Let $\kappa, \lambda$ be two infinite cardinals such that for all infinite $\mu, \mu^\kappa = \mu^\lambda$. Is it the case that $\kappa =\lambda$ ?
First of all, clearly if the generalized continuum hypothesis holds, then the answer is yes (just take $\mu = 2^\kappa$, if $\kappa \leq \lambda$).
If we don't assume GCH, then it is well-known that $\mu = 2^\kappa$ is not enough to answer. I was thinking that maybe evaluating at some specifc cardinals such as $\kappa, 2^\kappa, \aleph_\kappa, \beth_\kappa$ could help, but so far nothing has given me an answer.
It is also possible of course that it's consistent that $\kappa \neq \lambda$, although that would be surprising to me (a bit, with rime you get used to this stuff I guess); if that's the case can we even choose any reasonable$\kappa, \lambda$ ? (e.g. is it consistent that $\kappa = \aleph_0, \lambda = \aleph_1$ ?)
Best Answer
Yes.
Suppose that $\kappa<\lambda$, take $\mu=\beth_{\kappa^+}$, then $$\beth_{\kappa^+}^\kappa = \beth_{\kappa^+} <\beth_{\kappa^+}^{\kappa^+} \leq \beth_{\kappa^+}^\lambda.$$