% This is lecture no. 1 of MH3100/: Real Analysis I.

\chapter{Axiom of Completeness}%
\label{cha:axiom_of_completeness}

\section{Defining the Real Numbers}%
\label{sec:defining_the_real_numbers}

We have the natural numbers defined by:
\[
\mathbb{N}\vcentcolon=\left\{1, 2, 3, \dotsc, n, n+1, \dotsc\right\}
.\]
We can add and multiply numbers in this set and compare them.
Then, we have the set of all integers:
\[
\mathbb{Z}\vcentcolon= \left\{\dotsc, -i-1, -i,\dotsc,-3,-2,-1,0,1,2,3,\dotsc,i,i+1,\dotsc\right\}
.\]
We can add, multiply and subtract integers, and compare them.
Following that, we have $\mathbb{Q}$, the set of \textbf{all} rational numbers:
\[
\mathbb{Q}\vcentcolon = \left\{\frac{i}{n}\vcentcolon(i,n)\in\mathbb{Z}\times\mathbb{N}, |i|\ \text{and}\ n\ \text{have no common divisor}\ > 1\right\}
.\]
We can add, multiply, subtract, and divide two rational numbers, and compare them.
However, when it comes to the real numbers, although we may understand that it contains all the rational numbers, we also know it contains more numbers.
These are known as \textbf{irrational numbers}.
Both $\pi$, $\sqrt{2}$ and $e$ are irrational.
How this property is defined requires more analysis.

\section{Axiom of Completeness}%
\label{sec:axiom_of_completeness}

\begin{definition}
	Upper Bound and Lower Bound.
	A set $A\subseteq\mathbb{R}$ is said to be \textbf{bounded above} if there exists $b\in\mathbb{R}$ such that:
	\[
	\forall a\in A,\ a\leqslant b
	.\]
	The number $b$ is called an \textbf{upper bound} for $A$.
	We similarly define a set to be \textbf{bounded below} if there exists a \textbf{lower bound}.
	If $A$ is both bounded above and below, then we say that it is \textbf{bounded}.
\end{definition}

\begin{definition}
	Supremum and Infimum.
	A number $s\in\mathbb{R}$ is said to be the \textbf{supremum} of a set $A\subseteq\mathbb{R}$ if
	\begin{enumerate}
		\item $s$ is an upper bound for $A$; and
		\item $s$ is not greater than any upper bound for $A$.
	\end{enumerate}
	We write $s=\sup A$ to mean that $s$ is the \textbf{supremum} of $A$.
	We similarly define the  \textbf{infimum} of  a set and the notation $k =\inf A$.
\end{definition}

\begin{definition}
	Maximum and Minimum.
	A number $m\in\mathbb{R}$ is said to be the \textbf{maximum} of a set $A\subseteq\mathbb{R}$ if $m\in A$ and $m$ is an upper bound for $A$.
	We write $m = \max A$ to mean that $m$ is the \textbf{maximum} of $A$.
	We similarly define the \textbf{minimum} of a set and the notation $c =\min A$.
\end{definition}

It is important to know that the maximum/minimum of $A$ (if it exists) is always the supremum/infimum of $A$.
With this, we come to the axiom.

\begin{mdframed}
	\begin{center}
		\textbf{Axiom of Completeness}
	\end{center}
	Every \textbf{nonempty} set of real numbers that is \textbf{bounded above} has a \textbf{supremum}.
\end{mdframed}

Note that the empty set $\varnothing$ does not have a supremum because every real number is an upper bound for $\varnothing$.
Also note that a set $A\subseteq\mathbb{R}$ that is not bounded above does not have a supremum because $A$ does not have any upper bound.

\begin{thrm}
	Uniqueness of the Real Numbers.
	Every ordered field that satisfies the \textit{Axiom of Completeness} is isomorphic to $\mathbb{R}$.
\end{thrm}

Furthermore, we have the epsilon criterion for determining if a number is a supremum.
\begin{thrm}
	$\varepsilon$-Criterion for Suprema and Infima.
	For any set $A\subseteq\mathbb{R}$ and any number $s\in\mathbb{R}$, the number $s=\sup A$ if and only if $s$ is an \textbf{upper bound} for $A$ and
	\[
	\forall\varepsilon > 0,\ \exists a\in A\backepsilon s - \varepsilon <a
	.\]
	A similar criterion holds for $k = \inf a$.
\end{thrm}

\begin{proof}
	Proof via contrapositive.
	Recall that $\sup A$ is an upper bound that is no greater than any upper bound.
	Then:
	\begin{align*}
		\exists\varepsilon>0\backepsilon\forall a\in A\quad s-\varepsilon\geqslant a \iff & \exists\varepsilon > 0\backepsilon s-\varepsilon\ \text{is an upper bound for}\ A\\
		\iff & \exists b<s\quad b\ \text{is an upper bound for}\ A\\
		\iff & \exists\ \text{upper bound for}\ A<s.
	\end{align*}
\end{proof}
