# isomorphism of Dedekind complete ordered fields

In the last chapter of Spivak's Calculus, there is a proof that complete ordered fields are unique up to isomorphism. I find the first steps in it somewhat suspicious. Specifically, I believe he is treating numbers as rational numbers which are at best isomorphic to them.

(Here $\mathbb{R}$ is the field of Dedekind cuts. Also: Spivak is going to denote addition in $F$ with $\oplus$, in contrast to addition in $\mathbb{R}$, denoted by $+$.)

Verbatim:

Theorem: If $F$ is a complete ordered field, then $F$ is isomorphic to $\mathbb{R}$.

Proof: Since two fields are defined to be isomorphic if there is an isomorphism between them, we must actually construct a function $f$ from $\mathbb{R}$ to $F$ which is an isomorphism. We begin by defining $f$ on the integers as follows:

$$f(0) = \mathbf{0}$$ $$f(n) = \underbrace{ \mathbf{1} \oplus \text{ } ... \text{ } \oplus \mathbf{1}}_{n \text{ times}} \text{, }\text{ for n > 0}$$ $$f(n) = -( \underbrace{ \mathbf{1} \oplus \text{ } ... \text{ } \oplus \mathbf{1}}_{|n| \text{ times}}) \text{, }\text{ for n < 0.}$$

It is easy to check that

$$f(m + n) = f(m) \oplus f(n)$$ $$f(m \cdot n) = f(m) \odot f(n)$$

for all integers $m$ and $n$, and it is convenient to denote $f(n)$ by $\bf{n}$. We then >define $f$ on the rational numbers by $$f(\frac{m}{n}) = \bf{\frac{m}{n}} = \bf{m \odot n^{-1}}$$

(notice that the $n$-fold sum $\mathbf{1} \oplus \text{ } ... \text{ } \oplus \mathbf{1} \neq \bf{0}$ if $n>0$, since $F$ is an ordered field). This definition makes sense because if $\frac{m}{n} = \frac{k}{l}$, then $ml = nk$, so $\bf{m} \odot \bf{l} = \bf{k} \odot \bf{n}$, so $\bf{m} \odot \bf{n^{-1}} = \bf{k} \odot \bf{l ^ {-1}}.$ It is easy to check that

$$f(r_1 + r_2) = f(r_1) \oplus f(r_2)$$ $$f(r_1 \cdot r_2) = f(r_1) \odot f(r_2)$$

for all rational numbers $r_1$ and $r_2$ and that $f(r_1) \prec f(r_2)$ if $r_1 < r_2$.

My problem with this is - well, actually, I have two problems with this. The first one is,

\begin{align} \text{what does } \underbrace{ \mathbf{1} \oplus \text{ } ... \text{ } \oplus \mathbf{1}}_{n \text{ times}} \text{ mean?} \end{align}

I get that you can say, "look, $1x = x$, $2x = x+x$, $3x = (x+x)+x$, and so on." But hold it, I say: "and so on" sounds like induction. And the $n$ we referred to isn't really an integer (much less a positive integer in $\mathbb{N}$), but rather a member of $\mathbb{R}$. So, how can we induct on its value?

My other question needs still more introduction! If $z$ is an "integer" $\{t \in \mathbb{Q}: t<n\} \in \mathbb{R}$, we say $z \in \mathbb{Z_R}$. By $\mathrm{quot}_{\mathbb{R}}$, we will denote the set of products $\{m \cdot n^{-1} \in \mathbb{R}: m, n \in \mathbb{Z_R} \land n \neq 0 \}$. Finally, put $\Theta_x = \{f(y) \in F: y < x, y \in \mathrm{quot_\mathbb{R}} \}$. In these terms, Spivak subsequently claims things like

"Given $x, y \in \mathbb{R}$, it is clear that $x<y \implies \Theta_x \subset \Theta_y$."

(He really does say "clear". This would be clear if $\mathrm{quot_\mathbb{R}}$ were equal to $\mathbb{Q}$; then it would just require us to say, "if $x$ and $y$ are Dedekind cuts, then the set $X$ of rationals contained in $x$ is a subset of those rationals $Y$ contained in $y$; thus, $f(X) \subset f(Y)$". But you can't freakin' do that!)

His basic goal is to say that $\phi: x \rightarrow \sup \Theta_x$ is an isomorphism (he claims that $\phi$ agrees with $f$ wherever both are defined, i.e. on $\mathrm{quot}_{\mathbb{R}}$).

