# Peano axioms proof attempt

Given that $$K$$ is an ordered field satisfying the least upper bound property and $$1$$ as the multiplicative identity, the set of natural numbers $$\mathbb{N}_K$$ in $$K$$ is defined as: $$1 \in \mathbb{N}_K$$ and $$N + 1 \in \mathbb{N}_K$$ if $$N \in \mathbb{N}_K$$. In this case, $$N = 1 + ... + 1 \}$$ $$N$$ times. Show that $$\mathbb{N}_K$$ satisfies the Peano axioms.

My attempt:

Let $$S(N) = N + 1$$. We now go ahead and prove each axiom:

1. For all natural numbers $$N \in \mathbb{N}_K$$, $$S(N) \in \mathbb{N}_K$$.

Assume that $$N \in \mathbb{N}_K$$. As $$1, N \in \mathbb{N}_K$$, we have that $$N + 1 \in \mathbb{N}_K$$. Therefore, $$S(N) \in \mathbb{N}_K$$.

1. For all $$N \in \mathbb{N}_K$$, $$S(N) \neq 1$$.

Assume that $$1 < N \in \mathbb{N}_K$$ and $$S(N) = 1$$. We get $$S(N) = 1 \implies N + 1 = 1 \implies N < 1$$. Contradiction.

1. $$N = N' \iff S(N) = S(N')$$.

Assume that $$N, N', S(N)$$, and $$S(N')$$ are natural numbers in $$\mathbb{N}_K$$. We get that

$$N = N' \iff N + 1 = N' + 1 \iff S(N) = S(N')$$

We conclude that $$\mathbb{N}_K$$ satisfies the Peano axioms.

QED.

Is this proof correct? I am new to analysis and Peano axioms. So, can somebody correct me if I am wrong? Any assistance much appreciated.

You should, however, spell out some parts. Namely, you can use subtraction since $$K$$ is a field, so
1. $$N+1=1\implies N=0$$ and in an ordered field we indeed have $$0<1$$ and therefore $$0<1+1+\dots+1=N$$ for any $$N\in\Bbb N_K$$, i.e. $$K$$ must have characteristic $$0$$.
2. $$N+1=N'+1\implies N=N'$$ by subtracting $$1$$ from both sides.
• Thank you for replying! Do I need to prove that if $S \subset \mathbb{N}_K$ then $S = N$? Jan 29, 2021 at 1:03