Lesson 0007: Ordered Fields
Let $(F,+,\times)$ be a field and $(F,\preceq)$ a totally ordered set.
For all $a,b\in F$ we define
$$ \begin{aligned}
(a,b) &=\{c\in F: a\prec c\prec b\} \\
(a,b] &= \{c\in F: a\prec c\preceq b\} \\
[a,b) &= \{c\in F: a\preceq c\prec b\} \\
[a,b] &= \{c\in F: a\preceq c\preceq \}
\end{aligned}$$
Any of the above is called a bounded interval, and $a$ is called its left end-point, and $b$ is called its right end-point. End-points are called exclusive if they are delimited by a curved parenthesis. They are called inclusive if it is delimited by a "square bracket".
We define $\infty$ and $-\infty$ as new objects, and define $F^* = F\cup \{\infty,-\infty\}$. We then extend the ordering relation as follows, for every $x\in F^*$.
$$ \begin{aligned}
-\infty \preceq x\preceq \infty
\end{aligned}$$
We define $(F^\ast,\preceq)$ as the extended order of $F$.
We define the unbounded intervals,
$$\begin{aligned}
(-\infty,a) &= \{c\in F: c\prec a\} \\
(-\infty,a] &= \{c\in F: c\preceq a\} \\
(a,\infty) &= \{ c\in F: a\prec c \} \\
[a,\infty) &= \{ c\in F: a\preceq c\} \\
(-\infty,\infty) &= F
\end{aligned}$$
If an interval's end-points are all exclusive, or it has no end-points, then it is called an open interval. If its end-points are all inclusive, or it has no end-points, then it is called a closed interval.
If $X\subseteq F$ then we define the subset of positives,
$$ X^+ = \{a\in X: 0\prec a \}$$
and the subset of negatives
$$ X^- = \{a\in X: a\prec 0\} $$
and the subset of nonnegatives
$$ X^{\succeq 0} = \{a\in X: 0\preceq a\} $$
and the subset of nonpositives
$$ X^{\preceq 0} = \{a\in X: a\preceq 0\}$$
The set of all positive numbers is $F^+$ and the set of all negative numbers is $F^-$.
We say that $\preceq$ is compatible with addition if $\forall a,b,c\in F$
$$ a\preceq b \Rightarrow a+c\preceq b+c $$
We say that $\preceq$ is compatible with multiplication if $\forall a,b\in F$ and $c\in F^{\succeq 0}$,
$$ a\preceq b \Rightarrow ac\preceq bc $$
If $\preceq$ is compatible with both addition and multiplication then we say that $(F,+,\times,\preceq)$ is an ordered field.
Throughout this section, assume $(F,+,\times,\preceq)$ is an ordered field, and $X,Y\subseteq F$ nonempty subsets.
Also let $a,b\in F$ and assume $0\preceq b$.
We say that $\prec$ is compatible with addition if $\forall a,b,c\in F$
$$ a\prec b \Rightarrow a+c\prec b+c $$
We say that $\prec$ is compatible with multiplication if $\forall a,b\in F$ and $c\in F^+$,
$$ a\prec b\Rightarrow ac\prec bc $$
We define the notations:
$$\begin{aligned}
a+X &= \{a+p: p\in X\} \\
aX &= \{ap: p\in X\} \\
X+Y &= \{p+q: p\in X,q\in Y\}\\
XY &= \{ pq: p\in X, q\in Y \}
\end{aligned}$$
Also $-X = (-1)X$.
We define the element $\overline n = \overbrace{1+1+\cdots + 1}^n$, for each $n\in \Bbb N$. Similarly $\overline{-n} = \overbrace{-1-1-\cdots -1}^n$ for each $n\in \Bbb N$. And $\overline 0 = 0\in F$.
For each $m,n\in \Bbb Z$ with $n\ne 0$, we define $\overline{\frac{m}n} = \frac{\overline m}{\overline n}$.
We then define the field of fractions,
$$Q = \{\overline{\frac m n}: m,n\in \Bbb Z, n\ne 0\}$$
Note that we will begin to -- somewhat unofficially -- refer to elements in an ordered field as "numbers".
We do not make the definition of the word "number" official, because it is neither universally accepted nor is it terribly important that we do so.
But if any object is an element of an ordered field, we may refer to it as a number.
Number Theorems
Theorem: Ord. field simult. comp.
If $a\preceq b$ and $c\preceq d$ then $a+b\preceq c+d$.
