Examples

Since $R\{\xi,\xi\}$ is true for every cardinal $\xi$, what must be proved is that, for every set $E$ of cardinals the relation “$\xi\in E$ and $\eta\in E$ and $R\{\xi,\eta\}$” is a well-ordering relation on $E$. Consider the set $A=\bigcup_{\xi\in E}\xi$.

Every cardinal $\xi\in E$ is then a subset of $A$. By § 2, Theorem 1 there exists a well-ordering relation on $A$, which we shall denote by $\xi\leq\eta$, and every subset of $A$ is equipotent to a segment of $A$ (§ 2, no. 5, Theorem 3, Corollary x). For every cardinal $\xi\in E$ consider the set of segments of $A$ which are equipotent to $\xi$; this set of segments is not empty and therefore has a least element (§ 2, no. 1, Proposition 2); let $\varphi(\xi)$ denote this least element. The relation

“$\xi\in E$ and $\eta\in E$ and $\xi$ is equipotent to a subset of $\eta$”

is then equivalent to

“$\xi\in E$ and $\eta\in E$ and $\varphi(\xi)\subset\varphi(\eta)$”.

For clearly the second relation implies the first. Conversely, if $\xi$ is equipotent to a subset of $\varphi(\eta)$, we cannot have $\varphi(\eta)\subset\varphi(\xi)$ and $\varphi(\eta)\neq\varphi(\xi)$; otherwise there would exist a segment of $\varphi(\eta)$ equipotent to $\xi$ (§ 2, no. 5, Theorem 3, Corollary 3), contrary to the definition of $\varphi(\xi)$. Since the set of segments of $A$ is well-ordered by inclusion (§ 2, no. 1, Proposition 2), the theorem follows.

¶ We shall denote the relation $R\{\xi,\eta\}$ by $\xi\leq\eta$. A set $X$ is equipotent to a subset of a set $Y$ if and only if $\operatorname{Card}(X)\leq\operatorname{Card}(Y)$.

Clearly we have $0\leq\xi$ for every cardinal $\xi$, and $1\leq\xi$ for every cardinal $\xi\neq0$.

There exists a set E containing all the sets $\mathfrak{a}_\iota$ (e.g., the sum of these sets (Chapter II, § 4, no. 8)), whence $\mathfrak{a}_\iota \leqslant \mathfrak{a} =$ Card (E) for all $\iota \in I$. The set F of cardinals $\leqslant \mathfrak{a}$ is well-ordered and contains all the $\mathfrak{a}_\iota$, and therefore the family $(\mathfrak{a}_\iota)_{\iota \in I}$ has a least upper bound $\mathfrak{b}$ in F. Let $\mathfrak{c}$ be a cardinal $\geqslant \mathfrak{a}_\iota$ for all $\iota \in I$; if $\mathfrak{c} < \mathfrak{b} \leqslant \mathfrak{a}$, then $\mathfrak{c} \in F$, and the inequality $\mathfrak{a}_\iota \leqslant \mathfrak{c}$ contradicts the definition of the least upper bound of the family $(\mathfrak{a}_\iota)$ in the ordered set F; hence the result.

¶ By abuse of language, the cardinal $\mathfrak{b}$ is called the least upper bound of the family $(\mathfrak{a}_\iota)_{\iota \in I}$ and is denoted by $\sup_{\iota \in I} \mathfrak{a}_\iota$.

For there exists a section $s$ associated with $f$ (Chapter II, § 3, no. 8, Proposition 8) and $s$ is an injection of Y into X.

For there exists a bijection of P (resp. S) onto the product (resp. sum) of the sets $(\mathfrak{a}_\iota)$ (Chapter II, § 4, no. 8, Proposition 10, and § 5, no. 7, Corollary to Proposition 11).

For there exists a mapping of the sum S of the $\mathrm{E}_\iota$ onto the union of the $\mathrm{E}_\iota$ (Chapter II, § 4, no. 8); the Corollary therefore follows from Propositions 3 and 4.

(b) Let $(\mathfrak{a}_\iota)_{\iota \in \mathrm{I}}$ be a family of cardinals and let $(\mathrm{J}_\lambda)_{\lambda \in \mathrm{L}}$ be a partition of I. Then

("associativity of the sum and product").

(c) Let $((\mathfrak{a}_{\lambda\iota})_{\iota \in \mathrm{J}_\lambda})_{\lambda \in \mathrm{L}}$ be a family (indexed by L) of families of cardinals, and let $\mathrm{I} = \coprod_{\lambda \in \mathrm{L}} \mathrm{J}_\lambda$. Then

("distributivity of product over sum").

The relations in (a) follow from the analogous formulae for the union and product of sets, for the fact that $f$ is a bijection implies that if $(\mathrm{X}_\iota)_{\iota \in \mathrm{I}}$ is a family of mutually disjoint sets, the elements of the family $(\mathrm{X}_{f(x)})_{x \in \mathrm{K}}$ are also mutually disjoint (cf. Chapter II, § 4, no. 1, Proposition 1 and § 5, no. 3, Proposition 4).

