Hausdorff Maximal Principle/Proof 3

Proof
Let $\struct {\CC, \subseteq}$ be the set of all chains in $P$ ordered by inclusion.

By Set of Chains is Chain Complete for Inclusion, $\CC$ is a chain complete ordered set.

Now define $f: \CC \to \CC$ as follows:


 * If $C$ is a maximal chain then $\map f C = C$.
 * Otherwise $f$ chooses arbitrarily, using the axiom of choice, some chain $D$ which strictly contains $C$.

By construction, $\map f C \supseteq C$.

By the above, $\CC$ is chain complete.

Therefore the Bourbaki-Witt Fixed Point Theorem applies and $f$ has a fixed point $\map F M = M$.

But by the above construction, the only fixed points of $f$ are maximal chains.

Therefore $M$ is a maximal chain.

The use of the axiom of choice in the construction of $f$ is crucial.