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chapter-2/absolute-value-and-the-real-line.tex
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\section{Absolute Value and the Real Line}
\begin{definition}
The \textbf{absolute value} of a real number $a$, denoted by $|a|$, is defined by
\[|a|:=\begin{cases}
a & \text{if } a>0, \\
0 & \text{if } a=0, \\
-a & \text{if } a<0.
\end{cases}\]
\end{definition}
\begin{theorem}
\begin{enumerate}
\item[]
\item $|ab|=|a||b|$ for all $a,b \in \R$.
\item $|a|^2 = a^2$ for all $a \in \R$.
\item If $c \geq 0$, then $|a| \leq c$ if and only if $-c \leq a \leq c$.
\item $-|a|\leq a \leq |a|$ for all $a \in \R$.
\end{enumerate}
\end{theorem}
\begin{theorem}[\textbf{Triangle Inequality}]
If $a,b \in \R$, then $|a+b| \leq |a| + |b|$.
\end{theorem}
\begin{corollary}
If $a,b \in \R$, then
\begin{enumerate}
\item $\left| |a|-|b| \right| \leq |a-b|$,
\item $|a-b| \leq |a| + |b|$.
\end{enumerate}
\end{corollary}
\begin{corollary}
If $a_1, a_2, \dots, a_n$ are any real numbers, then
\[|a_1 + a_2 + \dots + a_n| \leq |a_1| + |a_2| + \dots + |a_n|\]
\end{corollary}
\begin{definition}
Let $a \in \R$ and $\varepsilon > 0$. Then the $\varepsilon$-\textbf{neighborhood} of $a$ is the set $V_\varepsilon(a):=\{x \in \R : |x-a| < \varepsilon\}$.
\end{definition}
chapter-2/applications-of-the-supremum-property.tex
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\section{Applications of the Supremum Property}
\begin{theorem}[\textbf{Archimedian Property}]
If $x \in \R$, then there exists $n_x \in \N$ such that $x \leq n_x$.
\end{theorem}
\begin{corollary}
If $S:= \{1/n : n \in \N\}$, then $\inf S = 0$.
\end{corollary}
\begin{corollary}
If $t >0$, there exists $n_t \in \N$ such that $0 < 1/n_t < t$.
\end{corollary}
\begin{corollary}
If $y>0$, there exists $n_y \in \N$ such that $n_y -1 \leq y \leq n_y$.
\end{corollary}
\begin{theorem}
There exists a positive real number $x$ such that $x^2 = 2$.
\end{theorem}
\begin{theorem}[\textbf{The Density Theorem}]
If $x$ and $y$ are any real numbers with $x<y$, then there exists a rational number $r \in \Q$ such that $x < r < y$.
\end{theorem}
\begin{corollary}
If $x$ and $y$ are real numbers with $x < y$, then there exists an irrational number $z$ such that $x < z < y$.
\end{corollary}
chapter-2/chapter-2.tex
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\chapter{The Real Numbers}
\subimport{./}{the-algebraic-and-order-properties-of-R.tex}
\subimport{./}{absolute-value-and-the-real-line.tex}
\subimport{./}{the-completeness-property-of-R.tex}
\subimport{./}{applications-of-the-supremum-property.tex}
\subimport{./}{intervals.tex}
chapter-2/intervals.tex
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\section{Intervals}
\begin{definition}
If $a,b \in \R$ satisfy $a<b$, then the \textbf{open interval} determined by $a$ and $b$ is the set
\[(a,b):= \{x \in \R : a <x < b\}\]
The points $a$ and $b$ are called the \textbf{endpoints} of the interval.
\end{definition}
\begin{definition}
If both endpoints $a$ and $b$ are adjoined to an open interval, then we obtain the \textbf{closed interval} determined by $a$ and $b$; namely, the set
\[[a,b]:=\{x \in \R : a \leq x \leq b\}\]
\end{definition}
\begin{definition}
The two \textbf{half-open} (or \textbf{half-closed}) intervals determined by $a$ and $b$ are $[a,b)$, which includes the endpoint $a$, and $(a,b]$, which includes the endpoint $b$.
\end{definition}
\begin{definition}
The \textbf{length} of an interval $(a,b)$ is defined by $b-a$.
\end{definition}
\begin{theorem}[\textbf{Characterization Theorem}]
If $S$ is a subset of $\R$ that contains at least two points and has the property
\[\text{if}\ \ \ \ \ x,y \in S\ \ \ \ \ \text{and}\ \ \ \ \ x < y,\ \ \ \ \ \text{then}\ \ \ \ \ [x,y] \subseteq S,\]
then $S$ is an interval.
