Definition 5.4.3.1. Let $\kappa $ be a cardinal. We say that a set $S$ is *$\kappa $-small* if the cardinality $|S|$ is strictly smaller than $\kappa $.

### 5.4.3 Smallness of Sets

We now introduce some terminology which will be useful for measuring the sizes of various mathematical objects.

Example 5.4.3.2. Let $\aleph _0$ denote the first infinite cardinal (Example 5.4.2.10). Then a set $S$ is $\aleph _0$-small if and only if it is finite.

Example 5.4.3.3. Let $\aleph _1$ denote the first uncountable cardinal (Example 5.4.2.12). Then a set $S$ is $\aleph _1$-small if and only if it is countable.

Remark 5.4.3.4. Let $\kappa $ be a cardinal and let $T$ be a $\kappa $-small set. Then:

Any subset of $T$ is also $\kappa $-small (see Proposition 5.4.2.3).

The set $T$ is $\lambda $-small for every cardinal $\lambda \geq \kappa $.

For every surjective morphism of sets $T \twoheadrightarrow S$, the set $S$ is also $\kappa $-small.

Proposition 5.4.3.5. 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 5.4.2.9, 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 $\mathrm{Ord}_{ \leq \alpha }$ and $\mathrm{Ord}_{ \leq \beta }$. Our inductive hypothesis guarantees that $K$ has cardinality $< \lambda $, contradicting Corollary 5.4.2.5. $\square$

Corollary 5.4.3.6. Let $\kappa $ be an infinite cardinal. Then the collection of $\kappa $-small sets is closed under finite coproducts.

**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 5.4.3.5.
$\square$

We will need the following generalization of Corollary 5.4.3.6:

Proposition 5.4.3.7. Let $\kappa $ and $\lambda $ be cardinals, where $\lambda $ is infinite. The following conditions are equivalent:

- $(1)$
The cardinal $\kappa $ is strictly smaller than the cofinality $\mathrm{cf}(\lambda )$ (see Definition 5.4.1.28).

- $(2)$
Let $\{ T_{s} \} _{s \in S}$ be a collection of $\lambda $-small sets indexed by a set $S$ of cardinality $\leq \kappa $. Then the coproduct $\coprod _{s \in S} T_{s}$ is $\lambda $-small.

**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 5.4.1.34, 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 5.4.3.6, 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. $\square$

Corollary 5.4.3.8. Let $\lambda $ be an infinite cardinal. Then $\kappa = \mathrm{cf}(\lambda )$ is the smallest cardinal for which there exists a set $S$ of cardinality $\kappa $ and a collection of $\lambda $-small sets $\{ T_{s} \} _{s \in S}$, where the coproduct $\coprod _{s \in S} T_ s$ is not $\lambda $-small.

**Proof.**
Proposition 5.4.2.14 guarantees that $\kappa $ is a cardinal. The characterization is a restatement of Proposition 5.4.3.7.
$\square$

Corollary 5.4.3.9. Let $\lambda $ be an infinite cardinal and let $\kappa = \mathrm{cf}(\lambda )$ be its cofinality. Suppose we are given a collection of $\lambda $-small sets $\{ T_{s} \} _{s \in S}$. If the index set $S$ is $\kappa $-small, then coproduct $\coprod _{s \in S} T_ s$ is $\lambda $-small.

Definition 5.4.3.10 (Regular Cardinals). Let $\kappa $ be a cardinal. We say that $\kappa $ is *regular* if it is infinite and $\mathrm{cf}(\kappa ) = \kappa $. Here $\mathrm{cf}(\kappa )$ denotes the cofinality of $\kappa $ (Definition 5.4.1.28). We say that $\kappa $ is *singular* if it is infinite but not regular.

Remark 5.4.3.11. Let $\kappa $ be an infinite cardinal. Then $\kappa $ is regular if and only if the collection of $\kappa $-small sets is closed under $\kappa $-small coproducts (this is a special case of Corollary 5.4.3.8).

Example 5.4.3.12. Let $\aleph _0$ denote the first infinite cardinal (Example 5.4.2.10). Then $\aleph _0$ is regular: that is, the collection of finite sets is closed under finite coproducts.

Example 5.4.3.13 (Successor Cardinals). Let $\kappa $ be an infinite cardinal and let $\kappa ^{+}$ be its successor (Example 5.4.2.11). Then a set $S$ is $\kappa ^{+}$-small if and only if it has cardinality $\leq \kappa $. It follows that $\kappa ^{+}$ is a regular cardinal. That is, if $\{ T_ s \} _{s \in S}$ is a collection of sets of cardinality $\leq \kappa $ indexed by a set $S$ of cardinality $\leq \kappa $, then the disjoint union $\coprod _{s \in S} T_{s}$ also has cardinality $\leq \kappa $. To prove this, choose a collection of monomorphisms $\{ i_ s: T_ s \hookrightarrow T \} _{s \in S}$, where $T$ is a set of cardinality $\kappa $. We then obtain a monomorphism

where the set $S \times T$ has cardinality $\leq \kappa $ by virtue of Proposition 5.4.3.5.

