Exercise 11.6

Proof. Suppose that $A$ is a collection of sets of finite type. Let $B$ be a subcollection of $A$ that is simply ordered by $⊊$. Consider next any finite subset $B$ of $\bigcup <!--\; FUNCTION\; APPLICATION\; -->B$. Then, for every $b\in B$, $b\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->B$ so that we can choose a set $B_b\in B$ such that $b\in B_b$. Note that this does not require the choice axiom since we need to make only a finite number of choices. Then the set $B^{\prime }=\left\{B_b\mid b\in B\right\}$ is clearly a finite set of elements of $B$. Since $B$ is simply ordered by $⊊$, it follows that $B^{\prime }$ is as well and so has a largest element $C$ since it is finite.

Hence, for any $b\in B$, we have that $b\in B_b\subset C$ so that $b\in C$, and so $B$ is a finite subset of $C$. Since $C\in B$ and $B\subset A$, clearly $C\in A$. Since $A$ is of finite type and $B$ is a finite subset of $C$, it follows that $B\in A$ also. Since $B$ was an arbitrary finite subset of $\bigcup <!--\; FUNCTION\; APPLICATION\; -->B$, it then follows that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->B$ is also in $A$ since it is of finite type. It then follows from the Kuratowski lemma (Exercise 11.5) that $A$ has an element that is properly contained in no other element of $A$ as desired. □