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\title{Math 171 WIM 2}
\author{Lucas Garron}
\fancyhead[CO,CE]{Math 171 WIM 2 - Lucas Garron}
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\maketitle

In this paper, we will construct the set of real numbers $\R$ as a completion of the rational numbers $\Q$ that satisfies the 9 Real Axioms listed in Appendix A.\\

We presume that the integers $\Z$ are already defined so that they satisfy algebraic and order properties. We further presume we have a natural definition of rational numbers as pairs of integers written ${p\/q}$ with $q \neq 0$. In Appendix B, we list some of the properties we assume of the rational numbers and verify the {\bf F1-F6}, {\bf O1}, {\bf O2} for the rational numbers using the properties of the integers. Importantly, $S$ does not hold, and we can use that property to define the reals.

We would like to define the real numbers $\R$ as equivalence classes of Cauchy sequences of rational numbers by following the construction of metric space completions from lecture / the textbook. For completeness, we state the most important definitions and introduce some of our own.

{\definition A \emph{Cauchy sequence} is a sequence $\{a_n\}$ such that for any $\epsilon > 0$ there is $n \in \N$ for all $m, n > N$, the bound $d(a_m, a_n) < \epsilon$ holds.\\
}

For $\Q$, we assume that $\epsilon$ is rational, and $d$ is the usual metric $d(x, y) = |x - y|$.

{\definition $\mathcal{R}$ denotes the set of Cauchy sequences in $\Q$.\\}

Since a limit only makes sense when we are discussing a sequence that approaches an existing value, it would not make sense to define equality by saying that sequences converge to the same value whose existence has not necessarily been proved yet. However, we can define Cauchy sequences as ``getting close to each other'' just as their respective terms eventually get close to each other.

{\definition $\{x_n\} \sim \{y_n\}$ if $\underset{n \to \infty}{\lim}(x_n, y_n) = 0$\\}

{\definition $\R$ is the set of equivalence classes of the set $\mathcal{R}$ under the relation $\sim$.\\}

Now, let us define addition and multiplication. First, let us choose some convenient and powerfully concise notation for talking about these equivalences classes of sequences:

{\definition For any expression $X$ containing variables $a, b, c, ...$ corresponding to sequences $\{a_n\}$, $\{b_n\}$, $\{c_n\}$, ..., the shorthand expression $\{(X)_n\}$ represents the sequence

\[\{X_1, X_2, X_3, ...\}\]

where $X_i$ is $X$ with any variable in the expression $X$ is replaced by the $i$th of its corresponding sequence (e.g. $a$ by $a_i$). \\}

For example:

\[\{(a+2b)_n\} = \{a_1 + 2b_1, a_2 + 2b_2, a_3 + 2b_3, ...\}\]

Now, we introduce a shorthand for equivalence classes of a sequence we have a name for:

{\definition Given  sequence $\{(a)_n\}$, the shorthand $[a]$ represents the equivalence class of $\{(a)_n\}$ in $\mathcal{R}$. \\}

Note that we have to be careful about using the shorthand in reverse. If we have an element of $\mathcal{R}$, it can be written as $[a]$ for some sequence $\{a_n\}$, but it is not to create a definition based on $[a]$ unless the definition is the same for every sequence in the same class.

Arithmetic operations on equivalence classes are defined as follows.

{\definition $[a] + [b] = [a + b]$ (the equivalence class of $\{(a+b)_n\}$).\\}

{\definition $[a] \* [b] = [a \* b]$ (the equivalence class of $\{(a\*b)_n\}$).\\}

Equivalence class addition and multiplication can easily be seen to be well-defined by the properties of limits applied to the definition of $\sim$. We will show that these definitions satisfy the Real Axioms for the set of $\mathcal{R}$, demonstrating that the equivalence classes provide a satisfactory construction of $\R$.

\section*{Verifying the Real Axioms}

Armed with all these definitions, we would like to verify that the Real Axioms hold for $\R$. Much of this will rely on the fact that brackets act much like parentheses under our definitions.

