% This is lecture no. 8 of MH3100/: Real Analysis I.

\chapter{The Bolzano-Weierstrass Theorem}%
\label{cha:the_bolzano_weierstrass_theorem}

\section{The Bolzano-Weierstrass Theorem}%
\label{sec:the_bolzano_weierstrass_theorem}

\begin{thrm}
	Bolzano-Weierstrass Theorem.
	If a sequence $\left\{a_n\right\}$ is \textbf{bounded}, then it has a \textbf{convergent subsequence}.
\end{thrm}

\begin{proof}
	There is a number $M>0$ for which $\abs{a_n}\leqslant M$, $\forall n\in\mathbb{N}$.
	Define the nonempty set
	\[
		A\defas \left\{x\in\mathbb{R}\vcentcolon x<a_n\text{ for infinitely many }n\in\mathbb{N}\right\}
	.\]
	For each $x\in A$,
	\[
		\left(\exists n\in\mathbb{N}\suchthat x< a_n \leqslant M\right)\implies x < M
	.\]
	Therefore, $M$ is an upper bound for $A$.
	The Axiom of Completeness then asserts that $A$ has a supremum.
	Let $a$ denote the supremum of $A$, that is:
	\[
		a=\sup\left\{x\in\mathbb{R}\vcentcolon x<a_n\text{ for infinitely many }n\in\mathbb{N}\right\}
	.\]
	Let $\left\{\varepsilon_k\right\}$ denote a sequence of positive numbers that converges to 0.
	For each $k\in\mathbb{N}$, the $\varepsilon$-Criterion for supremum asserts that
	\[
	\exists x_k \in A\suchthat a-\varepsilon_k < x_k
	.\]
	For each $k\in\mathbb{N}$, since $a+\varepsilon_k>a$ and a is an upper bound for $A$, the number $a+\varepsilon_k$ is not in $A$.
	Therefore, for each $k\in\mathbb{N}$, the number of terms greater than $a+\varepsilon_k$ is not infinite.
	Hence, for each $k\in\mathbb{N}$, there are infinitely many terms $a_n$ satisfying
	\[
	a-\varepsilon_k < x_k < a_n < a + \varepsilon_k
	,\]
	which implies that
	\[
	\abs{a_n-a}\leqslant\varepsilon_k
	.\]
	Define a subsequence $\left\{a_{n_{k}}\right\}$ with the recurrence
	\begin{align*}
		a_{n_1} & \defas \text{the first term among those satisfying }\abs{a_n-a}\leqslant\varepsilon_1,\\
		a_{n_{k+1}} & \defas \text{the first term among those in the tail }\left\{a_n\right\}^{\infty}_{n=n_k+1} \text{ satisfying}\\
			    &\mathrel{\hphantom{\defas}}\abs{a_n-a}\leqslant\varepsilon_{k+1}.
	\end{align*}
	The definition of $0=\lim\varepsilon_k$ then implies
	\begin{align*}
		\forall\varepsilon > 0, & \quad\left\{\varepsilon_k\right\}\text{ eventually satisfies }\varepsilon_k < \varepsilon\\
		\implies\forall\varepsilon > 0, & \quad\left\{a_{n_{k}}\right\}\text{ eventually satisfies }\abs{a_{n_{k}}-a}\leqslant\varepsilon_k<\varepsilon,
	\end{align*}
	which is the definition of $a = \lim a_{n_{k}}$.
\end{proof}

\section{Divergence Criterion}%
\label{sec:divergence_criterion}

Recall theorem \ref{divcrit}:
\begin{mdframed}
	Divergence Criterion.
	A \textbf{bounded} sequence $\left\{a_n\right\}$ is \textbf{divergent} iff it has \textbf{two subsequences} $\left\{a_{n_{k}}\right\}$ and $\left\{a_{m_{k}}\right\}$ \textbf{converging to different limits}.
\end{mdframed}

\begin{proof}
	The Bolzano-Weierstrass Theorem asserts that the bounded sequence $\left\{a_n\right\}$ has a convergent subsequence $\left\{a_{n_{k}}\right\}$.
	Let $a$ denote the limit of $\left\{a_{n_{k}}\right\}$.
	Since $\left\{a_n\right\}$ does not converge to $a$, there is a number $\varepsilon > 0$ such that
	\[
	\forall N\in\mathbb{N},\ \exists n\in\mathbb{N}\suchthat n>N\implies\abs{a_n-a}\geqslant\varepsilon
	.\]
	Therefore, there are infinitely many terms of $\left\{a_n\right\}$ of a distance at least $\varepsilon$ from $a$.

	The subsequence formed by such terms remains bounded.
	The Bolzano-Weierstrass Theorem concludes that this bounded subsequence has a convergent subsequence $\left\{a_{m_{k}}\right\}$.

	Since this convergent subsequence is of a distance at least $\varepsilon$ from $a$, its limit is different from a.
	Therefore, the two subsequences $a_{n_{k}}$ and $\left\{a_{m_{k}}\right\}$ converge to different limits.
\end{proof}
