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 Answers
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.
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Does that mean any model $M$ of ZF is definable in LST (i.e. as the unique class of elements satisfying some formula $\phi$)? – MacRance Nov 13 '18 at 14:43
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A class is usually taken to mean a definable class. – Asaf Karagila Nov 13 '18 at 14:44
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Okay, that makes sense. And then by Gödel 2 ZF is unable to prove that $M$ is a set, right? – MacRance Nov 13 '18 at 14:51
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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. – Asaf Karagila Nov 13 '18 at 14:53
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Hold on, so when Kunen wrote "$M$ is a transitive model" he meant "$M$ is an inner model"? – MacRance Nov 13 '18 at 14:57
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Not necessarily. You can also assume that a transitive model exists. This is a stronger assumption compared to just a model, but there is nothing inherently wrong with it. – Asaf Karagila Nov 13 '18 at 15:08
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Okay, final question (hopefully): if I assume a transitive model exists (and hence invoke completeness), is this model also definable in LST? – MacRance Nov 13 '18 at 15:25
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It is unclear to me what you mean by "definable". With a parameter? Yes, every set is definable with itself as a parameter ($x={y\mid\phi(y,x)}$ where $\phi(y,x)$ is just $y\in x$). Without parameters? Not necessarily, but not every inner model is definable without parameters either. Sets are always classes, exactly because the explanation above. If that is what you're trying to understand. – Asaf Karagila Nov 13 '18 at 15:28
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Thank you for your help and patience. I'll phrase my confusion differently: if $M$ is a model of ZF (invoking completeness again), where does $M$ "live" then? – MacRance Nov 13 '18 at 15:37
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Inside a larger universe, of whatever meta-theory you use (which is usually taken as ZFC itself once again). – Asaf Karagila Nov 13 '18 at 15:37
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I hope I'm not returning to my original question now, but with $M$ given in the larger universe how does $M \cap V_{\alpha}$ make sense then? Where do such $x$ live that are in $V_{\alpha} \setminus M$? – MacRance Nov 13 '18 at 16:43
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How does any intersection of two sets make sense? The sets in $V_\alpha\setminus M$ live in $V$, the same universe that $M$ lives in. This much is true even if $M$ is an inner model, either $V=M$ or there will be some $\alpha$ such that $V_\alpha\setminus M$ is non-empty. – Asaf Karagila Nov 13 '18 at 16:55
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Okay, I think I finally got it, thank you so much! – MacRance Nov 13 '18 at 17:00
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@AsafKaragila This is irrelevant to the question, but I think "we are Platonistic creatures", though true when "we" means mathematicians, is not so clear when "we" means the whole human race. I suspect that the people who can add "3 apples + 2 apples" and can add "3 books +2 books" but have no idea about $\frac38+\frac28$ are simply unable or unwilling to regard "eighths" as actual entities like apples and books. – Andreas Blass Nov 19 '18 at 02:06
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@Andreas: To be fair, I am mixing the meaning of "Platonistic" in a very non-standard way. What I meant to say is that we, as a species, experience reality through what we hope is concrete existence in a specific shared universe. So it is more of taking the meaning of mathematical Platonism, and applying it to reality. – Asaf Karagila Nov 19 '18 at 08:14