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Assuming ZFC has a model, is there a model of ZFC such that in that model, ZFC has no model?

Also, assuming ZFC has a model, is there a model of ZFC such that in that model, ZFC is inconsistent?

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Yes.

Recall that by the Completeness Theorem, having a model and being consistent are the same thing. Also, by Incompleteness, ZFC doesn't prove its own consistency. Finally, ZFC can prove the Soundness Theorem - that an inconsistent theory has no models!

So - assuming ZFC has a model - ZFC is consistent. If ZFC is consistent, then ZFC can't prove "ZFC is consistent." By completeness, this means there's a model of ZFC satisfying "ZFC is inconsistent." Since ZFC proves the Soundness Theorem, this model must think that ZFC has no model!


Note that there's a subtlety to the claim "There is a model $M$ of ZFC not containing a model of ZFC." What that means is that there is an $M$ which satisfies ZFC+"There is no model of ZFC." However, since ZFC is not finitely axiomatizable, what $M$ thinks ZFC is will be different from what ZFC actually is! And in fact every model $M$ of ZFC contains a structure $N$ which classically is a model of ZFC, but within $M$ appears to not be a model of ZFC! Neat, huh? See Joel David Hamkins' answer to https://mathoverflow.net/questions/51754/clearing-misconceptions-defining-is-a-model-of-zfc-in-zfc.

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    $\begingroup$ The post by Joel is probably my favorite use of the Reflection theorem. $\endgroup$ – Asaf Karagila Jun 14 '16 at 22:32
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    $\begingroup$ @AsafKaragila Yup, it's pretty sweet. :D $\endgroup$ – Noah Schweber Jun 14 '16 at 22:32
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    $\begingroup$ Neat! Now I'm curious: Which axiom of ZFC could $N$ appear to violate within $M$? $\endgroup$ – dankness Jun 15 '16 at 9:11
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    $\begingroup$ @dankness: Yes, but the object in $M$ which $M$ "thinks" is the theory we denote by ZFC is not the actual theory ZFC from $V$. It contains more axioms, whose length is not an actual natural number, but rather an object that $M$ thinks of as a natural number. Similarly, the inconsistency, or proof of contradiction, need not be a finite proof. It can be of a non-standard length, or appeal to inference rules which are themselves not actual inference rules. $\endgroup$ – Asaf Karagila Jun 15 '16 at 13:17
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    $\begingroup$ @dankness: Yes, $N$ is a model of an $M$-finite fragment of $M$-ZFC; it just happened that as this fragment is taken to be a non-standard fragment, it includes all the standard axioms, so from the point of view of the universe, $N$ is a model of ZFC. $\endgroup$ – Asaf Karagila Jun 15 '16 at 13:33
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Noah gave the correct answer. It is true, that if $\sf ZFC$ has any models, then there are models of $\sf ZFC$ in which there are no models of $\sf ZFC$.

But let me point out an important piece of information here.

We formulate statements about consistency in terms of natural numbers. This is because the language of set theory is finite, so we can encode formulas as natural numbers in a very robust way. How robust? Well, like a computer-program robust. It's a mechanical algorithm to move on thing into the other.

Why am I bringing this up? Because it serves the point that natural numbers are important to the statement that something is consistent or inconsistent.

More specifically, we work in a universe $V$. In $V$, it is true that $\sf ZFC$ is consistent. Now, it is important to point out that by this we mean "what $V$ interprets as the predicate which defines '$\sf ZFC$' in its own natural numbers". But we can show that if $M$ is a model of $\sf ZFC$ inside $V$, and $M$ and $V$ agree on the notion of "natural number", then both have the same first-order theory for the natural numbers.

In particular, the predicate '$\sf ZFC$' and the statement of its consistency are absolute between $V$ and $M$. And since $V$ thinks that $\sf ZFC$ is consistent, so must $M$.

This means that if $M$ is a model of $\sf ZFC$ which thinks that $\sf ZFC$ is inconsistent two things must be true:

  1. $M$ has non-standard integers (compared to $V$).
  2. The interpretation of $\sf ZFC$ inside $M$ is different the one in $V$.

Namely, '$\sf ZFC$' can be seen as a predicate in the language of arithmetic, and since in $M$ we have more natural numbers, this predicate is interpreted to have more axioms than it should really have; and first-order logic have more inference rules; and proofs can suddenly be "too long" as far as $V$ is concerned. All these things allow $M$ to find out a proof of contradiction from $\sf ZFC$.

This is the reason why we can find such models. And models whose natural numbers agree with that of the universe are called $\omega$-models, and in such models it is necessarily the case that $\sf ZFC$ is consistent, simply because the natural numbers agree with the universe, so the interpretation of '$\sf ZFC$' and the rules of first-order logic, are the same as in $V$. Since in $V$ we know that $\sf ZFC$ is consistent, it must be consistent in any $\omega$-model.

So while $\sf ZFC$ proves that if there are models of $\sf ZFC$, then there are models of $\sf ZFC+\lnot\operatorname{Con}(ZFC)$; it is not the case that $\sf ZFC$ proves that if there are models of $\sf ZFC$, then there are $\omega$-models of $\sf ZFC$, as this would be a violation of the incompleteness theorem.

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    $\begingroup$ Maybe I'm being a bit thick, but I don't quite see how your last sentence is true; can you elaborate? $\endgroup$ – Noah Schweber Jun 14 '16 at 22:39
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    $\begingroup$ @AsafKaragila Yes. I don't immediately see why ZFC proving "if there's a model of ZFC, then there's an $\omega$-model of ZFC" violates Incompleteness - what if ZFC proves $\neg Con(ZFC)$, in which case it trivially proves that statement? $\endgroup$ – Noah Schweber Jun 14 '16 at 22:41
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    $\begingroup$ @AsafKaragila Let me rephrase my question. Can you prove "If ZFC is consistent, then ZFC doesn't prove "if ZFC has a model, it has an $\omega$-model"" without a correctness assumption on ZFC? $\endgroup$ – Noah Schweber Jun 14 '16 at 22:43
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    $\begingroup$ @AsafKaragila E.g. $\Sigma^0_1$ correctness; if you prefer, a soundness assumption. My point is that your argument can't be completely without an extra assumption, since T=ZFC+Con(ZFC)+"ZFC proves $\neg Con$(ZFC)" proves that (ZFC is consistent, but that ZFC proves that (if ZFC has a model, then ZFC proves that ZFC has an $\omega$-model)). $\endgroup$ – Noah Schweber Jun 14 '16 at 22:44
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    $\begingroup$ Hehe. I hadn't yet read Noah's answer. $\endgroup$ – Andrés E. Caicedo Jun 14 '16 at 22:47

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