Timeline for Does anyone still seriously doubt the consistency of $ZFC$?
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14 events
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| Dec 26, 2022 at 5:08 | comment | added | Noah Schweber | Of course this in no way should increase our confidence in $\mathsf{ZFC}$, I just think it's an interesting point; I don't know of any natural foundational-ish theory which $(i)$ doesn't have a "best possible" implementation of $\mathbb{N}$ but $(ii)$ doesn't involve actively destroying all possible such implementations (I'm thinking of Hamkins' multiverse with "well-foundedness mirage" here, IIRC). I guess this is unsurprising since we expect foundational-ish theories to be able to implement some amount of second-order logic, and "pre-theoretic $\mathbb{N}$" is second-order characterizable. | |
| Dec 26, 2022 at 5:06 | comment | added | Noah Schweber | @Burak That said, I think it's worth noting that the usual $\mathsf{ZFC}$-implementation of $\mathbb{N}$ is the best possible given $\mathsf{ZFC}$ as our "implementer," in the following sense: (much less than $\mathsf{ZFC}$ can prove that) given any $\mathcal{M}\models\mathsf{ZFC}$ and any (say) discrete ordered positive semiring $\mathcal{P}$ interpretable in $\mathcal{M}$, $\omega^\mathcal{M}$ is $\mathcal{M}$-definably isomorphic to an initial segment of $\mathcal{P}$. So we don't have a situation with many "candidate implementations" of $\mathbb{N}$ - there's only one plausible choice. | |
| Dec 25, 2022 at 20:27 | comment | added | Burak | @TimothyChow: Yes. The scenario I was suggesting is basically about the smallest inductive set containing "unexpected" elements because all inductive sets do. (I don't want to say "non-standard" because here we are talking about the natural numbers of the metatheory/background theory itself, unless you also have another theory in mind, the natural numbers of which you take as the "standard" one.) In this case it is perfectly possible that ZFC proves an assertion $\exists n \in \omega\ \varphi(n)$ but none of $\varphi(0),\varphi(1),...$ | |
| Dec 25, 2022 at 19:55 | comment | added | Timothy Chow | @Burak In the MO question that JDH linked to, Emil Jeřábek quoted a result that PA is faithfully interpretable in ZFC if and only if ZFC is $\Sigma_1$-sound. Is that the sort of thing you're referring to? In particular, your suggestion applies only if ZFC fails to even be $\Sigma_1$-sound? | |
| Dec 25, 2022 at 19:29 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Dec 25, 2022 at 18:59 | comment | added | Burak | @TimothyChow: That depends. If our expectation from Inf is to only create a set that we would consider as an "infinite" set, then I wouldn't say that Inf is problematic; because it is doing what it is supposed to. If our expectation from Inf is to indirectly allow the creation of the "correct" set that represents $\mathbb{N}$ somehow, then, yes, I'd blame it on Inf, in case of unsoundness. | |
| Dec 25, 2022 at 18:59 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
added 20 characters in body
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| Dec 25, 2022 at 16:23 | comment | added | Julia Williams | To add onto Joel's point in his first comment: the construction of $L$, whence we get the relative consistency of a strong form of choice, goes through in much weaker theories than ZF. E.g. Mathias has shown (dpmms.cam.ac.uk/~ardm/maclane.pdf) how to do it in Mac Lane set theory. So even if one is skeptical of Replacement or Powerset or whatever other strong axiom, we still know that AC is harmless when it comes to consistency. | |
| Dec 25, 2022 at 12:57 | comment | added | Joel David Hamkins | @Burak, that is very interesting. You are saying that arithmetic unsoundness might be the result of our using a wrong interpretation of arithmetic in set theory. We already know there are other wrong interpretations. See eg mathoverflow.net/q/423328/1946. | |
| Dec 25, 2022 at 12:43 | comment | added | Timothy Chow | @Burak That sounds a lot like a suggestion that the problem is the axiom of infinity, right? | |
| Dec 25, 2022 at 11:50 | comment | added | Burak | The possible arithmetic unsoundness of ZFC need not be due to a specific axiom but rather due to the whole apparatus/approach itself. In ZFC, the set $\mathbb{N}$ is defined as the smallest inductive set, so the "arithmetic assertions" are actually assertions about the smallest inductive set. It is possible (and actually, consistent once formalized) that the smallest inductive set does not capture our intuitive notion of the set of natural numbers (that is, the set supposed to consist of 0,S0,...) So it may be that the formalization/objectification of natural numbers as a set is problematic. | |
| Dec 25, 2022 at 4:08 | comment | added | Timothy Chow | Without seeing the actual proof of inconsistency, I'd be inclined not to throw out any of the axioms or axiom schemata lock, stock, and barrel; rather, I'd expect that the inconsistency could be addressed by a suitable restriction. Maybe Separation or Replacement applies only to certain types of formulae, or only certain sets have power sets. | |
| Dec 25, 2022 at 2:57 | comment | added | Joel David Hamkins | Regarding your third bullet, we already know by Gödel that Con(ZF) iff Con(ZFC) and more to the point, ZFC is conservative over ZF for arithmetic assertions. So the axiom of choice can't be the cause either for inconsistency or arithmetic unsoundness. | |
| Dec 25, 2022 at 2:39 | history | answered | Qiaochu Yuan | CC BY-SA 4.0 |