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I'm sorry to have to report a basic confusion about logic on my part, and apologise to those having to make an effort to put it right.

Suppose that it was possible to derive a contradiction from ZFC. Then ZFC should have no model. But apparently, there are some models of ZFC; perhaps they are not sets, but why care? I would bet that at least the correctness of the logical inference rules (which is what we need here) is provable for proper classes as well. Von Neumann gave a clear definition of his universe using ordinal induction. The axioms of ZFC are readily verifyable from this definition.

So how is it that we can't rule out that ZFC is inconsistent? Where is the error in my thought?

Cloudscape
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    How do you know that ZFC has a model? What do you need to assume to get that result? :) – Nico Apr 01 '23 at 12:52
  • I need to assume the existence of the empty set, as well as the validity of the construction process that yields all the other sets. (This construction process does use two elements of ZFC, namely union and power set existence.) – Cloudscape Apr 01 '23 at 12:57
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    No! The axioms of ZFC are famously not strong enough to form the set of all sets (which would be a model of ZFC). This is on purpose, because the existence of such a set would lead to inconsistencies. You can not construct within ZFC a model of ZFC. Try it out and see where you can't make progress. – Nico Apr 01 '23 at 13:00
  • I was talking about classes here... – Cloudscape Apr 01 '23 at 13:08
  • Also, I didn't state that the two ZFC axioms used were all that is needed for the construction process... – Cloudscape Apr 01 '23 at 13:09
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    Ahh, okay. Then you might want to look at this question: https://math.stackexchange.com/questions/1442008/mathrmnbg-proves-its-own-consistency . NBG does not prove the consistency of ZFC. – Nico Apr 01 '23 at 13:10
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    The "snake in the grass" is this claim: "The axioms of ZFC are readily verifyable from this definition." What do you mean by verifiable? Informally, using our intuition about sets? Anyone who doubts the consistency of ZFC certainly won't accept that argument. Formally, inside some other formal system? (Something stronger than NBG.) But then how do we know that system is not inconsistent. – Michael Weiss Apr 01 '23 at 16:21
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    A bit of history helps. Both Cantor and Frege proved lots of results in an informal set theory that was later found to be inconsistent. The hope of ZFC and NBG is that those problems have been dealt with, but maybe not, or maybe there are new ones. As another example: in 1940 Quine published an axiom system for set theory (ML), which was later shown to be inconsistent. (See https://plato.stanford.edu/entries/quine-nf/) – Michael Weiss Apr 01 '23 at 16:28
  • @MichaelWeiss The proof that the von Neumann universe satisfies ZFC would use transfinite induction. – Cloudscape Apr 01 '23 at 17:20
  • Well, yes--the informal argument that the cumulative hierarchy satisfies the ZFC axioms uses transfinite induction. Drake in Chapter 1 of Set Theory: An Introduction to Large Cardinals sketches this train of thought, going through each of the axioms. But it's not a proof of consistency, just a semi-philosophical justification for ZFC. – Michael Weiss Apr 01 '23 at 17:57
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    A proof of consistency, to be worth anything, would have to be based on something much more solid. Hilbert famously hoped that set theory could be shown consistent by a purely finitary proof. Thanks to Gödel, we know that hope is vain. Even without Gödel, if we could give a proof of Con(ZFC) inside ZFC (or inside NBG), it would be open to the objection that if you suspect ZFC/NBG might be inconsistent, you surely won't trust any proof inside it. – Michael Weiss Apr 01 '23 at 18:02
  • Perhaps you really have in mind a different question: why can't we carry out a proof of Con(ZFC) inside NBG? That has a purely mathematical answer, given in the article linked in Nico's second comment. – Michael Weiss Apr 01 '23 at 18:06
  • The correctness of the logical inference rules cannot be proved (or even stated in general) for class models. The problem is that you can't even define (uniformly, in a way that lets you quantify over formulas) what it means for a formula to be satisfied in a class model. See my answer at https://math.stackexchange.com/questions/4536515/why-cant-a-class-model-serve-as-a-witness-of-the-consistency-of-a-theory. – Eric Wofsey Apr 02 '23 at 03:25

1 Answers1

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Von Neumann gave a clear definition of his universe using ordinal induction

This is highly debatable. There are two rather extreme gaps here

  1. What are the ordinals?
  2. What is a power set?

that are arguably not sharp concepts$.^*$

If you nonetheless believe this is clear (and that furthermore it's clear that this object obeys the ZFC axioms), then I agree, that is a valid basis on which to infer the consistency of ZFC.

On the other hand, if you're under the impression that all is well not because of Platonist convictions, but merely because we can define the Von-Neumann universe in axiomatic set theory (or class-set theory), then we have to think clearly about whether we can actually formulate such a consistency proof ("$V$ is a model for ZFC, therefore ZFC is consistent") in the theory we're working in.

