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Proposition Let $\kappa $ be an infinite cardinal. Then the collection of $\kappa $-small sets is closed under finite products.

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, 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 $\square$