Example 5.4.3.14. Let $\aleph _1$ denote the first uncountable cardinal (Example 5.4.2.12). Then $\aleph _1$ is regular: that is, the collection of countable sets is closed under the formation of countable disjoint unions. This is a special case of Example 5.4.3.13, since $\aleph _1 = \aleph _0^{+}$.

Example 5.4.3.15. Let $(T, \leq )$ be a nonempty linearly ordered set with no largest element. Then the cofinality $\kappa = \mathrm{cf}(T)$ is a regular cardinal. To see this, choose a well-ordered set $(S, \leq )$ of order type $\kappa $ and a cofinal function $f: S \rightarrow T$. Proposition 5.4.2.14 guarantees that $\kappa $ is a cardinal, and Example 5.4.1.31 shows that $\kappa $ is infinite. If it is not regular, then there exists a cofinal map $g: R \rightarrow S$, where $(R, \leq )$ is a well-ordered set of order type $\alpha < \kappa $. This contradicts the definition of $\kappa = \mathrm{cf}(T)$, since the composite map $(f \circ g): R \rightarrow T$ is cofinal.

It will be convenient to introduce the following bit of nonstandard terminology:

Definition 5.4.3.16. Let $\lambda $ be an infinite cardinal. We let $\mathrm{ecf}(\lambda )$ denote the least cardinal $\kappa $ with the following property: there exists a set $S$ of cardinality $\kappa $ and a collection of $\lambda $-small sets $\{ T_{s} \} _{s \in S}$ for which the product $\prod _{s \in S} T_ s$ is not $\lambda $-small. We will refer to $\mathrm{ecf}(\lambda )$ as the *exponential cofinality of $\lambda $*.

Remark 5.4.3.17. Let $\lambda $ be an infinite cardinal. Then the exponential cofinality $\mathrm{ecf}(\lambda )$ satisfies $\aleph _0 \leq \mathrm{ecf}(\lambda ) \leq \mathrm{cf}(\lambda )$. In particular, we have $\mathrm{ecf}(\lambda ) \leq \lambda $. The inequality $\aleph _0 \leq \mathrm{ecf}(\lambda )$ is a reformulation of the fact that the collection of $\lambda $-small sets is closed under finite products (Proposition 5.4.3.5). To prove the other inequality, choose a set $S$ of cardinality $\mathrm{cf}(\lambda )$ and a collection of $\lambda $-small sets $\{ T_ s \} _{s \in S}$ for which the coproduct $T = \coprod _{s \in S} T_{s}$ is not $\lambda $-small. We now observe that $T$ can be identified with a subset of the product $\prod _{s \in S} ( T_ s \coprod \{ s\} )$. Since each of the sets $T_{s} \coprod \{ s\} $ is also $\lambda $-small, we obtain $\mathrm{ecf}(\lambda ) \leq \mathrm{cf}(\lambda )$.

Remark 5.4.3.18. Let $\kappa $ and $\lambda $ be infinite cardinals. Then $\kappa \leq \mathrm{ecf}(\lambda )$ if and only if the following condition is satisfied: for every collection of $\lambda $-small sets $\{ T_ s \} _{s \in S}$ indexed by a $\kappa $-small set $S$, the product $\prod _{s \in S} T_{s}$ is also $\lambda $-small.

Remark 5.4.3.19. 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 5.4.3.5 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 $.

Definition 5.4.3.20. Let $\kappa $ be an infinite cardinal. We say that $\kappa $ is *strongly inaccessible* if $\kappa = \mathrm{ecf}(\kappa )$. In other words, $\kappa $ is strongly inaccessible if the collection of $\kappa $-small sets is closed under the formation of $\kappa $-small products.

Example 5.4.3.21. Let $\aleph _0$ be the least infinite cardinal. Then $\aleph _0$ is strongly inaccessible. That is, the collection of finite sets is closed under finite products.

Remark 5.4.3.22. Let $\kappa $ be a strongly inaccessible cardinal. Then $\kappa $ is regular: this follows immediately from the inequality $\mathrm{ecf}(\kappa ) \leq \mathrm{cf}(\kappa )$ of Remark 5.4.3.17.

Warning 5.4.3.23. The existence of uncountable strongly inaccessible cardinals cannot be proven from the axioms of Zermelo-Fraenkel set theory (assuming those axioms are consistent).

Proposition 5.4.3.24. Let $\lambda $ be an infinite cardinal and let $\kappa = \mathrm{ecf}(\lambda )$ be the exponential cofinality of $\lambda $. Then $\kappa $ is a regular cardinal.

**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.
$\square$