In ZFC or weaker set theories, the answer is obviously not, on grounds of the incompleteness theorem, if nothing else. But it goes deeper than that since, although we can define $V$ as a proper class in ZF - foundation, we can't even express "$V$ is a model of ZF". This is because there's no way in general of formulating a satisfaction predicate for proper classes, and in fact, Tarski's theorem implies that it will always fail when $V$ is the whole universe. This even holds through in stronger theories, e.g. ZFC + inaccessible, where we can prove the consistency of ZFC via set-sized models but still can't internally make sense of the whole universe being a model of it.

So given the issues of with the first order approach, we would naturally want to attack this in a class-set theory. As mentioned in Asaf's answer linked in the comments, in NBG, we run into a similar issue as before that we can't construct a first-order satisfaction predicate. Whereas, in Morse-Kelley, we can construct one, and the argument "$V$ is a model for ZFC, therefore ZFC is consistent" goes through just fine there.

So, as with anything, whether we accept that ZFC is consistent is a function of what we're willing to assume.


On the matter of the "logical inference rules" you bring up, there's something interesting we can say. The fact that we can interpret formal proofs as relative to some set or class, and that under reasonable assumptions, they do correctly reflect inference rules, leads to relative consistency proofs. For instance, $V$ can be defined in ZF - foundation, but $V$ satisfies all of ZF. From this, we can deduce that if ZF - foundation is consistent, then ZF is consistent, since any proof of inconsistency from ZF could be carried out in ZF - foundation, relativized to $V.$ Similarly, the constructible universe $L$ can be defined in ZF, but it satisfies ZFC + GCH + Suslin trees exist, etc., establishing a bunch of equiconsitencies with ZF.

But if we just try to do it with $V$ in a ZF(C) background, though, we don't get anything, since we can prove that $V$ is the whole universe, anyway. So we just get the boring result that Con(ZF(C)) implies Con(ZF(C)).


$^*$Since this remark has attracted one perplexed comment and ostensibly a downvote from some other user, let me explain more precisely what I mean here. I don't intend to say that the Von-Neumann universe is not precisely definable in set theory... it is. What I am pushing back on is the implication in the question that the Von Neumann universe construction somehow defines the set theoretical universe.

Sure, it's easy to define what a subset is and the power set axiom guarantees the power set operation exists. And replacement guarantees the ordinals "go on forever" in some reasonable sense, or at least that they don't have an unnatural stopping point. But what the ordinals and power sets precisely are depends in an inpredicative way on an already existing universe of sets. And this is how it's generally taught in set theory books and courses: Von Neumann's hierarchy does not "construct" the universe of well-founded sets from the bottom up, it merely gives a nice stratification of things that were already there.

And I'm not making this assertion that these concepts are arguably vague out of left field (in fact I don't fully agree that they are)... this is a well-known matter of debate. See e.g. here.

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    What?! I know very little set theory but I am extremely confused when you say that ‘ordinal’ and ‘power set’ are not sharp concepts. Both have precise definitions and are frequently used in “normal” mathematics with these precise definitions in mind – FShrike Apr 01 '23 at 22:28
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    @FShrike I say "the ordinals" is arguably not a sharp concept. For instance, are there any inaccessible cardinals? We say they "never stop", but can we really claim to have a sharp mental picture of what this means in the same way many would claim to have a precise structural understanding of what the natural numbers look like? Likewise, what exactly is an "arbitrary subset" of the natural numbers? – spaceisdarkgreen Apr 01 '23 at 22:56
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    @FShrike Sure, we can define these things in set theory, but the standard semantics is that there's some universe of set theory that already exists that we're describing with these axioms on which we're basing our definitions. On analysis, we find the possibility of many different models of the axioms, with different heights of ordinals and different power sets for $\mathbb N.$ – spaceisdarkgreen Apr 01 '23 at 23:01
  • @FShrike I guess to be more concise, both concepts only make sense once we know what "the universe of sets" is, and that's all the "precise definitions" you have in mind could possibly promise. – spaceisdarkgreen Apr 01 '23 at 23:16
  • I see. So you’re saying that axiomatically they’re of course having exact definitions but Platonically we don’t have definite pictures of these, because no model exists that we can ever view in its entirety with our own eyes. What an ordinal is is clear, but what is captures by “(all) the ordinals” is murky – FShrike Apr 01 '23 at 23:18
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    @FShrike Yes, at least in the sense that I mean Platonically, not axiomatically. I think implicit in the question is the claim that the Von-Neumann universe is "a model that we can view with our own eyes"... the point of my first remark is that its tangibility takes for granted the tangibility of the ordinals/power sets, which in turn depends on the tangibility of the universe of sets... so there's some circularity there. I don't personally find it very problematic, but it's controversial and one should know what they're signing up for. – spaceisdarkgreen Apr 02 '23 at 00:04