Lemma 2

The set $\mathbf{N} \times \mathbf{N}$ is equipotent to $\mathbf{N}$.

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.