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Let $\lambda, \kappa$ be infinite cardinals. Does it follow from the inequality $\kappa\lt2^\lambda$ that $2^\kappa\lt2^{2^\lambda}$?

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The < sign seems not to work inside $LaTeX$, at least for me. \lt is recommended instead. –  Ross Millikan Apr 20 '12 at 22:20
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Cross posted: mathoverflow.net/questions/94694/cardinal-exponentation - if you cross post to MO please wait a day or two first. 30 minutes is simply unreasonable. –  Asaf Karagila Apr 20 '12 at 23:07
    
You might want to have a look at Easton's theorem. It is mentioned also in this MO thread: When 2^a = 2^b implies a=b (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?. –  Martin Sleziak Apr 21 '12 at 6:45
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up vote 5 down vote accepted

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

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