Also if $a,b,c,d$ are all nonnegative and $a\preceq b$ and $c\preceq d$, then $ac\preceq bd$.
Statement and proof.
Statement: Let $a,b,c,d\in F$ such that $a\preceq b$ and $c\preceq d$. Then $a+c\preceq b+d$.
Also if $a,b,c,d\in F^{\succeq 0}$ and still $a\preceq b$ and $c\preceq d$, then $ac\preceq bd$.
Proof:
$a+c\preceq b+d$.
By compatibility with addition, $a+c\preceq b+c$. Also by compatibility with addition $b+c\preceq b+d$.
By transitivity then $a+c\preceq b+d$.
$ac\preceq bd$.
Assuming $a,b,c,d\in F^{\succeq 0}$, then $ac\preceq bc$ by compatibility with multiplication. Also $bc\preceq bd$.
Therefore by transitivity $ac\preceq bd$.
$\Box$
Theorem: Ord. field strict prec. compat.
Strict precedence is compatible with addition and multiplication.
Statement and proof.
Statement: $\prec$ is compatible with addition and multiplication.
Proof:
Comptability with addition.
Let $a,b,c\in F$ and assume $a\prec b$.
Then $a\preceq b$ and by compatibility of $\preceq$ we have $a+c\preceq b+c$.
Assume for contradiction that $a+c=b+c$. Then $a=b$ which in inconsistent with $a\prec b$.
Therefore $a+c\ne b+c$ which now implies $a+c\prec b+c$, which demonstrates compatibility with addition.
Compatibility with multiplication.
Let $a,b\in F$ and $c\in F^+$. Assume $a\prec b$.
Then $a\preceq b$ and therefore $ac\preceq bc$.
Assume for contradiction that $ac=bc$. Since $c\ne 0$ then $ac/c = bc/c$ which simplifies to $a=b$.
But this is inconsistent with $a\prec b$.
$\Box$
Theorem: Ord. field zero prec. one
$0\prec 1$
Statement and proof.
Statement: $0\prec 1$.
Proof: Assume for contradiction that $1\preceq 0$.
By compatbility with addition, $0\preceq -1$. By compatibility with multiplication therefore, $0\cdot (-1)\preceq (-1)(-1)$.
This simplifies to $0\preceq 1$ and therefore by anti-symmetry, implies $0=1$. But it is an axiom of fields that $0\ne 1$.
$\Box$
Theorem: Ord. field prod. pos. neg.
A product of two positive or two negative numbers is positive. A product of a positive and negative number is negative.
Statement and proof.
Statement: Let $a,b \in F^+$ and $c,d\in F^-$.
Then $ab, cd \in F^+$ and $ac \in F^-$.
Proof: Since $a\in F^+$ then $0\prec a$. Since $b\in F^+$ then by compatibility with multiplication, $0b = 0 \prec ab$.
Since $c,d\prec 0$ then by compatibility with addition $0\prec -c,-d$. From the result above, $(-c)(-d) = cd\in F^+$.
Finally, since $c\prec 0$ then by compatibility with multiplication, $ac\prec 0$ so $ac\in F^-$.
$\Box$
Theorem: Ord. field integer powers
An ordered field follows familiar ordering relations for integer exponents.
Statement and proof.
Statement: Let $a,b,c,d\in F^+$ such that $c\prec 1\prec d$. Then
(a) $a\preceq b$ if and only if $a^n\preceq b^n$ for all $n\in\Bbb N$.
(b) $a\preceq b$ if and only if $\frac 1 b \preceq \frac 1 a$.
(c) $0\prec c^2 \prec c \prec 1 \prec \frac 1 c$.
(d) $0\prec \frac 1 d \prec 1 \prec d\prec d^2$.
Proof:
(a) $a\preceq b$ if and only if $a^n\preceq b^n$.
If $a\preceq b$ then $a^n\preceq b^n$, for all $n\in\Bbb N$.
Suppose $a\preceq b$. As a note which will be useful soon, since $a\in F^+$ then by compatibility, $a^2\preceq ab$ and $ab\preceq b^2$. Therefore $a^2\preceq b^2$.
Now for the base case for induction, with $n=1$, the result is trivial.
For the inductive case, let $1\le n$ be a natural number such that $a^n\preceq b^n$. Then
$$\begin{aligned}
a^{n+1} &= aa^n \\
&\preceq bb^n \\
&=b^{n+1}
\end{aligned}$$
This concludes the inductive case, and therefore the claim holds for all $n\in\Bbb N$.
