Suppose I have an ordered field $F$ and $1 \in F$. Then we define the set of natural numbers recursively, here we call it $\mathbb{N}_F$ in the field $F$ via
$1 ∈ \mathbb{N}_F$, and if $\textbf{n} \in \mathbb{N}_F $ then $\textbf{n}+1 ∈ \mathbb{N}_F$, and $\textbf{n} = 1+ …. +1$, where the right side has $n$ "ones".
I want to prove this recurrence relation satisfies the Peano axioms. These kinds of questions look so basic so I don't what tools should I use to prove it/… I just know $\mathbb{N}_F$ is defined as an element $1$ and some successors to complete the set.
$1.$For proving 1 is not the successor of any $k$ ∈ $\mathbb{N}_F$. I am thinking contradiction and showing that there is some $k ∈ \mathbb{N}_F$, and I have that $k+1=1$ (since its a successor). But I don't know what to state next, I know I cannot explicitly define $0$ since it's not an element of $\mathbb{N}_F$.
$2.$ For proving if $\textbf{n},\textbf{m} ∈ \mathbb{N}_F$, and $\textbf{m}+1=\textbf{n}+1$, then $\textbf{n}=\textbf{m}$, this also seems obvious for me but I don't know how to prove it rigorously, as I don't know how to get rid of the $1$ on each side. Maybe I have to apply the fact that its an ordered field but I don't know where to start
$3.$ For $S ⊆ \mathbb{N}_F$, I want to show that $S=\mathbb{N}_F$ if i) $1 ∈ S$ and ii) if $\textbf{n} ∈ S$, then $\textbf{n}+1∈S$
Here I know that $S⊆\mathbb{N}_F$ is already assumed and I only have to prove that $\mathbb{N}_F⊆S$, but its very obvious since $S$ is defined exactly like $\mathbb{N}_F$ but how do I write it formally?
Any inputs are appreciated!
Best Answer
First, you need to define the successor function: $s(n) = n+1$. (Then the simplest definition of $\mathbb{N}_F$ would be as the smallest subset of $F$ that contains $1$ and is closed under $s$. This gives you (3) just by definition.)
You condition (2) follows from the axioms of a field: Assuming that $m,n\in\mathbb{N}_F$ are such that $s(m)=s(n)$ translates into $m+1=n+1$ in $F$. Adding $-1$ (by definition: the additive inverse of $1\in F$) to each side of this equation gives $m=n$, as desired.
To show that $1\ne s(n)$ for all $n\in\mathbb{N}_F$, you need to use that the characteristic of $F$ is zero. Try it yourself.