Suppose Morse-Kelley set theory consists of class comprehension, class foundation, class extension, axiom of infinity , limitation of size, and the general continuum hypothesis. Can the axiom of powerset,that the power set of a set is a set, be proven in this system? Without gch of course it can’t.

If it can be proven then Morse-Kelley set theory with gch can be presented as class comprehension, class foundation, class extension, axiom of infinity, and the following axiom which combines the axiom of limitation of size and gch: let P be the power class of a class C. If S is a class such that |S|<|P| then there exists an element of P with the same cardinality as S.

Here only one axiom explicitly refers to sets, the axiom of infinity.


1 Answer 1


No, you cannot prove powerset even under $\mathsf{GCH}$, although it depends on how we state $\mathsf{GCH}$. Let me follow the following form of $\mathsf{GCH}$:

For a given set $a$, if the powerset of $a$ exists, then there is a bijection between the powerset $\mathcal{P}(a)$ and $|a|^+$.

The existence of $|a|^+$ requires more justification. It follows from Choice, Replacement, and the existence of $\mathcal{P}(a)$. Since I am unfamiliar with Limitation of size, let me replace it with the combination of Replacement and Global choice, which should imply Limitation of size:

  • Replacement: For every class function $F$ and a set $a$, the image set $F[a] = \{F(x)\mid x\in a\}$ exists.
  • Global choice: There is a class well-order of $V$.

Then we can see the following:

Claim. There is a model of Morse-Kelley set theory plus Global choice, Replacement, $\mathsf{GCH}$, and the negation of Powerset.

Proof. Let us work over $\mathsf{ZFC} + (V=L)$ and consider $L_{\omega_1}$. Since $L_{\omega_1} = H_{\omega_1}$, it is a model of second-order $\mathsf{ZFC}$ without powerset. Furthermore, since $L_{\omega_1}\vDash (V=L)$, it satisfies Global Choice, $L_{\omega_1}$ satisfies $\mathsf{GCH}$ vacuously (since every set in there is countable) but it does not satisfy Powerset.

  • $\begingroup$ As far as formulation of gch, I wanted to use a formulation of gch and limitation of size together that does not explicitly refer to sets: given any power class P( a class equal to the power class of some class) and any class S with |S| <|P|, S has the same cardinality as some member of P. I’m pretty sure the answer to my question is no also for this formulation. Does your proof work for this formulation of gch also? Actually I like that gch and limitation of size can be bundled together that natural way as limiting cardinalities of subclasses of power classes. $\endgroup$
    – M. Solomon
    Commented Nov 21, 2023 at 21:50
  • $\begingroup$ @M.Solomon I do not see why your GCH is really the GCH we commonly say, since it follows from the Global Choice that is compatible with the failure of the usual GCH, if my understanding on your power class is correct: I understand your power class as a class $C$ containing all subsets of $C$. Then you can see by induction that $V_\alpha\subseteq C$ for every ordinal $\alpha$, so $C=V$. If a class $S$ satisfies $|S|<|V|$, and if we have the Global Choice, then $S$ must be a set so it has the same cardinality with a member of $V$. $\endgroup$
    – Hanul Jeon
    Commented Nov 21, 2023 at 21:59
  • $\begingroup$ I do consider the power class of a set to be the class of all subsets of that class. My formulation of gch is Let P be the power class of a class C. If |S| < |P| then S has the cardinality of some element of P (in other words the cardinalities of the subclasses of P are already the cardinalities of the elements of P or the cardinality of P). The case where C is a set is the usual gch, right? The case where C=V gives the axiom of limitation of size. Am I missing something? $\endgroup$
    – M. Solomon
    Commented Nov 21, 2023 at 22:20
  • $\begingroup$ The issue is that unlike that the powerset of a set has a larger cardinality than the given set, the powerclass of a given class is not necessarily "larger" than the given set. A powerclass of a class is more similar to, something like, a collection of "small" subsets of a given set (e.g., For an inaccessible cardinal $\kappa$, the collection of all subsets of $\kappa$ of cardinality $<\kappa$, which is completely different from the genuine powerset of $\kappa$.) $\endgroup$
    – Hanul Jeon
    Commented Nov 21, 2023 at 22:26
  • $\begingroup$ Even your current formulation of GCH is valid over the model $L_{\omega_1}$ I suggested: Take any subset $X\subseteq L_{\omega_1}$, and consider the powerclass $P(X)$ of $X$ computed over $L_{\omega_1}$, which is equal to the set of all countable subsets of $X$, whose cardinality is $\omega_1$. If $|S|<|P(X)|$ holds over $(L_{\omega_1}, \mathcal{P}(L_{\omega_1}))$, then $|S|<|P(X)| = \omega_1$ holds over $L$. This implies $S$ is countable, so it has the same cardinality with some member of $P(X)$. $\endgroup$
    – Hanul Jeon
    Commented Nov 21, 2023 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.