$\def\N{\mathbf{N}}$ The peano axioms:
- $0\in\N$
- $n\in\N \implies S(n)\in\N$
- $\forall n\in\N, S(n)\neq0$
- $\forall n,m\in\N, n=m \iff S(n)=S(m)$
- $[(0\in X) \wedge(\forall n\in\N, n\in X \implies S(n)\in X)]\implies X\supseteq\mathbf{N}$
I understand how the first 4 axioms alone are not enough to specify the natural numbers "as we know them." But I have yet to see a proper proof that, with the help of the 5th axiom of induction, this locks us into one and only one set, i.e. "our" natural numbers.
In particular, say we have a set $A=\{0,S(0),S(S(0)),\ldots\} \cup \{\omega,S(\omega),S(S(\omega)),\ldots\}$ Where $\omega$ behaves like another zero, in that it has no predecessor. (and also $\omega\neq0$). Using all 5 axioms, prove that this cannot be the set of all natural numbers.
Proof attempt. $A$ has the property that $0\in A$ and $n\in A \implies S(n)\in A$. Then by axiom (5), we can say $A\supseteq\N$. In particular, $\omega\in\N,S(\omega)\in\N$ and so on. I don't see how we have reached a contradiction yet. And I don't know how to proceed.
All in all, I'm of the opinion that instead of axiom 5, this axiom is simpler and should be equivalent.
5*. Let $m\in\N$. If for all $n\in\N$, we have $S(n)\neq m$, then $m=0.$
In other words, $0$ is the only number without a predecessor. Shouldn't we able to derive induction from this? Also the proof from above becomes very simple with this axiom instead of the induction axiom, since the existence of $\omega$ contradicts this axiom directly.
The Question. Can Axiom 5* replace Axiom 5 as the fifth axiom? If it cannot, how do we prove that the set $A$ defined as above is not the natural numbers?
EDIT: User Akiva Weinberger has shown that we cannot replace axiom 5 with axiom 5* in their comment to this post. It thus remains to prove $A$ as defined above is not the natural numbers, using the axiom of induction.