6.5  Assume the Axioms or Build a Model

Section 6.3 showed that two complete ordered fields are always linked, and by exactly one order-preserving field isomorphism (Theorem 6.3.1). This last section asks what that theorem means for the rest of the series. There are two honest ways to begin analysis: assume the axioms of a complete ordered field, or build a model of them. We compare the two, show what an isomorphism carries from one model to another and what it does not, and list what Analysis on the Real Line takes from this volume.

6.5.1. Two ways to begin analysis

(A) Assume the axioms. Declare that $\mathbb{R}$ is a set with an addition, a multiplication and a relation $<$ satisfying the field axioms (F1)–(F9) and $1\ne 0$ of Definition 2.3.1, the order axioms (O1)–(O4) of Definition 2.3.2, and completeness, Definition 5.6.1: every non-empty subset that is bounded above has a supremum. Then prove theorems from these axioms and nothing else. Analysis on the Real Line, Chapter 1 begins this way. It assumes an ordered field containing $\mathbb{Q}$ in which the least upper bound axiom holds, and every theorem of that volume about real numbers is proved from this assumption, with the logic and set theory of the earlier volumes.

(B) Build a model. Start from the axioms of set theory and construct the objects one by one. This volume did that: the natural numbers as the set $\omega$ in Chapter 1, the integers and rationals as classes of pairs in Chapter 2, and the real numbers as classes of Cauchy sequences of rationals in Chapter 5. The result is a complete ordered field (Theorem 5.6.3), and no property of it was assumed.

Route (A) is shorter, and it is the route of most analysis courses. On its own, though, it leaves two questions open.

Route (B) answers (Q1). The Cauchy model of Chapter 5 is a complete ordered field, built inside the set theory of Chapter 1. So if a contradiction could be derived from the axioms of a complete ordered field, the same derivation, run in the Cauchy model, would be a contradiction about one particular set built in that set theory. In short, the axioms are consistent if the set theory of Chapter 1 is. That is the most a construction can give: it moves the question of consistency from the axioms of $\mathbb{R}$ to the axioms of set theory.

Route (B) alone does not answer (Q2). A second builder who uses Dedekind cuts, as the optional Chapter 7 does, obtains a different set, whose elements are sets of rationals rather than classes of sequences. The answer to (Q2) is the theorem of Section 6.3, and the rest of this section explains why it is the right answer.

6.5.2. What uniqueness adds

Throughout this part, $F$ and $L$ are complete ordered fields and $\phi :F\to L$ is the order-preserving field isomorphism of Theorem 6.3.1. By definition (Definition 6.1.1), $\phi$ is a bijection that respects $+$, $\cdot$, $1$ and $<$. Everything else we need follows in a line or two. Within this section, (P1)–(P6) name the six facts below; they are not the positive-set clauses (P1)–(P2) of Proposition 2.3.3 used in Section 6.4.

The map acts on larger objects through their elements. A subset $A\subseteq F$ goes to its image $\phi (A)=\{\phi (a):a\in A\}$; a sequence $(a_n)$ in $F$ goes to $(\phi (a_n))$; a function $f:F\to F$ goes to $\phi \circ f\circ {\phi }^{-1}:L\to L$. Because $\phi$ is a bijection, $x\in A$ if and only if $\phi (x)\in \phi (A)$, and every subset, sequence or function of $L$ arises in this way from exactly one of $F$.

An isomorphism is a renaming: it changes what the elements are called and keeps every relation between them.

Here is the renaming at work on a statement with three quantifiers.

Example: limits are carried across. Let $(a_n)$ be a sequence in $F$ and $a\in F$. Then $a_n\to a$ in $F$ (Definition 4.3.2) if and only if $\phi (a_n)\to \phi (a)$ in $L$.

