The Kelley-Morse set theory can be thought as the "full-secondorderification of $\sf ZFC$", where we switch from sets to classes and allow the comprehension schema to include quantifiers on class variables.


As in the usual case of class-set theory, sets are exactly those classes which are elements of other classes. So we can consider this as a one-sort theory and have a definable predicate $\mathrm{Set}(x)\iff\exists y(x\in y)$.


The "standard" models of Kelley-Morse set theory are $V\_{\kappa+1}$ where $\kappa$ is a strongly inaccessible cardinal, and then $\mathrm{Set}$ is interpreted exactly as $V\_\kappa$.


But we can use the Lowenheim-Skolem theorem to obtain a countable model of Kelley-Morse. The trick here is that we keep "enough" classes to satisfy the impredicative comprehension schema, but the model is still countable.


My question is as follows:



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> Let $M$ be a model of Kelley-Morse What sort of limitations do we have on the cardinality of $M$ and $\mathrm{Set}^M$? Specifically, can we have a model with countably many sets and uncountably many classes? Does the answer change if we assume that $\mathrm{Set}^M$ is transitive?
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Uncountable models of Kelley-Morse set theory with only a countable number of sets
The first relevant theorem is the following classical result:


**Theorem A.** (Mostowski, Keisler) *If $M$ is a countable model of Kelley-Morse + Choice Scheme, then there is an elementary extension $M^{\*}$ of $M$ such that $\mathrm{Set}^{M}= \mathrm{Set}^{M^{\*}}$ and $M^{\*}$ has cardinality $\aleph\_1$.* 


In the above, the Choice Scheme consists of universal generalizations of statements of the following form, where the parameters in $\phi$ are suppressed:


$\forall x \exists Y \phi (x,Y) \rightarrow \exists Z \forall x \phi(x,(Z)\_x)$,


where $x$ range over sets, capital letters range over classes, and $(Z)\_x$ [read as: the $x$-th slice of $Z$] denotes $\{t: (x,t)\in Z\}$).


The theorem holds for models of Second Order Arithmetic, with the same proof that works for Kelley-Morse. The proof uses omitting types and elementary chains; see, e.g., Chapter 28 of Keisler's monograph *Model Theory for Infinitary Logic*. It is well-known that Kelley-Morse plus the Choice Scheme is bi-interpretable with with the extension of $ZFC^-$ ($ZFC$ without powerset, where $ZF$ is formulated using the collection and separation schemes) obtained by adding "the last cardinal exists, and it is inaccessible".


On the other hand;


**Theorem B.** (Jensen's Gap-1 Theorem). *Assuming $V=L$, every first order theory that admits a two-cardinal model of the form $(\aleph\_{1}, \aleph\_{0})$ admits a two-cardinal model of the form $(\kappa^{+},\kappa)$ for any prescribed infinite cardinal $\kappa$.*


By putting Theorems A & B together with the Loewenheim-Skolem theorem, we obtain the following corollary; in what follows, the "two-cardinal type" of a model $M$ of Kelley-Morse, is the ordered pair of cardinals $(|M|, |\mathrm{Set}^{M}|)$.


**Corollary.** *It is consistent with $ZFC$ that the following holds: for every consistent extension $T$ of Kelley-Morse + Choice Scheme (in the same language), the "two-cardinal" types of models of $T$ are precisely of the form $(\kappa, \kappa)$ or $(2^{\kappa}, \kappa)$, where $\kappa$ ranges over infinite cardinals.*


**Postscript.** I suspect that, using forcing, one can build a model of Kelley-Morse with countably many sets and continuumly many proper classes outright in $ZFC$. For models of Goedel-Bernays theory of classes, this was shown by Matt Kaufmann in [this paper.](http://matwbn.icm.edu.pl/ksiazki/fm/fm121/fm121121.pdf)


