Proof of Properties of the Canonical Map from the Natural Numbers to an Ordered Field

Throughout, $S$ denotes the successor map of Natural Numbers, so that $S(n)=n+1$ by statement 1 of Arithmetic of Addition on the Natural Numbers; each appeal to Principle of Induction for the Natural Numbers is to a subset $A\subseteq\mathbb{N}$ containing $1$ and closed under $S$. Finite sums and their properties are those of Finite Sum Notation in a Field and Properties of Finite Sums. For $n\in\mathbb{N}$, $[n]$ is the initial segment determined by $n$ and $u^{n}:[n]\to F$ is the constant family with value $1$ used in The Canonical Map from the Natural Numbers to a Field.

1. Base and step. By claim 1 of Properties of Finite Sums, $\iota_{F}(1)=\sum_{k=1}^{1}u^{1}_{k}=u^{1}_{1}=1$.

Let $n\in\mathbb{N}$. Since $n\le S(n)$ by statements 1 and 5 of Properties of the Order on the Natural Numbers, every $k\in[n]$ lies in $[S(n)]$, and the restriction of $u^{S(n)}$ to $[n]$ is the constant family $u^{n}$. Claim 1 of Properties of Finite Sums therefore gives

$$\iota_{F}(S(n))=\sum_{k=1}^{S(n)}u^{S(n)}_{k}=\Bigl(\sum_{k=1}^{n}u^{n}_{k}\Bigr)+u^{S(n)}_{S(n)}=\iota_{F}(n)+1 .$$

2. The unit is a lower bound. Let $A=\{n\in\mathbb{N}:1\le\iota_{F}(n)\}$. By part 1 and reflexivity of $\le$ we have $1\in A$. Suppose $n\in A$. Statement 6 of Elementary Order Arithmetic in an Ordered Field gives $0<1$, so statement 1 of the same result gives $\iota_{F}(n)+0<\iota_{F}(n)+1$; since $\iota_{F}(n)+0=\iota_{F}(n)$ by Additive Cancellation and Elementary Additive Identities in a Field, part 1 yields $\iota_{F}(n)<\iota_{F}(S(n))$. Combining $1\le\iota_{F}(n)$ with this by mixed transitivity, statement 2 of Elementary Order Arithmetic in an Ordered Field, gives $1<\iota_{F}(S(n))$ and in particular $1\le\iota_{F}(S(n))$, so $S(n)\in A$. By induction $A=\mathbb{N}$.

3. Positivity. Let $n\in\mathbb{N}$. From $0<1$ and $1\le\iota_{F}(n)$, mixed transitivity gives $0<\iota_{F}(n)$; in particular $\iota_{F}(n)\ne 0$. Statement 7 of Elementary Order Arithmetic in an Ordered Field then says that $\iota_{F}(n)^{-1}$ exists and satisfies $0<\iota_{F}(n)^{-1}$.

4. Additivity. Fix $m\in\mathbb{N}$ and let

$$A=\{n\in\mathbb{N}:\iota_{F}(m+n)=\iota_{F}(m)+\iota_{F}(n)\}.$$

Since $m+1=S(m)$, part 1 gives $\iota_{F}(m+1)=\iota_{F}(m)+1=\iota_{F}(m)+\iota_{F}(1)$, so $1\in A$. Suppose $n\in A$. Statements 4 and 2 of Arithmetic of Addition on the Natural Numbers give $m+S(n)=S(n)+m=S(n+m)=S(m+n)$, so by part 1 and the induction hypothesis

$$\iota_{F}(m+S(n))=\iota_{F}(m+n)+1=\bigl(\iota_{F}(m)+\iota_{F}(n)\bigr)+1=\iota_{F}(m)+\bigl(\iota_{F}(n)+1\bigr)=\iota_{F}(m)+\iota_{F}(S(n)),$$

using the associativity of addition in the field $F$. Hence $S(n)\in A$, and by induction $A=\mathbb{N}$.

5. Multiplicativity. Fix $m\in\mathbb{N}$ and let

$$A=\{n\in\mathbb{N}:\iota_{F}(mn)=\iota_{F}(m)\,\iota_{F}(n)\}.$$

Identity 3 of Natural Numbers gives $m\cdot 1=m$, and $\iota_{F}(m)\,\iota_{F}(1)=\iota_{F}(m)\cdot 1=\iota_{F}(m)$ by part 1 and the multiplicative identity of $F$; hence $1\in A$. Suppose $n\in A$. Identity 4 of Natural Numbers gives $m\cdot S(n)=mn+m$, so part 4 and the induction hypothesis give

$$\iota_{F}(m\cdot S(n))=\iota_{F}(mn)+\iota_{F}(m)=\iota_{F}(m)\,\iota_{F}(n)+\iota_{F}(m)\cdot 1=\iota_{F}(m)\bigl(\iota_{F}(n)+1\bigr)=\iota_{F}(m)\,\iota_{F}(S(n)),$$

using the distributivity of multiplication over addition in $F$. Hence $S(n)\in A$, and by induction $A=\mathbb{N}$.

6. Strict monotonicity. Let $m,n\in\mathbb{N}$ with $m<n$. By statement 7 of Properties of the Order on the Natural Numbers there is $k\in\mathbb{N}$ with $n=m+k$, so part 4 gives $\iota_{F}(n)=\iota_{F}(m)+\iota_{F}(k)$. By part 3 we have $0<\iota_{F}(k)$, so statement 1 of Elementary Order Arithmetic in an Ordered Field gives $\iota_{F}(m)+0<\iota_{F}(m)+\iota_{F}(k)$, and $\iota_{F}(m)+0=\iota_{F}(m)$ by Additive Cancellation and Elementary Additive Identities in a Field. Hence $\iota_{F}(m)<\iota_{F}(n)$.

7. Injectivity. We prove the contrapositive. Let $m,n\in\mathbb{N}$ with $m\ne n$. By trichotomy, statement 3 of Properties of the Order on the Natural Numbers, either $m<n$ or $n<m$. In the first case part 6 gives $\iota_{F}(m)<\iota_{F}(n)$ and in the second it gives $\iota_{F}(n)<\iota_{F}(m)$; in either case $\iota_{F}(m)\ne\iota_{F}(n)$.