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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

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up vote 13 down vote accepted

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
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@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
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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

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