Recall that a cardinal $\kappa$ is countably closed if $\nu^{\aleph_0}<\kappa$ for every $\nu<\kappa$. I wanted to get some kind of an intuition of countably closed cardinals as well as cardinals satisfying $\kappa^{<\kappa}=\kappa$.
For $\kappa^{<\kappa}=\kappa$ we first note that $\kappa$ must be regular, as otherwise $\kappa\geq\kappa^{\text{cof}(\kappa)}>\kappa$ by König's theorem. Furthermore if $\kappa=\lambda^+$ then $\kappa^{<\kappa}=\kappa^\lambda=\lambda^\lambda+\kappa=\kappa$ by the Hausdorff formula, so here $\kappa^{<\kappa}=\kappa$ precisely if $\textsf{CH}$ holds at $\lambda$. If $\kappa$ is limit we get that $\kappa=\kappa^{<\kappa}=\sup_{\lambda<\kappa}2^\lambda$, so that either $\kappa$ is inaccessible or $2^\lambda=\kappa$ for a tail of $\lambda<\kappa$. Summing up, $\kappa^{<\kappa}=\kappa$ holds iff
- $\kappa=\lambda^+$ with $\textsf{CH}_\lambda$; or
- $\kappa$ is inaccessible; or
- $\kappa$ is weakly inaccessible with $2^\lambda=\kappa$ for a tail of $\lambda<\kappa$ (this is possible: see Asaf's comment below).
Is there any similar characterisation of the countably closed cardinals? By Andreas' comment below it's not in general the case that weakly inaccessible cardinals are countably closed. In the special case of $\kappa=\aleph_\omega$ we have that $\aleph_n^{\aleph_0}=\aleph_n+\mathfrak c$, making $\aleph_\omega$ countably closed iff $\mathfrak c<\aleph_\omega$. But for $\aleph_{\omega_1}$ we also require that $\aleph_\omega^{\aleph_0}<\aleph_{\omega_1}$ (note that $\aleph_\omega^{\aleph_0}>\aleph_\omega$ by König's theorem and trivially $\mathfrak c\leq\aleph_\omega^{\aleph_0}$), so it seems like for arbitrary $\kappa$ we're simply winding up with the definition again.