Axiom of Choice implies Hausdorff's Maximal Principle/Proof 3

Theorem
Let $\left({\mathcal P, \preceq}\right)$ be an ordered set.

Then there exists a maximal chain in $\mathcal P$.

Proof
Let $\left({\mathcal C, \subseteq}\right)$ be the set of all chains in $P$ ordered by inclusion.

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

Now define $f: \mathcal C \to \mathcal C$ as follows:


 * If $C$ is a maximal chain then $f \left({C}\right) = C$.
 * Otherwise $f$ chooses arbitrarily, using the axiom of choice, some chain $D$ which strictly contains $C$.

By construction, $f \left({C}\right) \supseteq C$.

By the above, $\mathcal C$ is chain complete.

Therefore the Bourbaki-Witt Fixed Point Theorem applies and $f$ has a fixed point $F \left({M}\right) = M$.

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.