The converse direction holds trivially, setting $n=1$.
(b) $a\preceq b$ if and only if $\frac 1 b \preceq \frac 1 a$.
Since $a,b\in F^+$ then by compatibility, if $a\preceq b$ then multiplying both sides by $\frac 1 {ab}$ and simplifying, we obtain $\frac 1 b\preceq \frac 1 a$.
For the converse, assume $\frac 1 b\preceq \frac 1 a$ and multiply by $ab$.
(c) $0\prec c^2 \prec c \prec 1 \prec \frac 1 c$.
Since $0\prec c$ we have by compatibility $0c = 0 \prec cc = c^2$.
Since $c\prec 1$ then by compatibility $c^2\prec c$.
Finally, since $c\in F^+$ then $\prec c^{-1}$ exists. Since $1 = cc^{-1}$, if $c^{-1}$ were negative then $1$ would be negative, which we know it isn't. Therefore $0\prec \frac 1 c$. Now because $c\prec 1$ then by compatibility, $1\prec \frac 1 c$.
(d) $0\prec \frac 1 d \prec 1 \prec d\prec d^2$.
The proofs of these statements are nearly the same as for (c).
$\Box$
Set Bound Thereoms
Note that the theorems below all have dual theorems, meaning that anything stated in terms of an upper bound (either a simple upper bound, max, or sup) has a corresponding theorem stated in terms of lower bounds.
In general, stating the dual theorem involves toggling upper bounds (maxima, suprema) to lower bounds (minima, infima), and reversing the direction of inequalities. Therefore it is left to the reader to derive the dual theorems and their proofs.
Theorem: Ord. field dual.
Upper bounds correspond to negative lower bounds of negative sets.
Statement and proof.
Statement: (a) $\alpha\in UB_X \quad \Leftrightarrow \quad -\alpha\in LB_{-X}$. Therefore $UB_X = -LB_{-X}$.
(b) $\alpha=\max (X) \quad\Leftrightarrow \quad -\alpha = \min(-X)$.
(c) $\alpha=\sup(X) \quad\Leftrightarrow\quad -\alpha=\inf(-X)$.
Proof:
(a) $\alpha\in UB_X \quad \Leftrightarrow \quad -\alpha\in LB_{-X}$
By definition, then compatibility with addition, then computation, and then definition,
$$\begin{aligned}
\alpha\in UB_X &\Leftrightarrow \forall p\in X: p\preceq \alpha \\
&\Leftrightarrow \forall p\in X:p-p-\alpha\preceq \alpha-p-\alpha \\
&\Leftrightarrow \forall p\in X:-\alpha\preceq -p \\
&\Leftrightarrow -\alpha\in LB_{-X}
\end{aligned}$$
From the above, it is immediate that $UB_X = -LB_{-X}$.
(b) $\alpha=\max(X) \quad \Leftrightarrow \quad -\alpha=\min(-X)$
This follows from (a) and the fact that $p\in X \Leftrightarrow -p\in -X$ for all $p\in X$.
(c) $\alpha=\sup(X) \quad \Leftrightarrow \quad -\alpha=\min(-X)$
Note that since $UB_X = -LB_{-X}$ it is easy to then show that $-UB_X = LB_{-X}$. Therefore
$$\begin{aligned}
\alpha = \sup(X) = \min(UB_X) &\Leftrightarrow -\alpha = \max(-UB_X) = \max(LB_{-X}) = \inf(-X)
\end{aligned}$$
$\Box$
Theorem: Ord. field alg. sup.
The algebra laws of suprema and infima apply.
Statement and proof.
Statement: If $X, Y$ both have suprema then all following suprema exist and satify these equations.
$$\begin{aligned}
\sup(a+X) &= a+\sup(X) \\
\sup(bX) &= b\sup(X) \\
\sup(X+Y) &= \sup(X)+\sup(Y)\\
\text{ if } X = X^{\succeq 0}, Y = Y^{\succeq 0} &\text{ then } \sup(XY) = \sup(X)\sup(Y)
\end{aligned}$$
Proof:
$\sup(a+X) = a+\sup(X)$
Let $a+p\in a+X$, and observe that since $p\preceq \sup(X)$ therefore $a+p\preceq a+\sup(X)$. This demonstrates both that $UB_{a+X}\ne \emptyset$ and $a+\sup(X)\in UB_{a+X}$.
