% !TEX root =  Coeur.tex
\chapter{Intervals}\label{lect:Intervals}


\section{Absolute Value}\label{sec:abs}
The fact that negation reverses inequalities leads us to the following
\begin{definition}\label{dfn:abs}
	The \deffont{absolute value} of any number $x$ is defined to be
	\begin{equation*}
		\abs{x}=\max\single{x,-x}.
	\end{equation*}
\end{definition}
That is, the absolute value of any number is either the number itself or its negative, whichever is higher:
\begin{equation*}
	\abs{x}=
            \begin{cases}
               x & \text{if }x\geq0, \\
                -x  & \text{otherwise}.
            \end{cases}
\end{equation*}
Thus, the absolute value of $x$ is a non-negative number, which expresses the length of the interval with
endpoints $0$ and $x$.  This can also be interpreted as the \deffont{size} of the number $x$.  
We shall take some care to distinguish  saying that
$x$ is \emph{less} than $y$--or $y$ is \emph{higher} than $x$ ($x<y$)--from saying that 
$x$ is \emph{smaller} than $y$--or $y$ is \emph{larger} than $x$ ($\abs{x}<\abs{y}$).

For many reasons (not least of which is that inequalities among positive numbers are simpler
than inequalities among numbers that can sometimes be negative) we will very often be trying to establish
inequalities among \emph{absolute values} rather than numbers in general.  To this end, it is useful
to note the following 
\begin{remark}\label{rmk:abs}
Every number, as well as its negative, is less than or equal to its absolute value:
\begin{equation*}
	x\leq\abs{x}\text{ and }-x\leq\abs{x},
\end{equation*}
and for any other number $y$, 
\begin{quote}
	$\abs{x}\leq y$ if and only if both $x\leq y$ and $-x\leq y$.
\end{quote}
\end{remark}
(After all, the \emph{higher} of two numbers is less than $y$ precisely if 
each of the two numbers individually is less than $y$.)
 
We shall make frequent use of three basic properties of the absolute value, codified in the following
\begin{prop}\label{prop:abs}
	The absolute value function satisfies:
	\begin{description}
		\item[Positive-Definite:] For all $x$, $\abs{x}\geq0$, and $\abs{x}=0$ precisely when $x=0$.
		\item[Scaling:] For any $x$ and $y$, $\abs{xy}=\abs{x}\abs{y}$.
		\item[Triangle Inequality:] For any $x$ and $y$, $\abs{x+y}\leq\abs{x}+\abs{y}$.
	\end{description}
\end{prop}
The first property is an immediate consequences of the definition, but the other two require proof.
\begin{enumerate}
	\item
            \begin{proofof}{scaling}
            \begin{description}
            	\item[Case 1]
            	If $x$ and $y$ have the same sign, then $xy$ is non-negative, and hence equals \abs{xy}.
            	In this case, if $x$ and $y$ are both positive, then also $xy=\abs{x}\abs{y}$,
            	while if both are negative, then $\abs{x}\abs{y}=(-x)(-y)=xy=\abs{xy}$
            	
            	\item{Case 2}
            	If they have opposite signs, then $xy$ is negative, so $\abs{xy}=-xy$.
            	Now, either $\abs{x}=-x$ or $\abs{y}=-y$, \emph{but not both}, so $\abs{x}\abs{y}=-xy$ as well. 
            \end{description}
            \end{proofof}
            
         \item
            \begin{proofof}{triangle inequality}
            	We need to show that for any two numbers $x$ and $y$,
            	\begin{equation*}
            		\max\single{(x+y),-(x+y)}\leq \abs{x}+\abs{y}.
            	\end{equation*}
            	Applying \refer{rmk}{abs} to $x$ and $y$ individually,  we have the four inequalities
            	\begin{align*}
            		x\leq\abs{x}&\text{ and }
            		-x\leq\abs{x}\\
            		y\leq\abs{y}&\text{ and }
            		-y\leq\abs{y};
            	\end{align*}
            	adding the first and third inequality (and using the basic fact about adding unequals) yields
            	\begin{equation*}
            		x+y\leq \abs{x}+\abs{y}
            	\end{equation*}
            	while adding the second and fourth inequality yields
            	\begin{equation*}
            		-x-y\leq \abs{x}+\abs{y}
            	\end{equation*}
            	or $-(x+y)\leq\abs{x}+\abs{y}$.
            	These two observations let us apply \refer{rmk}{abs} again, this time to $(x+y)$, to get
            	\begin{equation*}
            		\abs{x+y}\leq\abs{x}+\abs{y}
            	\end{equation*}
            	which is the triangle inequality.
            \end{proofof}
\end{enumerate}
The absolute value \abs{x} of a number can be interpreted as the distance from $x$ to the origin, $0$.
More generally, we can measure the distance between two distinct numbers, $x$ and $y$, as 
\begin{equation*}
	\dist{x}{y}=\abs{x-y}
\end{equation*}
From this point of view, the triangle inequality for the absolute value translates to the statement\\
\textbf{Triangle Inequality for Distance:}\\\fbox{For any three points $x,y,z$, $\dist{x}{y}\leq\dist{x}{z}+\dist{z}{y}$.}