\end{theorem}
\begin{theorem}[\textbf{Nested Intervals Property}]
If $I_n=[a_n,b_n],\ n \in \N$, is a nested sequence of closed bounded intervals, then there exists a number $\xi \in \R$ such that $\xi \in I_n$ for all $n \in \N$.
\end{theorem}
\begin{theorem}
If $I_n :=[a_n,b_n],\ n \in \N$, is a nested sequence of closed, bounded intervals such that the lengths $b_n-a_n$ of $I_n$ satisfy
\[\inf \{b_n - a_n : n \in \N\}=0,\]
then the number $\xi$ contained in $I_n$ for all $n \in \N$ is unique.
\end{theorem}
\begin{theorem}
The set $\R$ of real numbers is not countable.
\end{theorem}
\begin{theorem}
The unit interval $[0,1] := \{x \in \R : 0 \leq x \leq 1\}$ is not countable.
\end{theorem}
chapter-2/the-algebraic-and-order-properties-of-R.tex
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\section{The Algebraic and Order Properties of $\R$}
\begin{theorem}[\textbf{Algebraic Properties of $\R$}]
On the set $\R$ of real numbers there are two binary operations, denoted by $+$ and $\cdot$ and called \textbf{addition} and \textbf{multiplication}, respectively. These operations satisfy the following properties:
\begin{enumerate}
\item[(A1)] $a+b=b+a\ \forall\ a,b \in \R$. (\textit{commutative property of addition});
\item[(A2)] $(a+b)+c=a+(b+c)\ \forall\ a,b,c \in \R$ (\textit{associative property of addition});
\item[(A3)] There exists and element $0$ in $\R$ such that $0+a=a$ and $a+0=a$ for all $a \in \R$ (\textit{existence of a zero element});
\item[(A4)] for each $a \in \R$ there exists and element $-a \in \R$ such that $a + (-a)=0$ and $(-a) + a=0$ (\textit{existence of negative elements});
\item[(M1)] $a \cdot b=b \cdot a\ \forall\ a,b \in \R$ (\textit{commutative property of multiplication});
\item[(M2)] $(a \cdot b) \cdot c = a \cdot (b \cdot c)\ \forall\ a,b,c \in \R$ (\textit{associative property of multiplication});
\item[(M3)] There exists an element $1 \in \R$ \textit{distinct from} $0$ such that $1 \cdot a=a$ and $a \cdot 1 = a\ \forall\ a \in \R$ (\textit{existence of a unit element});
\item[(M4)] for each $a \neq 0 \in \R$, there exists an element $1/a \in \R$ such that $a \cdot (1/a) = 1$ and $(1/a) \cdot a = 1$ (\textit{existence of reciprocals});
\item[(D)]$a \cdot (b+c)=(a \cdot b) + (a \cdot c)$ and $(b+c)\cdot a = (b \cdot a) + (c \cdot a)\ \forall\ a,b,c \in \R$ (\textit{distributive property of multiplication over addition}).
\end{enumerate}
\end{theorem}
\begin{theorem}
\begin{enumerate}
\item[]
\item If $z$ and $a$ are elements in $\R$ with $z+a=a$, then $z=0$.
\item If $u$ and $b \neq 0$ are elements in $\R$ with $u \cdot b=b$, then $u=1$.
\item If $a \in \R$, then $a \cdot 0=0$.
\end{enumerate}
\end{theorem}
\begin{theorem}
\begin{enumerate}
\item[]
\item If $a \neq 0$ and $b \in \R$ are such that $a \cdot b = 1$, then $b = 1/a$.
\item If $a \cdot b = 0$, then either $a=0$ or $b=0$.
\end{enumerate}
\end{theorem}
\begin{theorem}
There does not exists a rational number $r$ such that $r^2=2$.
\end{theorem}
\begin{definition}[\textbf{The Order Properties of $\R$}]
There is a nonempty subset $\mathbb{P}$ of $\R$, called the set of \textbf{positive real numbers}, that satisfies the following properties:
\begin{enumerate}
\item If $a,b$ belong to $\mathbb{P}$, then $a+b$ belongs to $\mathbb{P}$.
\item If $a,b$ belong to $\mathbb{P}$, then $ab$ belongs to $\mathbb{P}$.
