Theorem 1

Every set $\mathfrak{S}$ of subsets of a set $E$ which is of finite character has a maximal element (when ordered by inclusion).

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.