Set of Cuts forms Ordered Field

Theorem

Let $\CC$ denote the set of cuts.

Let $\struct {\CC, +, \times, \le}$ denote the ordered structure formed from $\CC$ and:

the operation $+$ of addition of cuts

the operation $\times$ of multiplication of cuts

the ordering $\le$ of cuts.

Then $\struct {\CC, +, \times, \le}$ is an ordered field.

Proof

First we show that $\struct {\CC, +, \times}$ is a field by demonstrating that it fulfils the field axioms:

The properties of a field are as follows.

For a given field $\struct {F, +, \circ}$, these statements hold true:

\((\text A 0)\)	$:$		Closure under addition	\(\ds \forall x, y \in F:\)	\(\ds x + y \in F \)
\((\text A 1)\)	$:$		Associativity of addition	\(\ds \forall x, y, z \in F:\)	\(\ds \paren {x + y} + z = x + \paren {y + z} \)
\((\text A 2)\)	$:$		Commutativity of addition	\(\ds \forall x, y \in F:\)	\(\ds x + y = y + x \)
\((\text A 3)\)	$:$		Identity element for addition	\(\ds \exists 0_F \in F: \forall x \in F:\)	\(\ds x + 0_F = x = 0_F + x \)				$0_F$ is called the zero
\((\text A 4)\)	$:$		Inverse elements for addition	\(\ds \forall x \in F: \exists x' \in F:\)	\(\ds x + x' = 0_F = x' + x \)				$x'$ is called a negative element
\((\text M 0)\)	$:$		Closure under product	\(\ds \forall x, y \in F:\)	\(\ds x \circ y \in F \)
\((\text M 1)\)	$:$		Associativity of product	\(\ds \forall x, y, z \in F:\)	\(\ds \paren {x \circ y} \circ z = x \circ \paren {y \circ z} \)
\((\text M 2)\)	$:$		Commutativity of product	\(\ds \forall x, y \in F:\)	\(\ds x \circ y = y \circ x \)
\((\text M 3)\)	$:$		Identity element for product	\(\ds \exists 1_F \in F, 1_F \ne 0_F: \forall x \in F:\)	\(\ds x \circ 1_F = x = 1_F \circ x \)				$1_F$ is called the unity
\((\text M 4)\)	$:$		Inverse elements for product	\(\ds \forall x \in F^*: \exists x^{-1} \in F^*:\)	\(\ds x \circ x^{-1} = 1_F = x^{-1} \circ x \)
\((\text D)\)	$:$		Product is distributive over addition	\(\ds \forall x, y, z \in F:\)	\(\ds x \circ \paren {y + z} = \paren {x \circ y} + \paren {x \circ z} \)

These are called the field axioms.

It has been established from Set of Cuts under Addition forms Abelian Group that $\struct {\CC, +}$ forms an abelian group.

Thus $\text A 0$ through to $\text A 4$ are fulfilled.

It remains to verify that $\struct {\CC, +, \times}$ fulfils the remaining field axioms.

Field Axiom $\text M0$: Closure under Product

From Product of Cuts is Cut:

$\forall \alpha, \beta \in \CC: \alpha \beta \in \CC$

Thus $\struct {\CC, \times}$ is closed.

$\Box$

Field Axiom $\text M1$: Associativity of Product

From Multiplication of Cuts is Associative:

$\paren {\alpha \beta} \gamma = \alpha \paren {\beta \gamma}$

Thus the operation $\times$ of multiplication of cuts is associative on $\struct {\CC, \times}$.

$\Box$

Field Axiom $\text M2$: Commutativity of Product

From Multiplication of Cuts is Commutative:

$\alpha \beta = \beta \alpha$

Thus the operation $\times$ of multiplication of cuts is commutative on $\struct {\CC, \times}$.

$\Box$

Field Axiom $\text M3$: Identity for Product

Consider the rational cut $1^*$ associated with the (rational) number $1$:

$1^* = \set {r \in \Q: r < 1}$

From Cut Associated with 1 is Identity for Multiplication of Cuts:

$\alpha \times 1^* = \alpha$

for all $\alpha \in \CC$.

From Multiplication of Cuts is Commutative it follows that:

$1^* \times \alpha = \alpha$

That is, $1^*$ is the identity element of $\struct {\CC, \times}$.

$\Box$

Field Axiom $\text M4$: Inverses for Product

We have that $1^*$ is the identity element of $\struct {\CC, \times}$.

Let $\alpha \ne 0^*$.

From Existence of Unique Inverse Element for Multiplication of Cuts, there exists a unique cut $\dfrac 1 \alpha$ such that:

$\alpha \times \dfrac 1 \alpha = 1^*$

From Multiplication of Cuts is Commutative it follows that:

$\dfrac 1 \alpha \times \alpha = 1^*$

Thus every element $\alpha$ of $\struct {\CC, \times}$ such that $\alpha \ne 0^*$ has an inverse $\dfrac 1 \alpha$.

$\Box$

Field Axiom $\text D$: Distributivity of Product over Addition

From Multiplication of Cuts Distributes over Addition:

$\alpha \paren {\beta + \gamma} = \alpha \beta + \alpha \gamma$

for all $\alpha$, $\beta$ and $\gamma$ in $\CC$.

$\Box$

All the field axioms are thus seen to be fulfilled, and so $\struct {\CC, +, \times}$ is a field.

$\Box$

Finally it is shown that $<$ is a total ordering on $\struct {\CC, +, \times}$:

This is demonstrated in Ordering on Cuts is Total.

From Properties of Ordered Field, this is all that is required to be shown.

$\blacksquare$

Sources

• 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Real Numbers