Do sets, whose power sets have the same cardinality, have the same cardinality?

Is it generally true that if $|P(A)|=|P(B)|$ then $|A|=|B|$? Why? Thanks.

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A very closely related question on MathOverflow: mathoverflow.net/questions/17152/… – Jonas Meyer Mar 27 '11 at 21:48
And also this one: mathoverflow.net/questions/67473/… – Martin Sleziak Nov 19 '15 at 14:21

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 $2^\kappa=2^\lambda$.

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Surely you need the generalized continuum hypothesis, not just CH itself? – Chris Eagle Mar 27 '11 at 21:54
@Chris: Of course, thanks for noticing it. :) – Apostolos Mar 27 '11 at 21:56
You mean you have to assume the greater continuum hypothesis. – Ross Millikan Mar 27 '11 at 22:09
@Ross: I'm not sure I get your question but you don't need to assume the generalized continuum hypothesis. Easton's Theorem is enough to show that the sentence is independent from the axioms of ZFC. I just used the generalized continuum hypothesis as a more concrete example (and since it's the first thing that came to my mind when I read the question). – Apostolos Mar 27 '11 at 22:18
Easton's theorem is overkill here. Cohen's original model for ZFC + $\neg$CH had $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$. – Andreas Blass Nov 7 '13 at 2:07