Sets question, without Zorn's lemma

Question

Is there any proof to $|P(A)|=|P(B)| \Longrightarrow |A|=|B|$ that doesn't rely on Zorn's lemma (which means, without using the fact that $|A|\neq|B| \Longrightarrow |A|<|B|$ or $|A|>|B|$ ) ?

Thank you!

Answer 1

Even with Zorn's Lemma, one cannot (under the usual assumption that ZF is consistent) prove that if two power sets have the same cardinality, then the sets have the same cardinality.

Answer 2

One can not prove that $|\mathcal P(A)|=|\mathcal P(B)|$ implies $|A|=|B|$, even assuming Zorn's lemma (or equivalently, the axiom of choice).

It is consistent with choice (i.e., Zorn's lemma) that $2^{\aleph_0}=\aleph_{\omega+1}$ and, for every ordinal $\alpha$ that is $0$ or limit and, for every natural number $n$, we have $2^{\aleph_{\alpha+n+1}}=\aleph_{\alpha+\omega+1}$. This gives us that for any infinite $\kappa$ there are infinitely many different sizes $\lambda$ such that $2^{\kappa^+}=2^\lambda$. For example, this gives us that $\mathcal P(\aleph_0),\mathcal P(\aleph_1),\dots,\mathcal P(\aleph_\omega)$ all have the same size $\aleph_{\omega+1}$, while $\mathcal P(\aleph_{\omega+1}),\mathcal P(\aleph_{\omega+2}),\dots,\mathcal P(\aleph_{\omega+\omega})$ all have the same size $\aleph_{\omega+\omega+1}$, etc.

Similarly, one can also arrange that for any infinite $\kappa$ there is a different $\lambda$ with $2^\kappa=2^\lambda$.

On the other hand, for any positive integer $n$, it is consistent with choice that for any infinite cardinal $\kappa$, we have $2^\kappa=\kappa^{+n}$ (the $n$-th successor of $\kappa$). In this case, we have $|\mathcal P(A)|=|\mathcal P(B)|$ implies $|A|=|B|$. The case $n=1$ is the $\mathsf{GCH}$.

(Small technical note: For $n>1$, the result in the paragraph above requires large cardinals. We can avoid this by requiring, for example, that if $\kappa$ is a limit cardinal then $2^\kappa=\kappa^+$, while if it is a successor cardinal, then $2^\kappa=\kappa^{++}$. This can be achieved by a standard Easton forcing over a model of $\mathsf{GCH}$, and no large cardinals are required.)

Perhaps you are interested in whether there is a condition that does not imply choice but implies your statement or, more directly, whether your statement already implies choice. This is open. It was asked on MO, here. The statement (that Asaf Karagila calls $\mathsf{ICF}$, injective continuum function) implies $\mathsf{dBS}$, the dual Bernstein-Schroder theorem, in which I've been interested myself. This direct consequence of choice states that for any two sets $A$ and $B$, if there is a surjection from $A$ onto $B$ and a surjection from $B$ onto $A$, then $A$ and $B$ have the same size. It is also open whether $\mathsf{dBS}$ implies choice, see here.

The two links above give about all the results I'm aware of relating $\mathsf{ICF}$ and choice. As Asaf points out in the comments, $\mathsf{ICF}$ has also been called $\mathsf{WPH}$, the weak power hypothesis, and was also considered by Tarski. The reference

Azriel Lévy. The Fraenkel-Mostowski method for independence proofs in set theory, in The Theory of Models, J. Addison, L. Henkin, and A. Tarski, eds., North Holland, Amsterdam, 1965, pp. 135-157,

indicates that Tarski had already observed that $\mathsf{ICF}$ implies $\mathsf{dBS}$. Lévy's paper also asks whether $\mathsf{dSB}$ implies choice.

(Actually, it appears your question had already been asked on this site. See here and here.)