Set Theory – Does Limitation of Size Imply Axiom of Powerset in Morse-Kelly?

set-theory

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.

Best Answer

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.

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