# 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
Don't use $\Rightarrow$ together with the word "if". That makes the entire implication the "if" clause of a statement (for which you have no consequent). To each worded "if" there should be a worded "then". Also, your use of ASCII would easily have been confused with you asking if $|P(A)|=|P(B)|\geq|A|=|B|$ was possible. –  Arturo Magidin Mar 27 '11 at 21:58

## 1 Answer

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
In your first line you say the OP's question is true if you assume CH, but that only solves it for $\aleph_0$. If you want it generally true, don't you need GCH? Your later statement that without GCH you can't be sure is correct. –  Ross Millikan Mar 27 '11 at 22:25