Comment by raincole

You're right... with a catch.

What you described doesn't rule out {0,1,2... Q-1,Q,Q+1...}, because you only defined how to yield new natural number, but not exclude things that are not yielded that way from N (the set of all natural numbers).

Now, our intuition is to add this missing part into our axioms, right?

So instead:

> 0 is a natural number, and if n is a natural number then so is S(n)

We say:

> For any X⊆N, if 0 ∈ X, and for every n ∈ X, S(n) ∈ X, then X=N.

This is a perfect valid axiom. And it does rule out the nonstandard shit: for a set N' that looks like {0,1,2... Q-1,Q,Q+1...}, we can get X = {0,1,2...}, which is a subset of N'. According to this axiom, if N'=N then X=N', but it clearly doesn't because Q∈X while ~(Q∈N'). Therefore, N' isn't N.

However, this axiom is not included in the commonly accepted Peano Arithmetic! The reason is that this uses second-order logic, and Peano Arithmetic is a first-order theory.

The above axiom effectively defines a predicate, f(X), which accepts a set as input and returns whether the set is N. This is second-order logic.

Peano Arithmetic, being first-order logic, doesn't have such predicate. This is why we can't rule out these nonstandard {0,1,2... Q-1,Q,Q+1...}.

When it comes to ZFC, it's more complicate as in ZFC, 'natural numbers' are ordinals of sets. But ZFC is written in first-order logic as well, and it's known that an axiomatic system written in first-order logic will have nonstandard models. Even if you can rule out {0,1,2... Q-1,Q,Q+1...} by defining PA in ZFC in some unusual way or adding new axioms to ZFC, as long as it's still a first-order theory, it will have 0 (if inconsistent) or multiple (if consistent) models[0].

[0]: https://en.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skolem_...