# Are cardinals with the same “continuum function” equal?

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 $$\kappa = \aleph_0, \lambda = \aleph_1$$ ?)

• $\kappa$ and $\lambda$ could be different finite (nonzero) cardinals.... – Henning Makholm Aug 8 at 11:05
• @Henning: Valid remark, of course. I suspect that the intention was to consider infinite cardinals only, though. – Asaf Karagila Aug 8 at 13:08
• @AsafKaragila: Yes, hence only a comment. – Henning Makholm Aug 8 at 13:35
• @Henning : you are of course right. Initially I had put no restrictions on $\mu$, which of course took care of the finite case, but I realized I was actually interested in infinite $\mu$ – Max Aug 9 at 1:22

Suppose that $$\kappa<\lambda$$, take $$\mu=\beth_{\kappa^+}$$, then $$\beth_{\kappa^+}^\kappa = \beth_{\kappa^+} <\beth_{\kappa^+}^{\kappa^+} \leq \beth_{\kappa^+}^\lambda.$$
• @Max Both the equality and the inequality follow from the fact that it's a strong limit with cofinality $\kappa^+$ (so yes I guess that boils down to Konig for the inequality). – spaceisdarkgreen Aug 9 at 1:43