\subsection*{F1 (Commutativity)}

This is straightforward to show via properties of limits (expressed using our shorthand sequence notation):

\[[a] + [b] = [a+b] = [b+a] = [b] + [a]\]

Similarly for multiplication:

\[[a] \* [b] = [a+b] = [b+a] = [b] \* [a]\]


\subsection*{F2 (Associativity)}

\[[a] + ([b] + [c]) = a + [b+c] = [a+(b+c)] = [(a+b)+c] = [a+b]+c = ([a]+[b])+c\]
\[[a] \* ([b] \* [c]) = a \* [b\*c] = [a\*(b\*c)] = [(a\*b)\*c] = [a\*b]\*c = ([a]\*[b])\*c\]


\subsection*{F3 (Distributivity)}

\[[a]\*([b] + [c]) = [a]\*[b+c] = [a\*(b+c)] = [ab + ac] = [ab] + [ac] = [a][b] + [a][c]\]


\subsection*{F4 (Identity)}

According to our definitions, $[0]$ denotes the equivalence class of $\{z_n\} = \{(0)_n\} = \{0, 0, 0, ...\}$.

\[[0] + [a] = [z] \* [a] = [z + a] = \{(z + a)_n\} = \{(0 + a)_n\} = [a]\]

Thus, $0$ does indeed satisfy the requirement for an additive identity.

$[1]$ denotes the equivalence class of $\{e_n\} = \{(1)_n\} = \{1, 1, 1, ...\}$. It is the multiplicatie identity for $\mathcal{R}$:

\[[1] \* [a] = [e] \* [a] = [e \* a] = \{(e \* a)_n\} = \{(1 \* a)_n\} = [a]\]

$[0]$ is also clearly distinct from $[1]$. 


\subsection*{F5 (Additive Inverse)}
If we define $-[a] = [-a] = \{(-a)_n\}$, then:

\[[a] + (-[a]) = [a] + \{(-a)_n\} = \{(a-a)_n\} = \{(0)_n\} = [0]\]

\subsection*{\emph{Not} F6 (Multiplicative Inverse)}
We will put this on hold in order to prove {\bf O1} first.

\subsection*{O1 (Existence of Sign)}

For any Cauchy sequence $\{a_n\}$, we define:

\[\{a_n\} \in P_0 \iff \{a_n\} \not \in [0]\t{ and }\E N \st a_n > 0\t{ for all }n > N\]

This definition satisfies the condition that exactly one of $\{a_n\} \in P_0, -\{a_n\} \in P_0, \{a_n\} = [0]$ holds for a given Cauchy sequence. We would like to extend this definition to equivalence classes:

\[P = \{r \in \mathcal{R}: a_n \in P_0 \t{ for every }[a] \in r\}\]

In order for this to be a consistent definition, we need to show that $\{a_n\} = \{b_n\}$ only if they ``have the same sign''. There are three cases:

\begin{itemize}
\item $\{a_n\} \in [0]$. Then $\{b_n\} \in [0]$.

\item $\{a_n\} \in P_0$. For any given $\epsilon > 0$, there is some $N$ such that \begin{itemize}
\item $a_N > 0$
\item $|(a_m) - (a_n)| < \epsilon/2$ for all $m, n \geq N$
\item $|(a_n) - (b_n)| < \epsilon/2$ for all $n \geq N$
\end{itemize}

Letting $\epsilon' = a_N/2$, there is an $N'$ such that the corresponding statements hold for $\epsilon'$. This means that $|b_n| > |a_N| - |a_N - a_n| - |a_n - b_n| > 2\epsilon' - \epsilon' - \epsilon' > 0$ for all $n > N'$. Thus, $\{b_n\} \in P_0$

\item $-\{a_n\} \in P_0$. In this case, we can apply the argument on $\{a_n\} \in P_0$ to show that $-\{b_n\} = -\{b_n\} \in P_0$
\end{itemize}

Thus, the sign of every equivalence class is determined consistently by all its elements, and the set $P$ has the desired properties. This means that we can use notation like $[a] > [b]$ for $[a]-[b] \in P$.

