Let A and B infinite sets and |A|<=|B|. Is that equivalent to |P(A)|<=|P(B)| in ZF? P denotes the powerset. I think it is equivalent, but I dont know exactly how to prove it.
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1Not even in $\sf ZFC$. This question has been asked more than several times. – Asaf Karagila Jun 28 '14 at 15:57
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Note that the case of equality is sufficient here, since now $\kappa<\lambda$ and $2^\kappa=2^\lambda$. Then indeed we have that $\kappa\leq\lambda$, but we also have that $\lambda\nleq\kappa$, but $2^\kappa=2^\lambda$. – Asaf Karagila Jun 28 '14 at 16:00
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Oops! Thanks @Asaf, for making explicit what already was explicit (yet I somehow failed to catch!) – amWhy Jun 28 '14 at 16:02
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Thanks Asaf for your help! – user160628 Jun 28 '14 at 16:25