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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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    No biggie... :-)2012-12-02

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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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    @MettaWorldPeace: You’re welcome.2012-12-02
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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.2012-12-02