Tukey's Lemma/Formulation 1

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

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

Proof
Let $\NN \subseteq \FF$ be a chain.

We will show that $\bigcup \NN \in \FF$.

Let $F$ be a finite subset of $\bigcup \NN$.

By the definitions of subset and of union, each element of $F$ is an element of at least one element of $\NN$.

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

Then $c \sqbrk F$ is a finite subset of $\NN$.

From Finite Totally Ordered Set is Well-Ordered, $f \sqbrk F$ has a greatest element $P \in \NN \subseteq \FF$.

Then:
 * $F$ is a finite subset of $P$

and:
 * $P \in \FF$

Since $\FF$ has finite character:
 * $F \in \FF$

We have thus shown that every finite subset of $\bigcup \NN$ is in $\FF$.

Since $\FF$ is of finite character:
 * $\bigcup \NN \in \FF$

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