For ($\Rightarrow$), let $\epsilon \in L_{>0}$, and put ${\epsilon }^{\prime }:={\phi }^{-1}(\epsilon )$. Then ${\epsilon }^{\prime }>0$ by (P3), since $\phi ({\epsilon }^{\prime })=\epsilon >0$. Since $a_n\to a$, there is $N\in \mathbb{N}$, depending on ${\epsilon }^{\prime }$ and so on $\epsilon$, with $|a_n-a|<{\epsilon }^{\prime }$ for every $n\ge N$. For such $n$, by (P1), (P4) and (P3), $$|\phi (a_n)-\phi (a)|=\phi (|a_n-a|)<\phi ({\epsilon }^{\prime })=\epsilon .$$ So $\phi (a_n)\to \phi (a)$. For ($\Leftarrow$), apply ($\Rightarrow$) to ${\phi }^{-1}$, which is allowed by (P6).

The step that needs $\phi$ to be onto is the first one: the tolerance $\epsilon$ is an element of $L$, and the hypothesis speaks only about tolerances in $F$. Onto is also what carries “converges to some limit” back, since that limit must have a preimage. The map $\kappa ={\iota }_{\mathbb{R}}:\mathbb{Q}\to \mathbb{R}$ (Theorem 5.4.1 (d)) is an order-preserving field map that is not onto, and it fails here: the Newton sequence $(x_n)$ has no limit in $\mathbb{Q}$ (Theorem 4.6.1), while $(\kappa (x_n))$ converges in $\mathbb{R}$ (Theorem 5.4.5) to $r=[(x_n)]$, which is not $\kappa$ of any rational (Example (The Class of the Newton Sequence)).

The example shows the general pattern. Take a statement about a complete ordered field that is phrased with $+$, $\cdot$, $<$, elements, and subsets, sequences and functions built from elements, with quantifiers ranging over these. Rename every object by $\phi$. Each basic piece keeps its truth value: an equation because $\phi$ respects $+$ and $\cdot$ and is injective, with (P1), (P2), (P4) and (P5) for terms that contain $-$, ${}^{-1}$, $|\cdot |$ or rationals; an inequality by (P3) as well; and a membership because $\phi$ is a bijection, as noted above. Quantifiers match as well: “for every element of $F$” matches “for every element of $L$” because $\phi$ is onto; the same holds for subsets, sequences and functions. So the statement is true in $F$ exactly when it is true in $L$. We do not state this as a theorem. Making “statement” precise is the business of formal languages, which Logic and Proof, Chapter 5 begins with its first-order formulas, and the proof would be an induction on how statements are built. A particular instance is checked as the example was: one renaming for each object, and one preservation fact for each basic piece.

Theorems need no uniqueness; truth does. One consequence of route (A) holds for any list of axioms, with no uniqueness theorem: a statement proved from the axioms is true in every structure satisfying them, since the proof can be run in each. The ordered-field axioms alone have the models $\mathbb{Q}$, $\mathbb{Q}(\sqrt{2})$ (Definition 6.4.2) and $\mathbb{R}(x)$ (Definition 6.4.1), and every result proved from those axioms, such as the rules of Proposition 2.3.5, holds in all three. Yet the three disagree about statements the axioms do not settle:

• “some $x$ has $x\cdot x=1+1$” is true in $\mathbb{Q}(\sqrt{2})$ and false in $\mathbb{Q}$ (Theorem 2.6.1);
• “some $y$ lies above $n\cdot 1$ for every $n\in \mathbb{N}$” is true in $\mathbb{R}(x)$ (witness $y=x$, Proposition 6.4.4) and false in $\mathbb{Q}$ (Theorem 2.5.3).

For complete ordered fields no such disagreement can occur. Every statement of the kind described above has the same truth value in all complete ordered fields, whether or not anybody has a proof of it from the axioms. That is what uniqueness adds to route (A): its axioms describe one structure, not a family of structures that agree only on what the axioms prove.

6.5.3. What is not carried across

The renaming keeps relations between elements. It does not keep what the elements are made of. In the Cauchy model, the zero is $0_{\mathbb{R}}=[(0{)}_n]$ (Definition 5.2.2), a class that contains the constant sequence $(0{)}_n$; so the zero of $\mathbb{R}$ is not the empty set. In another complete ordered field the zero can be the empty set, because the elements of a model can be chosen almost at will.