This is the same as the triangle inequality for absolute value, applied to
\begin{equation*}
	\abs{x-y}=\abs{(x-z)+(z-y)}.
\end{equation*}
Geometrically, if we draw the collapsed triangle whose vertices are the three points, it is the statement that
each side of the triangle is no longer than the sum of the other two.

\section{Intervals}\label{sec:intervals}
While subsets of \Reals{} come in many different varieties, one variety of subset which occurs very frequently
(for example, as the domain of a function) is the \emph{interval}:

\begin{definition}\label{dfn:interval}
	An \deffontcustom{interval}{interval!definition} is a set $I$ of real numbers with the property that for any two elements $x,y\in I$,
	every point between $x$ and $y$ is also an element of $I$.
\end{definition}
We have a special notation for intervals, which distinguishes several subvarieties:
\begin{itemize}
	\item The \deffontcustom{closed interval}{closed!interval}\index{interval!closed} 
	with endpoints $a\leq b$ is
	\begin{equation*}
		\boldsymbol{[a,b]}=\setbld{x\in\Reals}{a\leq x\text{ and }x\leq b}.
	\end{equation*}

	\item The \deffontcustom{open interval}{open!interval in \Reals}\index{interval!open} 
	with endpoints $a\leq b$ is
	\begin{equation*}
		\boldsymbol{(a,b)}=\setbld{x\in\Reals}{a< x\text{ and }x< b}.
	\end{equation*}

	\item We can also define \textbf{half-open} intervals:
	\begin{align*}
		\boldsymbol{(a,b]}&=\setbld{x\in\Reals}{a< x\text{ and }x\leq b}\\
		\boldsymbol{[a,b)}&=\setbld{x\in\Reals}{a\leq x\text{ and }x< b}.
	\end{align*}

\end{itemize}
All of these sets consist of the points ``between $a$ and $b$'' ; the distinction between them hinges on whether 
$a$ \resp{$b$} is included in, or excluded from, the set.  This distinction may seem a bit esoteric, 
but as we shall see it can often play a crucial role in our study of real numbers.
An interval specified in any of these ways as the collection of points between two specified real numbers
$a,b\in\Reals$ is called a \deffont{bounded interval} and the points $a$ and $b$ are its \deffont{endpoints}.
When sketching intervals, we mark an \emph{included} endpoint by a filled-in dot, and an \emph{excluded} one by 
a ``hollow'' dot:

\begin{figure}[htbp]
\begin{center}

\subfloat[Closed interval \clintab]{
\begin{pspicture}(-2,-0.5)(2,0.5)
	\psline[linestyle=dashed](-2,0)(-1,0)
	\psline[linewidth=1.5pt, arrows=*-*](-1,0)(1,0)
	\psline[linestyle=dashed](1,0)(2,0)
	
	\uput[u](-1,0){$a$}
	\uput[u](1,0){$b$}	
\end{pspicture}
}
\subfloat[Open interval \opint{a}{b}]{
\begin{pspicture}(-2,-0.5)(2,0.5)
	\psline[linestyle=dashed](-2,0)(-1,0)
	\psline[linewidth=1.5pt, arrows=o-o](-1,0)(1,0)
	\psline[linestyle=dashed](1,0)(2,0)
	
	\uput[u](-1,0){$a$}
	\uput[u](1,0){$b$}	
\end{pspicture}
}
\subfloat[Half-open Interval \lopint{a}{b}]{
\begin{pspicture}(-2,-0.5)(2,0.5)
	\psline[linestyle=dashed](-2,0)(-1,0)
	\psline[linewidth=1.5pt, arrows=o-*](-1,0)(1,0)
	\psline[linestyle=dashed](1,0)(2,0)
	
	\uput[u](-1,0){$a$}
	\uput[u](1,0){$b$}	
\end{pspicture}
}
\caption{Bounded Intervals}
\label{fig:bdint}
\end{center}
\end{figure}