\item If $a$ belongs to $\R$, then exactly one of the following holds:
\[a \in \mathbb{P},\ \ \ \ a=0,\ \ \ \ -a \in \mathbb{P}\]
(This condition is usually called the \textbf{Trichotomy Property}.)
\end{enumerate}
\end{definition}
\begin{definition}
Let $a,b$ be elements of $\R$.
\begin{enumerate}
\item If $a-b \in \mathbb{P}$, then we write $a >b$ or $b < a$.
\item If $a-b \in \mathbb{P} \cup \{0\}$, then we write $a \geq b$ or $b \leq a$.
\end{enumerate}
\end{definition}
\begin{theorem}
Let $a,b,c$ be any elements of $\R$.
\begin{enumerate}
\item If $a>b$ and $b>c$, then $a>c$.
\item If $a>b$, then $a+c>b+c$.
\item If $a>b$ and $c>0$, then $ca>cb$.
\\If $a>b$ and $c<0$, then $ca<cb$.
\end{enumerate}
\end{theorem}
\begin{theorem}
\begin{enumerate}
\item[]
\item If $a \in \R$ and $a \neq 0$, then $a^2>0$.
\item $1 >0$.
\item If $n \in \N$, then $n >0$
\end{enumerate}
\end{theorem}
\begin{theorem}
If $a \in \R$ is such that $0 \leq a < \varepsilon$ for every $\varepsilon>0$, then $a=0$.
\end{theorem}
\begin{theorem}
If $ab>0$, then either
\begin{enumerate}
\item $a>0$ and $b>0$, or
\item $a<0$ and $b<0$.
\end{enumerate}
\end{theorem}
\begin{corollary}
If $ab <0$, then either
\begin{enumerate}
\item $a<0$ and $b>0$, or
\item $a>0$ and $b<0$.
\end{enumerate}
\end{corollary}
\begin{definition}[\textbf{Bernoulli's Inequality}]
If $x>-1$, then
\[(1+x)^n \geq 1+nx\ \forall\ n \in \N\]
\end{definition}
chapter-2/the-completeness-property-of-R.tex
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\section{The Completeness Property of $\R$}
\begin{definition}
Let $S$ be a nonempty subset of $\R$.
\begin{enumerate}
\item The set $S$ is said to be \textbf{bounded above} if there exists a number $u \in \R$ such that $s \leq u$ for all $s \in S$. Each such number $u$ is called an \textbf{upper bound} of $S$.
\item The set $S$ is said to be \textbf{bounded below} if there exists a number $w \in \R$ such that $w \leq s$ for all $s \in S$. Each such number $w$ is called a \textbf{lower bound} of $S$.
\item A set is said to be \textbf{bounded} if it is both bounded above and bounded below. A set is said to be \textbf{unbounded} if it is not bounded.
\end{enumerate}
\end{definition}
\begin{definition}
Let $S$ be a nonempty subset of $\R$.
\begin{enumerate}
\item If $S$ is bounded above, then a number $u$ is said to be a \textbf{supremum} (or a \textbf{least upper bound}) of $S$ if it satisfies the conditions:
\begin{enumerate}
\item $u$ is an upper bound of $S$, and
\item if $v$ is any upper bound of $S$, then $u \leq v$.
\end{enumerate}
\item If $S$ is bounded below, then a number $w$ is said to be an \textbf{infimum} (or a \textbf{greatest lower bound}) of $S$ if it satisfies the conditions:
\begin{enumerate}
\item $w$ is a lower bound of $S$, and
\item if $t$ is any lower bound of $S$, then $t \leq w$.
\end{enumerate}
\end{enumerate}
\end{definition}
\begin{lemma}
A number $u$ is the supremum of a nonempty subset $S$ of $\R$ if and only if $u$ satisfies the conditions:
\begin{enumerate}
\item $s \leq u$ for all $s \in S$,
\item if $v < u$, then there exists $s' \in S$ such that $v < s'$.
\end{enumerate}
\end{lemma}
\begin{lemma}
An upper bound $u$ of a nonempty set $S$ in $\R$ is the supremum of $S$ if and only if for every $\varepsilon > 0$ there exists an $s_\varepsilon \in S$ such that $u - \varepsilon < s_\varepsilon$.
\end{lemma}
\begin{theorem}[\textbf{The Completeness Property of $\R$}]
Every nonempty set of real numbers that has an upper bound also has a supremum in $\R$. (This property is also called the \textbf{Supremum Property of $\R$}).
\end{theorem}