We shall denote by $\mathbf{N}$ the set of integers (also called "the set of natural integers" when necessary to avoid ambiguity). The cardinal of $\mathbf{N}$ is denoted by $\aleph_0$. Whenever $\mathbf{N}$ is considered as an ordered set, it is always the ordering (called the usual ordering) defined in § 3, no. 2 that is under consideration, unless the contrary is expressly stated.

We shall use the following two lemmas :

There exists a well-ordering relation on E (§ 2, no. 3, Theorem 1), which we shall denote by $x \leqslant y$. The hypothesis implies that the well-ordered set E cannot be isomorphic to a segment of $\mathbf{N}$ distinct from $\mathbf{N}$, for such a segment is of the form $[0, n]$ (§ 2, no. 1, Proposition 1) and is therefore finite. It follows that $\mathbf{N}$ is isomorphic to a segment of E (§ 2, no. 3, Theorem 3), whence the result.

Since $\mathbf{N} \times \mathbf{N}$ contains the set $\{0\} \times \mathbf{N}$, which is equipotent to $\mathbf{N}$, we have $\mathrm{Card}(\mathbf{N}) \leqslant \mathrm{Card}(\mathbf{N} \times \mathbf{N})$. To complete the proof it is enough to define an injection $f$ of $\mathbf{N} \times \mathbf{N}$ into $\mathbf{N}$. For this purpose we note that there exists an injection $\varphi$ of $\mathbf{N}$ into the set of mappings of $\mathbf{N}$ into $\mathrm{I} = \{0, 1\}$, obtained as follows : if $r$ is the least integer such that $n > 2^r$, and if $\sum_{k=0}^{r-1} \varepsilon_k 2^{r-k-1}$ is the dyadic expansion of $n$ (§ 5, no. 7), $\varphi(n)$ is defined to be the sequence $(u_m)_{m \in \mathbf{N}}$ such that $u_m = \varepsilon_{r-m-1}$ for $m < r$ and $u_m = 0$ for $m \geqslant r$. Proposition 8 of § 5, no. 7 shows that $\varphi$ is injective. For each pair $(n, n') \in \mathbf{N} \times \mathbf{N}$ we define $f(n, n')$ as follows : if $\varphi(n) = (u_m)$ and $\varphi(n') = (v_m)$, let $f(n, \; n')$ be the integer $s$ such that $\varphi(s) = w_m$, where $w_{2m} = u_m$ and $w_{2m+1} = v_m$ for all $m \in \mathbf{N}$. It is clear that the relation $f(n, \; n') = f(n_1, \; n_1')$ implies $\varphi(n) = \varphi(n_1)$ and $\varphi(n') = \varphi(n_1')$; hence $(n, n') = (n_1, n_1')$, and therefore $f$ is injective.

¶ We come now to the proof of Theorem 2. Let E be a set such that $\mathrm{Card}(\mathrm{E}) = \mathfrak{a}$. Let D be a subset of E equipotent to $\mathbf{N}$ (Lemma 1). Then there exists a bijection $\psi_0$ of D onto $\mathrm{D} \times \mathrm{D}$ (Lemma 2). Let $\mathfrak{M}$ be the set of pairs $(\mathrm{X}, \psi)$, where X is a subset of E containing D and $\psi$ is a bijection of X onto $\mathrm{X} \times \mathrm{X}$ which extends $\psi_0$. Order the set $\mathfrak{M}$ by means of the relation

between $(\mathrm{X}, \psi)$ and $(\mathrm{X}', \psi')$. Then it is immediately seen that $\mathfrak{M}$ is inductive (cf. § 2, no. 4, Example 2). Hence, by Zorn's Lemma (§ 2, no. 4, Theorem 2), $\mathfrak{M}$ has a maximal element $(\mathrm{F}, f)$. We shall show that $\mathrm{Card}\,(\mathrm{F}) = \mathfrak{a}$, which will suffice to prove the theorem. If $\mathrm{Card}\,(\mathrm{F})$ is not equal to $\mathfrak{a}$, let $\mathrm{Card}\,(\mathrm{F}) = \mathfrak{b} < \mathfrak{a}$. Then $\mathfrak{b} = \mathfrak{b}^2$ and $\mathfrak{b}$ is infinite, and $\mathfrak{b} \leqslant 2\mathfrak{b} \leqslant 3\mathfrak{b} \leqslant \mathfrak{b}^2 = \mathfrak{b}$ (§ 3, no. 6, Proposition 14); hence $2\mathfrak{b} = \mathfrak{b}$ and $3\mathfrak{b} = \mathfrak{b}$. From the hypothesis $\mathfrak{b} < \mathfrak{a}$ it follows that $\mathrm{Card}\,(\mathrm{E} - \mathrm{F}) > \mathfrak{b}$, for otherwise we should have $\mathrm{Card}\,(\mathrm{E}) \leqslant 2\mathfrak{b} = \mathfrak{b}$, and we have assumed that $\mathfrak{b} < \mathrm{Card}\,(\mathrm{E})$. Hence there is a subset $\mathrm{Y} \subset \mathrm{E} - \mathrm{F}$ equipotent to F. Let $\mathrm{Z} = \mathrm{F} \cup \mathrm{Y}$; we shall show that there exists a bijection $g$ of Z onto $\mathrm{Z} \times \mathrm{Z}$ which extends $f$. We have

and the four products on the right-hand side are mutually disjoint. Since F and Y are equipotent, we have

whence

There is therefore a bijection $f_1$ of Y onto the set

the mapping $g$ of Z into Z $\times$ Z, which is equal to $f$ on F and to $f_1$ on Y, is therefore a bijection which extends $f$, contrary to the definition of $f$. Hence $\mathrm{Card}(\mathrm{F}) = \mathfrak{a}$, and the proof is complete.

