Exercise 8.1.6

Proof. $\text{(→)}$ Suppose Zorn’s Lemma and let $A$ be an arbitrary system of sets of finite character. Suppose that $B$ is any subset of $A$ that is linearly ordered by $\subseteq$ and let $C$ be any finite subset of $\bigcup <!--\; FUNCTION\; APPLICATION\; -->B$. Now, for each $x\in C$ there is a set $X_x\in B$ such that $x\in X_x$, since $x\in C\subseteq \bigcup <!--\; FUNCTION\; APPLICATION\; -->B$. Clearly the set $D=\left\{X_x\mid x\in C\right\}$ is a subset of $B$ so that $D$ is also linearly ordered by $\subseteq$. Also clearly $D$ is finite since $C$ is. Hence $D$ has a $\subseteq$-greatest element $X$. Note that the Axiom of Choice is not needed in selecting the set $X_x$ for each $x\in C$ since we are only making a finite number of choices. So consider any $x\in C$ so that $x\in X_x\subseteq X$. Hence $x\in X$ so that $C\subseteq X$ since $x$ was arbitrary. We also have that $X\in B\subseteq A$ so $X\in A$. Therefore $C$ is a finite subset of $X$, which is an element of $A$, so that $C$ is also in $A$ since $A$ has finite character. Since $C$ was an arbitrary finite subset of $\bigcup <!--\; FUNCTION\; APPLICATION\; -->B$ and $C\in A$ it follows that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->B\in A$. Hence, since $B$ was an arbitrary linearly ordered (by $\subseteq$) subset of $A$, we have by Exercise 8.1.5 and Zorn’s Lemma that $A$ has a $\subseteq$-maximal element as desired.

$\text{(←)}$ Now suppose that every system of sets of finite character has a $\subseteq$-maximal element. Let $(A,\le )$ be any ordered set and let $C$ be the set of all chains of $(A,\le )$. Now suppose that $X\in C$ and let $Y$ be any finite subset of $X$. Clearly since $X\in C$, it is linearly ordered by $\le$ so that $Y$ is as well since $Y\subseteq X$. Hence $Y\in C$. Now let $X^{\prime }$ be any set such that every finite subset of $X^{\prime }$ is in $C$. Consider any $x,y\in X^{\prime }$. Then $\left\{x,y\right\}$ is clearly a finite subset of $X^{\prime }$ so that it is in $C$ and therefore a $\le$-chain. Hence $x$ and $y$ are comparable in $\le$, which shows that $X^{\prime }$ itself is a $\le$-chain since $x$ and $y$ were arbitrary. Hence $X^{\prime }\in C$. Thus we have just shown that $X\in C$ if and only if every finite subset of $X$ is in $C$ so that $C$ has finite character by definition. Therefore, by the initial supposition, $C$ has a $\subseteq$ maximal element. Since again $C$ is the set of all chains of the arbitrary $(A,\le )$, Zorn’s Lemma follows from Exercise 8.1.4. □