Complete Ordered Fields
1. Fields Defined
A triple \((F,+,\cdot)\) where \(F\) is a set and \(+,\cdot\) are binary operations \(+,\cdot : F \times F \rightarrow F\) is called a field when the following hypotheses are satisfied:
- (Commutativity) For any elements \(x\) and \(y\) of \(F\),\[ x + y = y + x \mbox{ and } x \cdot y = y \cdot x. \]
- (Associativity) For any elements \(x,y\) and \(z\) of \(F\),\[{}(x + y ) + z = x + (y+z){}\]\[{}\text{ and } (x \cdot y) \cdot z = x \cdot (y \cdot z).{}\]
- (Identity Elements) There exist elements \(0,1 \in F\) with \(0 \neq 1\) such that, for all \(x \in F\),\[ 0 + x = x \mbox{ and } 1 \cdot x = x. \]
- (Inverse Elements) For any \(x \in F\), there exists \(-x \in F\) such that\[ x + (-x) = 0. \]Moreover, for any \(x \in F \setminus \{0\}\), there exists \(x^{-1} \in F\) such that\[ x \cdot x^{-1} = 1. \]
- (Distributivity) For any \(x,y,z \in F\),\[ (x + y) \cdot z = (x \cdot z) + (y \cdot z). \]
Essentially all of the familiar facts from arithmetic follow from these basic hypotheses. For example, \(0 \cdot a = 0\) for any \(a \in F\). One possible derivation goes like this:
\[{}0 {}\]
\[{}= (0 \cdot a) + (-(0\cdot a)){}\]
\[{}= ((0 + 0) \cdot a) + (-(0\cdot a)){}\]
\[{}= ((0 \cdot a) + (0 \cdot a)) + (-(0\cdot a)){}\]
\[{}= (0 \cdot a) + ( (0 \cdot a) + (-(0\cdot a))){}\]
\[{}= (0 \cdot a) + 0 = 0 + (0 \cdot a) = 0 \cdot a,{}\]
where the hypotheses used are additive inverses, additive identity, distributivity, additive associativity, additive inverses, additive commutativity, and additive identity properties, respectively. Likewise, for any \(a \in F\), any additive inverse \(-a\) of \(a\) must equal \((-1) \cdot a\) for any additive inverse of \(1\):
\[{}-a{}\]
\[{}= 0 + (-a){}\]
\[{}= (0 \cdot a) + (-a){}\]
\[{}= ((1 + (-1)) \cdot a) + (-a){}\]
\[{}= ((1 \cdot a) + ((-1) \cdot a)) + (-a) {}\]
\[{}= (a + ((-1) \cdot a)) + (-a){}\]
\[{}= (((-1) \cdot a)+ a) + (-a){}\]
\[{}= ((-1) \cdot a) + (a + (-a)) {}\]
\[{}= ((-1) \cdot a) + 0{}\]
\[{}= 0 + ((-1) \cdot a){}\]
\[{}= (-1) \cdot a.{}\]
Here we have used additive identity, \(0 = 0 \cdot a\), additive inverses, distributivity, multiplicative identity, additive commutativity, additive associativity, additive inverses, additive commutativity, and additive identity, respectively. (Note that the field hypotheses do not directly insist that there be only one additive inverse. But one can establish that, in fact, additive and multiplicative inverses are unique.)
2. Ordered Fields
An ordered field is a collection \((F,+,\cdot,<)\) consisting of a set \(F\), binary operations \(+,\cdot\) on \(F\) and a binary relation \(<\) on \(F\) such that \((F,+,\cdot)\) is a field and \(<\) satisfies the following hypotheses:
- (Trichotomy) For any \(x,y \in F\), exactly one of the three statements\[ x < y, \qquad x = y, \qquad y < x \]is true.
- (Transitivity) For any \(x,y,z \in F\),\[ x < y \mbox{ and } y < z \mbox{ imply } x < z. \]
- (Addition) For any \(x,y,z \in F\)\[ x < y \mbox{ implies } x + z < y + z. \]
- (Multiplication) For any \(x,y \in F\) and any \(c \in F\) with \(c > 0\),\[ x < y \mbox{ implies } c \cdot x < c \cdot y. \]
The most important consequence of these order axioms is that \(0 < 1\). (Any \(a \in F\) with \(0 < a\) is, of course, called positive. When \(a < 0\), \(a\) is called negative.)
Theorem
In any ordered field, \(0 < 1\).
Proof
By trichotomy and the fact that \(0 \neq 1\) (as is assumed for any field), either \(0 < 1\) or \(1 < 0\). Since \(0 < 1\) is the conclusion we expect, let us consider the possibility that \(1 < 0\) is instead what one obtains from the trichotomy law. In this case \(1 + (-1) < 0 + (-1)\), meaning \(0 < (-1)\). Consequently \((-1) \cdot 0 < (-1) \cdot (-1)\), giving \(0 < 1\). This contradicts our assumption that \(1 < 0\) (by order trichotomy) so the assumption must have been false and indeed it must be true that \(0 < 1\).
A second important consequence of the ordered field axioms is that when \(c < 0\),
\[ x < y \Rightarrow c \cdot y < c \cdot x. \]
This conclusion may be reached as follows. When \(c < 0\), \(c + (-c) < (-c)\), so \(0 < (-1) \cdot c\). Thus
\[ x < y \Rightarrow ((-1) \cdot c) \cdot x < ((-1) \cdot c) \cdot y. \]
But associativity of multiplication and the equality \(-z = (-1) \cdot z\) gives that
\[ x < y \Rightarrow - (c \cdot x) < - (c \cdot y). \]
Finally, adding \((c \cdot x) + (c \cdot y)\) and using additive commutativity, additive identity, and additive inverse properties leads to the desired conclusion that
\[ x < y \Rightarrow c \cdot y < c \cdot x. \]
3. Completeness
Within the real numbers, every nonempty set which is bounded above has a least upper bound, which is called its supremum. Similarly, every nonempty set which is bounded below has a greatest lower bound, called its infimum.
Exercises
- Suppose that \(F\) is an ordered field and \(a,b,c,d \in F\) satisfy \(a \leq b\) and \(c \leq d\). Prove that \(a + c \leq b + d\).
- In the setup from the previous exercise, prove that \(a + c = b + d\) if and only if \(a = b\) and \(c = d\).
- If \(F\) is an ordered field and \(a,b,c,d \in F\) satisfy \(0 \leq a \leq b\) and \(0 \leq c \leq d\). Prove that \(ac \leq bd\).