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: