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\noindent
{\bf Math 3220}
\hfill {\Large \bf Axioms which Determine $\mathbb{R}$} \hfill 
{\bf Intro. to Real Analysis}

\vspace{0.1in}

Without getting too philosophical, much like the natural numbers, for the real numbers it does not matter so much how we formally construct $\mathbb{R}$. What really matters is the formal properties that the real numbers possess. What makes the real numbers stand out is that they are the only {\bf complete ordered field} (up to an appropriate equivalence). What does this mean?

First, the real numbers, denoted $\mathbb{R}$, form a {\bf field}. This means that $\mathbb{R}$ is equipped with two operations: addition is denoted $a+b$ and multiplication is denoted $ab$ for any $a,b \in \mathbb{R}$. To be a field the following properties must hold:

\begin{description} \setlength{\itemsep}{0pt}
\item[Closure:] For all $a,b \in \mathbb{R}$, $a+b, ab \in \mathbb{R}$.
\item[Associativity:] For all $a,b,c \in \mathbb{R}$, $(a+b)+c=a+(b+c)$ and $(ab)c=a(bc)$.
\item[Identity:] There are special elements $0,1 \in \mathbb{R}$ such that $0 \not= 1$ and for all $a \in \mathbb{R}$, $0+a=a=a+0$ and $1a=a=a1$.
\item[Inverses:] For each $a \in \mathbb{R}$ there is some $-a \in \mathbb{R}$ such that $(-a)+a=0=a+(-a)$. Also, when $a\not=0$, there exists $a^{-1} \in \mathbb{R}$ such that $a^{-1}a=1=aa^{-1}$.
\item[Commutativity:] For all $a,b \in \mathbb{R}$, $a+b=b+a$ and $ab=ba$.
\item[Distribution:] For all $a,b,c \in \mathbb{R}$, $(a+b)c=ac+bc$ and $a(b+c)=ab+ac$. 
\end{description}

Next, $\mathbb{R}$ is {\bf totally ordered}. This means that $\mathbb{R}$ is equipped with a relation, denoted $\leq$, such that the following properties hold:

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\item[Reflexive:] For all $a \in \mathbb{R}$, $a \leq a$.
\item[Antisymmetric:] For all $a,b \in \mathbb{R}$, if $a \leq b$ and $b \leq a$, then $a=b$.
\item[Transitive:] For all $a,b,c \in \mathbb{R}$, if $a \leq b$ and $b \leq c$, then $a \leq c$.
\item[Totally Comparability:] For every $a,b \in \mathbb{R}$ we have either $a \leq b$ or $b \leq a$,
\end{description}

To make $\mathbb{R}$ an {\bf ordered field} we must require that its ordering is compatible with $\mathbb{R}$'s field operations:

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\item[Compatibility:] For all $a,b,c \in \mathbb{R}$, if $ a\leq b$, then $a+c \leq b+c$ and if $0 \leq a$ and $0 \leq b$, then $0 \leq ab$.
\end{description}

Maybe it is surprising that all of our familiar algebraic equality and inequality manipulations follow from this list of properties. Things such as ``negative times negative is positive'' and ``$a \leq b$ implies $cb \leq ca$ when $c \leq 0$'' and ``$0 \leq a^2$'' follow from these axioms.

Notice that the rational numbers $\mathbb{Q} = \{ a/b \;|\; a,b \in \mathbb{Z} \mbox{ and } b\not=0 \}$ is an ordered field while $\mathbb{C}$ is not (we cannot impose an order on the complex numbers that plays nicely with its arithmetic: $i^2=-1<0$ whereas in an ordered field $x^2 \geq 0$ for all $x$). Completeness is what sets the reals apart from all other ordered fields.

