Hausdorff Maximal Principle and Axiom of Choice I need to show that Hausdorff Maximal Principle is equivalent to the Axiom of Choice. Suggested is to use Tukeys Lemma. 
So far I have that Hausdorff Maximal Principle states that whenever < is a strict partial order of a set A, there is a maximal chain C $\subseteq$ A. 
and
Tukeys Lemma states that $F \subseteq P(A)$ is of finite character if and only if for all $X \subseteq A : X \in F$ iff every finite subset of X is in $F$. 
This is coming from my book Kunen Foundations of Mathematics. 
I also understand a chain to be a totally ordered set, if for all x,y, in X, either $x \leq y$ or $y \leq x$. A chain has at most one maximal element. 
I am really stuck on how to relate these two to each other. I think I have an idea of how to show equivalence from Tukey's Lemma to Axiom of Choice. But I am stuck when it comes to showing that the Hausdorff Maximal Principle is equivalent to Tukeys Lemma. Any help or references is greatly appreciated. I have also been reading up on this website:
http://web.science.mq.edu.au/~chris/sets/CHAP09%20Axiom%20of%20Choice.pdf
 A: I actually don't see how to use Tukey's lemma here.
To me, the most natural proof is along the following lines (leaving some gaps to fill in):


*

*AC implies HMP: Use AC in the form of Zorn. Given a partial order $P$, let $P^*$ be the partial order consisting of chains in $P$, ordered by inclusion. Then a maximal element in $P^*$ corresponds to a maximal chain in $P$.

*HMP implies AC: Fix a set $\mathcal{A}$ of nonempty disjoint sets $A_i$ ($i\in I$); using $HMP$, I'll show that $\mathcal{A}$ has a choice function. Let $P$ be the partial order whose elements are partial choice functions: maps $p$ with domain $J_p\subseteq I$ satisfying $p(j)\in A_j$ for all $j\in J_p$. We order $P$ by extension: $p\ge q$ if $J_p\supseteq J_q$ and $p\upharpoonright J_q=q$. Then given a maximal chain $C\subseteq P$, let $q=\bigcup C$ be the union of the functions in $C$. Clearly $q\in P$, and $q\ge p$ for all $p\in C$; and if $dom(q)\not=I$, then we can extend $q$ to get a strictly larger element $q'$ of $P$, which will contradict the assumption that $C$ is maximal. So $q$ is in fact a full choice function, and we're done.
A: The key point here is that being a chain is a property of finite character. $C$ is a chain if and only if every one of its finite subsets is a chain. 
So Hausdorff's Maximality Principle follows easily from Tukey's Lemma. 
In the other direction, given a family of finite character, pick a maximal chain and show the union of that chain is a maximal member of your family, here the trick is that if $C$ is the chain and $X$ is its union, then a finite subset of $X$ is a subset of some element of $C$. 