¶ The relations in (b) are immediate consequences of the associativity formulae for unions and products (Chapter II, § 4, no. 2, Proposition 2 and § 5, no. 5, Proposition 7) and the distributivity of intersection over union (Chapter II, § 5, no. 6, Proposition 8), which shows that if $(\mathrm{X}_\iota)_{\iota \in \mathrm{I}}$ is a family of mutually disjoint sets, then the elements of the family

are also mutually disjoint.

¶ Finally, (c) follows from the distributivity of the product over union and intersection (Chapter II, § 5, no. 6, Proposition 9 and Corollary 1).

¶ Let $\mathfrak{a}$ and $\mathfrak{b}$ be two cardinals. If I is a set consisting of two distinct elements (e.g., the cardinal 2), there exists a mapping $f$ of I onto $\{\mathfrak{a}, \mathfrak{b}\}$ which defines a family of cardinals. The sum and product of this family depend only on $\mathfrak{a}$ and $\mathfrak{b}$ (by reason of Proposition 5(a)); these cardinals are called respectively the sum and the product of $\mathfrak{a}$ and $\mathfrak{b}$, and are denoted by $\mathfrak{a} + \mathfrak{b}$ and $\mathfrak{a}\mathfrak{b}$. Similarly for the sum and product of three or more cardinals. Proposition 5 then implies the following corollary :

(1)

(2) (3)

The proposition is obvious as regards the sum, for the sum $\mathrm{S}_\mathrm{I}$ of the family of sets $(\mathfrak{a}_\iota)_{\iota \in \mathrm{I}}$ is equipotent to the union of the sum $\mathrm{S}_\mathrm{J}$ of the family $(\mathfrak{a}_\iota)_{\iota \in \mathrm{I}}$ and the empty set, and hence equipotent to $\mathrm{S}_\mathrm{J}$. The assertion concerning products follows from the fact that the projection $\mathrm{pr}_\mathrm{K}$ of the product set $\prod_{\iota \in \mathrm{I}} \mathfrak{a}_\iota$ onto the partial product $\prod_{\iota \in \mathrm{K}} \mathfrak{a}_\iota$ is a bijection (Chapter II, § 5, no. 5, Remark 1).

The second formula is a consequence of the fact that any set is the union of its one-element subsets. The first formula follows from the second by multiplying by $\mathfrak{a}$ and using Corollary 1.

This is merely a translation of the condition that a product set should be non-empty (Chapter II, § 5, no. 4, Proposition 5, Corollary 2).

Let $X = \mathfrak{a} + 1 = \mathfrak{b} + 1$. Then there exist subsets A, B of X with cardinals $\mathfrak{a}$, $\mathfrak{b}$, respectively, such that the complements $X - A, X - B$ each consist of a single element. Let $u$, $v$ be these elements. The intersection $C = A \cap B$ has as complement in X the set $\{u, v\}$. If $u = v$, then $A = B = C$, so that $\mathfrak{a} = \mathfrak{b}$. If $u \neq v$, then $A = C \cup \{v\}$, $B = C \cup \{u\}$, and therefore $\mathfrak{a} = 1 + \mathrm{Card}\ (C) = \mathfrak{b}$.

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The reader should beware of assuming that $\mathfrak{a} + \mathfrak{m} = \mathfrak{b} + \mathfrak{m}$ implies $\mathfrak{a} = \mathfrak{b}$ for all cardinals $\mathfrak{m}$ (cf. § 6); \* we shall see later, however, that this implication is true if $\mathfrak{m}$ is finite (§ 5, no. 2, Proposition 3, Corollary 4 and § 6, no. 3, Theorem 3, Corollary 4). \*

For there exists a bijection of $X^{Y}$ onto the set of mappings of $\mathfrak{b}$ into $\mathfrak{a}$ (Chapter II, § 5, no. 2, Proposition 2, Corollary).

This follows from the definition of the product of a family of sets as a set of functional graphs (Chapter II, § 5, no. 3).

Let S be the sum of the sets $\mathfrak{b}_{\iota}$, and put $\mathfrak{a}_{s} = \mathfrak{a}$ for all $s \in S$. Both sides of the equality to be proved are then equal to $\mathbf{P}_{s \in S} \mathfrak{a}_{s}$, by virtue of Proposition 10 and the associativity formula for products (no. 3, Proposition 5(b)).

Let $\mathfrak{a}_{\iota\beta} = \mathfrak{a}_\iota$ for each pair $(\iota, \beta) \in \mathbf{I} \times \mathfrak{b}$. Then, by associativity of the product, we have

Let $\mathfrak{b}_\gamma = \mathfrak{b}$ for all $\gamma \in \mathfrak{c}$. Then

by virtue of Corollary 1.

