# The reals (constructed as Dedekind cuts) form a field

The real numbers, when constructed as Dedekind cuts over the rationals, form a field.

We shall often write the one-sided Dedekind cut \((A, B)\) noteRecall: “one-sided” means that \(A\) has no greatest element. as simply \(\mathbf{A}\) (using bold face for Dedekind cuts); we can do this because if we already know \(A\) then \(B\) is completely determined. This will make our notation less messy.

The field structure, together with the total ordering on it, is as follows (where we write \(\mathbf{0}\) for the Dedekind cut \((\{ r \in \mathbb{Q} \mid r < 0\}, \{ r \in \mathbb{Q} \mid r \geq 0 \})\)):

\((A, B) + (C, D) = (A+C, B+D)\)

\(\mathbf{A} \leq \mathbf{C}\) if and only if everything in \(A\) lies in \(C\).

Multiplication is somewhat complicated.

If \(\mathbf{0} \leq \mathbf{A}\), then \(\mathbf{A} \times \mathbf{C} = \{ a c \mid a \in A, a > 0, c \in C \}\). we’ve missed out the complement in this notation, and can’t put set-builder sets in boldface

If \(\mathbf{A} < \mathbf{0}\) and \(\mathbf{0} \leq \mathbf{C}\), then \(\mathbf{A} \times \mathbf{C} = \{ a c \mid a \in A, c \in C, c > 0 \}\).

If \(\mathbf{A} < \mathbf{0}\) and \(\mathbf{C} < \mathbf{0}\), then \(\mathbf{A} \times \mathbf{C} = \{\} \) write down the form of the set

where \((A, B)\) is a one-sided Dedekind cut (so that \(A\) has no greatest element).

(Here, the “set sum” \(A+C\) is defined as “everything that can be made by adding one thing from \(A\) to one thing from \(C\)“: namely, \(\{ a+c \mid a \in A, c \in C \}\) in set builder notation; and \(A \times C\) is similarly \(\{ a \times c \mid a \in A, c \in C \}\).)

# Proof

## Well-definedness

We need to show firstly that these operations do in fact produce Dedekind cuts.

### Addition

Firstly, we need everything in \(A+C\) to be less than everything in \(B+D\). This is true: if \(a+c \in A+C\), and \(b+d \in B+D\), then since \(a < b\) and \(c < d\), we have \(a+c < b+d\).

Next, we need \(A+C\) and \(B+D\) together to contain all the rationals. This is true: this, it’s quite boring

Finally, we need \((A+C, B+D)\) to be one-sided: that is, \(A+C\) needs to have no top element, or equivalently, if \(a+c \in A+C\) then we can find a bigger \(a' + c'\) in \(A+C\). This is also true: if \(a+c\) is an element of \(A+C\), then we can find an element \(a'\) of \(A\) which is bigger than \(a\), and an element \(c'\) of \(C\) which is bigger than \(C\) (since both \(A\) and \(C\) have no top elements, because the respective Dedekind cuts are one-sided); then \(a' + c'\) is in \(A+C\) and is bigger than \(a+c\).

### Multiplication

this section

### Ordering

this section

## Additive commutative group structure

identity, associativity, inverse, commutativity

## Ring structure

multiplicative identity, associativity, distributivity

## Field structure

inverses

## Ordering on the field

a ⇐ b implies a+c ⇐ b+c, and 0 ⇐ a, 0 ⇐ b implies 0 ⇐ ab

Parents:

- Real number (as Dedekind cut)
A way to construct the real numbers that follows the intuition of filling in the gaps.