ZFC proves that $\kappa^{\mathrm{cf}(\kappa)} \leq \kappa^\kappa$ for all infinite cardinal numbers $\kappa$. Further, it is consistent with ZFC that we always have equality (e.g. assume GCH).

Question. For which cardinal numbers $\kappa$ is it consistent with ZFC that $\kappa^{\mathrm{cf}(\kappa)} < \kappa^\kappa$?

Clearly no such $\kappa$ can be regular, and $\kappa = \beth_\omega$ will not work, but that's all I know.

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    $\begingroup$ $\kappa$ cannot be strong limit (which is why $\beth_\omega$ does not have the property). $\endgroup$ May 2, 2014 at 5:42
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    $\begingroup$ On the other hand, we can have $2^{\aleph_0}=\aleph_1$ and $2^{\aleph_1}=\aleph_{\omega_4}$, but the first assumption already implies (by Shelah's pcf results) that $(\aleph_\omega)^{\aleph_0}<\aleph_{\omega_4}\le 2^{\aleph_\omega}$. $\endgroup$ May 2, 2014 at 5:49

2 Answers 2


(A very partial answer only, but too long for a comment.)

In light of your second comment under Monroe's answer, I interpret your question as follows:

Given a formula $\varphi(x)$, write $\kappa_\varphi$ for the least cardinal satisfying $\varphi$. ($\kappa_\varphi$ could be undefined, of course. But you are only concerned with formulas which are satisfied by a unique cardinal.)

Write $C(\kappa)$ to abbreviate $\kappa^{cf(\kappa)} < \kappa^{\kappa}$.

How can we find out if "ZFC   +   $\kappa_\varphi$ exists   +   $C(\kappa_\varphi)$" is consistent?

A sufficient condition is the following: $\varphi$ is absolute under cardinal-preserving extensions (so in particular, $\varphi$ may use ordinal arithmetic, cardinal successor, the $\aleph$-function, etc), ZFC proves that $\kappa_\varphi$ exists and is singular. (This is really just a reformulation of Monroe Eskew's answer.)

An example is the formula $\kappa=\aleph_\omega$.


Start with a model of GCH. Let $\kappa$ be singular. Add $\kappa^{++}$ Cohen subsets of $\mathrm{cf}(\kappa)^+$-- the forcing is $\mathrm{cf}(\kappa)^+$-closed and $\mathrm{cf}(\kappa)^{++}$-c.c. Then we'll have $\kappa^\kappa = \kappa^{++}$ but $\kappa^{\mathrm{cf}(\kappa)} = \kappa^+$ since we've added no $\mathrm{cf}(\kappa)$-sequences of ordinals.

  • $\begingroup$ What happens if $\kappa$ was a strong limit cardinal to begin with? $\endgroup$ May 3, 2014 at 1:31
  • $\begingroup$ This procedure destroys strong limits. $\endgroup$ May 3, 2014 at 1:36
  • $\begingroup$ 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? $\endgroup$ May 3, 2014 at 2:13
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    $\begingroup$ @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). $\endgroup$ May 3, 2014 at 2:25

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