\subsection*{F6 (Multiplicative Inverse)}
We would like to show that the definition $[a]\i = [a\i]$ satisfies $[a][a]\i = 1$ for all $[a] \neq [0]$.

We assume that $[a] \in P$ and note that the same argument applies to show that $(-[a])\i = [-a]\i$ satisfies the inverse condition if $-[a] \in P$.\\

To show that $\{(a\i)_n\}$ exists: for any given (rational) $1 > \epsilon > 0$ we would like to find $N$ such that $|(a_m)\i - (a_n)\i| \leq \epsilon$ for all $m, n > N$.

We would like to find a term $a_N$ and a good bound $e$ such that all the $a_i$ (from some point on) are contained in $[a_N - e, a_N + e]$. When we take multiplicative inverses of numbers in this range, they need to map to a sufficiently small interval around $(a_N)\i$, satisfying the Cauchy condition for this arbitrary value of $\epsilon'$. We need to find a value $1 > e > 1$ such that

\begin{enumerate}
\item the interval contains only positive rationals. We need to ensure that $N'$ is a value such that $a_n > 0$ for all $n > N'$ (which must exist because $\{a_n\} \in P$), and that $e < a_{N'}$.
\item the multiplicative inverse map of this interval satisfies the Cauchy condition for $[a]\i$.

Denote $B = a_{N'} - e$, $C = a_{N'} + e$, which must both be positive (i.e. $e < a_{N'}$, which we already need). In order for $C\i < a_n < B\i$ for all $n > N'$, we need that
\[B\i - C\i < \epsilon\]
\[\imp {A+e - (A-e) \/ A^2 - e^2} < \epsilon \imp 2e < \epsilon\*(A^2 - e^2) \imp \epsilon e^2 + 2e < \epsilon \* (a_{N'})^2\]

We assumed that $\epsilon, e < 1$, so this holds when $e < \epsilon \* (a_{N'})^2 / 3$.

For the second condition, we need:

\[a_N\i + \epsilon' > C\i > a_N\i > B\i > A\i - \epsilon'\]

This means that

\[{1\/A\i + \epsilon'} < B < a_N < C < {1\/A\i - \epsilon'}\]

Solving for the conditions gives us that $e < {\epsilon' a_N \/ 1 - \epsilon' a_N}$ and $e < {\epsilon' a_{N'} \/ 1 + \epsilon' a_{N'}}$. (The second condition is stronger.)

\end{enumerate}

If we take $e < a_{N'}$ and $e < {\epsilon' a_{N'} \/ 1 + \epsilon' a_{N'}}$, then there is a value $N$ satisfying the Cauchy condition for $[a]$ and $e$, which also ensures that $|(a_m)\i - (a_n)\i| \leq \epsilon$ for all $m, n > N$.

Since his argument can be adapted to any initial rational $\epsilon > 0$, the multiplicative inverse sequence $\{(a\i)_n\}$ converges and exists. Since $[a\*a\i] = [1]$, it must be the case that $[a][a\i] = 1$, as desired.

\subsection*{O2 (Sign Rule)}

$[a], [b] \in P$ means that there is an $N$ such that $a_n, b_n > 0$ for all $n > N$. This means that $a_n\*b_n, a_n + b_n \in P$ for all $n>0$, so $[a][b], [a]+[b] \in P$

\subsection*{S (Supremum)}

{\definition $B \in \mathcal{R}$ is an upper bound for the set $S \subset \mathcal{R}$ if $s < B$ for all $s \in S$. \\}

Suppose $S$ is a non-empty subset of $\mathcal{R}$ with an upper bound. We need to show that there is a least upper bound. This means that we need to demonstrate the existence of an equivalence class $B$ that is an upper bound on $S$, such that there is no equivalence class $B'$ with $B - B' \in P$ that is also an upper bound on $S$.

