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


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

| cite | improve this answer | |
  • 5
    $\begingroup$ Easton's theorem is overkill here. Cohen's original model for ZFC + $\neg$CH had $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$. $\endgroup$ – Andreas Blass Nov 7 '13 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.