Is ZFC+Con(ZFC) powerful enough to show there isn't any standard model of ZFC? What you think about it?

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    $\begingroup$ Are you sure that the question you stated is the one you really wanted to ask? What you're asking is, as long as there exists some model of ZFC, does it follow that no models are standard? Maybe you meant to ask, as long as there exists some model of ZFC, does it follow that some model is standard? $\endgroup$ – Timothy Chow Mar 16 '13 at 20:31
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    $\begingroup$ A clarification: I believe Jaykov is referring to the content of a posting he made to the Foundations of Mathematics email list, about a purported proof that ZFC has no well-founded models, which led to a discussion about models of ZFC. With Jaykov's permission, I can post a link to the original email (the FOM archives are public and searchable, so in principle anyone can find it, but I would rather not link it without permission), which contains a link to the purported proof. If this is correct, I would like to point out that MO is not for testing conjectures or checking original work. $\endgroup$ – Noah Schweber Mar 16 '13 at 20:45
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    $\begingroup$ Jaykov, as I explain in my answer, if this were true, it would refute all large cardinal axioms. Are you claiming to have such a proof? $\endgroup$ – Joel David Hamkins Mar 16 '13 at 21:24
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    $\begingroup$ No, because the axioms of ZFC + Con(ZFC) are true and there are standard models of ZFC. (Of course, I'm not saying that these things are provable in ZFC, only that they are true.) $\endgroup$ – Andreas Blass Mar 17 '13 at 0:17
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    $\begingroup$ Noah, thanks for the clarification. I'm on FOM and I saw that posting, but I had forgotten the author's name. I agree with you that in light of that context, this question is not appropriate for MO, and I have just voted to close as "off topic." $\endgroup$ – Timothy Chow Mar 18 '13 at 14:21

Many set theorists define that a model $\langle M,E\rangle$ of set theory is standard to mean that the set membership relation $E$ of the model is the actual set membership relation $\in$, restricted to objects in $M$.

The existence of such a standard model of ZFC is equivalent to the existence of a well-founded model of ZFC, since every well-founded model is isomorphic to its transitive collapse, via the Mostowski collapse. The hypothesis that there is such a transitive model of ZFC should be considered a very weak large cardinal axiom. For example, the transitive model of ZFC hypothesis appears near the bottom of the large cardinal hierarchy in Cantor's Attic.

In particular, nearly all the usual large cardinal axioms imply the existence of a standard model of ZFC, and so very few set theorists want or expect ZFC to rule them out. Since we think that large cardinals are consistent with ZFC, we also expect that it is consistent with ZFC that there are standard models of ZFC.

For example, if $\kappa$ is an inaccessible cardinal, then the set $H_\kappa$ of all sets having hereditary size less than $\kappa$ forms a transitive model of ZFC. Under the Grothendieck Universe Axiom, there will be a proper class of such inaccessible cardinals, and so the entire set-theoretic universe will be the union of such standard set models of ZFC.

Indeed, I would say much more: it is a fundamental part of the reflection paradigm, the view that truths in the whole set-theoretic universe are increasingly found in proper initial segments of it, that we should expect many initial segments of the universe to be ZFC models, and these will be standard models. This is the kind of philosophical justification that one often hears for various large cardinal hypotheses.

Meanwhile, it is easy to see that if ZFC is consistent, then it is consistent with ZFC that there is no standard model of ZFC. To prove this, take any model of ZFC. If it has no transitive models of ZFC, then we're done. If it does have a transitive model, then it will have an $\in$-minimal transitive model of ZFC, and this will be a model of ZFC having no transitive models of ZFC.

Furthermore, since every transitive model of ZFC has exactly the same arithmetic truths as the ambient universe, it follows that if there is a transitive model of ZFC, then there is one having no transitive model of ZFC, in which Con(ZFC) still holds. In particular, ZFC+Con(ZFC), if consistent, cannot prove the existence of a standard model of ZFC. One may iterate this to the consistency hierarchy of models of ZFC. Basically, no amount of iterating the Con operator will get you to the existence of a standard model of ZFC, and this seems relevant for your revised question.

Finally, let me point out that if ZFC is inconsistent, then of course it is strong enough to prove any assertion, including the one you ask about, and so we cannot say for sure that ZFC doesn't prove your assertion without going beyond ZFC in our axioms.