For there exists a unique mapping of $\varnothing$ into any given set (namely, the mapping whose graph is the empty set); the set of mappings of a set consisting of a single element into an arbitrary set X is equipotent to X (Chapter II, § 5, no. 3); there exists a unique mapping of an arbitrary set into a set consisting of a single element; and, finally, there is no mapping of a non-empty set into $\varnothing$.

¶ Note in particular that $0^0 = 1$.

Let $\alpha$ and $\beta$ be the elements of the cardinal 2. For each subset Y of X let $f_{\mathrm{Y}}$ be the mapping of X into 2 defined by $f_{\mathrm{Y}}(x) = \alpha$ if $x \in \mathrm{Y}$ and $f_{\mathrm{Y}}(x) = \beta$ if $x \in \mathrm{X} - \mathrm{Y}$. Let $u$ be the mapping $\mathrm{Y} \to f_{\mathrm{Y}}$ of $\mathfrak{P}(\mathrm{X})$ into $2^{\mathrm{X}}$. Conversely, with each mapping $g$ of X into 2 let us associate the subset $\overset{-1}{g}(\alpha)$ of X, and let $v$ be the mapping $g \to \overset{-1}{g}(\alpha)$ of $2^{\mathrm{X}}$ into $\mathfrak{P}(\mathrm{X})$. It is clear that $u \circ v$ and $v \circ u$ are the identity mappings of $2^{\mathrm{X}}$ and $\mathfrak{P}(\mathrm{X})$; hence $u$ and $v$ are bijections (Chapter II, § 3, no. 8, Proposition 8, Corollary) and therefore Card $(\mathfrak{P}(\mathrm{X})) = 2^{\mathfrak{a}}$.

For the relation $\mathfrak{a} \geqslant \mathfrak{b}$ means that there exists a subset B of $\mathfrak{a}$ which is equipotent to $\mathfrak{b}$ (no. 2), i.e., $\mathfrak{a}$ is equipotent to the set which is the sum of $\mathfrak{b}$ and a set $c$.

☡

If $\mathfrak{a} \geqslant \mathfrak{b}$, there usually exist many cardinals $\mathfrak{c}$ such that $\mathfrak{a} = \mathfrak{b} + \mathfrak{c}$ (cf. § 6); in general, therefore, it is not possible to define the "difference" $\mathfrak{a} - \mathfrak{b}$ of two such cardinals (cf. § 5, no. 2).

The second inequality follows from the inclusion relations between products of sets (Chapter II, § 5, no. 4, Proposition 6, Corollary 3). As to the first inequality, if a set E is the union of a family $(A_\iota)_{\iota \in I}$ of mutually disjoint subsets and if $B_\iota \subset A_\iota$ for all $\iota \in I$, then the $B_\iota$ are also mutually disjoint and $\bigcup_\iota B_\iota \subset \bigcup_\iota A_\iota$ (Chapter II, § 4, no. 2).

Put $\mathfrak{b}_\iota = \mathfrak{a}_\iota$ if $\iota \in J$, and $\mathfrak{b}_\iota = 0$ (resp. $\mathfrak{b}_\iota = 1$) if $\iota \in I - J$. Then apply Proposition 14, observing that the relation $\mathfrak{a} \neq 0$ implies $\mathfrak{a} \geqslant 1$.

For $\mathfrak{a}^\mathfrak{b} \leqslant \mathfrak{{a}'}^\mathfrak{b}$ by Propositions 10 and 14, and $\mathfrak{{a}'}^\mathfrak{b} \leqslant \mathfrak{{a}'}^{\mathfrak{b}'}$ by Proposition 10 and Corollary 1 to Proposition 14.

We have $\mathrm{Card}(\mathfrak{P}(\mathfrak{a})) = 2^\mathfrak{a}$ (no. 5, Proposition 12). The mapping $x \to \{x\}$ $(x \in \mathfrak{a})$ is an injection of $\mathfrak{a}$ into $\mathfrak{P}(\mathfrak{a})$, whence $\mathfrak{a} \leqslant 2^\mathfrak{a}$. Hence it is enough to show that $\mathfrak{a} \neq 2^\mathfrak{a}$, i.e., that for every mapping $f$ of $\mathfrak{a}$ into $\mathfrak{P}(\mathfrak{a})$, the image $f(\mathfrak{a})$ is distinct from $\mathfrak{P}(\mathfrak{a})$. Let X be the set of all $x \in \mathfrak{a}$ such that $x \notin f(x)$. If $x \in X$, we have $x \notin f(x)$, whence $f(x) \neq X$; if $x \in \mathfrak{a} - X$, we have $x \in f(x)$ and $x \notin X$, whence $f(x) = X$. This shows that $X \notin f(\mathfrak{a})$ and proves the theorem.

If U were such a set, there would exist a set S, the sum of the family of sets $(X)_{X \in U}$, so that every cardinal is equipotent to a subset of S. In particular, let $\mathfrak{S} = \mathrm{Card}(S)$; since $2^\mathfrak{S}$ is a cardinal, we would have $2^\mathfrak{S} \leqslant \mathfrak{S}$, in contradiction to Theorem 2.