\refer{dfn}{interval} encompasses, in addition to intervals defined by two inequalities, intervals defined by a 
single inequality, like the set of strictly positive numbers noted before \refer{dfn}{interval}.  
We adapt our notation to these sets by use of the symbols $\infty$ and $-\infty$, which we can think of as the 
right and left ``ends'' of the number line, so \emph{every} real number $x$ satisfies the two inequalities
$-\infty<x$ and $x<\infty$.  This device lets us write \opint{0}{\infty} for the set of strictly positive numbers,
and \lopint{-\infty}{0} for the set of non-positive numbers--we can even write
\begin{equation*}
	\Reals=\opint{-\infty}{\infty}
\end{equation*}
for the interval defined by no inequalities at all.  It is important to remember that 
$\pm\infty$ \textit{are not numbers}--we cannot perform arithmetic operations on them in a meaningful way--and by
convention we never write them as ``included'' endpoints of an interval. 
We refer to an interval 
%whose endpoints are both real numbers as \deffont{bounded intervals} and any interval
with at least one ``infinite endpoint'' as an \deffont{unbounded interval}.
%
%\begin{figure}[htbp]
%\begin{center}
%
%\subfloat[Closed interval \clintab]{
%\begin{pspicture}(-2,-0.5)(2,0.5)
%	\psline[linestyle=dashed](-2,0)(-1,0)
%	\psline[linewidth=1.5pt, arrows=*-*](-1,0)(1,0)
%	\psline[linestyle=dashed](1,0)(2,0)
%	
%	\uput[u](-1,0){$a$}
%	\uput[u](1,0){$b$}	
%\end{pspicture}
%}
%\subfloat[Open interval \opint{a}{b}]{
%\begin{pspicture}(-2,-0.5)(2,0.5)
%	\psline[linestyle=dashed](-2,0)(-1,0)
%	\psline[linewidth=1.5pt, arrows=o-o](-1,0)(1,0)
%	\psline[linestyle=dashed](1,0)(2,0)
%	
%	\uput[u](-1,0){$a$}
%	\uput[u](1,0){$b$}	
%\end{pspicture}
%}
%\subfloat[Half-open Interval \lopint{a}{b}]{
%\begin{pspicture}(-2,-0.5)(2,0.5)
%	\psline[linestyle=dashed](-2,0)(-1,0)
%	\psline[linewidth=1.5pt, arrows=o-*](-1,0)(1,0)
%	\psline[linestyle=dashed](1,0)(2,0)
%	
%	\uput[u](-1,0){$a$}
%	\uput[u](1,0){$b$}	
%\end{pspicture}
%}
%
%\caption{Bounded Intervals}
%\label{fig:bdint}
%\end{center}
%\end{figure}

\begin{figure}[htbp]
\begin{center}
\subfloat[\opint{a}{\infty}]{
\begin{pspicture}(-2,-0.5)(2,0.5)
	\psline[linestyle=dashed](-2,0)(-1,0)
	\psline[linewidth=1.5pt, arrows=o->](-1,0)(2,0)
%	\psline[linestyle=dashed](1,0)(2,0)
	
	\uput[u](-1,0){$a$}
%	\uput[u](1,0){$b$}	
\end{pspicture}
}
\subfloat[\lopint{-\infty}{b}]{
\begin{pspicture}(-2,-0.5)(2,0.5)
%	\psline[linestyle=dashed](-2,0)(-1,0)
	\psline[linewidth=1.5pt, arrows=<-*](-2,0)(1,0)
	\psline[linestyle=dashed](1,0)(2,0)
	
%	\uput[u](-1,0){$a$}
	\uput[u](1,0){$b$}	
\end{pspicture}
}
\subfloat[$\opint{-\infty}{\infty}=\Reals$]{
\begin{pspicture}(-2,-0.5)(2,0.5)
%	\psline[linestyle=dashed](-2,0)(-1,0)
	\psline[linewidth=1.5pt, arrows=<->](-2,0)(2,0)
%	\psline[linestyle=dashed](1,0)(2,0)
	
%	\uput[u](-1,0){$a$}
%	\uput[u](1,0){$b$}	
\end{pspicture}
}
\caption{Unbounded Intervals}
\label{fig:unbdint}
\end{center}
\end{figure}

%\section*{Exercises for Lecture 1}
%\addcontentsline{toc}{section}{Exercises}
\LectExHead
\begin{enumerate}
%	\item Show that if $m\in\Nat$ then $m^{2}$ is even if and only if $m$ is even.
%	\textit{Note that there are two things to prove:} if $m$ is even, then so is $m^{2}$, 
%	and if $m^{2}$ is even then $m$ is even.
		