So, this has all left me wondering,

\begin{align} \text{how do I navigate his multiple meanings for rationals?} \end{align}

I mean, I really don't know how to prove any of his claims without playing fast and loose with how $\mathbb{N}, \mathbb{Q}$ and $\mathrm{quot}_{\mathbb{R}}$ relate.

Can somebody give me some ideas/hints?

(A little note here: when I think of $\mathbb{N}$, formally I think of the minimal successor set, satisfying the Peano axioms. I'm willing to accept $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$ as god-given, but $\{t \in \mathbb{Q}: t<n\} \in \mathbb{R}$ is very different from $n \in \mathbb{N}$. I do know the obvious isomorphism between them.)

I get that you can say, "look, $1x = x$, $2x = x+x$, $3x = (x+x)+x$, and so on." But hold it, I say: "and so on" sounds like induction. And the $n$ we referred to isn't really an integer (much less a positive integer in $\mathbb{N}$), but rather a member of $\mathbb{R}$. So, how can we induct on its value?

You might or might not want to consider $\mathbb{N}$ as a subset of $\mathbb{R}$. If you don't want to do this, then in any case there is a natural inclusion $i : \mathbb{N} \hookrightarrow \mathbb{R}$, and then you can replace each '$n$' in your inductive proof by '$i(n)$', so that what you have is really induction on $\mathbb{N}$.

But actually, this isn't necessary. You can perform induction on any well-founded relation. A relation $R \subset X \times X$ is well-founded if for each non-empty subset $Y \subset X$ there is an element $y \in Y$ with no $R$-predecessor in $Y$; that is, there is no $z \in Y$ with $(z,y) \in R$. For a well-founded relation $R$ on a set $X$, we have the principle of $R$-induction. This states that for a property $\phi$ of elements of $X$, if for each $x$, whenever $\phi$ holds for all $R$-predecessors of $x$, it also holds for $x$ then $\phi$ holds for all $x \in X$.

Mathematical induction on $\mathbb{N}$ is an example of this. For strong induction, the relation $R$ is given by $(x,y) \in R$ if and only if $x<y$. For weak induction, the relation is given by $(x,y) \in R$ if and only if $x+1=y$. But we can transfer this relation from $\mathbb{N}$ to $i(\mathbb{N}) \subseteq \mathbb{R}$ easily, so there's no problem performing induction on 'the natural numbers in $\mathbb{R}$' rather than 'the natural numbers themselves'.

How do I navigate his multiple meanings for rationals?

The idea is again the same. You have the integers $\mathbb{Z}$ and the 'integers' $\mathbb{Z}_{\mathbb{R}}$, and an inclusion $i_{\mathbb{Z}} : \mathbb{Z} \hookrightarrow \mathbb{R}$ taking each integer $n$ to the corresponding Dedekind cut in $\mathbb{Z}_{\mathbb{R}}$. You also have the rationals $\mathbb{Q}$ and the 'rationals' $\text{quot}_{\mathbb{R}}$, and an includion $i_{\mathbb{Q}} : \mathbb{Q} \hookrightarrow \text{quot}_{\mathbb{R}}$ taking each rational number $q$ to the corresponding quotient of Dedekind cuts. All the properties of the integers/rationals transfer across under these identifications, and so you can argue about the subsets of $\mathbb{R}$ by using arguments based on the actual sets of integers and rationals without any problem.

Moral of the story: Stop worrying about whether a 'integer' or 'rational' in $\mathbb{R}$ really is a integer or rational number or not. Whether or not you view the integers and rationals as subsets of $\mathbb{R}$ doesn't matter because when we embed them isomorphically into $\mathbb{R}$, all their properties transfer across. I mean, a real number isn't really a subset of $\mathbb{Q}$ anyway, is it?

