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Here $2^A$ is the set of all functions from $A$ to $\{0,1\}$.

This is evidently true if one of $A$, $B$ is finite. To me it seems like it should be true for infinite $A$, $B$, but a proof is oddly hard to come by. Is there a quick proof or counterexample?

Axesilo
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    It is independent from ZFC. See here: http://math.stackexchange.com/questions/639112/is-2-alpha-2-beta-rightarrow-alpha-beta-a-sf-zfc-independence-result – Jack D'Aurizio Oct 16 '15 at 18:03
  • Thanks, you're right that this question is a duplicate--I didn't see that one. – Axesilo Oct 16 '15 at 18:05

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The problem you stated above is independent of ZFC.
Schema of proof:
The statement is true if $GCH$ holds then if $|P(A)|= \beth _{\alpha+1}$ and $|P(B)|= \beth _{\alpha+1}$ then $|A|=|B|=\beth _{\alpha}$, where $P(C)$ denotes the power set of $C$
The statement is false if $MA+ \lnot CH$ holds then for any infinite cardinal $\kappa$ lesser than $2^{ \aleph _0}$ we have $2^\kappa =2^{\aleph _0}$, so $2^{\aleph _1} =2^{\aleph _0}$.

M. Stawiski
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  • Thanks for your answer. If I understand you correctly, you're saying it depends on whether or not you accept the continuum hypothesis? – Axesilo Oct 16 '15 at 18:07
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    No, generalized continuum hypothesis (there is no set of cardinality strictly between $\kappa$ and and its power set) implies your statement but it's not equivalent to it. Continuum hypothesis alone doesn't proove nor disproove this fact. If continuum hypothesis holds we just need to search for counterexamples "deeper". You can find an interesting example that shows if certain large cardinal exists (supercompact) then there it is consistent with ZFC that for every $2^\kappa$ there exists exactly one (besides $\kappa$) $\lambda$ such that $2^\kappa=2^\lambda$. – M. Stawiski Oct 16 '15 at 18:11