Tukey's Lemma/Formulation 1

Theorem
Let $S$ be a non-empty set of finite character.

Then $S$ has an element which is maximal with respect to the subset relation.

Proof
Let $C \subseteq S$ be a chain.

We will show that $\ds \bigcup C \in S$.

Let $x$ be a finite subset of $\ds \bigcup C$.

By the definitions of subset and of union, each element of $x$ is an element of at least one element of $C$.

By the Principle of Finite Choice, there is a mapping $c: x \to C$ such that:
 * $\forall a \in x: a \in \map c x$

Then $\map c x$ is a finite subset of $C$.

From Finite Totally Ordered Set is Well-Ordered, $\map c x$ has a greatest element $m \in C \subseteq S$.

Then:
 * $x$ is a finite subset of $m$

and:
 * $m \in S$

Since $S$ has finite character:
 * $x \in S$

We have thus shown that every finite subset of $\ds \bigcup C$ is in $S$.

Since $S$ is of finite character:
 * $\ds \bigcup C \in S$

Thus by Zorn's Lemma, $S$ has a maximal element.