Let $\mathsf{MK^-}$ be the theory "$\mathsf{MK}-\text{Foundation}-\text{Limitation of size}-\text{Union}+\text{Subsets}$", where $\mathsf{MK}$ is Morse-Kelley set theory with axioms mentioned in:


Of course axioms of pairing and union and one direction of axiom of limitation of size are known to be redundant with this specific formulation of $\mathsf{MK}$.

Removal of axiom of limitation of size and putting the axiom of subsets (the axiom asserting that every subclass of a set is a set) instead of it would make it possible for some classes to be equinumerous to sets and yet not being sets, since it is known that the axiom of subsets does not imply the assertion that every class that is equinumerous to a set is a set. So is it possible to have a model $M$ of $\mathsf{MK^-}$ such that we have all of the followings?

$M \vdash \exists P (\text {$P$ is a proper class} \wedge P < V)$
$M \vdash \forall P (\text{$P$ is a proper class} \wedge P < V \implies \exists x\in V (\text{$P$ is equinumerous to x}))$

where "<" denotes "strict subnumerousity" defined in the customary manner; and $V$ is the class of all sets. Given that, it is clear that $M$ cannot satisfy closure of replacement over sets. The general context of this question is about size of proper classes and to what extent that can be shared with size of sets in absence of Replacement and global choice.

Also related to this is the following question:

If the above is possible then can we add the following:

$M \vdash \forall P (\text{$P$ is a proper class} \wedge P < V \implies \exists x \in V (x=\{\{y\}| y \in P\}))$

The special context of those questions raised when I was investigating an alternative to the axiom of limitation of size of $\mathsf{MK}$. A version that I've lately posted to FOM thread is referred to by the following link:


This version proves all axioms of Empty set, Pairing, Power, Separation (subsets), Replacement, Limitation of size, Global choice and Set union, so it is more powerful than the usual limitation of size axiom.

I've been trying to further simplify this axiom to the following:

$\forall x (x \in V \iff U(x) < V)$

However I couldn't prove Replacement nor union, so I was left with the above situation (i.e., theory $\mathsf{MK^-}$); so if the possibilities that I've asked about specifically above were inconsistent then this would mean that this axiom can be simplified to the above version.

  • $\begingroup$ Could you clarify the question? Any model of KM itself has a bijection of V with Ord by the global choice axiom, and so any class that is strictly smaller than V is in fact bijective with an ordinal and hence is a set. So the property seems to be true (vacuously) in any model of KM, without needing to remove any axioms at all. $\endgroup$ Commented Sep 15, 2016 at 12:22
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    $\begingroup$ My point is that every model of KM itself is also a model of that weak fragment, since nothing in your theory prevents this. I guess you intended to ask for a model with some additional properties, such as not having global choice etc.? $\endgroup$ Commented Sep 15, 2016 at 13:39
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    $\begingroup$ I'm still a little confused about what you want. If there is a proper class that is smaller than $V$, then by the meaning of "proper" class, it is not a set. $\endgroup$ Commented Sep 15, 2016 at 15:56
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    $\begingroup$ Could you state clearly the axioms that are in your theory, rather than which are not? There are several different equivalent formulations of Kelley-Morse, and some of them, for example, do not use the so-called Limitation of size axiom. $\endgroup$ Commented Sep 20, 2016 at 16:19
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    $\begingroup$ In general, the operation of removing an axiom from a theory is not well-defined up to logical equivalence. An interesting case is the removal of the power set axiom from ZFC: jdh.hamkins.org/what-is-the-theory-zfc-without-power-set $\endgroup$ Commented Sep 20, 2016 at 17:34

3 Answers 3


I believe the following gives a positive answer to your first question, modulo large cardinals:

Let $\kappa$ be measurable, and consider a Prikry-generic extension of the universe $V[G]$. Look at $\mathcal{M}=(V_{\kappa+1})^{V[G]}$. We can view this as a model of $MK^-$ by taking the proper classes to be exactly those elements of $\mathcal{M}$ of rank $\kappa$. In particular, the generic $G$ - which is a cofinal map from $\omega$ to $\kappa$ - is a "proper class" in the sense of $\mathcal{M}$, yet equinumerous with the $\mathcal{M}$-set $\omega$.

I strongly suspect that the measurable is unnecessary here - I'm just too lazy to think up the details for a ZFC-example (maybe something closer to Namba forcing?). The reason I'm using Prikry here is because Prikry forcing doesn't add any bounded-rank subsets of $\kappa$, which gives a fast (but probably unnecessary) proof that $\mathcal{M}$ is in fact a model of $MK^-$.

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    $\begingroup$ Are you sure there isn't a typo in how you defined the proper classes? Also, wouldn't that work with the first worldly cardinal without the need for forcing? $\endgroup$
    – Asaf Karagila
    Commented Sep 21, 2016 at 4:33
  • $\begingroup$ I don't think we need to go that far just to answer the first two conditions. I'll post an answer to this effect in ZFC $\endgroup$ Commented Sep 21, 2016 at 12:21
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    $\begingroup$ @AsafKaragila Derp. Fixed. And yes, it would work with just a worldly (but I really like Prikry forcing :P). $\endgroup$ Commented Sep 22, 2016 at 13:34
  • $\begingroup$ Who doesn't? Prikry forcing is an epitome of awesomeness! $\endgroup$
    – Asaf Karagila
    Commented Sep 24, 2016 at 15:58