  • $\begingroup$ Hi Joel. Do the statements ‘The consistency of $ \mathsf{ZFC} $ does not imply the existence of a standard model of $ \mathsf{ZFC} $’ and ‘The existence of a standard model of $ \mathsf{ZFC} $ is strictly stronger than the consistency of $ \mathsf{ZFC} $’ mean $$ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \mathsf{SM}, $$ or do they mean $$ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \text{Con}(\mathsf{ZFC} + \mathsf{SM})? $$ $\endgroup$ – Leonard Jul 21 '13 at 8:56
  • $\begingroup$ These statements are what I see in most standard texts on set theory, and they somehow seem to mean $$ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \mathsf{SM}, $$ which I am unable to prove. I had a discussion with Asaf over at Math StackExchange about the proof of $$ \mathsf{ZFC} \nvdash \text{Con}(\mathsf{ZFC}) \rightarrow \text{Con}(\mathsf{ZFC} + \mathsf{SM}), $$ which we can rewrite as $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \nvdash \text{Con}(\mathsf{ZFC} + \mathsf{SM}). $$ $\endgroup$ – Leonard Jul 21 '13 at 9:17
  • $\begingroup$ Proof Suppose instead that $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \vdash \text{Con}(\mathsf{ZFC} + \mathsf{SM}). $$ Then as $$ \mathsf{ZFC} \vdash \text{Con}(\mathsf{ZFC} + \mathsf{SM}) \rightarrow \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})), $$ we would obtain $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \vdash \text{Con}(\mathsf{ZFC} + \text{Con}(\mathsf{ZFC})), $$ which contradicts Gödel’s Second Incompleteness Theorem if $ \mathsf{ZF} + \text{Con}(\mathsf{ZFC}) $ is consistent. $ \blacksquare $ $\endgroup$ – Leonard Jul 21 '13 at 9:27
  • $\begingroup$ The problem is: How can I remove the ‘if $ \mathsf{ZF} + \text{Con}(\mathsf{ZFC}) $ is consistent’ clause so that I get a proof of $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \nvdash \text{Con}(\mathsf{ZFC} + \mathsf{SM})? $$ Thank you for your time! $\endgroup$ – Leonard Jul 21 '13 at 9:30
  • $\begingroup$ @Leonard, you can't remove that assumption, since if $\text{ZF}+\text{Con}(\text{ZFC})$ is inconsistent, then $\text{ZF}\vdash\text{Con}(\text{ZFC})\to\text{anything}$. All nonprovability results take place under a consistency assumption, and you shouldn't think that we haven't proved the final non-provability result, since we have proved it under the relevant consistency assumption. Note that any assertion of the form $\text{ZFC}+\text{Con}(\text{ZFC})\not\vdash\psi$ outright implies that $\text{ZFC}+\text{Con}(\text{ZFC})$ is consistent. $\endgroup$ – Joel David Hamkins Jul 21 '13 at 12:11

Joel has explained that the existence of standard models and a negative answer to the question follow from large cardinal axioms. He also mentioned, in the context of philosophical justification for large cardinal axioms, "the reflection paradigm, the view that truths in the whole set-theoretic universe are increasingly found in proper initial segments of it". Let me add that the existence of standard models follows from (ZFC plus) mere acceptance (in an appropriate sense) of the very notion of "truths in the whole set-theoretic universe". More precisely, suppose we add to the language of ZFC an additional predicate symbol Sat, intended to denote satisfaction, in the whole universe, of formulas of the original language of ZFC. Suppose we add to the ZFC axioms the clauses that specify Sat by induction on formulas. And suppose we also extend the axiom scheme of replacement by allowing formulas of the enlarged language. Then, by formalizing a downward Löwenheim-Skolem argument, we can prove that ZFC has standard models, in fact ones of the form $V_\kappa$. (We actually get more, namely $V_\kappa$'s that are elementary submodels of the universe, with respect to the original language of ZFC.) In other words, for the limited sort of "truths in the whole set-theoretic universe" needed in proving the existence of standard models of ZFC, the reflection principle mentioned by Joel becomes provable, once one incorporates the relevant notion of truth into the theory.

  • $\begingroup$ Very good. And it may be worthwhile pointing out that your theory is a proper part of Kelly-Morse set theory, which proves the existence of such satisfaction predicates not only with respect to first-order truth, but also with respect to satisfaction for assertions involving any given class predicate parameter. So one also gets Sat relative to Sat, and so on transfinitely. $\endgroup$ – Joel David Hamkins Mar 17 '13 at 0:44

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