• Could you give me an example (maybe, one of the cases Spivak wants to prove)? – Chris Jan 2 '13 at 20:36
• An example of what? A complete ordered field? – Clive Newstead Jan 2 '13 at 20:41
• Maybe, inducting using the natural inclusion? – Chris Jan 2 '13 at 20:47
• Ok sure. Say you have $\mathbb{R}$. The natural inclusion $i_{\mathbb{N}} : \mathbb{N} \hookrightarrow \mathbb{R}$ is given by $i_{\mathbb{N}}(n) = n$, where the $n$ on the right-hand side is the element of $\mathbb{R}$ which is '$n$ units along the number line'. If you're viewing $\mathbb{R}$ as the set of Dedekind cuts of $\mathbb{Q}$, then you could set $i_{\mathbb{N}}(n) = \{ q \in \mathbb{Q}\, :\, q < n \}$. – Clive Newstead Jan 2 '13 at 20:51
• Another complete ordered field is $F = \{ (x,2x)\, :\, x \in \mathbb{R} \} \subseteq \mathbb{R}^2$, which is the line through the origin with gradient $2$. The isomorphism $\mathbb{R} \to F$ is given by $x \mapsto (x,2x)$. We also have copies of $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ inside $F$ given by, for instance, $\mathbb{N}_F = \{ (n, 2n)\, :\, n \in \mathbb{N}_{\mathbb{R}} \}$, where I've used $\mathbb{N}_{\mathbb{R}}$ to denote the copy of $\mathbb{N}$ in $\mathbb{R}, whatever that is. – Clive Newstead Jan 2 '13 at 20:52 ‘I do know the obvious isomorphism between them.’ And presumably you also know the isomorphism between$\operatorname{quot}_{\Bbb R}$and$\Bbb Q$, whatever your particular definition of$\Bbb Q$may be. Such isomorphisms are the answer to all of your question. Spivak simply identifies$\Bbb N$with its image in$\Bbb R$and$\Bbb Q$with$\operatorname{quot}_{\Bbb R}$in order to reduce notational clutter. If you really wish to do so, you can introduce names for the various isomorphisms and rewrite all of the statements that are bothering you in formally correct fashion. It is perhaps worthwhile to carry out such an exercise once, but in practice the more careful version simply obscures the real idea. Added: ‘Specifically, I believe he is treating numbers as rational numbers which are at best isomorphic to them.’ Although I haven’t a copy of the book to hand, I’m pretty sure that at this point Spivak takes it for granted that$\Bbb N\subseteq\Bbb Q\subseteq\Bbb R$. No, this isn’t the$\Bbb N$from which you built$\Bbb Z$as a set of equivalence classes of ordered pairs; but that$\Bbb N$has an isomorphic copy in$\Bbb Z$, and isomorphic means that as far as algebraic and order properties are concerned the two are interchangeable. That$\Bbb Z$and it’s isomorphic copy of$\Bbb N$aren’t subsets of the$\Bbb Q$that you constructed from it, but they have isomorphic copies in that$\Bbb Q$, which again are interchangeable with their originals as far as algebraic and order properties are concerned. And that$\Bbb Q$and its$\Bbb Z$and$\Bbb N$aren’t subsets of the$\Bbb R$that you construct via Dedekind cuts (which isn’t the one that you construct via equivalence classes of Cauchy sequences, or the one that can be constructed by Conway’s surreal number procedure), but they have isomorphic copies in that$\Bbb R$, to which the same boring refrain applies. And neither of these versions of$\Bbb Q$is literally identical to the field of quotients of the copy of$\Bbb N$that lives in this$\Bbb R$, but both are isomorphic to it. And at this point, when the goal is to prove that up to isomorphism$\Bbb R$is the unique complete ordered field, none of these formal details of construction matter: you’re entitled to use all of the standard properties of$\Bbb R$, and to think of real numbers as just real numbers, not as special sets of rational numbers. • @BrianMScott I worked part of this out in Rudin last year; I saw how$\mathbb{Q}$is isomorphic to a subfield of$\mathbb{R}$. I assume that it's pretty similar for$\mathbb{N}$. I genuinely do not see how to go about here; otherwise I wouldn't have spent the past hour typesetting. (i.e., it isn't true that "if I really wish to do so", I can; I don't see how to, and that's why I asked.) – Chris Jan 2 '13 at 20:10 • @user1296727: You have some isomorphism$h:\Bbb N\to\Bbb Q$, and you have an isomorphism$e:\Bbb Q\to\Bbb R$, so you have an isomorphism$e\circ h:\Bbb N\to\Bbb R$. Given those, I honestly don’t see what the difficulty is. – Brian M. Scott Jan 2 '13 at 20:12 • Alright, and I took that for granted. But how can I use it? The difficulty is, there are several instances where specific properties of those structures are used, and I don't know how to show they hold just as well in their isomorphisms. Like induction on the range of$e \circ h$. – Chris Jan 2 '13 at 20:13 • @user1296727: Does the example that I just added help? For induction on$\operatorname{ran}e\circ h$, just use$(e\circ h)^{-1}$to transfer statements to$\Bbb N$, do the induction there, and use the isomorphism to transfer back to the copy of$\Bbb N$in$\Bbb R$. These are isomorphisms: they preserve everything that you care about. – Brian M. Scott Jan 2 '13 at 20:19 • I don't think that's right. I think you want the isomorphism mapping$\mathrm{quot}_{\mathbb{R}} \rightarrow \mathbb{Q}\$. More to the point: could you tell me a bit about how this makes things work? – Chris Jan 2 '13 at 20:20