We say that $\mathbb{R}$ is a {\bf complete ordered field} since it is an ordered field with the following property: Suppose $X$ is a non-empty subset of $\mathbb{R}$. In addition, suppose that there is some $M \in \mathbb{R}$ such that $x \leq M$ for all $x \in X$ (i.e., $X$ is bounded above). Then there exists a number $s = \mathrm{sup}(X)$ (called the {\it supremum} or {\it least upper bound}) such that $x \leq s$ for all $x \in X$ and given any $m \in \mathbb{R}$ such that $x \leq m$ for all $x \in X$, we have $s \leq m$ (i.e., the supremum is the least possible upper bound of $X$). Briefly, we require:

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\item[Completeness:] Let $X$ be a non-empty subset of $\mathbb{R}$. If $X$ is bounded above, then $\mathrm{sup}(X)$ exists.
\end{description}

\THM {}[All Complete Ordered Fields are Isomorphic] Let $\mathbb{F}$ be a complete ordered field. Then there exists an order preserving isomorphism between $\mathbb{R}$ and $\mathbb{F}$. In particular, there is an invertible map $\varphi:\mathbb{R} \to \mathbb{F}$ such that $\varphi(a+b)=\varphi(a)+\varphi(b)$, $\varphi(ab)=\varphi(a)\varphi(b)$, and $a \leq b$ implies $\varphi(a) \leq \varphi(b)$ for all $a,b \in \mathbb{R}$. 

Note that you need to be careful with the term ``complete''. It can mean different things in different contexts. For example, the complex numbers are {\it algebraically complete} (every non-constant complex polynomial has a complex root). This is a different kind of completeness. This is made a little trickier by the fact that the completeness of the real numbers can be characterized in a bunch of different ways. 

First, a quick technical note, any complete ordered field (i.e., $\mathbb{R}$) has the following property:

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\item[Archimedean Property:] For all $x \in \mathbb{R}$ such that $0 \leq x$ there is a positive integer $n$ such that $x \leq n$.
\end{description}

\newpage
Although far from exhaustive, here are a few properties equivalent to completeness (in an Arichimedean field)\dots

\vspace{-0.05in}

\THM Given an ordered field with the Archimedean property,\\ \force\hfill any one of the following statements could replace the completeness axiom:

\vspace{-0.05in}

\begin{itemize} \setlength{\itemsep}{-2.5pt}
\item A supremum (least upper bound) exists for any non-empty set that is bounded above.
\item An infimum (greatest lower bound) exists for any non-empty set that is bounded below.
\item Every Cauchy sequence converges.
\item Every monotone sequence converges.
\item Bolzano-Weierstrauss: Every bounded sequence contains an accumulation point.
\item Every bounded sequence has a convergent subsequence.
\item Given a nested sequence of intervals: $[a_1,b_1] \supseteq [a_2,b_2] \supseteq \cdots$ such that $\displaystyle \lim_{n \to \infty} (b_n-a_n) = 0$, their intersection

\vspace{-0.1in}

contains exactly one real number: $\bigcap\limits_{n=1}^\infty [a_n,b_n] = \{ x\}$.
\item Every continuous function which takes on positive and negative values has a root.
\item Continuous functions satisfy the conclusion of the intermediate value theorem.
\item Every decimal expression: $\pm N.d_1d_2d_3\dots$ (where $N \in \mathbb{Z}_{\geq 0}$ and $d_j \in \{0,1,\dots,9\}$) represents a real number.
\end{itemize}

%\vspace{-0.05in}

While these properties determine how the real numbers must behave, this leaves a dangling issue of {\it do such a system of real numbers actually exist?} Of course the answer is ``Yes!''
There are several approaches to building the real numbers. One could use decimal expansions, but it's incredibly messy to try state exactly how arithmetic is done and then prove that various properties hold. The two standard methods for constructing the real numbers (both with advantages and disadvantages) are sketched out below. Both constructions assume we have already constructed the rational numbers $\mathbb{Q}$.

\vspace{-0.05in}

\begin{description}\setlength{\itemsep}{-1pt}
\item[Dedekind Cuts:] Let $\alpha \subseteq \mathbb{Q}$ such that $\alpha$ is non-empty and $\alpha \not= \mathbb{Q}$. Suppose that given $x \in \alpha$, $y \in \mathbb{Q}$, and $y<x$, then $y \in \alpha$ (if $x$ belongs to $\alpha$ so does everything below $x$). Also, if $x \not\in \alpha$, $y \in \mathbb{Q}$, and $x<y$, then $y \not\in \alpha$ (if $x$ does not belong to $\alpha$ neither does anything above $x$). Such a set $\alpha$ is called a (Dedekind) {\it cut}. Let $\mathbb{R} = \{ \alpha \subseteq \mathbb{Q} \;|\; \alpha \mbox{ is  a cut} \}$.

