A while ago I was interested in the grounded Martin’s axiom, a principle that said that $V$ was a ccc forcing extension of some inner model $W$ and Martin’s axiom held in $V$ but only for posets from $W$ (see this post or my dissertation for more information). I was implicitly only considering set forcing extensions when formulating this axiom, but the question arose whether one could also allow class forcing.

So what is really being asked is whether there is (or can be) a ccc class forcing that is somehow not equivalent to any set forcing. For the class forcing to make sense from the point of view of the axiom, we should also require that it is tame (i.e. preserves ZFC). Omitting either the chain condition or the preservation of ZFC gives rise to simple counterexamples: the forcing $\mathrm{Add}(\omega,\mathrm{Ord})$ is ccc but does not preserve ZFC, and the forcing to add a Cohen subset to each infinite cardinal is a tame class forcing not equivalent to any set forcing, but which does not have a bounded chain condition.

Recently I realized that something like a combination of the intermediate model theorem and Bukovský’s theorem gives an answer to this question. I was then coerced (by people who shall remain unnamed) to write down the argument.

Theorem.(GBC) Let $\kappa$ be a cardinal and suppose that $\mathbb{P}$ is a $\kappa$-cc tame class forcing notion. Then $\mathbb{P}$ is equivalent to a set forcing.

Proof: Since $\mathbb{P}$ is tame, we may as well assume that it is a set-complete Boolean algebra. Since $\mathbb{P}$ is $\kappa$-cc and tame, there is some cardinal $\lambda$ such that $\mathbb{P}\Vdash 2^\kappa\leq\lambda$. We can therefore fix a $\mathbb{P}$-name $\dot{A}$ for a subset of $\lambda$ such that it is forced that $\dot{A}$ codes all the subsets of $\kappa$ in the extension. Now let $\mathbb{Q}$ be the complete subalgebra generated by the elements $\Vert\check{\alpha}\in\dot{A}\Vert$ for $\alpha<\lambda$. Note that $\mathbb{Q}$ is generated by a set in a $\kappa$-cc complete Boolean algebra, so it itself is a set. Note that we have arranged matters so that $\mathbb{P}\Vdash \mathcal{P}(\kappa)\in V[\dot{G}\cap\mathbb{Q}]$. Now let $\dot{\mathbb{R}}=\mathbb{P}/\mathbb{Q}$ name the quotient algebra and observe that $\dot{\mathbb{R}}$ is forced to be $\kappa$-cc, but it cannot add any subsets to $\kappa$, since all of those are already there in $V^{\mathbb{Q}}$. It follows that $\mathbb{R}$ cannot add any sets at all, by an argument I described in an earlier post on Bukovský’s theorem. We can thus conclude that $\mathbb{P}$ and $\mathbb{Q}$ are equivalent. $\Box$

To be honest, I expect that there is a combinatorial argument showing that there simply are no proper class tame $\kappa$-cc separative forcing notions, superseding the above result, but it is not clear to me how it would go. Tameness, in particular, is crucial, because of the $\mathrm{Add}(\omega,\mathrm{Ord})$ example above, but as far as I know, we do not have a good combinatorial description of which pretame forcing notions are tame. One might try instead to analyze the quotient forcing $\mathbb{R}$ above and show that it is a set. We showed that it is trivial as a forcing notion, but right now I do not see a reason that it could not be a proper class.