0.1 Appendix: existence and uniqueness of a complete ordered field

This appendix is devoted to the proofs of [link] and [link] , which together assert that there exists a unique complete ordered field.Our construction of this field will follow the ideas of Dedekind, which he presented in the late 1800's.

By a Dedekind cut, or simply a cut, we will mean a pair $(A,B)$ of nonempty (not necessarily disjoint) subsets of the set $Q$ of rational numbers for which the following two conditions hold.

1.   $A\cup B=Q.$ That is, every rational number is in one or the other of these two sets.
2. For every element $a\in A$ and every element $b\in B,$ $A\le b.$ That is, every element of $A$ is less than or equal to every element of $B.$

Recall that when we define the rational numbers as quotients (ordered pairs) of integers, we faced the problem that two different quotients determine the same rational number, e.g., $2/3\equiv 6/9.$ There is a similar equivalence among Dedekind cuts.

Two Dedekind cuts $(A_1,b_1)$ and $(A_2,B_2)$ are called equivalent if $a_1\le b_2$ for all $a_1\in A_1$ and all $b_2\in B_2,$ and $a_2\le b_1$ for all $a_2\in A_2$ and all $b_1\in B_1.$ In such a case, we write $(A_1,B_1)\equiv (A_2,B_2).$

There are three relatively simple-sounding and believable properties of cuts, and we present them in the next theorem.It may be surprising that the proof seems to be more difficult than might have been expected.

Let $(A,B)$ be a Dedekind cut. Then

1. If $a\in A$ and $a^{\text{'}}<a,$ then $a^{\text{'}}\in A.$
2. If $b\in B$ and $b^{\text{'}}>b,$ then $b^{\text{'}}\in B.$
3. Let $ϵ$ be a positive rational number. Then there exists an $a\in A$ and a $b\in B$ such that $b-a<ϵ.$

Suppose $a$ is an element of $A,$ and let $a^{\text{'}}<a$ be given. By way of contradiction suppose that $a^{\text{'}}$ does not belong to $A.$ Then, by Condition (1) of the definition of a cut, it must be that $a^{\text{'}}\in B.$ But then, by Condition (2) of the definition of a cut, we must have that $a\le a^{\text{'}},$ and this is a contradiction, because $a^{\text{'}}<a.$ This proves part (1). Part (2) is proved in a similar manner.

To prove part (3), let the rational number $ϵ>0$ be given, and set $r=ϵ/2.$ Choose an element $a_0\in A$ and an element $b_0\in B.$ Such elements exist, because $A$ and $B$ are nonempty sets. Choose a natural number $N$ such that $a_0+Nr>b_0.$ Such a natural number $N$ must exist. For instance, just choose $N$ to be larger than the rational number $(b_0-a_0)/r.$ Now define a sequence $\left\{a_k\right\}$ of rational numbers by $a_k=a_0+kr,$ and let $K$ be the first natural number for which $a_K\in B.$ Obviously, such a number exists, and in fact $K$ must be less than or equal to $N.$ Now, $a_{K-1}$ is not in $B,$ so it must be in $A.$ Set $a=A_{K-1}$ and $b=A_K.$ Clearly, $a\in A,$ $b\in B,$ and