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

  1. $\kappa=\lambda^+$ with $\textsf{CH}_\lambda$; or
  2. $\kappa$ is inaccessible; or
  3. $\kappa$ 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.

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    $\begingroup$ (1) The second paragraph makes little to no sense some of it grammatically, and some of it notationally. (2) Weakly inaccessible means regular limit, but not strong limit, so the three options you mention do not cover all the options, I imagine you mean something closer to "strong limit"? (3) To your question about weakly inaccessible, if you start with GCH+inaccessible, then add inaccessibly many Cohen reals, you get exactly the third scenario. $\endgroup$ – Asaf Karagila Aug 1 '17 at 22:16
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    $\begingroup$ (Or do I need to break your other leg? :P) $\endgroup$ – Asaf Karagila Aug 1 '17 at 22:17
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    $\begingroup$ Weakly inaccessible cardinals need not be countably closed. More precisely, if the existence of inaccessible cardinals is consistent with ZFC, then so is $\mathfrak c$ is weakly inaccessible. (The model is the same as in the last part of Asaf's first comment.) $\endgroup$ – Andreas Blass Aug 1 '17 at 23:48
  • $\begingroup$ For the case of $\aleph_{\omega_1}$ you need to require that all the limit ordinals have this property. In other words, you need that for all $\lambda<\kappa$ whose cofinality is countalbe, $\lambda^{\aleph_0}<\kappa$. But this is just rewriting the definitions again. $\endgroup$ – Asaf Karagila Aug 2 '17 at 9:10
  • $\begingroup$ Also, the problem with "Is there something" is that it's a $\Sigma^0_1$ question, and not a $\Delta^0_1$ question, so it's kinda hard to answer this. $\endgroup$ – Asaf Karagila Aug 2 '17 at 9:10

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