From Hausdorff maximal principle to well-ordering principle

Question

How to prove the well-ordering principle if we have known the Hausdorff maximal principle?

Answer 1

Let $A$ be a non-empty set, and let $P$ be the collection of all partial well-orders, i.e. all the well orders of subsets of $A$. We order by me extension, i.e. $R<S$ iff $R$ is a proper initial segment of $S$.

By Hausdorff Maximality Principle we have a maximal chain, its union is a well-ordering of $A$. Denote this order by $R$. It should be fairly clear how to show that the domain of $R$ is $A$ (Hint: it follows from the maximality of the chain).

Suppose $\varnothing\neq B\subseteq A$, then $B$ is a subset of the domain of $R$, so there exists some $S$ which is a well-order of a subset of $A$ whose domain contains at least one point from $B$. Since $S$ is a well-order and $B\cap\operatorname{dom}(S)$ is non-empty there exists a minimal element there. Now you have to show that this minimal element is actually $R$-minimal (Hint: it follows from the definition of $<$ on the collection of well-orders).