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.


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.


1 Answer 1


It's basically correct.

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.

  • $\begingroup$ Thank you for replying! Do I need to prove that if $S \subset \mathbb{N}_K$ then $S = N$? $\endgroup$ Jan 29, 2021 at 1:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .