I think I read somewhere the following.

If a first-order sentence $\varphi$ in the language of set theory holds for every well-founded model of ZFC, then nonetheless:

  • $\varphi$ may fail for a non-well-founded model;
  • in other words, $\varphi$ needn't be a theorem of ZFC.

What is an example of such a $\varphi$?

  • 5
    $\begingroup$ You can take as $\varphi$ the statement "$\mathsf{ZFC}$ is consistent". $\endgroup$ May 21 '14 at 5:13
  • $\begingroup$ "Well-founded model" is surprisingly difficult to search for in this context; all results I've found are about logic programming. $\endgroup$ May 21 '14 at 5:20
  • 1
    $\begingroup$ @user2357112 It may be easier to find references to "transitive model". $\endgroup$ May 21 '14 at 5:21

Every statement which is in its essence a true [first-order] number theoretic statement in the universe must be true for every well-founded model. The most striking example for these statements are consistency of various theories.$\DeclareMathOperator{\con}{con}$

For example, if there are well-founded models of $\sf ZFC$, then $\con\sf(ZFC)$ holds. It follows that every well-founded model satisfies $\con\sf(ZFC)$. Similarly if there is a model with an inaccessible cardinal, then $\con\sf(ZFC+I)$ holds, so it must hold in every well-founded model, and if there is a model with a proper class of supercompact cardinals, then in every well-founded model it is true that there is a model with a proper class of supercompact cardinals.

On the other hand, if there is a model of $\sf ZFC$ then there is a model for $\sf ZFC+\lnot\con(ZFC)$. And this model is necessarily not well-founded, and in fact not even an $\omega$-model (meaning: it has non-standard integers).

  • $\begingroup$ Hmm okay. I think I get it. So would it be fair to say that every well-founded model of ZFC is also an $\omega$-model? I did not realize well-foundedness was so very strong. $\endgroup$ May 21 '14 at 8:17
  • $\begingroup$ Of course. If it is well-founded we can collapse it, and its ordinals are real ordinals. In particular its $\omega$ is the real $\omega$. $\endgroup$
    – Asaf Karagila
    May 21 '14 at 8:52

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