# Given $\kappa$ infinite cardinal, $\{\lambda \in \text{Card } | \ \lambda^\kappa>\lambda\}$ is a proper class

I'm trying to do the following exercise

Show that for any infinite cardinal $$\kappa$$ the classes $$\{\lambda \in \text{Card }| \ \lambda^\kappa = \lambda\}$$ and $$\{\lambda \in \text{Card }| \ \lambda^\kappa > \lambda\}$$ are proper.

I think it suffices to show that both classes are unbounded in Card (which is proper).

• For the first class $$\{\lambda \in \text{Card }| \ \lambda^\kappa = \lambda\}$$, I'd define the following functional: \begin{align}f : \text{Card} &\longrightarrow \text{Card} \\ \lambda &\longmapsto \lambda^\kappa \end{align} this function is increasing and continuous, therefore the class of the fixed points is unbounded.
• Regarding the second class $$\{\lambda \in \text{Card }| \ \lambda^\kappa > \lambda\}$$, which is the complement of the previous class, I would use the fact that $$\kappa^{\text{cof}(\kappa)}>\kappa$$ for every infinite cardinal $$\kappa$$. Therefore if I consider a cardinal $$\lambda$$ s.t. $$\text{cof}(\lambda) = \text{cof}(\kappa)$$ I'd have $$\lambda^\kappa \ge\lambda^{\text{cof}(\kappa)}=\lambda^{\text{cof}(\lambda)}>\lambda$$ So I'd proceed proving that $$\{\lambda \in \text{Card }| \ \text{cof}(\lambda) = \text{cof}(\kappa)\}$$ is unbounded (1).

I have some doubts about the correctness of my approach to the second part of the exercise. It seems a bit too convoluted.

If the approach is correct, I'd prove (1) by noticing that given an ordinal $$\alpha$$ and a regular cardinal $$\kappa$$, then $$\aleph_{\alpha+\kappa} > \aleph_\alpha \text{ and }\text{cof}(\aleph_{\alpha+\kappa})=\kappa$$ where the sum in the index of $$\aleph$$ is meant as an ordinal sum. Is this correct regardless of the main exercise? Thanks

• You should use \mid instead of \ |\ , by the way. Feb 9 '20 at 11:52

The function $$\lambda\mapsto\lambda^\kappa$$ is not continuous. Exactly because limits of $$\kappa$$-cofinality will witness that.
Instead, note that for every $$\mu$$, $$\lambda=\mu^\kappa$$ satisfies $$\lambda^\kappa=\lambda$$.
For the second part, the approach is indeed the correct one. We essentially want to show there is a proper class of cardinals of cofinality $$\kappa$$. Indeed, your suggestion is the correct one.
However, one can combine the two into a single approach. Note that $$\beth_{\alpha+\kappa}^\kappa>\beth_{\alpha+\kappa}$$, by the same argument as you pointed. But now notice that $$\lambda=\beth_{\alpha+\kappa}^+$$ satisfies $$\lambda^\kappa=\lambda$$.