An answer to the first question is: Take a model $\mathcal{M}$ of $\text{ZF +GCH}$, the stage $V_{w+w+1}$ of $\mathcal{M}$ will satisfy all axioms of $\text{MK}^-$ plus the first two conditions of the first question: the stage $V_{w+w}$ would interpret the class $V$ of all sets of $\text{MK}^-$, a set of $\text{MK}^-$ is intepreted as an element of $V_{w+w}$, while a proper class of $\text{MK}^-$ is a subset of $V_{w+w}$ that is not an element of $V_{w+w}$; now we have $|V_{w+w}|=\aleph_w$, and since the Generalized Continuum Hypothesis "$\text{GCH}$" is equivalent to $\beth_\alpha=\aleph_\alpha$ for every infinite ordinal $\alpha$, then every cardinality less than the cardinality of $V_{w+w}$ would be either $n$ or $\aleph_{w+n}$ for a natural $n$, all of which are cardinalities of elements of $V_{w+w}$, and thus cardinalities of sets, that a proper class exists that is strictly subnumerous to $V$ is witnessed by the set of all $V_i$ stages with $i<w+w$, since it is a subset of $V_{w+w}$ and it is not an element of $V_{w+w}$ and its cardinality is $\aleph_0 <\aleph_w$. The rest of axioms of $\text{MK}^-$ (Extensionality, Class comprehension axioms, Pairing, Power, Infinity and Subsets) are interpreted straightforwardly.


The stage $(V_{w+w+1})^L$ would serve as a model of $MK^-$ as far as the first two conditions are concerned, but it would fail the third condition. Classes would be interpreted as elements of $(V_{w+w+1})^L$, Sets are elements of $(V_{w+w})^L$ and proper classes are elements of $(V_{w+w+1})^L$ that are not elements of $(V_{w+w})^L$. $(V_{w+w})^L$ interprets $V$. Clearly it is provable within $(V_{w+w+1})^L$ that for each $n=0,1,2,..,w$ we have $(V_{w+n})^L < (V_{w+n+1})^L$ where < is strict subnumerousity, since each $(V_{i+1})^L$ is the constructible power set of $(V_i)^L$ and $L$ is a model of $ZFC$, now the largest cardinality "as seen inside $(V_{w+w+1})^L$" is the cardinality of $(V_{w+w})^L$, and because $L$ satisfies the general continuum hypothesis then we can define cardinality within $(V_{w+w+1})^L$ as:

$|x|=y \iff \exists F (F:y\to x \wedge \text{F is a bijection})\wedge (\text {y is a finite von Neumann} \lor \exists i (\text {i is infinite} \wedge y=V_i^L))$.

In English the cardinality of a set $x$ is the finite von Neumann ordinal that is equinumerous with $x$ or the infinite stage of the $(V_{w+w+1})^L$ that is equinumerous with $x$. So we define $\aleph_n$ as $(V_{w+n})^L$ for each $n=0,1,2,3,...,w$.

Now because $L$ satisfies the Generalized Continuum Hypothesis so it is provable within $L$ that all cardinals strictly smaller than $\aleph_w$ are either $n$ or $\aleph_n$ for some natural $n$ and each one of those is the cardinality of some element of $(V_{w+w})^L$ and each one of those is itself an element of $(V_{w+w})^L$. The set of all $V_i^L$ stages in $(V_{w+w})^L$ is a proper class of cardinality $\aleph_0$ which is strictly smaller than $\aleph_w$, and every proper class that is provable within $(V_{w+w+1})^L$ to be strictly smaller than $\aleph_w$ would be provable within $(V_{w+w+1})^L$ to be of cardinality $n$ or of cardinality $\aleph_n$ for a natural $n$, and so equinumerous to a set. So $(V_{w+w+1})^L$ satisfies $MK^-$ plus the first two conditions. However the third condition cannot be met since the set of all singletons of elements of any proper class would be a proper class too. And so this remains a partial answer.

The same argument would hold in any model of $ZF+GCH$, the stage $V_{w+w+1}$ of the cumulative hierarchy of that model would be a model that satisfies all axioms of $MK^-$ plus the first two conditions of the question. However it still fails the third condition. So it is also a partial answer.

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    $\begingroup$ (1) $2\omega=\omega$, you probably mean $\omega2$; (2) $L_{\omega2+1}$ is countable, it knows about no infinite cardinal except $\omega$ itself. $\endgroup$
    – Asaf Karagila
    Commented Sep 21, 2016 at 13:50
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    $\begingroup$ No. $2\omega=\omega$ and $\omega+\omega=\omega2$, the latter of which is indeed greater than $\omega$. Secondly, there are only countably many formulas in the language of set theory, and it follows that for an infinite $\alpha$, $|\alpha|=|L_\alpha|$. Unless your $2w$ is some uncountable ordinal, you're just wrong. I'm not saying that your idea cannot work, just that the way you present it contains several mistakes. $\endgroup$
    – Asaf Karagila
    Commented Sep 21, 2016 at 14:16
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    $\begingroup$ The number of formulas is relevant in that the definition of the stages of L involves taking definable subsets of the previous stages. Asaf is perfectly right, $|L_\alpha|=|\alpha|$ for infinite ordinals $\alpha$, and standard ordinal multiplication makes $2\cdot\omega=\omega$ (the well order you get by putting $\omega$ many pairs after each other is still $\omega$). $\endgroup$ Commented Sep 21, 2016 at 15:27
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    $\begingroup$ I believe you are confusing $L_\alpha $ with $(V_\alpha)^L $; the $\alpha $th level of the $L $ hierarchy is vastly smaller in general than what $L $ thinks is the $\alpha $th level of the cumulative hierarchy. In particular, your first-paragraph sentence beginning "clearly it is provable" is completely false as written. $\endgroup$ Commented Sep 22, 2016 at 3:53
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    $\begingroup$ @zuhair see my comment, the answer is still incorrect. $\endgroup$ Commented Sep 22, 2016 at 3:56

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