to be \textbf{Cauchy} if
	\[
	\forall\varepsilon>0,\ \exists N\in\mathbb{N}\suchthat n>N \implies \abs{a_n-a_N}<\varepsilon
	.\]
	Equivalently, a sequence $\left\{a_n\right\}$ is \textbf{Cauchy} if
	\[
	\forall\varepsilon>0,\ \exists N\in\mathbb{N}\suchthat\forall n,m\in\mathbb{N}\quad n,m>N\implies \abs{a_n-a_m}<\varepsilon
	.\]
\end{definition}

\section{Cauchy Criterion}%
\label{sec:cauchy_criterion}

\begin{thrm}
	Cauchy Criterion.
	A sequence $\left\{a_n\right\}$ is \textbf{convergent} if and only if it is \textbf{Cauchy}.
\end{thrm}

\begin{proof}
	To prove that a convergent sequence is Cauchy, let $a$ denote the limit of $\left\{a_n\right\}$.
	The definition of $a=\lim a_n$ implies that $\forall\varepsilon>0$, $\exists N\in\mathbb{N}\suchthat\forall n\in\mathbb{N}$ 
	\[
	n>N\implies\abs{a_n-a}<\frac{\varepsilon}{2}
	.\]
	Therefore, $\forall\varepsilon>0$, $\exists N^{\prime}=N+1\in\mathbb{N}\suchthat\forall n\in\mathbb{N}$ 
	\begin{align*}
		n>N^{\prime}>N\implies \abs{a_n-a_{N^{\prime}}} & \leqslant \abs{a_n-a} + \abs{a-a_{N^{\prime}}}\\
								& < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}\\
								& = \varepsilon.
	\end{align*}
	This is the definition of $\left\{a_n\right\}$ being Cauchy.
	To prove that a Cauchy Sequence is convergent, let $\left\{\varepsilon_n\right\}$ denote a sequence of positive number that converges to 0.
	The definition of $\left\{a_n\right\}$ being Cauchy implies that for each $m\in\mathbb{N}$, $\exists N_m\in\mathbb{N}\suchthat$ 
	\begin{align*}
		\forall n\in\mathbb{N}\quad n>N_m & \implies \abs{a_n-a_{N_m}}<\varepsilon_m\\
						  & \implies a_{N_m}-\varepsilon_m<a_n<a_{N_m}+\varepsilon_m.
	\end{align*}
	This defines a sequence $\left\{N_m\right\}$ of natural numbers.
	For each $m\in\mathbb{N}$, define
	\begin{align*}
		c_m & \defas \max\left\{a_{N_1}-\varepsilon_1,a_{N_2}-\varepsilon_2,\dotsc,a_{N_m}-\varepsilon_m\right\}\\
		\intertext{and}
		d_m & \defas \min\left\{a_{N_1}-\varepsilon_1,a_{N_2}-\varepsilon_2,\dotsc,a_{N_m}-\varepsilon_m\right\}
	\end{align*}
	Then the sequences $\left\{c_m\right\}$ and $\left\{d_m\right\}$ satisfy the conditions $c_1\leqslant c_2\leqslant c_3\leqslant \dotsb$ and $d_1\geqslant d_2\geqslant d_3\geqslant \dotsb$.
	The definition of $\left\{N_m\right\}$ implies that for each $m\in\mathbb{N}$, we have that $\forall k\in\left\{1,2,\dotsc,m\right\}$:
	\[
	\max\left\{N_1,N_2,\dotsc,N_m\right\}\geqslant N_k \implies a_{N_k}-\varepsilon_k < a_{\max\left\{N_1,N_2,\dotsc,N_m\right\}}<a_{N_k} + \varepsilon_k
	.\]
	Therefore, the definitions of $\left\{c_m\right\}$ and $\left\{d_m\right\}$ imply that for each $m\in\mathbb{N}$
	\[
	c_m <a_{N_k}-\varepsilon_k < a_{\max\left\{N_1,N_2,\dotsc,N_m\right\}}<a_{N_k} + \varepsilon_k<d_m
	.\]
	Therefore, the sequences $\left\{c_m\right\}$ and $\left\{d_m\right\}$ further satisfy
	\[
	c_1\leqslant c_2\leqslant c_3 \leqslant\dotsb\leqslant d_3\leqslant d_2\leqslant d_1
	.\]
	This allows us to define a nested sequence of closed and bounded intervals $I_1\supseteq I_2\supseteq I_3\supseteq\dotsb$, where $I_m\defas[c_m,d_m]$ for each $m\in\mathbb{N}$.
	The Nested Interval Property asserts that there is a number $a\in\mathbb{R}$ satisfying
	\[
	c_1\leqslant c_2\leqslant c_3 \leqslant\dotsb\leqslant a\leqslant\dotsb\leqslant d_3\leqslant d_2\leqslant d_1
	.\]
	The definition for $\lim\varepsilon_m=0$ implies
	\[
	\forall\varepsilon>0,\ \exists m\in\mathbb{N}\suchthat \varepsilon_m<\frac{\varepsilon}{2}
	\]
	and
	\[
	a_{N_m}-\varepsilon_m\leqslant c_m\leqslant a\leqslant d_m\leqslant a_{N_m}+\varepsilon_m
	.\]
	Therefore, together with the definition of $\left\{N_m\right\}$, $\forall\varepsilon>0$, $\exists N_m\in\left\{N_k\right\}\subseteq\mathbb{N}\suchthat\forall n\in\mathbb{N}$:
	\begin{align*}
		n > N_m & \implies \abs{a_n-a_{N_m}} < \varepsilon_m\\
			&
		\begin{aligned}
			{} \implies \abs{a_n-a} & \leqslant \abs{a_n-a_{N_m}} + \abs{a_{N_m}-a}\\
						& <\varepsilon_m + \varepsilon_m\\
						& = 2\varepsilon_m\\
						& < \varepsilon,
		\end{aligned}
	\end{align*}
	which is the definition of $a=\lim a_n$.
\end{proof}
