Does forcing need a countable transitive model or can forcing be done using any model $M$ of $\mathsf{ZFC}$?

Also: what about standard vs. non-standard $M$? (here standard just means that the relation on $M$ is the actual $\in$) Then: can forcing be done on a non-standard model? (The answer to which intuitively seems to be yes)

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    $\begingroup$ I think for definable notions of forcing, one does not even need a model of ZFC, via the technique of boolean-valued models... $\endgroup$
    – Zhen Lin
    May 1, 2013 at 10:05
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    $\begingroup$ Forcing can be applied within any model of $\mathsf{ZFC}$, standard or otherwise, countable, or proper sized. $\endgroup$ May 1, 2013 at 22:54

1 Answer 1


For the first part: if you prefer to view forcing as being done over ctm's, then the fact that your models are countable guarantees the existence of filters for any family of dense sets for your poset (Rasiowa-Sikorski lemma).

The fact is though that one does not need ctm's in order to force. We can view forcing as a purely syntactic notion (introduce a forcing language, show it satisfies some properties, and then argue metamathematically to show consistency). This is how Kunen does forcing.

Or, like how Zhen Lin says, we can take Boolean-valued models as a method of forcing (Jech forces this way), but I'm not sure anyone actually forces this way 'in the real world' - apparently this is really unwieldy, and it's much easier to talk about posets instead.

(disclaimer: just a masters student, take everything i say with a pinch of salt)

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    $\begingroup$ In almost all the cases I have seen thus far, no one bothers with these aspects. We just take a countable transitive model of $\sf ZFC$ and get it over with. The fact that we know how to translate the same arguments into a setting without CTM's is what allows us to use them whenever we want. $\endgroup$
    – Asaf Karagila
    May 1, 2013 at 17:50
  • $\begingroup$ Yeah, it should be said that actually no-one really cares how you force, just that you can do it. $\endgroup$
    – Kris
    May 1, 2013 at 22:58
  • $\begingroup$ @AsafKaragila, isn't it necessary then to add to ZF some axiom that standard models exist? $\endgroup$
    – Alexey
    Sep 4, 2015 at 13:31
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    $\begingroup$ @Alexey: If you want them, yes. But you can do without. You can prove that every finite fragment of $\sf ZF$ has a countable transitive model. So if you can prove that every finite list of axioms of $\sf ZF$ is consistent with the statement, then $\sf ZF$ does not prove its negation (note that this is a meta-theorem; but all forcing arguments "internal to the universe" are meta-theorems). You can also use Boolean-valued models; of Feferman's extension which is equiconsistent with $\sf ZF$, that there is a countable transitive set which is an elementary submodel of the universe [...] $\endgroup$
    – Asaf Karagila
    Sep 4, 2015 at 13:49
  • $\begingroup$ [...] But the fact of elementarity is a meta-fact, so you do not increase the consistency strength. It's quite clever and nice. But in any case. You can use that countable model to force over. Then you prove that it is a model of $\sf ZF$ (but this proof is a meta-theorem, again, since you can only prove the axioms hold one by one, not all together; or at least not necessarily all together). $\endgroup$
    – Asaf Karagila
    Sep 4, 2015 at 13:50

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