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.)
Best Answer
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'.
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?