The following statement [cannot](http://mathforum.org/kb/message.jspa?messageID=182924) be proven in $\mathsf{ZFC}$:



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> > (S) : If $A, B$ are sets with $|A| < |B|$, then $2^{|A|} = |{\cal P}(A)| < |{\cal P}(B)| = 2^{|B|}$.
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Obviously, $\mathsf{GCH}$ implies (S). Does (S) imply $\mathsf{GCH}$ too?



Is injectivity of $2^{(\ldots)}$ weaker than $\mathsf{GCH}$?
No, take Merimovich's model in which $2^\kappa=\kappa^{+3}$ for all cardinals $\kappa.$


Merimovich, Carmi [A power function with a fixed finite gap everywhere](http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=9156282&fileId=S0022481200005272). J. Symbolic Logic 72 (2007), no. 2, 361–417.


Let me add that the statement "the power function is injective" is known as weak GCH, which has important applications. See [Why the weak GCH is true!](http://homepages.math.uic.edu/~jbaldwin/asl07.pdf) .


