Let $\lambda, \kappa$ be infinite cardinals. Does it follow from the inequality $\kappa\lt2^\lambda$ that $2^\kappa\lt2^{2^\lambda}$?
Cardinal exponentation
2
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set-theory
cardinals
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0You might want to have a look at [Easton's theorem](http://en.wikipedia.org/wiki/Easton%27s_theorem). It is mentioned also in this MO thread: [When 2^a = 2^b implies a=b (a,b cardinals)](http://mathoverflow.net/questions/17152/when-2a-2b-implies-ab-a-b-cardinals) and in the same question on this site [Does $2^X \cong 2^Y$ imply $X \cong Y$ without assuming the axiom of choice?](http://math.stackexchange.com/questions/74477/does-2x-cong-2y-imply-x-cong-y-without-assuming-the-axiom-of-choice). – 2012-04-21
1 Answers
6
No.
Suppose $2^{\aleph_0}=\aleph_2$ and $2^{\aleph_1}=2^{\aleph_2}=\aleph_3$.
Let $\lambda=\aleph_0$ and $\kappa=\aleph_1$, now we have $2^\kappa=2^{2^\lambda}$.