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

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    $\begingroup$ $\kappa$ and $\lambda$ could be different finite (nonzero) cardinals.... $\endgroup$
    – hmakholm left over Monica
    Aug 8, 2019 at 11:05
  • $\begingroup$ @Henning: Valid remark, of course. I suspect that the intention was to consider infinite cardinals only, though. $\endgroup$
    – Asaf Karagila ♦
    Aug 8, 2019 at 13:08
  • $\begingroup$ @AsafKaragila: Yes, hence only a comment. $\endgroup$
    – hmakholm left over Monica
    Aug 8, 2019 at 13:35
  • $\begingroup$ @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 $\mu$ $\endgroup$
    – Maxime Ramzi
    Aug 9, 2019 at 1:22

1 Answer 1

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Yes.

Suppose that $\kappa<\lambda$, take $\mu=\beth_{\kappa^+}$, then $$\beth_{\kappa^+}^\kappa = \beth_{\kappa^+} <\beth_{\kappa^+}^{\kappa^+} \leq \beth_{\kappa^+}^\lambda.$$

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  • $\begingroup$ I assume the strixt inequality follows from König's lemma ? I'll have to look into the cardinal arithmetic of the first inequality for myself ! Thank you ! $\endgroup$
    – Maxime Ramzi
    Aug 9, 2019 at 1:24
  • $\begingroup$ @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). $\endgroup$
    – spaceisdarkgreen
    Aug 9, 2019 at 1:43
  • $\begingroup$ @spaceisdarkgreen : yes I realized that the equality is easy to prove by elementary considerations anyway $\endgroup$
    – Maxime Ramzi
    Aug 9, 2019 at 10:18

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