Is it generally true that if $|P(A)|=|P(B)|$ then $|A|=|B|$? Why? Thanks.
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5A very closely related question on MathOverflow: http://mathoverflow.net/questions/17152/when-2a-2b-implies-ab-a-b-cardinals – Jonas Meyer Mar 27 '11 at 21:48
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1And also this one: http://mathoverflow.net/questions/67473/equality-of-cardinality-of-power-set – Martin Sleziak Nov 19 '15 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 $2^\kappa=2^\lambda$.
Martin Sleziak
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12Easton'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 07 '13 at 02:07