11.2.3 Dedekind reals are Dedekind complete

We obtained $\mathbb{R}_{\mathsf{d}}$ as the type of Dedekind cuts on $\mathbb{Q}$ . But we could have instead started with any archimedean ordered field $F$ and constructed Dedekind cuts on $F$ . These would again form an archimedean ordered field $\bar{F}$ , the Dedekind completion of $F$ , with $F$ contained as a subfield. What happens if we apply this construction to $\mathbb{R}_{\mathsf{d}}$ , do we get even more real numbers? The answer is negative. In fact, we shall prove a stronger result: $\mathbb{R}_{\mathsf{d}}$ is final.

Say that an ordered field $F$ is admissible for $\Omega$ when the strict order $<$ on $F$ is a map ${<}:F\to F\to\Omega$ .

Theorem 11.2.1.

Every archimedean ordered field which is admissible for $\Omega$ is a subfield of $\mathbb{R}_{\mathsf{d}}$ .

Proof.

Let $F$ be an archimedean ordered field. For every $x : F$ define $L,U:\mathbb{Q}\to\Omega$ by

L_{x}(q):\!\!\equiv(q<x)\qquad\text{and}\qquad U_{x}(q):\!\!\equiv(x<q).

(We have just used the assumption that $F$ is admissible for $\Omega$ .) Then $(L_{x},U_{x})$ is a Dedekind cut. Indeed, the cuts are inhabited and rounded because $F$ is archimedean and $<$ is transitive, disjoint because $<$ is irreflexive, and located because $<$ is a weak linear order. Let $e:F\to\mathbb{R}_{\mathsf{d}}$ be the map $e(x):\!\!\equiv(L_{x},U_{x})$ .

We claim that $e$ is a field embedding which preserves and reflects the order. First of all, notice that $e(q)=q$ for a rational number $q$ . Next we have the equivalences, for all $x, y : F$ ,

x<y\Leftrightarrow(\exists(q:\mathbb{Q}).\,x<q<y)\Leftrightarrow(\exists(q:% \mathbb{Q}).\,U_{x}(q)\land L_{y}(q))\Leftrightarrow e(x)<e(y),

so $e$ indeed preserves and reflects the order. That $e(x+y)=e(x)+e(y)$ holds because, for all $q:\mathbb{Q}$ ,

q<x+y\Leftrightarrow\exists(r,s:\mathbb{Q}).\,r<x\land s<y\land q=r+s.

The implication from right to left is obvious. For the other direction, if $q<x+y$ then there merely exists $r:\mathbb{Q}$ such that $q-y<r<x$ , and by taking $s:\!\!\equiv q-r$ we get the desired $r$ and $s$ . We leave preservation of multiplication by $e$ as an exercise. ∎

To establish that the Dedekind cuts on $\mathbb{R}_{\mathsf{d}}$ do not give us anything new, we need just one more lemma.

Lemma 11.2.2.

If $F$ is admissible for $\Omega$ then so is its Dedekind completion.

Proof.

Let $\bar{F}$ be the Dedekind completion of $F$ . The strict order on $\bar{F}$ is defined by

((L,U)<(L^{\prime},U^{\prime})):\!\!\equiv\exists(q:\mathbb{Q}).\,U(q)\land L^% {\prime}(q).

Since $U(q)$ and $L^{\prime}(q)$ are elements of $\Omega$ , the lemma holds as long as $\Omega$ is closed under conjunctions and countable existentials, which we assumed from the outset. ∎

Corollary 11.2.3.

The Dedekind reals are Dedekind complete: for every real-valued Dedekind cut $(L,U)$ there is a unique $x:\mathbb{R}_{\mathsf{d}}$ such that $L(y)=(y<x)$ and $U(y)=(x<y)$ for all $y:\mathbb{R}_{\mathsf{d}}$ .

Proof.

By \autoreflem:cuts-preserve-admissibility the Dedekind completion $\bar{\mathbb{R}}_{\mathsf{d}}$ of $\mathbb{R}_{\mathsf{d}}$ is admissible for $\Omega$ , so by \autorefRD-final-field we have an embedding $\bar{\mathbb{R}}_{\mathsf{d}}\to\mathbb{R}_{\mathsf{d}}$ , as well as an embedding $\mathbb{R}_{\mathsf{d}}\to\bar{\mathbb{R}}_{\mathsf{d}}$ . But these embeddings must be isomorphisms, because their compositions are order-preserving field homomorphisms which fix the dense subfield $\mathbb{Q}$ , which means that they are the identity. The corollary now follows immediately from the fact that $\bar{\mathbb{R}}_{\mathsf{d}}\to\mathbb{R}_{\mathsf{d}}$ is an isomorphism. ∎

Title	11.2.3 Dedekind reals are Dedekind complete
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