For the relations $\alpha = \mathfrak{b}$ and $\alpha + 1 = \mathfrak{b} + 1$ between cardinals $\alpha$ and $\mathfrak{b}$ are equivalent (§ 3, no. 4, Proposition 8); the relations $\alpha \ne \alpha + 1$ and $\alpha + 1 \ne (\alpha + 1) + 1$ are therefore equivalent.

¶ It is clear that $0 \ne 1$; hence $0$ is an integer. It follows that $1$ and $2$ are integers. The cardinals $2 + 1$ and $(2 + 1) + 1$ are integers, denoted by $3$ and $4$, respectively.

If $\alpha \le n$, there exists a cardinal $\mathfrak{b}$ such that $n = \alpha + \mathfrak{b}$ (§ 3, no. 6, Proposition 13). Then $(\alpha + 1) + \mathfrak{b} = (\alpha + \mathfrak{b}) + 1 = n + 1$ (§ 3, no. 3, Proposition 5, Corollary); and since $n \ne n + 1$, we have

Hence $\alpha + 1 \ne \alpha$, which means that $\alpha$ is an integer. If $n \ne 0$, we have $n \ge 1$ (§ 3, no. 2), and therefore there exists a unique cardinal $m$ such that $n = m + 1$ (§ 3, no. 6, Proposition 13 and no. 4, Proposition 8). Since $m \leqslant n$, $m$ is an integer, from what has already been proved. Finally, if an integer $a$ is such that $a < n$, we have $n = a + b$, with $b \neq 0$ (§ 3, no. 6, Proposition 13); since $b$ is an integer, we have $b = c + 1$ and $n = m + 1 = (a + c) + 1$. It follows that $m = a + c$ (§ 3, no. 4, Proposition 8), hence $a \leqslant m$. Conversely, if $a \leqslant m$, we have

and if $a = n = m + 1$, we would have $a > m$, contrary to hypothesis.

For X is contained in the complement X$'$ of a subset of E consisting of a single element; we have $\mathrm{Card}\,(\mathrm{X}) \leqslant \mathrm{Card}\,(\mathrm{X}')$ and $\mathrm{Card}\,(\mathrm{E}) = \mathrm{Card}\,(\mathrm{X}') + 1$, hence (Proposition 2) $\mathrm{Card}\,(\mathrm{X}') < \mathrm{Card}\,(\mathrm{E})$ and a fortiori $\mathrm{Card}\,(\mathrm{X}) < \mathrm{Card}\,(\mathrm{E})$.

Definition 1 shows that, conversely, if E is a set such that

for every subset X of E such that X $\neq$ E, then E is finite.

For $\mathrm{Card}\ (f(\mathrm{E})) \leqslant \mathrm{Card}\ (\mathrm{E})$ (§ 3, no. 2, Proposition 3).

(a) $f$ is an injection; (b) $f$ is a surjection; (c) $f$ is a bijection.

It is enough to prove that (a) and (b) are equivalent. If $f$ is injective, then $\mathrm{Card}(f(\mathrm{E})) = \mathrm{Card}\,(\mathrm{E}) = \mathrm{Card}\,(\mathrm{F})$, whence $f(\mathrm{E}) = \mathrm{F}$ (Corollary 2). If $f$ is not injective, let $x$ and $x'$ be two elements of E such that $x \neq x'$ and $f(x) = f(x')$. Then, putting $\mathrm{E}' = \mathrm{E} - \{x\}$, we have $f(\mathrm{E}') = f(\mathrm{E})$, whence $\mathrm{Card}(f(\mathrm{E})) \leqslant \mathrm{Card}\,(\mathrm{E}') < \mathrm{Card}\,(\mathrm{E})$ by virtue of Corollary 2; but since $\mathrm{Card}\,(\mathrm{F}) = \mathrm{Card}\,(\mathrm{E})$, it follows that $f(\mathrm{E}) \neq \mathrm{F}$.

The proof is by induction on the number $n$ of elements of the subset under consideration. The result is trivial for $n = 1$. Let X be a subset of $n + 1$ elements of E (with $n \geqslant 1$), and put $\mathrm{X} = \mathrm{Y} \cup \{x\}$, where Y has $n$ elements and is therefore not empty. The inductive hypothesis implies the existence of an upper bound (resp. least upper bound, resp. greatest element) $y$ of Y. Since E is right directed (resp. a lattice, resp. totally ordered), $\{x, y\}$ has an upper bound (resp. least upper bound, resp. greatest element) which is evidently an upper bound (resp. least upper bound, resp. greatest element) of X.

For such a set is inductive by Corollary 1 (cf. § 2, no. 4, Theorem 2).

By Theorem 2 of § 2, no. 4 it is enough to show that $\mathfrak{S}$ is inductive. To do this we shall show that if $\mathfrak{G}$ is any subset of $\mathfrak{S}$ which is totally ordered by inclusion, the union $X$ of the sets of $\mathfrak{G}$ belongs to $\mathfrak{S}$ (§ 2, no. 4, Theorem 2, Corollary 2). Since $\mathfrak{S}$ is of finite character, it is enough to show that every finite subset $Y$ of $X$ belongs to $\mathfrak{S}$. Now, for each $y \in Y$, there exists a set $Z_y \in \mathfrak{G}$ such that $y \in Z_y$. Since the set of sets $Z_y$ $(y \in Y)$ is finite and totally ordered by inclusion, it has a greatest element $S$ (no. 4, Corollary 1 to Proposition 3); in other words, there exists a set $S \in \mathfrak{G}$ such that $Y \subset S$. But since $S \in \mathfrak{S}$ and since $Y$ is a finite subset of $S$, we have $Y \in \mathfrak{S}$ because $\mathfrak{S}$ is of finite character; this completes the proof.