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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

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    $\begingroup$ You should use \mid instead of \ |\ , by the way. $\endgroup$
    – Asaf Karagila
    Commented Feb 9, 2020 at 11:52

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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 $\lambda^\kappa=\lambda$.

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