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.
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.
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).
No, not even closely.
Let $\kappa$ be of countable cofinality and $\kappa>\frak c$. We have that ${\frak c}^{\aleph_0}=\frak c$, but $\kappa^{\aleph_0}=\kappa^{\operatorname{cf}(\kappa)}>\kappa$.