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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.

I am new to analysis and was not really taught about the Peano axioms and can't seem to find anything on the web about exactly what I need to prove in this problem. Can someone please briefly list and/or describe the axioms that need to be shown in order to complete the above proof? Any assistance is much appreciated.

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  • $\begingroup$ You can find the list here. To show that any of those axioms are satisfied by $\mathbb{N}_K$ you'll want to use the fact that $K$ is a completely ordered field, i.e., going back to the field axioms, etc.. $\endgroup$
    – Hayden
    Jan 28, 2021 at 23:37
  • $\begingroup$ @Hayden It starts with "$0$ is a natural number" however in the book I use, it suggests that they start from $1$? $\endgroup$ Jan 28, 2021 at 23:56
  • $\begingroup$ Yeah, whether $0$ is considered a natural number is a bit up to personal preference. Within the first-order axiomatization, replace the first axiom with $\forall x ~(S(x) \neq 1)$, the third axiom with $\forall x ~(x+1 = S(x))$, the fifth axiom with $\forall x ~(x \cdot 1 = x)$, and in the induction axiom schema, replace $0$ with $1$. Although I haven't checked the details carefully, any model of this modified system of axioms should arise as the set of all nonzero elements of a model of the original system, and vice versa. $\endgroup$
    – Hayden
    Jan 29, 2021 at 3:01

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Suppose you are allowed to sum numbers, and if you are in the natural numbers, this is quite natural. And define the operation $S(x) = x+1$ to be it, then we have to believe:

  1. $S(x) = S(y) $, meaning that our $x$ equals $y$, so $S$ is injective;

  2. 1 is the only element, that doesn't succed any other, then there is no $x$ in natural numbers such that $S(x) = 1$;

  3. Suppose there is a subset of natural numbers with 1 inside it, then if in this subset there is another, then it's because $S(x) \in X$, and so, this subset is the natural numbers set.

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  • $\begingroup$ Are these the only three things to be shown? $\endgroup$ Jan 28, 2021 at 23:58
  • $\begingroup$ First, I want to know why $N= 1+1+1+ ... +1$ in this case above as you described? $\endgroup$ Jan 29, 2021 at 0:17

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