My question is essentially in the title. I'm reading some notes about the proof of consistency of ZFC assuming the consistency of ZF. The author first assumes theres is a model for ZF-Foundation and proves that the von Neumann universe is then a model for ZF.

Subsequently the author constructs the constructible universe L and proves that it is a model for the axiom of constructibility V=L and subsequently for AC and GCH.

As far as I understand why L believes in V=L, it's not clear to me whether V is a model of V=L or not.

To be a model for a sentence means that the sentence is true or false under an interpretation, so it seems to me that there should be a definite answer to that, i.e. either V is a model of V=L or it is not.

I was unable to find any reference for that in the literature. Would anyone care to help me with that?

(In particular if V believed in V=L then it would already be a model for ZFC. Or is the answer that we simply cannot deduce $\langle V,\in \rangle\models V=L$ nor V doesn't model V=L? But if so, is it a theorem than we cannot prove any of the above (from ZFC) or we simply don't know?)

  • 3
    $\begingroup$ Could you perhaps introduce some line breaks? $\endgroup$ – Asaf Karagila Feb 11 '15 at 12:50

Every model of $\sf ZF$ is a von Neumann universe, namely it has a decomposition into a hierarchy of $V_\alpha$'s, such that $V_0=\varnothing$, $V_{\alpha+1}$ is the power set (in $V$) of $V_\alpha$, and at limit ordinals we take limits.

Now, in order to show that $V=L$ is not a theorem of $\sf ZF$ you need to exhibit models where $V\neq L$ holds. For this you need to know two things:

  1. If $M$ and $N$ are two models of $\sf ZFC$ with the same ordinals, then $L$ as constructed in $M$ and in $N$ is the same model. Namely, $L$ depends only on the ordinals of the universe.

  2. If $\sf ZFC$ is consistent then $\sf ZFC+\lnot CH$ is consistent. Since $V=L$ implies $\sf CH$, this means that $V=L$ is not provable from $\sf ZFC$. The same can be deduced by the theorem that $\sf ZF+\lnot AC$ is consistent if and only if $\sf ZFC$ is consistent, since $L$ satisfies $\sf AC$. For this you need to know what forcing is, which is generally something much more advanced than just the construction of $L$.

To sum up, every model of $\sf ZF$ is a von Neumann universe, but not every model of $\sf ZF$ satisfies $V=L$.

  • $\begingroup$ So I take my V to be $V(0)=\emptyset$ and then define it recursively by taking power sets, that doesn't model $V=L$? $\endgroup$ – settheoryguest Feb 11 '15 at 13:32
  • $\begingroup$ Not necessarily. It might, though. $\endgroup$ – Asaf Karagila Feb 11 '15 at 13:32
  • $\begingroup$ So here is when it is not clear to me, I see $V$ as a one well defined thing, I don't understand how it can model the constructibilty axiom or not. What that does depend on? $\endgroup$ – settheoryguest Feb 11 '15 at 13:35
  • $\begingroup$ But $V$ is not a unique object in all the universe. EVERY universe of $\sf ZF$ sees itself as a von Neumann universe. $L$ sees itself as a von Neumann universe in which $V=L$ is true, and models which satisfy $\sf ZF+\lnot AC$ are von Neumann universes in which $V=L$ is false. $\endgroup$ – Asaf Karagila Feb 11 '15 at 13:42
  • $\begingroup$ In the notes I'm referring to, $V$ was constructed from a model $V^{\ast}$ of ZF-Foundation (which was just assumed to exist without out explicitly describing what it is) by $V(0)=\emptyset$ and then taking power sets. How is this not unique? Or should I understand it this way - whether $V$ constructed in this way believes in $V=L$ or not depends on the model $V^{\ast}$ which I started with and about which I don't know anything, so I can't argue neither in favour nor against $\langle V,\in\rangle\models V=L$? $\endgroup$ – settheoryguest Feb 11 '15 at 14:04

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