Kerodon

Remark 4.7.3.19. Let $\kappa $ be an infinite cardinal. Then there are arbitrarily large regular cardinals $\lambda $ satisfying $\mathrm{ecf}(\lambda ) > \kappa $. To see this, it will suffice (by enlarging $\kappa $) to show that there exists some regular cardinal $\lambda $ of exponential cofinality $\geq \kappa $. Let $S$ be a set of cardinality $\kappa $ and let $2^{\kappa }$ denote the cardinality of the power set $P(S) = \{ S_0: S_0 \subseteq S\} $. Proposition 4.7.3.5 implies that the product $S \times S$ also has cardinality $\kappa $, so that $P(S \times S) \simeq {\prod }_{s \in S} P(S)$ also has cardinality $2^{\kappa }$. It follows that the collection of sets of cardinality $\leq 2^{\kappa }$ is closed under the formation of products indexed by sets of cardinality $\leq \kappa $, so that $\lambda = ( 2^{\kappa } )^{+}$ has exponential cofinality $> \kappa $.