By induction on $n$.

Let $\mathfrak{b}$ denote the product and let $n$ be the number of elements in I. Then $\mathfrak{b} \leqslant \mathfrak{a}^n = \mathfrak{a}$ (§ 3, no. 6, Proposition 14). On the other hand, since $\mathfrak{a}_i \geqslant 1$ for all $i \in \mathrm{I}$, we have $\mathfrak{b} \geqslant \mathfrak{a}$ (§ 3, no. 6, Proposition 14).

Let $\mathfrak{b}$ be the cardinal of I; then we have $\sum\limits_{\iota \in \mathrm{I}} \mathfrak{a}_\iota \leqslant \mathfrak{a}\mathfrak{b} \leqslant \mathfrak{a}^2 = \mathfrak{a}$ (§ 3, no. 6, Proposition 14), and $\sum\limits_{\iota \in \mathrm{I}} \mathfrak{a}_\iota \geqslant \mathfrak{a}_\varkappa$ for all $\varkappa \in \mathrm{I}$.

This follows directly from Corollaries 2 and 3.

The first assertion is obvious. The others follow from the Corollaries to Theorem 2, no. 3.

¶ We have proved (no. 3, Lemma 1) that if $\mathfrak{a}$ is any infinite cardinal, then $\mathrm{Card}(\mathbf{N}) \leqslant \mathfrak{a}$. This has the following consequences :

For $\mathrm{Card}(\mathrm{E}) \leqslant \mathrm{Card}(\mathbf{N})$ by definition; and $\mathrm{Card}(\mathbf{N}) \leqslant \mathrm{Card}(\mathrm{E})$ since E is infinite.

For $\mathrm{Card}(\mathrm{E}) = \mathrm{Card}(\mathrm{E})\ \mathrm{Card}(\mathbf{N})$ (no. 3, Theorem 2, Corollary 4).

For the sets $\overset{-1}{f}(y)$ $(y \in \mathrm{F})$ form a partition of E; hence

and $\mathrm{Card}(\mathrm{F}) \leqslant \mathrm{Card}(\mathrm{E})$ by Proposition 3 of § 3, no. 2.

For each integer $n$, let $\mathfrak{F}_n$ denote the set of subsets of E which have $n$ elements. For each $\mathrm{X} \in \mathfrak{F}_n$ there exists a bijection of $[1, n]$ onto X. Hence the cardinal of $\mathfrak{F}_n$ is at most equal to that of the set of mappings of $[1, n]$ into E, i.e., to $\mathrm{Card}(\mathrm{E}^n) = \mathrm{Card}(\mathrm{E})$ (no. 3, Theorem 2, Corollary 1). Therefore

On the other hand, since $x \to \{x\}$ is an injective mapping of E into $\mathfrak{F}(\mathrm{E})$, we have $\mathrm{Card}(\mathrm{E}) \leqslant \mathrm{Card}(\mathfrak{F}(\mathrm{E}))$.

(a) Every non-empty subset of E has a maximal element. (b) Every increasing sequence $(x_n)$ of elements of E is stationary.

We show first that (a) implies (b). Let X be the set of elements of the sequence $(x_n)$, and let $x_m$ be a maximal element of X. If $n \geqslant m$, we have by hypothesis $x_n \geqslant x_m$, and therefore $x_n = x_m$ by the maximality of $x_m$. Conversely, suppose that there exists a non-empty subset A of E which has no maximal element. For each $x \in A$, let $T_x$ be the set of all $y \in A$ such that $y > x$. By assumption, $T_x \neq \emptyset$ for all $x \in A$; hence there exists a mapping $f$ of A into A such that $f(x) > x$ for all $x \in A$ (Chapter II, § 5, no. 4, Proposition 6). If $a \in A$, then the sequence $(x_n)_{n \in \mathbf{N}}$ defined inductively by the conditions $x_0 = a$, $x_{n+1} = f(x_n)$ is evidently increasing and not stationary.

For to say that E is well-ordered is equivalent to saying that every non-empty subset of E has a minimal element (§ 1, no. 10, Proposition 10), and the assertion therefore follows from Proposition 6.

For every finite ordered set has a maximal element (§ 4, no. 4, Proposition 3, Corollary 2).

¶ An ordered set E which satisfies the equivalent conditions of Proposition 6 is sometimes said to be Noetherian.

Indeed, suppose E $\neq$ F; then E $-$ F has a maximal element $b$. By definition we have $x \in F$ for all $x > b$; but this implies $b \in F$, which is absurd.