Take the 2-minute tour ×  Is there any example that for cardinal
numbers $\kappa < \lambda$, we have $2^\kappa = 2^\lambda$?  My
guess is that it only depends on whether GCH holds. Is it true?
share|improve this question I just noticed the exact wording of
the question was not answered by my answer previously. I added
more. –  Asaf Karagila Nov 28 '12 at 2:02 Thank you very much.
I'm so amazed. It seems to me that you ponder upon these
problems like crazy :) –  Metta World Peace Nov 28 '12 at 2:10 I
do (and I am). I mean, seriously. 4am. I am going to sleep! (I
have to give a talk tomorrow afternoon, and I won't have time to
doze off around noon.) –  Asaf Karagila Nov 28 '12 at 2:11 What
should I say?! I don't want to sound like Cee lo in a talent
show. But what you have been doing do amaze and impress me。 – 
Metta World Peace Nov 28 '12 at 5:59 add comment  1 Answer  up
vote 4 down vote accepted  This is independent of ZFC. It is
consist that there are no cardinals, for example if GCH holds.
Note that $\lam