Kerodon

Proof. We first note that the collection of finite sets is closed under finite products. It will therefore suffice to show that, for every infinite cardinal $\lambda $, the following condition is satisfied:

$(\ast _{\lambda })$

If $S$ and $T$ are sets of cardinality $\leq \lambda $, then the product $S \times T$ has cardinality $\leq \lambda $.

By virtue of Remark 4.7.2.9, we may assume that condition $(\ast _{\mu } )$ is satisfied for every cardinal $\mu < \lambda $. Without loss of generality, we may assume that $S = \mathrm{Ord}_{< \lambda } = T$, where $\mathrm{Ord}_{< \lambda }$ denotes the collection of ordinals smaller than $\lambda $. Given a pair of elements $(\alpha ,\beta ), (\alpha ', \beta ') \in S \times T$, let us write $(\alpha ', \beta ') \preceq (\alpha , \beta )$ if either $\mathrm{max}( \alpha ', \beta ') < \mathrm{max}( \alpha , \beta )$, or $\mathrm{max}( \alpha ', \beta ') = \mathrm{max}( \alpha , \beta )$ and $\alpha ' < \alpha $, or $\mathrm{max}( \alpha ', \beta ' ) = \mathrm{max}(\alpha , \beta )$ and $\alpha ' = \alpha $ and $\beta ' \leq \beta $. The relation $\preceq $ defines a well-ordering of the set $S \times T$. To prove $(\ast _{\lambda })$, it will suffice to show this well ordering has order type $\leq \lambda $. Assume otherwise. Then there exists an element $(\alpha ,\beta ) \in S \times T$ such that $\lambda $ is the order type of the initial segment $K = \{ (\alpha ',\beta ') \in S \times T: (\alpha ', \beta ') \prec (\alpha , \beta ) \} $. Note that $K$ is a subset of the product $\mathrm{Ord}_{ \leq \gamma }$ and $\mathrm{Ord}_{ \leq \gamma }$, where $\gamma = \mathrm{max}( \alpha , \beta )$. Our inductive hypothesis guarantees that $K$ has cardinality $< \lambda $, contradicting Corollary 4.7.2.5. $\square$