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This problem is originated from the experience that I was trying to prove 5.19 and 5.20 on Page 60 of Set Theory, Jech(2006).

It seems to be right with AC, but I don't know how to prove it.

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I had just finished writing the answer when you hit delete. But since you found the duplicate I suppose it's better this way. :-) –  Asaf Karagila Dec 2 '12 at 15:18
Apologies for imposing a deadweight loss on you. I shall be more careful next time. –  Metta World Peace Dec 2 '12 at 15:20
No biggie... :-) –  Asaf Karagila Dec 2 '12 at 15:21

2 Answers 2

up vote 3 down vote accepted

No, it does not: if $\mu>\kappa$ has cofinality to $\lambda$, then $\mu^\lambda>\mu$ by König’s theorem (which does require the axiom of choice).

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Thank you. I got it. –  Metta World Peace Dec 2 '12 at 11:36
@MettaWorldPeace: You’re welcome. –  Brian M. Scott Dec 2 '12 at 11:44

No, not even closely.

Let $\kappa$ be of countable cofinality and $\kappa>\frak c$. We have that ${\frak c}^{\aleph_0}=\frak c$, but $\kappa^{\aleph_0}=\kappa^{\operatorname{cf}(\kappa)}>\kappa$.

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Thank you. As you have told me in another problem, "$\operatorname{cf}{\kappa}$ is not monotone with respect to $\kappa$", I shouldn't expect cardinals behave like that. –  Metta World Peace Dec 2 '12 at 11:36

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