Is the following statement consistent with ZF (obviously, this is not true in ZFC)?



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> There are cardinals $\kappa, \lambda$ such that $\kappa \neq \lambda$ and $2^\kappa = 2^\lambda$.
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Can the powersets of two different cardinals be of the same cardinality?
Yes, this is possible, even with ZFC. In Cohen's model, where the continuum hypothesis fails, he gets ZFC plus $2^\omega=2^{\omega\_1}$. This statement is known as [Luzin's hypothesis](https://www.encyclopediaofmath.org/index.php/Luzin_hypothesis), an alternative to CH.


More generally, the situation for regular cardinals is described by [Easton's theorem](https://en.wikipedia.org/wiki/Easton%27s_theorem), which says that if ZFC plus the GCH holds, then for any function $E:\kappa\mapsto E(\kappa)$ on the regular cardinals satisfying the properties that $\kappa\leq\lambda\to E(\kappa)\leq E(\lambda)$ and $\kappa<\operatorname{cof}(E(\kappa))$, then there is a forcing extension in which $2^\kappa=E(\kappa)$. 


In particular, one can easily make functions $E$ that will thereby ensure ZFC plus many instances of $2^\kappa=2^\lambda$. 


