% This is lecture no. 17 of MH3100/: Real Analysis I.

\chapter{Nested Set Property}%
\label{cha:nested_set_property}

\section{Nested Set Property}%
\label{sec:nested_set_property}

\begin{thrm}
	Nested Set Property.
	If $K_n\subseteq\mathbb{R}$ is a nonempty, closed, and bounded set fore each $n\in\mathbb{N}$ for which $K_n\supseteq K_{n+1}$ for all $n\in\mathbb{N}$, then the resulting \textbf{nested sequence of closed and bounded sets} 
	\[
	K_1\supseteq K_2\supseteq K_3\supseteq K_4\supseteq\dotsb
	\]
	has a \textbf{nonempty intersection}, that is, there is some number $a\in\mathbb{R}$ with $a\in K_n$ for all $n\in\mathbb{N}$
\end{thrm}

\begin{note}
	The Nested Interval Property is a special case of the Nested Set Property where each closed and bounded set $K_n$ is a closed and bounded interval $[a_n,b_n]$.
\end{note}

\begin{proof}
	For each $n\in\mathbb{N}$, since $K_n$ is nonempty, there is some number $a_n\in K_n$.
	since $K_1\supseteq K_n$ for every $n\in\mathbb{N}$, this forms a sequence $\left\{a_n\right\}$ contained in $K_1$.
	The Sequential Compactness of Closed and Bounded Sets then gives a convergent subsequence $\left\{a_{n_{k}}\right\}$ with its limit $a$ in $K_1$.
	For each $n\in\mathbb{N}$:
	\begin{align*}
		\forall k\in\mathbb{N}\quad k > n & \implies n_k > n_n \geqslant n\\
					      & \implies K_{n_{k}}\subseteq K_n\\
					      & \implies a_{n_{k}}\in K_{n_{k}}\subseteq K_n\\
		\therefore\ &\left\{a_{n_{k}}\right\}\text{ is eventually contained in }K_n.
	\end{align*}
	Therefore, for each $n\in\mathbb{N}$, the Sequential Criterion for Closed Sets asserts that the limit $a$ is in $K_n$.
\end{proof}
