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

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

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

Proof
Let $\mathcal{C}$ be the set of all chains in $X$ ordered by inclusion.

This is clearly a poset. In fact it is a chain complete poset - every chain in it has a least upper bound.

Proof: Let $\mathcal{A} \subseteq \mathcal{C}$ be a chain. Then $A = \bigcup \mathcal{A} \in \mathcal{C}$, because if $x, y \in A$ then they must both be in some element of $\mathcal{A}$ (as $\mathcal{A}$ is a chain - find an element containing x, an element containing y, the larger of the two contains them both) and thus $x \leq y$ or $y \leq x$. So this is a chain, and clearly it must be the least upper bound (because it is the least upper bound in the power set poset).

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


 * if C is maximal then f(C) = C.
 * else f chooses arbitrarily (i.e. using the axiom of choice) some chain D which strictly contains C.

By construction, $f(C) \supseteq C$. By the above, $\mathcal{C}$ is chain complete. Therefore the Bourbaki Witt Fixed Point Theorem applies and $f$ has a fixed point $F(M) = M \supseteq T$. But the only fixed points of $f$ are maximal chains, therefore $M$ is a maximal chain containing $T$.