Given an ordinal $\alpha$ we will denote by $V_\alpha$ the $\alpha$-stage of the Von Neumann hierarchy for the set-theoretic universe. An exercise from Kunen's book says that ZFC can't prove the existence of an ordinal $\alpha$ such that $V_\alpha$ is a model of ZFC. This is clear by virtue of Gödel's Incompleteness Theorem but Kunen proposes a different approach, pointing out in a hint that if the existence of such an ordinal is provable, then taking $\alpha$ to be least such that ${\rm ZFC}\vdash "V_\alpha\vDash {\rm ZFC}"$, there would exist another ordinal $\beta\in \alpha$ such that ${\rm ZFC}\vdash "V_\beta\vDash {\rm ZFC}"$.

Could someone explain to me why would there be such a $\beta$? Hints or any comments would be much appreciated. Thanks in advance.

  • 1
    $\begingroup$ I tried to clean up your prose a little, but what you wrote is not quite what Kunen said. If ZFC proves that some $V_\alpha$ is a model of ZFC, then (ZFC proves that) there is a least such $\alpha$. This is not the same as saying that ZFC proves that $V_\alpha$ is a model of ZFC, since claiming the latter in particular indicates that $\alpha$ is a definable term, while the former is only claiming the existence of some ordinal, but not its definability. Similarly with the ordinal $\beta$. (Yes, the least such $\alpha$ is definable, but this is just a happy accident.) $\endgroup$ – Andrés E. Caicedo Jan 16 '16 at 3:35
  • $\begingroup$ Thank you @AndrésCaicedo! But I don't understand that subtlety about the definability of $\alpha$... $\endgroup$ – Ergonvi Jan 16 '16 at 14:15
  • 2
    $\begingroup$ ZFC only proves sentences (formulas without free variables). You can write a sentence saying "$\omega_1$ is regular" because $\omega_1$ is definable. Your formula would actually say something like "the first uncountable ordinal is regular". The statement "$V_\alpha\vdash{\rm ZFC}$", on the other hand, is not a sentence, but a formula $\psi(\alpha)$ that may be true of some ordinals $\alpha$, and false of others. $\endgroup$ – Andrés E. Caicedo Jan 16 '16 at 15:27
  • $\begingroup$ So, does definable mean definable without parameters? $\endgroup$ – Ergonvi Jan 16 '16 at 15:37
  • 1
    $\begingroup$ That's the sense in which I am using the term, yes. $\endgroup$ – Andrés E. Caicedo Jan 16 '16 at 15:41

Suppose that ZFC proves the existence of such an $\alpha$, and take the least such $\alpha$. Then $V_\alpha$ is a model of ZFC, so $V_\alpha\vDash\text{"There exists an ordinal $\beta$ such that $V_\beta$ is a model of ZFC"}$. So there is some $\beta\in V_\alpha$ such that $$V_\alpha\vDash\text{"$\beta$ is an ordinal and $V_\beta$ is a model of ZFC"}.$$ If we knew that that last statement in quotes was absolute for $V_\alpha$, then we would conclude that $\beta$ is an ordinal and $V_\beta$ is a model of ZFC. And since $\beta\in V_\alpha$, $\beta<\alpha$, so this contradicts the minimality of $\alpha$. So what you need to prove is that "$\beta$ is an ordinal and $V_\beta$ is a model of ZFC" is absolute for $V_\alpha$.

(By the way, your phrasing "taking $\alpha$ the less ordinal such that $ZFC\vdash V_\alpha\vDash ZFC$" doesn't make much sense. We don't want the least $\alpha$ such that ZFC proves that $V_\alpha$ is a model of ZFC (that doesn't make sense, because ZFC can't even express the sentence "$V_\alpha$ is a model of ZFC" for any particular ordinal $\alpha$ unless $\alpha$ is definable). What we want is just the least $\alpha$ such that $V_\alpha\vDash ZFC$, without reference to what ZFC can prove.)

  • $\begingroup$ Thank you @EricWofsey! In this regard, it is known that $V_\alpha\vDash \phi$ ($\phi$ an axiom from ZFC) can be formalized in ZFC by virtue of Tarski's definability. But, are there some way to formalize inside ZFC when a set is a model of the whole ZFC? I'm thinking that the problem with it is that we can't quantify over an infinite set of formulas. $\endgroup$ – Ergonvi Jan 16 '16 at 14:30
  • $\begingroup$ If you encode formulas in the language of set theory as (say) natural numbers, you can use recursion on the complexity of $\phi$ to express "$V_\alpha\vDash\phi$" as a formula in the language of set theory with a free variable for $\phi$ (note that this does not work if you replace $V_\alpha$ by a proper class, since the recursive definition involves quantifying over functions from arbitrary finite sets to $V_\alpha$). So you're allowed to quantify over $\phi$, and so you can express the statement $\forall\phi\in ZFC (V_\alpha\vDash\phi)$. $\endgroup$ – Eric Wofsey Jan 16 '16 at 18:04
  • $\begingroup$ OK, so for example if you supose that the languaje of set theory is bild up by members $V_\omega$ then a formula is simply a finite subset of $V_\omega$ and then we can view the collection of all axioms of ZFC as a subset of $V_\omega$? On the other hand, then why you said that ZFC can't express the sentence "$V_\alpha$ is a model of ZFC"? $\endgroup$ – Ergonvi Jan 16 '16 at 18:28
  • $\begingroup$ ZFC can only express the sentence "$V_\alpha$ is a model of ZFC" if it can refer to $V_\alpha$ at all. It can only do so if either $\alpha$ is treated as a free variable, or if the particular value of $\alpha$ you are trying to use happens to be definable. This is the same point as Andres Caicedo was making in his comments on the question. (Also, this is not relevant to your reply, but I misspoke a bit: the recursion involves quantifying not just over functions from finite sets to $V_\alpha$, but over classes of such functions.) $\endgroup$ – Eric Wofsey Jan 16 '16 at 18:36
  • $\begingroup$ OK. Thank you very much. But, another question. Why if $\alpha$ is definable then ZFC can express that statement? And moreover, which ordinals are not definable in ZFC? $\endgroup$ – Ergonvi Jan 16 '16 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.