Now let $\alpha\in UB_{a+X}$ and $p\in X$. Since $a+p\preceq \alpha$ then $p\preceq\alpha-a$ so $\alpha-a\in UB_X$ and then $\sup(X)\preceq \alpha-a$.
Therefore $a+\sup(X)\preceq \alpha$ which now shows $a+\sup(X)=\sup(a+X)$.
$\sup(bX) = b\sup(X)$.
Let $bp\in bX$, and observe that since $p\preceq\sup(X)$ and since $0\prec b$ therefore $bp\preceq b\sup(X)$. This shows that $UB_{bX}\ne \emptyset$ and $b\sup(X)\in UB_{bX}$.
Now let $\alpha\in UB_{bX}$ and $p\in X$. Since $bp \preceq \alpha$, and because $0\prec b$, then $p\preceq \alpha/b$. So $\alpha/b \in UB_X$ and therefore $\sup(X)\preceq \alpha/b$.
Therefore $b\sup(X) \preceq \alpha$ which shows that $b\sup(X)=\sup(a+X)$.
$\sup(X+Y) = \sup(X)+\sup(Y)$.
Let $a\in X, b\in Y$ so that $a+b\in X+Y$. Then $a+b\preceq \sup(X)+\sup(Y)$, so $\sup(X)+\sup(Y)\in UB_{X+Y}$.
Now let $\alpha\in UB_{X+Y}$, and $a\in X,b\in Y$. Then $a+b\preceq \alpha$ therefore $b\preceq \alpha-a$.
This shows that, with $a$ fixed, $\alpha-a\in UB_Y$. Therefore $\sup(Y)\preceq \alpha-a$.
Then $a\preceq \alpha-\sup(Y)$ for all $a\in F$. So $\alpha-\sup(Y)\in UB_X$, and so $\sup(X)\preceq \alpha-\sup(Y)$.
Therefore $\sup(X)+\sup(Y)\preceq \alpha$, which then shows $\sup(X)+\sup(Y) = \sup(X+Y)$.
$\sup(XY) = \sup(X)\sup(Y)$ if $X=X^{\succeq 0}, Y=Y^{\succeq 0}$.
Because $X$ is nonnegative, its supremum can only be 0 or positive.
Consider the case that $\sup(X) = 0$. In that case, $X = \{0\}$ and therefore $XY = \{0\}$, in which case it trivially follows that $\sup(XY)=0=\sup(X)\sup(Y)$.
If $\sup(Y)=0$ the same result follows. Therefore, through the remainder of this proof, assume that neither suprema is zero, and therefore both are positive.
Let $a\in X,b\in Y$ so that $ab\in XY$. Then $ab \preceq \sup(X)\sup(Y)$, therefore $\sup(X)\sup(Y)\in UB_{XY}$.
Now let $\alpha\in UB_{XY}$. Since neither $\sup(X)$ nor $\sup(Y)$ is zero, both $X$ and $Y$ have positive elements. Let $a\in X, b\in Y$ be positive elements.
Then $ab\preceq \alpha$, and therefore $a\preceq \alpha/b$. Moreover $0\prec \alpha/b$, therefore $\alpha/b\in UB_X$ so $\sup(X)\preceq \alpha/b$.
So $b\preceq \alpha/\sup(X)$ and therefore $\alpha/\sup(X)\in UB_Y$. Then $\sup(Y)\preceq \alpha/\sup(X)$.
Therefore $\sup(X)\sup(Y)\preceq \alpha$ which shows that $\sup(X)\sup(Y) = \sup(XY)$.
$\Box$
Theorem: Ord. field func. sum sup.
$\sup(f+g)\preceq \sup(f)+\sup(g)$
Statement and proof.
Statement: If $Z$ is any set and $f,g:Z\to F$ are functions such that $\sup(f)$ and $\sup(g)$ each exist, then $\sup(f+g)\preceq \sup(f)+\sup(g)$.
Proof: Let $r\in Z$. Then $f(r)\preceq \sup(f)$ and $g(r)\preceq \sup(g)$.
Therefore $f(r)+g(r)=(f+g)(r)\preceq \sup(f)+\sup(g)$ which shows $\sup(f)+\sup(g)\in UB_{\text{Im}(f+g)}$. Then $\sup(f+g)\preceq\sup(f)+\sup(g)$.
$\Box$