Exercise 11.7

Proof. Following the hint, suppose that the set $A$ has strict partial order $\prec$ and let $A$ be the collection of all subsets of $A$ that are simply ordered by $\prec$. We show that $A$ has finite type, i.e. that a subset $B\subset A$ is in $A$ if and only if every finite subset of $B$ is.

$\text{(⇒)}$ Suppose that $B\subset A$ is in $A$ so that it is simply ordered by $\prec$. Clearly, any finite subset of $B$ is also simply ordered by $\prec$ so that it is also in $A$, which shows the result.

$\text{(⇐)}$ Now suppose that $B\subset A$ and that every finite subset of $B$ is in $A$. Now consider two distinct elements $x$ and $y$ of $B$. Clearly, then the set $\left\{x,y\right\}$ is a finite subset of $B$ and hence is in $A$. Then this means that $\left\{x,y\right\}$ is simply ordered by $\prec$ so that clearly $x$ and $y$ are comparable. Since $x$ and $y$ were arbitrary this shows that $B$ is simply ordered by $\prec$ and hence $B\in A$.

We have thus shown that $A$ is of the finite type so that it has a set $C$ such that is properly contained in no other element of $A$. Since $C\in A$, it is simply ordered by $\prec$. It is also maximal since, if $D$ is any proper superset of $C$ then it cannot be that $D$ is simply ordered for then we would have $D\in A$ and $C⊊D$, which would contradict the definition of $C$. Hence $C$ is the maximal simply ordered subset of $A$ that shows the maximum principle. □