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Omega rule for sets$``\omega^{ Set}"$: if $\{\varphi_0(y),\varphi_1(y), \varphi_2(y),...\}$ is the set of all formulas in the language of $ZF$ in one free variable symbol $``y"$, and if $\psi(x)$ is a formula in one free variable symbol $``x"$, then

From: for each $n=0,1,2,3,...; \text{ we have }\forall x \ [x=\{y|\varphi_n(y)\} \to \psi(x) ]$

We Infer: $\forall x \ [\psi(x)]$

Question: Is $ZF + \omega^{Set}$ complete?

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Zuhair
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1 Answers1

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No. In fact, the set of theorems of $ZF + \omega^{Set}$ is exactly the same as the set of theorems of $ZF+V=HOD$, which is an ordinary recursively axiomatizable first-order theory and is therefore incomplete (assuming ZF is consistent).

First, note that every definable set is ordinal-definable (here and below, "definable" always means without parameters). Your "$\omega$-rule" then implies that every set is ordinal-definable, so $ZF + \omega^{Set}$ proves $V=HOD$.

Conversely, I claim that every model $M$ of $ZF+V=HOD$ is a model of $ZF + \omega^{Set}$, and so $ZF + \omega^{Set}$ cannot prove any theorems which are not theorems of $ZF+V=HOD$. To see this, note that first that since $M\vDash V=HOD$ there is a definable (internal) well-ordering of $M$. Now suppose $\forall x \ [\psi(x)]$ is false in $M$ for some formula $\psi$. Then there is a definable element $a\in M$ such that $M\vDash\neg\psi(a)$, namely the least such $a$ with respect to the definable well-ordering of $M$. Say $a$ is defined by the formula $\varphi$. Then $a=\{y\mid\varphi_n(y)\}$ where $\varphi_n(y)$ is $\forall x(\varphi(x)\rightarrow y\in x)$. Thus $M\vDash\neg\forall x \ [x=\{y|\varphi_n(y)\} \to \psi(x) ]$. Since $\psi$ was arbitrary, your $\omega$-rule is sound in $M$ and $M$ is a model of $ZF + \omega^{Set}$.

Eric Wofsey
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  • but ZF interprets PA, and here with $\omega^{Set}$, we have this theory interpreting PA+omega rule, which is complete for arithmetic, I don't think any model of ZF satisfying your qualifications would interpret PA+omega rule. So I think this answer is not correct as far as the details given is concerned? – Zuhair Feb 05 '19 at 23:40
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    How does your theory interpret PA+omega rule, exactly? Having an omega rule with respect to all definable sets is quite different from having an omega rule with respect to all standard natural numbers. – Eric Wofsey Feb 05 '19 at 23:47
  • I'm not sure really, but it should do. If you have a non-standard natural then you must be able to violate $\omega^{Set}$, it just cannot be the case that PA + omega rule be stronger than this theory? – Zuhair Feb 05 '19 at 23:55
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    Note that if you similarly added an omega rule for PA that used all definable numbers instead of just the standard natural numbers, you would get a theory that is no stronger than PA, since any nonempty definable subset of a model of PA has a definable element (its least element). – Eric Wofsey Feb 06 '19 at 00:59
  • @Zuhair "If you have a non-standard natural then you must be able to violate $\omega^{Set}$" Why? Having a nonstandard natural doesn't mean that there is an undefinable element in the model. The issue is simple: $(1)$ Any model of ZF in which every element is parameter-freely definable automatically satisfies $\omega^{Set}$. $(2)$ In any model $M$ of ZF + V=HOD, the substructure $Def(M)$ of parameter-freely definable elements of $M$ is an elementary substructure of $M$, hence is a model of ZF in which every element is definable. – Noah Schweber Feb 06 '19 at 01:01
  • So any model of ZF + V=HOD has an elementary submodel consisting of only parameter-freely definable elements. This means that this elementary submodel satisfies $\omega^{Set}$, and hence by elementarity so does the whole original model. – Noah Schweber Feb 06 '19 at 01:10
  • @EricWofsey, I see your point. Thanks – Zuhair Feb 06 '19 at 01:11
  • @NoahSchweber, hmm.. I see your point. Thanks – Zuhair Feb 06 '19 at 01:11
  • @Zuhair Take for example a (necessarily!) non-$\omega$-model $M$ of ZF + V=HOD + $\neg$Con(ZF+V=HOD). Then every element of $Def(M)$ is parameter-freely definable in $Def(M)$, but by elementarity $Def(M)$ is definitely not an $\omega$-model since it sees a contradiction in ZF+V=HOD. – Noah Schweber Feb 06 '19 at 01:12
  • Incidentally, the use of the submodel of definable elements of $M$ turns out to be unnecessary and kind of a distraction from the main point--you can make the argument directly in $M$ using its definable well-ordering. I've edited the answer to do so. – Eric Wofsey Feb 06 '19 at 01:14
  • Sure, but ... elementary submodels! They're the best! – Noah Schweber Feb 06 '19 at 01:15
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    @Zuhair I strongly recommend (again) the Hamkins/Linetsky/Reitz paper on pointwise-definable models of ZF - I think it's a good source for dispelling certain reasonable-but-incorrect intuitions. – Noah Schweber Feb 06 '19 at 01:17