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If $M \in V$ is a transitive, countable model of ZFC and $G$ a generic filter on a forcing poset $P \in M$, then we can construct the forcing extension M[G] of M. It is the smallest transitive model of ZFC such that $M[G] \supset M$ and $G \in M[G]$. Is it possible to do the same if we assume that the ground model $M \subset V$ is a proper class?

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  • $\begingroup$ Yes we can. Are you familiar with relative constructibility? $\endgroup$
    – Asaf Karagila
    Commented Sep 6, 2017 at 11:37
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    $\begingroup$ Note that there is an issue that arises when thinking about forcing over proper classes, but it's not the one you're talking about; namely, whereas a generic filter over $M$ exists whenever $M$ is countable, there's no reason for this to be true for $M$ uncountable, let alone a proper class. E.g. the statement "For every poset $\mathbb{P}$ in $L$, there is a $\mathbb{P}$-generic-over-$L$ filter" is not provable in ZFC since it implies $V\not=L$; and the statement "For every poset $\mathbb{P}$ in $V$, there is a $\mathbb{P}$-generic-over-$V$ filter" is of course outright disprovable in ZFC. $\endgroup$ Commented Sep 6, 2017 at 12:59
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    $\begingroup$ @Noah: That is outright disprovable, since that would imply that there is a generic for every $\operatorname{Col}(\omega,\alpha)$ from $L$. Which would imply that all ordinals in $L$ are countable in $V$, which is of course preposterous. Of course, your statement can be remedied by limiting the cardinality of the forcing posets or something like that. $\endgroup$
    – Asaf Karagila
    Commented Sep 6, 2017 at 16:00
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    $\begingroup$ @NoahSchweber Assuming enough large cardinals (in $V$), a version of that statement holds, with $\mathbb P$ reasonably small (say, smaller than the first $L$-indiscernible). More is true, as you can prove the existence of $L$-generics for certain proper class-sized forcing posets. $\endgroup$ Commented Sep 6, 2017 at 16:46
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    $\begingroup$ Now, one should probably also emphasize that regardless of whether we are discussing a proper class ground model or not, extensions always exist as Boolean-valued models. $\endgroup$ Commented Sep 6, 2017 at 16:47

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The most beautiful point about forcing, is that ultimately, this is all something you can do internally, granted that there are generic filters (which you can sometimes prove to exist).

More specifically, given a class $M\subseteq V$, and some $\Bbb P\in M$, if there is some $G\subseteq\Bbb P$ such that $G$ is an $M$-generic filter over $M$, then $M[G]$ is the least transitive class satisfying all the axioms of $\sf ZF$ such that $M$ is a subclass of $M[G]$ and $G\in M[G]$.

It even follows that:

  1. $M[G]=\bigcup\{L[x,G]\mid x\in M\}$, and
  2. $M$ is a class of $M[G]$, granted that $M$ was a model of $\sf ZFC$ (in which case $M]G]$ also satisfies choice).

There are some caveats, though:

  1. The forcing theorem is not a theorem, but rather a schema, or a meta-theorem. But this is the same situation as when working with countable transitive models of $\sf ZFC$. In the set-model case, this is because we prove the theorem about $M$ in $V$, and $V$ is the meta-theoretic universe as far as $M$ is concerned.

  2. The existence of generic filters is no longer a theorem. This, however, also happens when we force over uncountable models. For example, if $M$ is a transitive model which contains all the reals (in particular $M$ is uncountable), then there are no generic filters for any forcing in $M$ which "should add reals". Simply because all the reals are already in $M$, so we can't add anything new.

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  • $\begingroup$ That's a very good answer! On the other hand, there are posets $\mathbb{P}$ (for example if $\left| \mathbb{P} \right| = 1$) in $L$ such that there is a $\mathbb{P}$-generic-over-$L$ filter $G \in V$. Of course we have that $G \in L$, if $V = L$. Assuming $V \models ZFC + \lnot GCH$ (which implies that $V \neq L$), is there a forcing extension $L \left[ G \right]$ of $L$ such that $L \left[ G \right] \models \lnot GCH$? More generally, are there any nontrivial forcing extensions of $L$ if we assume only $V \neq L$? $\endgroup$
    – User_754
    Commented Sep 7, 2017 at 11:33
  • $\begingroup$ Not necessarily. It is consistent that $V=L[x]$ for some set $x$ which is not generic over $L$, and for every set, either $V=L[y]$ of $y\in L$. In such universe we do not have any nontrivial generic extensions of $L$. For the GCH question, I don't have an example off-hand, but I think that the answer is negative. $\endgroup$
    – Asaf Karagila
    Commented Sep 7, 2017 at 11:38
  • $\begingroup$ How difficult is it to prove this consistency? $\endgroup$
    – User_754
    Commented Sep 9, 2017 at 13:01
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    $\begingroup$ Quite. It requires Jensen's coding the universe into a real, but in a variant (I think due to Sy Friedman) that results in a "minimal real". There's a lot of technical precursor to fully understanding these things. $\endgroup$
    – Asaf Karagila
    Commented Sep 9, 2017 at 13:03

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