In $ \mathsf{ZFC} $, is it true that $ \text{cf}(\kappa) < \text{cf}(2^{\kappa}) $ for all cardinals $ \kappa $?

Question

Question. In $ \mathsf{ZFC} $, is it true that $ \text{cf}(\kappa) < \text{cf}(2^{\kappa}) $ for all cardinals $ \kappa $?

I am particularly interested in the case when $ \kappa = \mathfrak{c} \stackrel{\text{def}}{=} 2^{\aleph_{0}} $.

Answer 1

A particular case of Koenig's theorem states that if $\kappa$ is an infinite cardinal, then $\kappa^{\mathrm{cf}(\kappa)}>\kappa$. Applying this result when $\kappa=2^\lambda$ gives us that $\mathrm{cf}(2^\lambda)>\lambda\ge\mathrm{cf}(\lambda)$, because $$ (2^\lambda)^\lambda=2^{(\lambda\times \lambda)}=2^\lambda. $$

This result uses choice, of course. It seems problematic to define cofinality for non-well-orderable cardinals, but note that it is consistent (with the absence of choice) that the reals are a countable union of countable sets, which can be reasonably interpreted as saying that $\mathrm{cf}(\mathfrak c)=\omega$.