# Are there natural numbers that are not the descendant of 0?

Based on the Peano Axioms (wich are a way to correctly absolutely define the set of natural numbers – correct me if i’m wrong) it is possible to construct a set of symbols that doesn’t quite look the way i imagine the natural numbers:

If there is a circle of other symbols next to the infinite row of known natural numbers, doesn’t this also fit all the requirements?

So are there multiple unequal sets of natural numbers?

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The axiom of induction fails for the structure above.
– mjqxxxx
Oct 20 at 18:10

1

The peano axioms aren’t so much definitions as… axioms. Definitions are new words for things that already exist. When working in the Peano axioms, one assumes there exists a structure satisfying those axioms… It’s different. More like axioms for set theory or Euclidean geometry.
– Noah Olander
2 days ago

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The induction axiom ensures that N\Bbb N cannot contain a cycle like your a,b,ca,b,c cycle. It says that if

0∈A0\in A, and
for each n∈Nn\in\Bbb N, n∈An\in A implies that n+1∈An+1\in A,

then A=NA=\Bbb N. Take AA to be everything in your diagram except a,ba,b, and cc; this AA satisfies both of these requirements, yet itâ€™s not the whole set shown in your diagram. Thus, the set in your diagram doesnâ€™t satisfy the Peano axioms, and indeed they characterize N\Bbb N.

However, the induction axiom cannot be expressed in first-order logic, and there are structures other than N\Bbb N that satisfy the first-order counterpart of the Peano axioms, though they still donâ€™t contain cycles. All of them are linearly ordered and consist of a copy of the standard N\Bbb N followed by copies of Z\Bbb Z (so that everything except 00 has a unique immediate predecessor). There are restrictions on how these copies of Z\Bbb Z can be ordered relative to one another. For instance, the only possibility for a countable non-standard model looks like N\Bbb N followed by Q×Z\Bbb Q\times\Bbb Z ordered lexicographically.

How does N\mathbb{N} followed by copies of Z\mathbb{Z} satisfy the induction axiom? It seems I should be able to take AA to be the copy of N\mathbb{N}.
– arkeet
Oct 20 at 18:22

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@arkeet: I need to revise that slightly. It doesnâ€™t satisfy the second-order induction axiom that I stated; it does satisfy the first-order axiom schema that replaces the induction axiom if you try to give a corresponding first-order axiomatization.
– Brian M. Scott
Oct 20 at 18:29

The induction axiom is:

If P(0)P(0) holds, and if P(n+1)P(n + 1) follows from P(n)P(n), then PP is true of all values.

If you let S={a,b,c}S = \{a, b, c\}, and P(n)=n∉SP(n) = n \not \in S, then the above axiom fails.

P(0)P(0) is true, P(n+1)P(n + 1) follows from P(n)P(n) (it hold for n∈Sn \in S due to vacuous implication), but P doesn’t hold for all numbers in your structure (it fails for n∈Sn \in S).

So the induction axiom will fail for any structure with non-natural numbers, just let SS be the set of numbers not in the natural number chain.

Going by wikipedia. https://en.wikipedia.org/wiki/Peano_axioms

The first 8 axioms (particularly 8): For every natural number n, S(n) = 0 is false. That is, there is no natural number whose successor is 0.) seem to assure that the is the “number line” 0->1->2->3…. and that it is infinite and not circular. But they don’t say any thing about there not being any a->b->c->d…. or oom<->pah. which are completely distinct and separate from this set of number lines.

They do state that they must be distinct though. If S(S(S(….(a)))) = a then a ≠\ne S(S(…..(0))))). Or if b ∉\not \in 0->1->2->3…. then no predecessor or successor of b is either.

But 9) the induction axiom does imply there are not distinct threads.

If K is a set such that:

0 is in K, and
for every natural number n, n being in K implies that S(n) is in K,

then K contains every natural number.

As K = 0->1->2->3 … is such a set we have N⊂K⊂N\mathbb N \subset K \subset \mathbb N and 0->1->2->3…. are precisely the natural numbers and nothing else are.

This is not just theoretical. 1/2 ->1 1/2 -> 2 1/2 -> 3 1/2 ⊄N\not \subset \mathbb N is just such a separate string that does exist but are not the natural numbers.

I like to think (and I’m probably abusing some theory and I could be wrong) that 1-8 define the smallest thing the natural numbers can be. And the induction axiom restricts it to the smallest it can be. I may be naively abusing math though.