Is it generally true that if $|P(A)|=|P(B)|$ then $|A|=|B|$? Why? Thanks.
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5$\begingroup$ A very closely related question on MathOverflow: mathoverflow.net/questions/17152/… $\endgroup$– Jonas MeyerCommented Mar 27, 2011 at 21:48
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1$\begingroup$ And also this one: mathoverflow.net/questions/67473/… $\endgroup$– Martin SleziakCommented Nov 19, 2015 at 14:21
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Your question is undecidable in ZFC. If you assume the generalized continuum hypothesis then what you state is true. On the other hand Easton's theorem shows that if you have a function $F$ from the regular cardinals to cardinals such that $F(\kappa)>\kappa$, $\kappa\leq\lambda\Rightarrow F(\kappa)\leq F(\lambda)$ and $cf(F(\kappa))>\kappa$ then it's consistent that $2^\kappa=F(\kappa)$. This of course shows that it's consistent that we can have two cardinals $\kappa<\lambda$ such that $2^\kappa=2^\lambda$.
answered Mar 27, 2011 at 21:47
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12$\begingroup$ Easton's theorem is overkill here. Cohen's original model for ZFC + $\neg$CH had $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$. $\endgroup$ Commented Nov 7, 2013 at 2:07