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\newtheorem*{example}{Example}
\newtheorem*{exercise}{Exercise}
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\begin{document}
\title{Completeness (continued)}
\section{The Approximation Property}
The proof of the approximation property was by contradiction; we started off assuming the hypothesis of the theorem was true and the conclusion false, and arrived at a contradiction.
\par\vspace{0.2 cm}
We proved that the existence of a finite supremum for a set $E$ implies the existence, for any $\epsilon>0$, of 
and element of $E$ with the property that
\[
\sup E-\epsilon < a \leq \sup E
\vspace{0.2 cm}
\]
Actually, we can strengthen this to an if and only if theorem.  
\begin{theorem}
Suppose $s$ is an upper bound for a set $E$ with the property that, for any $\epsilon>0$, there exists an $a\in E$ 
with the property that
\[
s-\epsilon < a \leq s
\vspace{0.2 cm}
\]
Then $s=\sup E$.
\end{theorem}
\begin{proof}
This means that, no matter how small $\epsilon$ is chosen, $s-\epsilon$ is no longer an upper bound for $E$.  This is equivalent to saying that $s$ has to be the smallest upper bound for $E$, that is, if $u$ is any upper bound for $E$, it must be true that $u\geq s$.  Since $s$ is an upper bound for $E$ and, for any upper bound $u$ of $E$, $u\geq s$, 
by definition $s=\sup E$.
\end{proof} 
\par\vspace{0.3 cm}
Now we can say that $s=\sup E$ if and only if, for every $\epsilon>0$, there exists an $a\in E$ such that
\[
s-\epsilon < a \leq s
\vspace{0.2 cm}
\]
\par\vspace{0.3 cm}
Note that we did \textbf{not} prove that every set $E$ that has a finite upper bound has a supremum.  What we proved was that if $E$ has a finite upper bound that satisfies the approximation property, then that upper bound is $\sup E$.
\par\vspace{0.3 cm}

The approximation property useful for proving that a number $s$ is in fact the supremum of a set $E$, given that $E$ has a finite supremum.
\par\vspace{0.3 cm}
\begin{example}
Suppose 
\[
E=\left\{\frac{n-1}{n} : n\in\mathbb{N}\right\}
\vspace{0.2 cm}
\]
Prove that $\sup E=1$.
\begin{proof}
We know $(n-1)/n<1$ for every $n\in\mathbb{N}$, so $1$ is an upper bound for $E$.  Let $\epsilon>0$ be given. We need to find $a\in E$ such that
\[
1-\epsilon < a \leq 1
\vspace{0.2 cm}
\]
Since $(n-1)/n$ is always less than $1$, the inequality will hold as long as
\[
1-\epsilon < a\quad\mbox{or}\quad -\epsilon < a-1\quad\Rightarrow\quad \epsilon > 1-a
\vspace{0.2 cm}
\]
But if $a\in E$, then
\[
a=\frac{n-1}{n}\quad\mbox{for some}\; n\in\mathbb{N}
\vspace{0.2 cm}
\]
and we can write
\[
\epsilon > 1-a = 1-\frac{n-1}{n}=\frac{1}{n}
\vspace{0.2 cm}
\]
Multiplying both sides by $n$ and dividing by $\epsilon$, we obtain
\[
n > \frac{1}{\epsilon}
\vspace{0.2 cm}
\]
So as long as we take $n>1/\epsilon$, we can say that
\[
1-\epsilon < \frac{n-1}{n}\in E \leq 1
\vspace{0.2 cm}
\]
Since, given $\epsilon > 0$, we can always find such an $n$, we conclude that $1=\sup E$.
\end{proof}
\end{example}
\section{Suprema}
Loosely speaking, the original proof hinged on the fact that if we had a gap between $\sup E$ and some smaller number
\[
s_0=\sup E-\epsilon_0
\vspace{0.2 cm}
\]
that did not contain any elements of $E$, then $s_0$ would be an upper bound for $E$ that is smaller than $\sup E$.
\par\vspace{0.2 cm}
Note that by assuming there was \textbf{not} an $a\in E$ such that
\[
\sup E-\epsilon <a\leq \sup E
\vspace{0.2 cm}
\]
we implicitly assume that $\sup E\notin E$ (because of the $\leq$ condition).
\par\vspace{0.3 cm}
This brings up a subtle but important point.  If $\sup E$ belongs to $E$, it's OK to have a gap between $\sup E$ and 
the next smaller element of $E$, because the $\leq$ condition in the statement of the approximation property makes it true whenever $\sup E\in E$, regardless of the value of $\epsilon$.
\par\vspace{0.3 cm}
However, if $\sup E\notin E$, we \textbf{always} have to be able to find another element of $E$ in the open interval between $\sup E-\epsilon$ and $\sup E$.  This means that, in this case, there can be no gaps without violating the approximation property.
\par\vspace{0.3 cm}
\begin{theorem}If $E$ is a finite set, $\sup E$ exists and $\sup E\in E$.
\end{theorem}
\begin{proof}
By applying the Archimedean principle repeatedly with $a=1$, we can find an integer $n\in\mathbb{N}$ such that $n>a$ for every $a\in E$, so $E$ is bounded above, and therefore $\sup E$ exists by the completeness axiom.
\par\vspace{0.2 cm}
Now suppose $s=\sup E\notin E$.  By the approximation property, for any $\epsilon>0$, we can find an $a_0\in E$ such that
\[
s-\epsilon < a \leq s\quad\Rightarrow\quad s-\epsilon < a < s\quad\mbox{if}\; s\notin E
\vspace{0.2 cm}
\]
Now let $\epsilon_0=1$.  Together with the assumption that $s\notin E$, the approximation property guarantees that we can find $a_0\in E$ such that
\[
s-\epsilon_0 < a_0 < s
\]
Now let 
\[
\epsilon_1=\frac{s-a_0}{2}
\vspace{0.2 cm}
\]
Using the approximation property again in the same way we are guaranteed that there exists $a_1\in E$ such that
\[
s-\epsilon_1 < a_1 < s
\vspace{0.2 cm}
\]
Now let
\[
\epsilon_2=\frac{s-a_1}{2}
\vspace{0.2 cm}
\]
We can continue this process indefinitely, and each time we have to produce a different $a_n\in E$.  But since $E$ 
is finite, the process must end at some point when all elements of $E$ have been exhausted, resulting in a contradiction of the approximation property since for that $\epsilon$, we cannot find a corresponding $a\in E$ such that 
\[
s-\epsilon < a < s
\vspace{0.2 cm}
\]
\end{proof} 



\end{document}



