Timeline for Is the statement "All numbers are counting numbers" independent of $PA$?
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| S Jan 30, 2021 at 22:41 | history | edited | Michael Albanese | CC BY-SA 4.0 |
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| S Jan 30, 2021 at 22:41 | history | suggested | none | CC BY-SA 4.0 |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Dec 17, 2016 at 1:58 | comment | added | Thomas Benjamin | @CarlMummert: If there should be an induction axiom for $C$, need that induction axiom imply the existence of a completed infinity? That seems to be Nelson's primary concern. For what it's worth, Nelson, in his paper " Hilbert's Mistake", makes the following claim: "We cannot say, 'For all numbers $x$ there exists a numeral d such that $x$=$d$' since this is a category mistake conflating the formal with the genetic [genetic as pertaining to origins--my comment]." How does this claim pertain to your comments? | |
| Dec 16, 2016 at 12:41 | comment | added | Carl Mummert | @David Treumann: the deeper question is why $C(x)$ would not be subject to an induction axiom, of course. Nelson does manage to find a delicate balance, but the usual intuition that the natural numbers are the smallest set of a certain kind suggests that there should be an induction axiom for $C$ as well. | |
| Dec 16, 2016 at 12:39 | comment | added | Carl Mummert | @Gro-Tsen: of course, if we extend the language of PA with a unary relation symbol such as $C$ that is true for counting numbers, then we can write a single axiom "$C0 \land (\forall y)(Cy \to C(Sy))$" to express Nelson's axioms, and we could write a single first-order induction axiom involving $C$ that implies all naturals are counting numbers. So there is no essential use of second-order logic, just in case some other readers wonder about that. | |
| Dec 13, 2016 at 17:02 | comment | added | Gro-Tsen | @ThomasBenjamin In second order logic, I would probably write it as the induction axiom ($\forall P((P(0)\land\forall n(P(n)\Rightarrow P(n')))\Rightarrow\forall n(P(n))$), but I'm not sure what Nelson means so (especially given that he doesn't believe in the consistency of some things that seem obvious to me) maybe that's not right. But it can't be as simple as $\alpha$. | |
| Dec 13, 2016 at 15:27 | comment | added | Thomas Benjamin | Why the downvote? | |
| Dec 13, 2016 at 14:03 | comment | added | Thomas Benjamin | @Gro-Tsen: How then would you properly formalize the statement "Every number is a counting number"? | |
| Dec 13, 2016 at 13:53 | vote | accept | Thomas Benjamin | ||
| Dec 13, 2016 at 13:20 | answer | added | Joel David Hamkins | timeline score: 7 | |
| Dec 13, 2016 at 13:05 | comment | added | David Treumann | I love that article. Nelson is not defining a notion of counting number inside of the peano system, he is extending the peano system by introducing a new predicate C(x), calling it "x is a counting number," and subjecting it to a couple of axioms. But he does not subject C to anything like an induction axiom. Your proposition alpha is also true in this extension of the peano system (call it PA+C?), but you cannot use it to prove that for all x, C(x). | |
| Dec 13, 2016 at 13:04 | comment | added | Gro-Tsen | The statement $\alpha$ is a consequence of PA (it is trivially proved by induction), but it is definitely weaker than the statement "all natural numbers are counting numbers" (which is problematic to formalize except by using second-order logic); $\alpha$ just says that every nonzero number has a predecessor, not necessarily that numbers can be obtained inductively from zero and successor. | |
| Dec 13, 2016 at 12:33 | history | asked | Thomas Benjamin | CC BY-SA 3.0 |