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. $\square$

Proof. Let $\{ S_ i \} _{i \in I}$ be a finite collection of $\kappa $-small sets. Then the disjoint union ${\coprod }_{ i \in I} S_ i$ can be identified with a subset of the product ${\prod }_{i \in I} (S_ i \coprod \{ i \} )$, which is $\kappa $-small by virtue of Proposition 4.7.3.5. $\square$

Proof. Assume first that condition $(1)$ is satisfied. Let $\{ T_{s} \} _{s \in S}$ be a collection of $\lambda $-small sets indexed by a set $S$ of cardinality $\leq \kappa $; we wish to show that the coproduct $T = {\coprod }_{s \in S} T_{s}$ is $\lambda $-small. Using Theorem 4.7.1.34, we can choose a well-ordering $\leq _{S}$ on the set $S$, and a well-ordering $\leq _{s}$ on the set $T_{s}$ for each $s \in S$. For elements $t \in T_{s}$ and $t' \in T_{s'}$, write $t \leq _{T} t'$ if either $s <_{S} s'$, or $s = s'$ and $t \leq _{s} t'$. Then $\leq _{T}$ is a well-ordering of the set $T$. If $T$ is not $\lambda $-small, then it has an initial segment of order type $\lambda $. Passing to subsets, we may assume without loss of generality that $T$ itself has order type $\lambda $. Moreover, we may assume without loss of generality that each of the sets $T_{s}$ is nonempty, and therefore contains a smallest element $t_{s}$. We consider two cases:

• Suppose that $S$ contains a largest element $s$. In this case, we can write $T$ as the disjoint union of the initial segment $T' = {\coprod }_{s' < s} T_{s'}$ with the set $T_{s}$. Since $T_{s}$ is nonempty, $T'$ has order type smaller than $\lambda $, and is therefore $\lambda $-small. Applying Corollary 4.7.3.6, we deduce that $T = T' \coprod T_{s}$ is also $\lambda $-small.

• Suppose that $S$ does not have a largest element. In this case, the construction $(s \in S) \mapsto (t_ s \in T)$ is a cofinal function from $S$ to $T$. It follows that the order type of $(S, \leq _{S} )$ is greater than or equal to the cofinality $\mathrm{cf}(T) = \mathrm{cf}(\lambda )$, contradicting assumption $(1)$.

We now prove the reverse implication. Assume that condition $(2)$ is satisfied. Choose a well-ordering $(S, \leq _{S} )$ of order type $\mathrm{cf}(\lambda )$ and a cofinal map $f: S \rightarrow \mathrm{Ord}_{< \lambda }$. If $\kappa \geq \mathrm{cf}(\lambda )$, then condition $(2)$ implies that the disjoint union ${\coprod }_{s \in S} \mathrm{Ord}_{ < f(s) }$ is $\lambda $-small. Since $f$ is cofinal, the tautological map ${\coprod }_{s \in S} \mathrm{Ord}_{ < f(s) } \rightarrow \mathrm{Ord}_{< \lambda }$ is surjective. It follows that $\mathrm{Ord}_{< \lambda }$ is $\lambda $-small, which is a contradiction. $\square$

Proof. Proposition 4.7.2.14 guarantees that $\kappa $ is a cardinal. The characterization is a restatement of Proposition 4.7.3.7. $\square$

Proof. Suppose that $\kappa $ is not regular: that is, there is a collection of $\kappa $-small sets $\{ T_{s} \} _{s \in S}$ indexed by a $\kappa $-small set $S$ such that $T = {\coprod }_{s \in S} T_ s$ has cardinality $\geq \kappa $. Choose a collection of $\lambda $-small sets $\{ U_{t} \} _{t \in T}$ for which the product $U = {\prod }_{t \in T} U_ t$ is not $\lambda $-small. For each $s \in S$, let $U_{s}$ denote the product ${\prod }_{ t \in T_{s} } U_{t}$. Since $T_{s}$ is $\mathrm{ecf}(\lambda )$-small, the set $U_{s}$ is $\lambda $-small. Since $S$ is also $\mathrm{ecf}(\lambda )$-small, it follows that $U \simeq {\prod }_{s \in S} U_ s$ is also $\lambda $-small, which is a contradiction. $\square$