% This is lecture no. 16 of MH3100/: Real Analysis I.

\chapter{Compact Sets}%
\label{cha:compact_sets}

\section{Closed \& Bounded Sets}%
\label{sec:closed_&_bounded_sets}

\begin{thrm}
	Extrema of Closed and Bounded Sets.
	A \textbf{nonempty, closed, and bounded} set $K\subseteq\mathbb{R}$ has a \textbf{maximum} and a \textbf{minimum}.
\end{thrm}

\noindent
By using the fact that
\[
\max(-K) = -\min(K)
\]
where
\[
-K\defas\left\{-a\vcentcolon a\in K\right\} = \left\{a\vcentcolon -a\in K\right\}
\]
is compact when $K$ is compact, it is enough to prove that every nonempty closed and bounded set has a maximum.

\begin{proof}
	Existence of maximum.
	The Axiom of Completeness then asserts that $K$ has a supremum.
	The Closure Criterion for Closed Sets then asserts that $\overline{K}=K$.
	Since $\sup K\in\overline{K}$, we have $\sup K\in\overline{K}=K$, which implies that $\sup K=\max K$
\end{proof}

\section{Sequential Compactness of Closed \&\\ Bounded Sets}%
\label{sec:sequential_compactness_of_closed_and_bounded_sets}

\begin{thrm}
	Sequential Compactness of Closed and Bounded Sets.
	A set $K\subseteq\mathbb{R}$ is \textbf{closed and bounded} iff \textbf{every sequence} contained in the set has a \textbf{convergent subsequence} with its \textbf{limit in the set}.
\end{thrm}

\begin{definition}
	Sequentially Compact Sets.
	A set $K\subseteq\mathbb{R}$ is said to be \textbf{sequentially compact} if every sequence contained in $K$ has a convergent subsequence converging to a limit in $K$.
\end{definition}

\begin{itemize}
	\item Closedness:
		Every convergent sequence in $K$ has its limit in $K$.
	\item Sequential Compactness:
		Every sequence in $K$ has a convergent subsequence with a limit in $K$.
\end{itemize}

\begin{proof}
	Let $\left\{a_n\right\}$ be a sequence contained in the closed and bounded set $K$.
	The Bolzano-Weierstrass Theorem asserts that the bounded sequence $\left\{a_n\right\}$ has a convergent subsequence.
	The Sequential Criterion for Closed Sets then asserts that this convergent subsequence has its limit in $K$.

	Let $\left\{a_n\right\}$ be an arbitrary convergent sequence contained in $K$.
	Then $\left\{a_n\right\}$ has a convergent subsequence $\left\{a_{n_{k}}\right\}$ with a limit $a$ in $K$.
	The sequence $\left\{a_n\right\}$ shares the same limit $a\in K$.
	Since $\left\{a_n\right\}$ is arbitrary, the Sequential Criterion for Closed Sets asserts that $K$ is closed.

	Let $\left\{M_n\right\}$ denote a positive increasing ubounded sequence.
	The definition of the unboundedness of $K$ implies that $\forall n\in\mathbb{N}$ 
	\[
	\exists a_n\in K\suchthat\abs{a_n}>M_n
	.\]
	This forms a sequence $\left\{a_n\right\}$ contained in $K$.
	Let $\left\{a_{n_{k}}\right\}$ denote an arbitrary subsequence of the sequence $\left\{a_n\right\}$.
	The definition of the unboundedness of $\left\{M_n\right\}$ implies that
	\[
	\forall M>0,\exists n\in\mathbb{N}\suchthat M_n=\abs{M_n}>M
	.\]
	Therefore, $\forall M>0$, $\exists k = n\in\mathbb{N}$ such that
	\[
	\abs{a_{n_{k}}} > M_{n_{k}}\geqslant M_k = M_n > M
	,\]
	which is the definition of the unboundedness of $ \left\{a_{n_{k}}\right\}$; the Boundedness of Convergent Sequence then implies the divergence of $ \left\{a_{n_{k}}\right\}$.
	Since $ \left\{a_{n_{k}}\right\}$ is arbitrary, all subsequences of $ \left\{a_n\right\}$ are divergent.
\end{proof}
