Existence of model of ZF in which every uncountable cardinal is the cardinality of some power set?

Question

Does there exist a model of ZF in which any uncountable cardinal number is equal to the cardinality of the power set of some set ?

( In ZFC it is not possible as is shown by the answers to this cardinal numbers and the power-set , so I am asking only for a model of ZF )

Answer 1

No.

For two main reasons:

1. Even in the absence of choice $\aleph_\omega$ cannot be a power set of anything. The proof is the same usual proof that relies on Koenig's theorem, which may require choice, but it does not require choice if the power set is well-orderable.

   So in any case it is provable you will have uncountable cardinals which are not power sets of anything.

But those are just singular cardinals, let's assume every regular cardinal is a cardinal of a power set. Which is a segue to the second reason,

2. If $\aleph_1$ is a cardinality of a power set, then it is necessarily the continuum. Moving to $\aleph_2$, then it has to be the cardinal of a power set, so it has to be the power set of $\omega_1$. Continuing by induction we get that the power set of an ordinal is well-ordered, and that implies choice over $\sf ZF$.

Answer 2

Let me elaborate on Asaf's claim that $\aleph_\omega$ cannot be a power set. Suppose $|P(X)|=\aleph_\omega$ for some set $X$. Since $|X|<|P(X)|$, $|X|<\aleph_\omega$, and so $|X|=\aleph_n$ for some $n<\omega$. It would then follow that $$\aleph_\omega^{\aleph_0}\leq\aleph_\omega^{\aleph_n}=(2^{\aleph_n})^{\aleph_n}=2^{\aleph_n^2}=2^{\aleph_n}=\aleph_\omega.$$

But by König's theorem $\aleph_\omega^{\aleph_0}>\aleph_\omega$. While König's theorem in general involves the axiom of choice, choice is not needed here since all the sets involved in this application are alephs and hence come with a canonical well-ordering.