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It is known that Morse-Kelly (MK) set theory forms a metatheory for ZFC. For example:

MK proves Con(ZFC). In fact, Joel David Hamkins claims in his blog post "Kelly-Morse set theory implies Con(ZFC) and much more" that in MK "there is a transitive model of ZFC, and furthermore that the universe $V$ is the union of an elementary chain of elementary rank initial segments $V_\theta$ of $V$, each of which , in particular, is a transitive model of ZFC."

My question is this: If in MK there is a transitive model $M$ of ZFC, is there in MK the generic extension $M$[$G$], where the generic $G$ is definable in MK? Also, if so, is there a $M$[$G$] in MK in which cardinals are collapsed?

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If there is a transitive model of ZFC, then there is a smallest ordinal $\alpha$ such that $L_\alpha$ is a model of ZFC, and this is called the minimal transitive model of ZFC, because it is contained in all others. So we can define this model in the theory KM. Since it is countable in $L$, furthermore, there will be $L_\alpha$-generic filters $G\in L$ for any particular poset $\mathbb{P}\in L_\alpha$, and so there will be an $L$-least such $G$ in $L$ that is $L_\alpha$-generic. Thus, we can also define all kinds of forcing extensions $L_\alpha[G]$, for any particular $\mathbb{P}\in L_\alpha$, since all elements of $L_\alpha$ are definable there without parameters and hence also definable in KM. So we arrive in this way at numerous models $L_\alpha[G]$ that are definable in KM. By taking $\mathbb{P}$ to be the forcing to collapse $\omega_1$ to $\omega$ (from the perspective of $L_\alpha$), these extensions will also collapse cardinals of $L_\alpha$.

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  • $\begingroup$ @Prof Hamkins: In the proof of the statement of yours I quoted, is that model countable or uncountable, and in KM can you define models of ZFC of arbitrary cardinality? $\endgroup$
    – Thomas Benjamin
    Commented Dec 2, 2014 at 21:53
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    $\begingroup$ If there is any transitive model of ZFC, then there is a countable one, so you can assume that it is countable. But in fact, in KM you can define transitive models of ZFC of any particular cardinality, since because of the elementary tower $V_{\kappa_0}\prec V_{\kappa_1}\prec\cdots$, we'll have unboundedly many models of ZFC of the form $L_\beta$ for various ordinals $\beta$, unboundedly often. So we can define the $\alpha^{th}$ one of any particular kind or cardinality. $\endgroup$
    – Joel David Hamkins
    Commented Dec 2, 2014 at 21:55
  • $\begingroup$ @Prof Hamkins: Thanks for that clarification. $\endgroup$
    – Thomas Benjamin
    Commented Dec 2, 2014 at 21:57
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Well, if $M$ is a model of $\sf ZFC$, you can assume - by using the Skolem-Lowenheim theorem - that $M$ is countable. Then forcing works as it does in $\sf ZF$.

Under additional assumptions, you can get more. For example $M$ is a model with particular properties, $G$ is a generic filter for such and such forcing, $V$ satisfies such and such properties.

For example, if $V$ is a model in which $0^\#$ exists, then you can arrange for a model, even uncountable, to have cardinals which $V$ knows are countable. For example $L_{\omega_1}$ has many cardinals which in $V$ are countable. Therefore there are generic filters for the collapsing forcing in that model, already in $V$.

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  • $\begingroup$ You beat me again! (But I see also that, again, we make different points.) $\endgroup$
    – Joel David Hamkins
    Commented Dec 2, 2014 at 21:48
  • $\begingroup$ I was just checking to see between brushing my teeth and getting into bed, so I had the chance. I think that we make similar points here, even if not quite the same one. Namely, once you have a model of ZFC, you can assume it is countable and you don't even care what the metatheory is, as long that it is sufficiently strong to develop forcing, of course. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 2, 2014 at 21:54
  • $\begingroup$ Yes, that is of course the main point. $\endgroup$
    – Joel David Hamkins
    Commented Dec 2, 2014 at 21:56
  • $\begingroup$ @Prof. Hamkins: Could one then define the Naturalisitic account of forcing in KM? $\endgroup$
    – Thomas Benjamin
    Commented Dec 2, 2014 at 22:07

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