# An alternative axiom to the peano axiom of induction?

$$\def\N{\mathbf{N}}$$ The peano axioms:

1. $$0\in\N$$
2. $$n\in\N \implies S(n)\in\N$$
3. $$\forall n\in\N, S(n)\neq0$$
4. $$\forall n,m\in\N, n=m \iff S(n)=S(m)$$
5. $$[(0\in X) \wedge(\forall n\in\N, n\in X \implies S(n)\in X)]\implies X\supseteq\mathbf{N}$$

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.

• This is Dedekind's categoricity theorem. Nov 29, 2022 at 3:49
• As for replacing Axiom 5 with Axiom 5*, consider the set $\{0,1,2,\dots,x-2,x-1,x,x+1,x+2,\dots\}$, that is, the natural numbers union all polynomials of the form $x+k$ for $k\in\Bbb Z$. This should satisfy your axioms. Nov 29, 2022 at 3:50
• Thank you @AkivaWeinberger for the counter-example. It thus remains to prove rigorously the above proposition using the induction axiom. I have edited the post in light of your comment. Nov 29, 2022 at 4:17

$$5^*$$ cannot replace 5, as shown in the comments. However, $$5^*$$ clearly follows from 5, as we can take $$X = \{n \in \mathbb{N} \mid$$ If $$n$$ has no predecessor, then $$n = 0\}$$. We see easily that $$0 \in X$$ and that for all $$n \in X$$, $$S(n) \in X$$. It follows from 5 that $$X = \mathbb{N}$$, and thus the only number without a predecessor is $$0$$.
• Nice. To further ask, how does axiom 5 rule out Akiva Weinberger's example? i.e. $\{0,1,2,3\ldots\}\cup\{\ldots \omega-2,\omega-1,\omega,\omega+1,\omega+2\ldots\}$ where $\omega\notin\{0,1,2,3\ldots\}$. To go even further, how are we sure that no matter which counter-example I come up with, there will always be a way to prove via axiom 5 that it isn't the natural number set? Nov 29, 2022 at 10:35
• @user1020500 In that instance, take $X = \{0, 1, 2, \ldots\}$. In general, if you try to add numbers to get a set $N’$ with $\mathbb{N} \subseteq N’$, $N’$ won’t satisfy 5, since we can take $X = \mathbb{N}$. Nov 29, 2022 at 16:34
• @MarkSaving What if the set was $A = \{0,S(0),S(S(0)),...\} \cup \{a,b\}$ with $S(a) = b$ and $S(b) = a$. Would it be sufficient to show that $S = \{ n \in \mathbb{N} \mid S(n) \neq a\}$ is equal to $\mathbb{N}$? and thus that $A$ is not a subset of $\mathbb{N}$? Would such a proof rely on the assumption that $a \neq 0,S(0),....$? What about $A = \{0, S(0),S(S(0)),...\} \cup \{\star\}$ where $S(\star) = \star$? Jul 18 at 15:23
• @DiegoCimadom Suppose we have elements $a, b \in \mathbb{N}$ such that $a = S(b)$ and $b = S(a)$. Then show using 5 that $\mathbb{N} \subseteq \mathbb{N} \setminus \{a, b\}$. This is a contradiction. For your $\star$ example, apply the previous result with $a = b = \star$. Jul 18 at 20:59
• @MarkSaving Thanks for the reply. I think this is clearer than what I had. However, I feel like we are still relying on the assumption that $\star \neq 0$ and no successor of zero is equal to $\star$. Should I state these as well or are they implicit in the set theoretic notation $A = \{0, S(0),S(S(0)),...\} \cup \{\star\}$? Jul 19 at 10:04