If $S$ has a rational least upper bound, then we are done. Therefore, we can assume that $S$ does not have a rational least upper bound.

Let $\{q_i\}$ denote an enumeration of the rational numbers. Let $\{r_i\}$ denote the subsequence of $\{q_i\}$ where an element $q_i$ is in $\{r_i\}$ if $[(s - q_i)] > [0]$ for all $s \in S$, and $r_i < r_j$ for all $j < i$. Thus, $\{r_i\}$ is a strictly decreasing sequence.

$\{r_i\}$ is bounded below. Let $s$ be an element of $S$, and $\{s_i\} \in s$. We can find some $N$ such that $|s_N - s_n| < \epsilon/2$ and $s_n < r_n$ for all $n > N$. bounding $\{r_i\}$ below by $s_N - \epsilon/2$.

$\{r_i\}$ is Cauchy. If not, then there is some $\epsilon > 0$ such that we can find an (infinite) subsequence of $\{r_i\}$ with each term decreasing by at least $\epsilon$ -- which would mean that $\{r_i\}$ is not bounded below. Thus, $[r] \in \mathcal{R}$, and in fact $[r] - s > [0]$ for all $s \in S$, so $[r]$ is an upper bound for $S$.

We need to show that $[r]$ is a least upper bound for $S$. If it were not, there would be a sequence $\{t\}$ that is an upper bound for $S$ with $[t] < [r]$. This means that $[r-t] > 0$, so there is a rational $\epsilon > 0$ such that there is no $N$ with $|r_n - t_n| < \epsilon$ for all $n > N$. There is also an $N'$ such that $|r_i - r_j| < \epsilon/3$ and $|t_i - t_j| < \epsilon/3$ for $i, j > N'$. Letting $c = r_{N'} - \epsilon / 3$, $c < r_i$ but $t_i < c$ for all $i > N'$. This means that $c$ is a rational number satisfying $[c] > s$ for all $s \in S$, and $c < r_i$ for \emph{all} $i \geq 1$, contradicting the definition of $\{r_i\}$ (because $r_i$ does not contain $c$, but $c$ satisfies the conditions for being a term of $\{r_i\}$). Thus, $[r]$ is a least upper bound for $S$.

\section*{Conclusion}
Since all the Real Axioms hold for $\mathcal{R}$, we have constructed a definition of the real numbers as a completion of the rational numbers.
\newpage
\section*{Appendix A: Definitions of the Real Axioms}

In this paper, we say that a set $S$ satisfies the 9 Real Axioms if it satisfies the following

\begin{mylist}{}

\subsection*{Field Axioms}

\item[F1 (Commutativity)] $\forall\ a, b \in S:$ 
\[ a + b = b + a\ \ \ \ \ \ ,\ \ \ \ \ \ a \* b = b \* a \]

\item[F2 (Associativity)] $\forall\ a, b, c \in S:$ 
\[ a + (b + c) = (a + b) + c\ \ \ \ \ \ ,\ \ \ \ \ \ a \* (b \* c) = (a \* b) \* c \]

\item[F3 (Distributivity)] $\forall\ a, b, c \in S:$ 
\[ a \* (b + c) = a \* b + a \* c \]

\item[F4 (Identity)] $\E\ 0, 1$ such that
\[ 0 \neq 1 \]
and $\forall a \in S$:
\[ 0 + a = a + 0 \]
\[ 1 \* a = a \* 1 \]

\item[F5 (Additive Inverse)] $\forall\ a \in S \setminus \{0\}$, $\E a\i \in S$:
\[a \* a\i = 0\]

\item[F6 (Multiplicative Inverse)] $\forall\ a \in S$, $\E -a \in S$:
\[a + (-a) = 0\]

\subsection*{Order Axioms}

There is a subset $P \subset S$ such that the following two axioms hold:

