Timeline for For which cardinal numbers $\kappa$ is it consistent with ZFC that $\kappa^{\mathrm{cf}(\kappa)} < \kappa^\kappa$?
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| May 3, 2014 at 2:25 | comment | added | Andrés E. Caicedo | @user18921 $\kappa$ is no longer $\beth_\omega$ in the extension. It is still a singular cardinal of cofinality $\omega$, though. As I indicate in my previous comments, there is a more general pcf theoretic result behind the scenes, which is really what you should study, and Monroe's answer (very easily) verifies that the result is not just vacuously true (that is, its assumptions are consistent). | |
| May 3, 2014 at 2:13 | comment | added | goblin GONE | I'm just not really sure what to do with this. Superficially, your answer implies: "Let $\varphi(\kappa)$ denote a sentence in the language of $\{\in\}$ such that ZFC proves that there is a unique $\kappa$ such that $\varphi(\kappa)$, and furthermore that this $\kappa$ is a singular cardinal number. Then it is consistent with ZFC to assume that $\varphi(\kappa) \rightarrow \kappa^{\mathrm{cf}(\kappa)} < \kappa^\kappa.$" However, I surmise that this is not a consequence of your claim, because $\varphi(\kappa) := (\kappa = \beth_\omega)$ is a clear counterexample. So, what am I missing? | |
| May 3, 2014 at 1:36 | comment | added | Monroe Eskew | This procedure destroys strong limits. | |
| May 3, 2014 at 1:31 | comment | added | goblin GONE | What happens if $\kappa$ was a strong limit cardinal to begin with? | |
| May 3, 2014 at 1:16 | history | answered | Monroe Eskew | CC BY-SA 3.0 |