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This is a try to enforce having standard models of ZFC in a way similar to how Omega rule do it for PA. I made a failed try at the posting titled "Omega rule and standard models of ZFC?", here I'll present it in a different way. It appears that the heart of ZFC is axiom of Foundation, now since $\in-$induction proves Regularity, then I'll try to make an Epsilon $\omega-$rule version of it [much as the omega rule for arithmetic can be viewed as strengthening the usual induction schema of PA].

$Epsilon \ \omega-rule$: if $\{\phi_1(y), \phi_2(y), \phi_3(y),...\}$ is the set of ALL formulas in the first order language of ZFC in which only symbol $``y"$ occur free, and only free, and symbol $``x"$ never occur, and if $\psi(y)$ is a formula in the same language in which only symbol $``y"$ occurs free, and only free, and symbol $``x"$ never occur, and if $\psi(x)$ is the formula obtained from $\psi(y)$ by merely replacing all occurrences of symbol $``y"$ by symbol $``x"$; then

from: $for \ i=1,2,3,.... \\ \forall x [\forall y (y \in x \leftrightarrow \phi_i(y)) \to (\forall y \in x (\psi(y))\to \psi(x))]$

we infer:

$\forall x \psi(x)$

In English: if we have every parameter free definable set fulfilling the antecedent of $\in-$induction for a parameter free definable property after formula $\psi$, then all sets would satisfy that property.

Question: would ZFC formulized in a language extended with the above rule, have all of its models being standard models?

Zuhair
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    What dou you call a standard model? And btw, Axiom of Foundation is really not at the heart of ZFC. – Régis Jun 15 '18 at 20:47
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    @Régis Others would argue differently. Famously, Adrian Mathias said that set theory was the study of well-foundedness. – Andrés E. Caicedo Jun 15 '18 at 21:00
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    Some authors do not include AF in ZFC. And you do not need it to speak about well foundedness so my assertion does not contradict Mathias' one. Anyway that was a side comment. My real question was: what do you call "standard model"? – Régis Jun 15 '18 at 21:05
  • @Régis "Standard model" refers to a model which is well-founded (equivalently, whose ordinals are well-ordered; equivalently, which is isomorphic to a transitive set). See here. Note that there are (under reasonable assumptions) lots of non-isomorphic standard models of ZFC; this is in contrast with the arithmetic situation, where "standard model of arithmetic" means "structure isomorphic to the natural numbers." – Noah Schweber Jun 15 '18 at 21:19
  • To the OP: As with your previous question, I think your "$\phi_i$" should be "$\phi_i(y)$" in the statement of your rule. – Noah Schweber Jun 15 '18 at 21:21
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    @Régis No modern set theorist working on ZFC or any of the related standard theories excludes foundation from the standatd list of axioms, and certainly it is understood to be part of ZFC. – Andrés E. Caicedo Jun 15 '18 at 22:26
  • @Régis To elaborate on Andres' point, while certainly Foundation is omitted in some texts these are never texts in research set theory itself, but rather expository texts for students. As far as I am aware, in a research paper on set theory foundation is always assumed unless explicitly stated otherwise. – Noah Schweber Jun 16 '18 at 01:28
  • @AndrésE.Caicedo: I think it may be a question of perspective: If one considers ZFC primarily as a vehicle for formalizing ordinary mathematics, then it seems reasonable to consider Regularity to be a somewhat peripheral axiom: it is very rarely needed for that purpose. (And a useful amount of well-foundedness to study pops up independently of Reg through the well-ordering theorem). On the other hand "modern working set theorists" tend not to care a lot about that view of ZFC, possibly because there doesn't seem to be a lot left to say about it beyond "yes, this seems to work pretty well". – hmakholm left over Monica Jun 16 '18 at 20:02

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The answer is still no, for the same reasons:

  • Any pointwise definable model of set theory satisfies your principle trivially since everything is parameter-freely definable (so your principle is just $\in$-induction for such models, hence follows from ZFC as usual), and there are ill-founded pointwise definable models.

  • More generally, what you've written is still a computable infinitary formula, so by the Barwise-Kreisel compactness theorem (discussed in more detail in my answer to your previous question) either it drastically restricts the "height" of the models which satisfy it or there are non-well-founded models which satisfy it.

However, let me go one step further with that second bulletpoint, and point out something that didn't occur to me in my answer to your previous question. Note that any well-founded model of ZFC has height at least $\omega_1^{CK}$ (this is a good exercise), so we can in fact use the BKCT to show that if $\theta$ is a computable infinitary sentence consistent with ZFC then ZFC+$\theta$ has ill-founded models (any well-founded model of ZFC+$\theta$ has "too large rank" for $\theta$ to avoid having ill-founded models). So there is a real sense in which anything along the lines you propose is doomed.

Noah Schweber
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  • These ill-founded pointwise definable models, do they have ill-founded ordinals? – Zuhair Jun 16 '18 at 15:54
  • @Zuhair A model is ill-founded if and only if some of its ordinals are ill-founded, so yes. (Think about how rank and membership interact, and remember that rank is ordinal-indexed.) – Noah Schweber Jun 16 '18 at 16:00
  • But in the posting "what are standard models of ZF-Regularity?" the answer by Asaf was that it would be a model whose ordinals are well founded, but according to your equivalence above then this would only be a well founded model? I generally thought that one can have an ill founded model with all of its ordinals being well founded for example take a model of ZF that has ONE quine atom and in which the only non-well founded sets are those that has that quine atom in their transitive closures, the von Neumann ordinals are not affected? – Zuhair Jun 17 '18 at 05:49
  • @Zuhair By "a model" I mean "a model of ZF(C)," not ZF(C)-Regularity. You asked "would ZFC formulized in a language extended with the above rule, have all of its models being standard models?" not "would ZFC-Regularity formulized in a language extended with the above rule, have all of its models being standard models?", so that's the question I aimed my answer at. – Noah Schweber Jun 17 '18 at 13:08
  • @ah I see, yes of course, know I understand your prior comment. – Zuhair Jun 17 '18 at 17:48