\item[O1 (Existence of Sign)] $\forall\ a \in S$, exactly one of the following holds:
\begin{itemize}
\item $a \in P$
\item $a \not \in P$
\item $a = 0$
\end{itemize}

\item[O2 (Sign Rule)] $\forall\ a, b \in S$:
\[\Big(a, b \in P\Big) \imp \Big(ab, a+b \in P\Big)\]

\subsection*{Supremum Axiom}

\item[S (Supremum)] $\forall\ S \subset \R$:
\[\Big(S\t{ non-empty} \wedge S\t{ bounded above} \Big) \imp \Big(S\t{ has a least upper bound}\Big)\]

\end{mylist}


\newpage
\section*{Appendix B: Verifying the Real Axioms for $\Q$}

We presume that the rational numbers are defined as pairs of integers written ${p\/q}$ with $q \neq 0$, satisfying arithmetic properties/definitions including:

\[q \neq 0\]
\[x = {p\/q} \iff x\*q = p\]
\[{p\/q} + {r\/q} = {p+r\/q}\]
\[{p\/q} \* {s\/t} = {p\*s\/q\*t}\]
\\\\
We will verify axioms {\bf F1-F6}, {\bf O1}, {\bf O2} for $\Q$. (We will not verify $S$ because it is false for $\Q$.)

\begin{mylist}{}

\newcommand{\vertiff}[0]{\begin{turn}{90} $\iff$ \end{turn}}

\subsection*{Field Axioms}

\item[F1 (Commutativity)] $\forall\ a, b \in \Q$, let $a = {p\/q}, b = {s\/t}$. If we multiply through by all the denominators in the rational numbers of the expression, we essentially end up only having to verify the axiom for integers in the numerator.

First we note that ${x\/y} = {x\*z\/y\*z}$, then:
\[
a + b = {p\/q} + {s\/t} = {pt\/qt} + {qs\/qt} = {pt + qs\/qt}
=
{qs + pt\/qt} = {qs\/qt} + {pt\/qt} = {s\/t} + {p\/q} = b + a
\]


\item[F2 (Associativity)] We can essentially multiply through by the denominators again.
\item[F3 (Distributivity)] Multiply through by the denominators again.

\item[F4 (Identity)] $0$ and $1$ are also rational numbers, e.g. $0 = {0\/1}$, $1 = {1\/1}$. $0$ is still not equal to $1$, and $\forall\ {p\/q} \in \Q$:
\[ 0 + {p\/q} = {0 \/ q} + {p\/q} = {0 + p \/ q} = {p\/q} \]
\[ 1 \* {p\/q} = {1 \/ 1} \* {p\/q} = {1 \* p \/ 1 \* q} = {p\/q} \]

\item[F5 (Additive Inverse)] Let ${p\/q}^{-1} = {q\/p}$ and multiply through by the denominators again.

\item[F6 (Multiplicative Inverse)] Let $-{p\/q} = {-p\/q}$ and multiply through by the denominators again.
Define ${p\/q} = 0$ by $p=0$, and ${p\/q} \in P$ by $p\*q \in P$.

\subsection*{Order Axioms}

There is a subset $P \subset S$ such that the following two axioms hold:

\item[O1 (Existence of Sign)] Holds by definition.

\item[O2 (Sign Rule)] $\forall\ {p\/q}, {s\/t} \in \Q$:
\[\Big({p\/q}, {s\/t} \in P\Big) \imp \Big(pq, st \in P\Big) \imp \Big(pqst \in P\Big) \imp \Big({ps\/qt} \in P\Big)\]
The square $a^2 = a\*a$ of an integer $a$ is positive, so:
\[\Big({p\/q}, {s\/t} \in P\Big) \imp \Big(q^2, t^2, pq, st, pqt^2, stq^2, pqt^2 + stq^2, q^2t^2, {pqt^2 + stq^2 \/ q^2t^2}, {pt + sq \/ qt}\*\big({qt \/ qt}\big), {pt + sq \/ qt}, {p\/q} + {s\/t} \in P\Big)\]

\end{mylist}

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