We know that if $M$ is a model of ZFC, then taking $\mathcal{C}$ to be the collection of all classes definable in $M$ with set parameters, and taking $\in$ to be the obvious extension of $M$'s epsilon relation to $\mathcal{C}$, then $(M,\mathcal{C})$ is a model of BG. Moreover, we can add a uniform choice function through a class forcing that is $k$-closed for every regular $k$ and so adds no sets to the universe. The resulting model $(M,\mathcal{C}')$ of BGC extends $(M,\mathcal{C})$ and has the same sets and the same restriction of $\in$ to sets. So given a model $M$ of ZFC we can always construct a model of BGC. Does this fact holds for MK? That is, assuming $M\vDash$ ZFC, is it possible to obtain a model $(M,\mathcal{C})$ of MK? I think that previous argument does not work here. Indeed, if $\mathcal{C}$ is collection of all classes definable in $M$, then $(M,\mathcal{C})$ does not satisfy MK since Class-comprehension fails.
1 Answer
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No.
Note that $\sf MK$ can define the truth predicate for the class of sets. In particular that means that it proves the consistency of $\sf ZFC$, and in fact it implies there are worldly cardinals (and more) as well.
So if $M$ is a model of $\sf ZFC$ in which there are no worldly cardinals, for example, then it cannot be extended to a model of $\sf MK$ without adding sets.
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1$\begingroup$ Since the question never restricted attention to transitive models, the same argument provides another sort of counterexample: A model of ZFC + $\neg$Con(ZFC) can't be extended to a model of MK without adding natural numbers (and can't be end-extended to a model of ZFC at all). $\endgroup$ Commented Feb 27, 2021 at 15:54