Now one defines $\alpha \leq \beta$ if and only if $\alpha \subseteq \beta$. It isn't hard to show that this is a total order on $\mathbb{R}$ and it satisfies the completeness axiom (supremums of bounded sets exist).

Next, define $\alpha+\beta = \{ x+y \;|\; x \in \alpha \mbox{ and } y \in \beta\}$. It can be shown that $\alpha+\beta$ is a cut whenever $\alpha$ and $\beta$ are and that this operation satisfies all of the needed properties. It turns out that the set of negative rational numbers is a cut and is the $0$ in $\mathbb{R}$.

Multiplication is a little tricky. For positive cuts $\alpha, \beta > 0$, one defines $\alpha\beta = \{ q \in \mathbb{Q} \;|\; q \leq xy \mbox{ for some } x \in \alpha \mbox{ and } y \in \beta\}$. All other multiplications are defined from the positive case: $\alpha\beta = (-\alpha)(-\beta)$ for $\alpha,\beta <0$, $\alpha\beta = - [(-\alpha)\beta]$ for $\alpha<0$ and $\beta>0$, and $\alpha\beta = - [\alpha(-\beta)]$ for $\alpha>0$ and $\beta<0$. It turns out that $\{ q \in \mathbb{Q} \;|\; q <1\}$ is a cut which takes on the role of $1$ in $\mathbb{R}$.

With inequality, addition, and multiplication defined, it can be shown that all necessary properties hold so $\mathbb{R}$ is a complete ordered field. 
This construction was developed by Richard Dedekind in 1858 (published in 1872). 

\item[Cauchy Sequences:] Let $\{ a_n \}$ be a sequence of rational numbers. If given any $\epsilon \in \mathbb{Q}$ such that $\epsilon >0$ we have some positive integer $N$ such that $|a_n - a_m| < \epsilon$ for all $m,n \geq N$, we say $\{ a_n \}$ is a Cauchy sequence. Moreover, if for each $\epsilon >0$ there exists some $N$ such that $|a_n| < \epsilon$ for all $n \geq N$, we say that $\{ a_n \}$ converges to $0$. 

Let $\{ a_n \}$ and $\{ b_ n \}$ be Cauchy sequences of rational numbers. Define $\{a_n \} \sim \{b_n \}$ if and only if $\{ a_n-b_n \}$ converges to $0$. It isn't hard to show that
$\sim$ is an equivalence relation. Let $[\{a_n\}]$ be the equivalence class of some Cauchy sequence $\{ a_n \}$. Define $\mathbb{R} = \{ [\{ a_n \}] \;|\; \{a_n\} \mbox{ is a Cauchy sequence of rational numbers} \}$ (here the ``real numbers'' are equivalence classes of rational Cauchy sequences).

For $[\{ a_n\}], [\{b_n\}] \in \mathbb{R}$ one defines $[\{ a_n\}] + [\{b_n\}] = [\{ a_n+b_n\}]$ and $[\{ a_n\}]\cdot [\{b_n\}] = [\{ a_n\cdot b_n\}]$. One can show that these are
well-defined operations that turn $\mathbb{R}$ into a field. One can identify each rational number $q \in \mathbb{Q}$ with the equivalence class of the constant sequence: $q,q,q,\dots$ (constant sequences are Cauchy). 

Finally, define $[\{ a_n\}] < [\{b_n\}]$ if and only if for every $\epsilon>0$ there is some $N$ such that $a_n < b_n$ for all $n \geq N$ (i.e., eventually the terms in $\{a_n\}$ are below the terms in $\{b_n\}$). Again, one can show that this relation is well-defined and that together this inequality relation with addition and multiplication make $\mathbb{R}$ into a complete ordered field. 

This construction was somewhat anticipated by Bernard Bolzano (circa 1817)  but wasn't really fleshed out completely until work done by M\'eray, Heine, and Cantor (around 1869--1872).   For more detailed history see: \href{http://www-history.mcs.st-and.ac.uk/HistTopics/Real_numbers_2.html}{http://www-history.mcs.st-and.ac.uk/HistTopics/Real\underline{\;\;}numbers\underline{\;\;}2.html}
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