There are many papers that point out that the behavior of forcing is quite different with and without the AC.

I wonder how much of an impact this has on relative consistency proofs.


1 Answer 1


Sometimes, but not quite.

You don't need to assume $\sf AC$ in order to get forcing to work. You also don't need to assume $\sf AC$ in the meta-theory order to get forcing to work over countable models of set theory. You also don't need to assume $\sf AC$ in the meta-theory in order to do general independence results, since we can always resort to Boolean-valued models and the likes of that.

And so, for the general mechanism, you don't quite need the assumption of $\sf AC$ for things to work out.

However, you could be talking about independence results in the sense that we want to prove that something is independent by forcing. So, for example, given any model of $\sf ZFC$ it has a forcing extension which satisfies $\sf CH$ and another which does not.

In that sense, forcing will preserve $\sf AC$, and while we can force $\sf AC$ itself over some models of $\sf ZF$, we also know of some models where we simply cannot do it.

Finally, there are concepts that just don't have good definitions without $\sf AC$, for example cardinal characteristics of the continuum, and that requires further explorations and work before we can say something concrete about the independence and the techniques that we use.

But overall, if something is independent of $\sf ZFC$, it will be independent of $\sf ZF$, and in fact, of any weaker theory.

  • $\begingroup$ So, if we live inside a $ZF+\neg AC$ model, there will be some independence results that cannot be obtained by forcing? Can this be seen as AC giving forcing the power to go beyond Boolean-valued? $\endgroup$ Sep 6, 2022 at 12:47
  • $\begingroup$ The question is what exactly do you mean by independence results. Are those number theoretic statements, e.g., Con(ZF) implies Con(ZFC), or are those forcing constructions? If you work with "Let $M$ be a countable transitive model of ZFC, etc.", then AC in the universe has exactly no bearing on forcing over $M$. If, however, you want to construct extensions of the universe, internally, then some models of ZF cannot be extended to models of ZFC via forcing. But then again, no model of ZFC can be extended to a model of ZF with the negation of AC via forcing either. $\endgroup$
    – Asaf Karagila
    Sep 6, 2022 at 12:56
  • $\begingroup$ To get independence results about different ACs, one has to use the corresponding different-ACs+ZF as the ground models, which seems to sound reasonable. But then there is no only best ground model that can formalize all relatively consistent results when different ACs. $\endgroup$ Sep 6, 2022 at 13:45
  • $\begingroup$ Unfortunately, I don't really understand what you mean by that last comment. $\endgroup$
    – Asaf Karagila
    Sep 6, 2022 at 14:45
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    $\begingroup$ @Ember: The whole concept of ccc is a bit problematic in ZF, and proper forcing only makes sense under DC. $\endgroup$
    – Asaf Karagila
    Jan 26, 2023 at 16:49

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