There's a bit about models and absoluteness in particular that confuses me (I think my question is related to Noah Schweber's answer to Why cumulative hierarchy of Sets is not model of ZF). In Kunen's standard text on independence proofs (1980), there's a section about absoluteness (see pp. 128 in particular). Here, it is stated that for any transitive model $M$ of ZF we have $V_{\alpha}^M = M \cap V_{\alpha}$. Mathematically, the proof follows easily as $$V_{\alpha}^M = \{ x \in M\mid (\operatorname{rank}_V(x) < \alpha)^M) \} = \{ x \in M \mid \operatorname{rank}_v(x) < \alpha \} = M \cap V_{\alpha}.$$ But conceptually, I struggle to make sense of the term $V_{\alpha} \cap M$; does this tacitly assume the platonic view that there is a universe of sets which we denote by $V$ (if $M$ is a model of ZF, then $M$ models the existence of every $V_{\alpha}$, which appears contradictory to me)? How can a model $M$ of ZF only contain some elements of the universe of sets? Similarly, how would $M=V$ make sense in this context?


1 Answer 1


No, it is a shorthand of writing $\{x\in V_\alpha\mid M(x)\}$, where $M(x)$ is the predicate which defines $M$, here without parameters for clarity.

To your question how can $M$ only have "some of the sets", let me ask you how can $\Bbb N$ have "some of the numbers"? Or how can $\Bbb Q$ be a field and only have "some" of the real numbers? And so on and so on. Being a class-model of ZF(C) means that you satisfy the axioms with the sets that you include. It doesn't mean that you are everything. You just "think" that to be the case. Much like $\Bbb Q$ is "unaware" of $\sqrt2$ or $\pi$, and $\Bbb R$ is "unaware" of $i$.

The key point to take from this is that $V_\alpha$ is really a class, or in this case even a set, defined from the parameter $\alpha$. You can ask what is the interpretation of this definition inside $M$, or rather what do we get from the relativization of the formula to $M$ (or more accurately, to the formula defining $M$, after fixing the parameters). And the nice thing is that you just get "what you would expect to get". Which is the intersection.

I will say, however, that tacitly taking a Platonistic view for the sake of clarity is often a very productive approach towards mathematics. Especially if one needs to explain an argument to someone else. We are Platonistic creatures, so talking in terms of Platonistic arguments makes sense to us. If you are not a Platonist, you can always preface this by a remark that this is just a crutch, and not the proof itself.

  • $\begingroup$ Does that mean any model $M$ of ZF is definable in LST (i.e. as the unique class of elements satisfying some formula $\phi$)? $\endgroup$
    – MacRance
    Nov 13, 2018 at 14:43
  • $\begingroup$ A class is usually taken to mean a definable class. $\endgroup$
    – Asaf Karagila
    Nov 13, 2018 at 14:44
  • $\begingroup$ Okay, that makes sense. And then by Gödel 2 ZF is unable to prove that $M$ is a set, right? $\endgroup$
    – MacRance
    Nov 13, 2018 at 14:51
  • $\begingroup$ No, that has nothing to do with that. Class models, or inner models, always contain all the ordinals. They are provably not sets. The 2nd incompleteness theorem means that ZF cannot prove that there is a set model of ZF, unless it is inconsistent. Those are two different things. $\endgroup$
    – Asaf Karagila
    Nov 13, 2018 at 14:53
  • $\begingroup$ Hold on, so when Kunen wrote "$M$ is a transitive model" he meant "$M$ is an inner model"? $\endgroup$
    – MacRance
    Nov 13, 2018 at 14:57

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