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As a consequence of Godel's incompleteness theorem, no axiomatic system can be the definition of standard natural numbers because any axiomatic system of arithmetic will always be satisfied by non-standard models too. E.g.-

  1. Robinson's arithmetic may be satisfied by models whose elements are neither even nor odd

  2. Peano arithmetic may be satisfied by models whose elements do not satisfy Goodstein's theorem

The natural number system that humans intuitively know are either even or odd and they do satisfy Goodstein's theorem

But what exactly are the natural numbers that humans intuitively know? What's their mathematical definition? Since no axiomatic system (even ZFC) can be taken to define them, is there some non-axiomatic definition of them?

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  • $\begingroup$ Second-order arithmetic; see also this post $\endgroup$ Mar 8, 2023 at 11:59
  • $\begingroup$ @MauroALLEGRANZA But doesn't this just push the can one step down to road to "defining the unique model of the second-order arithmetic"? How do we define that model? $\endgroup$
    – Ryder Rude
    Mar 8, 2023 at 12:11
  • $\begingroup$ @MauroALLEGRANZA It says this in the accepted answer : "Anything we could do with full second-order semantics, we could do in our usual first-order semantics by just talking about the standard model explicitly". So there is still the problem of defining what the standard model is mathematically $\endgroup$
    – Ryder Rude
    Mar 8, 2023 at 12:16
  • $\begingroup$ By the Lowenheim-Skolem Theorem, there are nonstandard models of arithmetic that satisfy all of the same logical properties as (are logically identical to) the standard model. Incompleteness isn't the thing that's giving us nonstandard models. $\endgroup$
    – TomKern
    Mar 8, 2023 at 16:38
  • $\begingroup$ @TomKern Interesting. But then, in what sense are they different from the standard model, for them to be called "non-standard"? Also, you used the term "the standard model". What is the definition of the standard model? $\endgroup$
    – Ryder Rude
    Mar 8, 2023 at 17:03

1 Answer 1

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Your use of the word "intuitive" means that we're entering philosophical waters.

In ZFC, as you know, one can prove the formal version of the assertion "there is, up to isomorphism, only one model of the second-order Peano axioms". But as you are also aware, there are non-standard models of ZFC (assuming it's consistent), and these can have non-standard omegas.

Now to someone who adopts a certain kind of formalist attitude, there's no more to be said. For an intuitionist like Brouwer, on the other hand, our intuition of the natural numbers is basic, and not something that one can (or should even try) to formalize.

For most Platonists, it's simply assumed (an "article of faith") that the entities of mathematics exist in some "universe of abstractions". $\mathbb{N}$ is then a denizen of this universe. Our axiom systems, like PA and ZFC, capture some of the "facts" about this universe that we somehow know.

If you're asking for a precise mathematical definition of the natural numbers, I would say either "there isn't any", or "omega, working inside the ZFC axioms, or something along those lines".

Consider why you're sure that Goodstein's theorem holds for "the standard natural numbers". Surely it's because you believe in ZFC, in some sense. Or at least some fragment of ZFC. In other words, the power set of $\omega$ is ascribed some sort of reality.

Now consider ZFI, which is ZFC plus "there is an inaccessible cardinal". The consistency of ZFI does not follow from the consistency of ZFC. The statement Con(ZFI) is, however, a statement in PA (but of course not provable in PA). Are you sure it's true? Likewise for even stronger large cardinal axioms, or other extensions of ZFC that cannot be shown to be relatively consistent.

As it happens, John Baez and I carried on a lengthy discussion of these matters. Baez said:

Roughly, my dream is to show that “the” standard model is a much more nebulous notion than many seem to believe.

Along the way we considered the "ultra-finitists", who consider even such a notion as $2^{100}$ as unclear.

You'll find the conversation here. Post 5 and post 6 especially focus on your question.

One last comment: regardless of one's philosophical stance, if someone says "This is true of the standard natural numbers", the purely mathematical content of this claim usually amounts to "this can be proved for $\omega$ in ZFC" (or some fragment of ZFC). At least that's how I would understand their assertion.

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  • $\begingroup$ "Consider why you're sure that Goodstein's theorem holds for "the standard natural numbers". Surely it's because you believe in ZFC, in some sense." ← This is completely wrong. It's easy to prove Goodstein's theorem in 2nd-order arithmetic, or 3rd-order if you don't even want to do coding. One can easily believe either of these while being skeptical of the arithmetical soundness of ZFC. Nobody truly believes ZFC in any platonic sense, because it doesn't make platonic sense once you try to stuff unbounded Specification and Replacement in. $\endgroup$
    – user21820
    Oct 9, 2023 at 14:01
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    $\begingroup$ @user21820 "Nobody truly believes ZFC in any platonic sense". That's completely wrong. Plenty of people believe ZFC in full, plenty don't, plenty don't care that much, and on and on for all variations. As for your other point, I didn't want to get too far into the weeds, that's why I said "in some sense". If you trust 2nd order or 3rd order arithmetic, then you believe that the "mathematical universe" allows for quantification over "all sets of natural numbers", which is at least the beginnings of a Platonistic viewpoint. $\endgroup$ Oct 9, 2023 at 14:22
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    $\begingroup$ "...it doesn't make platonic sense once you try to stuff unbounded Specification and Replacement in." How is this based on anything except your personal philosophy of mathematics? To my mind, claiming that unrestricted quantification is fine and dandy for the first few levels of the hierarchy but not beyond is a much stranger position to take. But "you do you", as the saying goes. $\endgroup$ Oct 9, 2023 at 15:54
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    $\begingroup$ @user21820 So in other words, you've just confirmed the fact that this is simply a philosophical disagreement between you and some other mathematicians. Your assertion "Nobody truly believes ZFC in any platonic sense" except people who don't understand it, is on a par with "Nobody actually enjoys opera" or similar pronouncements. For the record, I am not a strong Platonist myself. I will mkae a small modification to my answer to address the one substantive part of your comments $\endgroup$ Oct 9, 2023 at 16:14
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    $\begingroup$ Just saw your last comment. Your "completely wrong" remark is, alas, completely wrong. The original post brought up ZFC, which is why I continued in that vein, with the qualifications "or something along those lines" and "in some sense". Sure, there are all sorts of further philosophical issues one could go into, but that doesn't invalidate what I wrote. $\endgroup$ Oct 9, 2023 at 16:24

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