\documentclass[12pt]{amsart}
\theoremstyle{plain}    
\newtheorem*{definition}{Definition}
\newtheorem*{example}{Example}
\newtheorem*{exercise}{Exercise}
\newtheorem*{conjecture}{Conjecture}
\newtheorem*{theorem}{Theorem}
\newtheorem*{lemma}{Lemma}
\newtheorem*{claim}{Claim}
\newtheorem*{remark}{Remark}
\begin{document}
\title{Completeness}
\section{Upper Bounds}
\begin{definition}[upper bound] Suppose $E\subset\mathbb{R}$ is a nonempty, proper subset of $\mathbb{R}$.  We say that $E$ is \textbf{bounded above} if and only if
\[
\mbox{there exists an } M\in\mathbb{R}\quad\mbox{such that}\quad a\leq M\quad\mbox{for all}\; a\in E
\vspace{0.2 cm}
\]
\end{definition}
\begin{example}
Suppose $E=\{x\in\mathbb{R}:1\leq x\leq 2\}$=$[1,2]$.  Then $2$ is an upper bound for $E$ because
\[
a\leq 2\quad\forall\quad a\in[1,2]
\vspace{0.2 cm}
\]
\end{example}
\begin{example}
Again suppose $E=\{x\in\mathbb{R}:1\leq x\leq 2\}$=$[1,2]$.  Then $4$ is an upper bound for $E$ because
\[
a\leq 4\quad\forall\quad a\in[1,2]
\vspace{0.2 cm}
\]
In fact, any number greater than or equal to $2$ is an upper bound for $E$.
\end{example}
\par\vspace{0.2 cm}
In the previous examples, both $4$ and $2$ are upper bounds for $E=[1,2]$, and as we noted any number greater than or equal to $2$ is also an upper bound.  In this case $2$ is special in the sense that not only is it an upper bound for $E$, but it is the \textit{smallest possible} upper bound for this set.  That is, for any upper bound $u$ of $E$, $2\leq u$.  This is the motivation for the following definition.
\begin{definition}[supremum] A number $s$ is called the \textit{supremum} of the set $E$ if and only if $s$ is an upper bound of $E$ and $s\leq M$ for all other upper bounds of $E$.
\par\vspace{0.2 cm}
In this case we say that $E$ has a \textbf{finite supremum} and write $s=\sup E$.
\end{definition}
\par\vspace{0.2 cm}
So the essential characteristics of $s=\sup E$ are:
\begin{itemize}
\item $s$ is an upper bound for $E$
\item $s$ is the smallest upper bound for $E$
\end{itemize}
\par\vspace{0.2 cm}
It is important to distinguish between $\sup E$ and $\max E$.  Recall that $\max E$ is the largest element of $E$, that is,
\[
\max E\;=\;\{m\in E\mbox : a\leq m\quad\forall a\in E\}
\vspace{0.2 cm}
\]
The difference between the maximum and an upper bound is that we do not require that an upper bound belong to $E$.  This means we also do not require the supremum to belong to the set $E$.  In fact, it may or may not belong to $E$.
\par\vspace{0.2 cm}
\begin{example}
Consider the set
\[
E=[1,2]=\{x\in\mathbb{R}:1\leq x\leq 2\}
\vspace{0.2 cm}
\]
Clearly $\max E$ and $\sup E$ both exist and are both equal to $2$.
\end{example}
\par\vspace{0.2 cm}
Now consider a slightly different set $E$:
\begin{example}
\[
E=(1,2)=\{x\in\mathbb{R}:1 < x < 2\}
\vspace{0.2 cm}
\]
$2$ is an upper bound for $E$, and no smaller number is, so $\sup E=2$.  However, $\max E$ does not exist.
\end{example}
\begin{claim}If $E=(1,2)=\{x\in\mathbb{R}:1<x<2\}$, then $\max E$ does not exist.
\end{claim}
\begin{proof} The proof is by contradiction.  Suppose that $\max E=m$ does exist.  Then by the definition of $\max$, $m\in E$ and
\[
a\leq m\quad\forall a\in E
\vspace{0.2 cm}
\]
From the definition of $E$, every $x\in E$ has the property that $1<x<2$, so $m\in E$ implies that $m<2$. This means that $0<2-m$ and therefore $0<\delta=(2-m)/2$, and we can write 
\[
m<m+\delta=m+\frac{2-m}{2}=\frac{m}{2}+1
\vspace{0.2 cm}
\]
$m\in E$, so by the definition of $E$, $m<2$, and therefore
\[
m<2\Rightarrow\frac{m}{2}<1\Rightarrow\frac{m}{2}+1<1+1=2
\vspace{0.2 cm}
\]
so $m+\delta$, which is larger than $m$, is also less than $2$ and therefore belongs to $E$, contradicting the hypothesis that $\max E=m$.  Since $m$ was an arbitrary choice, we arrive at such a contradiction for any $m\in E$, and therefore $\max E$ does not exist.
\end{proof}
\par\vspace{0.2 cm}
The following statements are easy to prove:
\begin{claim} If a set has one upper bound, it has infinitely many upper bounds.
\end{claim}
\begin{proof}
Suppose $m\in\mathbb{R}$ is an upper bound for $E\subset\mathbb{R}$.  Then $m+1\in\mathbb{R}$ is also an upper bound for $E$, as are $m+2,m+3,\ldots$. 
\end{proof}
\begin{claim} If a set has a supremum, it is unique.
\end{claim}
\begin{proof}
Suppose $s_1$ and $s_2$ are suprema of $E\subset E$.  By the definition of supremum, both $s_1$ and $s_2$ are upper bounds.  First consider $\sup E=s_1$ and $s_2$ an upper bound for $E$.  By the definition of supremum, $s_1\leq s_2$.  A similar argument assuming $\sup E=s_2$ implies that $s_2\leq s_1$.  Taken together,
\[
s_1\leq s_2\quad\mbox{and}\quad s_2\leq s_1\quad\Rightarrow\quad s_1=s_2
\vspace{0.2 cm}
\] 
\end{proof}
\begin{theorem}[Approximation Theorem for Suprema] If $E$ has a finite supremum and $\epsilon>0$ is any positive number, there is a point $a\in E$ such that
\[
\sup E-\epsilon<a\leq \sup E
\vspace{0.2 cm}
\]
\end{theorem}
We will prove this by contradiction, that is, we will assume the theorem is false and try to arrive at a contradiction.  The statement we start will will be the negation of the theorem, so starting with the theorem,
\[
\mbox{ if }s=\sup E\mbox{ exists },\quad\forall \epsilon>0,\quad\exists a\in E\mbox{ such that } s-\epsilon<a\leq s
\vspace{0.2 cm}
\]
We negate the consequent or right hand side of the implication.  Recall that the negation of a generalization
\[
\forall x P
\vspace{0.2 cm}
\]
is an existence statement:
\[
\exists x : \sim P
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\]
and the negation of an exitence statement
\[
\exists x : p
\]
is a generalization
\[
\forall x : \sim P
\]
we would write this as
\[
\exists \epsilon_0>0 : \forall a\in E, s-\epsilon_0>a\mbox{  or  } s<a
\]
Since the second part of the "or" cannot be true if $s=\sup E$, for the "or" to be true the first part must be true, 
that is,
\[
s-\epsilon_0\geq a\;\Rightarrow\;-\epsilon_0>a-s\geq0
\]
 
\begin{proof}
The proof is by contradiction.  Suppose the theorem is false.
\end{proof}
\end{document}