A model with any elements we like. Let $F$ be a complete ordered field, $X$ a set, and $\beta :F\to X$ a bijection. Define, for $u,v\in X$, $$\begin{array}{rl}u\oplus v& :=\beta ({\beta }^{-1}(u)+{\beta }^{-1}(v)),u\odot v:=\beta ({\beta }^{-1}(u){\beta }^{-1}(v)),\\ u\prec v& \text{if and only if}{\beta }^{-1}(u)<{\beta }^{-1}(v).\end{array}$$ Then $X$, with $\oplus$, $\odot$ and $\prec$, is a complete ordered field, and $\beta$ is an order-preserving field isomorphism. Indeed, with $u=\beta (a)$ and $v=\beta (b)$, the definitions read $\beta (a+b)=\beta (a)\oplus \beta (b)$, $\beta (ab)=\beta (a)\odot \beta (b)$, and $a<b$ if and only if $\beta (a)\prec \beta (b)$. Each axiom for $X$, rewritten in this way, becomes the same axiom for $F$: for (F1), $u\oplus v=\beta (a+b)=\beta (b+a)=v\oplus u$; the zero and one of $X$ are $\beta (0)$ and $\beta (1)$, and $\beta (1)\ne \beta (0)$ because $\beta$ is injective and $1\ne 0$ in $F$; (F2)–(F9) and (O1)–(O4) go the same way. For completeness, let $B\subseteq X$ be non-empty and bounded above by $w$. Then ${\beta }^{-1}(B)$ is non-empty and bounded above by ${\beta }^{-1}(w)$, so it has a supremum $s$ in $F$, and $\beta (s)$ is a supremum of $B$, because $\beta$ and ${\beta }^{-1}$ both preserve the order by the definition of $\prec$.

Apply this to the Cauchy model with $X:=(\mathbb{R}\setminus \{0_{\mathbb{R}}\})\cup \{\mathrm{\varnothing }\}$, $\beta (0_{\mathbb{R}}):=\mathrm{\varnothing }$ and $\beta (x):=x$ otherwise. Every element of $\mathbb{R}$ is a class $[a]$, which contains $a$ and so is not empty; hence $\mathrm{\varnothing }\notin \mathbb{R}$, and $\beta$ is a bijection. The zero of $X$ is $\beta (0_{\mathbb{R}})=\mathrm{\varnothing }$. So the statement “the zero element is the empty set” is false in $\mathbb{R}$ and true in $X$, although the two are isomorphic. The statement is about what an element is made of, and no renaming can be expected to keep that.

So “which construction is $\mathbb{R}$?” has no mathematical content. Every construction yields a complete ordered field; any two are linked by exactly one isomorphism; and that isomorphism decides, in the same way in both, every question posed with $+$, $\cdot$ and $<$. Questions such as “is $\sqrt{2}$ a set of rationals?” do depend on the model. In the Cauchy model, $\sqrt{2}$ is the element $r$ of Example (The Class of the Newton Sequence), a set of sequences; in the cut model of Chapter 7 it is the set of rationals below it. Nothing in analysis depends on the answer, and from Analysis on the Real Line on the question is never asked again.

6.5.4. What Analysis on the Real Line takes from here

The analysis volume uses route (A). The table matches each of its assumptions with the place where this series supplies or justifies it.

Some results are proved in both volumes, from different starting points. Analysis on the Real Line, Chapter 1 proves from its axioms that $\mathbb{N}$ is not bounded above, that $\mathbb{Q}$ is dense in $\mathbb{R}$, and that positive numbers have square roots. Here the same three facts were proved for an arbitrary complete ordered field: Theorem 6.1.2, Theorem 4.2.5 (which needs Archimedes, supplied by the first) and Proposition 6.1.4. The two volumes start from different places, and neither volume’s proofs cite the other’s.

This ends the main line of the volume. The optional Chapter 7 builds a second complete ordered field from Dedekind cuts; by Theorem 6.3.1 it is a copy of the Cauchy model, and the chapter can be read as a second look at the rationals-below sets of Section 6.1.

6.5.5. Exercises

A. Check your understanding

B. Practice

C. Going deeper

R. Review