[Math] Hausdorff Maximal Principle and Axiom of Choice

axiom-of-choiceset-theory

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

Best Answer

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.

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