	\item Each of the following statements is true 
	whenever all the letters represent positive real numbers.
	For each statement, either give an example (involving some negative numbers) for which it is false,
	or prove that it is true for \emph{all} real numbers:
	\begin{enumerate}
%		\item\subfixtwoacross[2.0in] {If $a<b$ then $\recip{a}>\recip{b}$.}%
%		{ If $a<b$ then $-a>-b$,}%
		\item\subfixtwoacross[2.0in] {$\abs{a-b}\leq\abs{a}+\abs{b}$.}%
		{$\abs{a-b}\geq\abs{a}-\abs{b}.$}
	\end{enumerate}
	
%	\item Suppose $0<x<y$ and $0<u<v$.
%%	\begin{align*}
%%		0<x&<y\\
%%		0<u&<v.
%%	\end{align*}
%	What is the relation between $\frac{x}{u}$ and $\frac{y}{v}$?

	\item What is closer to $1$: $\frac{3}{4}$ or $\frac{4}{3}$?  As a general rule, which positive
	real numbers are closer to $1$ than their reciprocal?  Justify your answer.
	
%	\item \textbf{Cauchy's mediant lemma:}  \cite[Thm 1, p. 13]{Cauchy:CoursEnglish}%
%	\footnote{
%            	The fraction appearing in the middle of the inequality in (a) is called
%            	 the \deffont{mediant} or \deffont{Farey sum} of the outer fractions.
%	 	If we consider all the reduced proper fractions with denominator less than a given bound,
%		listed in increasing order, then each entry is the mediant of its two immediate neighbors.
%	 	This was observed (without proof) by \Fareyref{}\cite{Farey};  Cauchy then published a 
%		proof of this \cite{Cauchy:1816}, crediting Farey with the idea of mediants.  However, it
%		seems this same observation had been used much earlier by \Harosref{} when preparing tables
%		giving the conversion of fractions to decimal form \cite{Haros}.  The corresponding quantity
%		in (b) is also called the mediant of the collection of fractions $\as{i}/\bs{i}$;  Cauchy was interested
%		in this as a kind of average.
%%            	 \Harosref{} used mediants to generate fractions with small denominators in \cite{Haros};
%%            	 \Fareyref{} commented on a special property of mediants \cite{Farey} which Cauchy then
%%            	 proved \cite{Cauchy:1816}, crediting Farey with the idea of mediants.
%	 }
%	\begin{enumerate}
%            	\item Show that if 
%            	\begin{equation*}
%            		0<\frac{\as{1}}{\bs{1}}\leq\frac{\as{2}}{\bs{2}}
%            	\end{equation*}
%            	Then
%            	\begin{equation*}
%            		\frac{\as{1}}{\bs{1}}\leq\frac{\as{1}+\as{2}}{\bs{1}+\bs{2}}\leq\frac{\as{2}}{\bs{2}}.
%            	\end{equation*}
%		\textit{Note that we defined a fraction to have a positve denominator (but a numerator of either sign}).
%%            	Is the assumption that both fractions are positive necessary for this to hold?
%	
%		\item Extend this (by induction on $n$) to a collection of $n$ positive fractions $\frac{\as{i}}{\bs{i}}$,
%		$i=1,\dots,n$:
%			\begin{equation*}
%				\min\single{\frac{\as{i}}{\bs{i}}}
%				\leq\frac{\as{1}+\cdots+\as{n}}{\bs{1}+\cdots+\bs{n}}
%				\leq\max\single{\frac{\as{i}}{\bs{i}}}
%			\end{equation*}
%	\end{enumerate}
%            	Is the assumption that both fractions are positive necessary for this to hold?
%%	\textit{Note that we defined a fraction to have a positve denominator (but a numerator of either sign}).
	
	\item If \pof{x} is a polynomial of degree $n$,
	then for any $a\in\Reals$, there are at most $n$ real numbers that satisfy the equation $\pof{x}=a$.
	So a \emph{solution} of such an equation is a list of \emph{all} (real) numbers that satisfy it.
	In general, an \emph{inequality} can be satisfied by infinitely many numbers, so such a list is impossible.
	But we could reasonably say that we have ``solved'' the inequality if we can express the collection of
	all the numbers that satisfy it as a disjoint union of intervals.
	
	The function 
	\begin{equation*}
		\fofx=x^{2}-x^{4}
	\end{equation*}
	has a local maximum at $x=\pm\recip{\sqrt{2}}$ and a local minimum at $x=0$.
	
	Solve each of the following:
	\begin{enumerate}
		\item \subfixthreeacross[1.2in]{$\fofx= -2$}%
		{$\fofx= 1$}%
    		{$\fofx= 0 $}%
    		\item \subfixthreeacross[1.2in]{$\fofx> 0$}%
		{$\fofx\leq 0 $}%
		{$\abs{\fofx}\leq2$}%
		\item{$\abs{\fofx}>2 $}
	\end{enumerate}
	
\end{enumerate}