Editing Hausdorff maximal principle (section)

== Proof 1==
The idea of the proof is essentially due to Zermelo and is to prove the following weak form of [[Zorn's lemma]], from the [[axiom of choice]].<ref>{{harvnb|Halmos|1960|loc=§ 16.}}</ref><ref>{{harvnb|Rudin|1986|loc=Appendix}}</ref>

:Let <math>F</math> be a nonempty set of subsets of some fixed set, ordered by set inclusion, such that (1) the union of each totally ordered subset of <math>F</math> is in <math>F</math> and (2) each subset of a set in <math>F</math> is in <math>F</math>. Then <math>F</math> has a maximal element.

(Zorn's lemma itself also follows from this weak form.) The maximal principle follows from the above since the set of all chains in <math>P</math> satisfies the above conditions.

By the axiom of choice, we have a function <math>f : \mathfrak{P}(P) - \{ \emptyset \} \to P</math> such that <math>f(S) \in S</math> for the power set <math>\mathfrak{P}(P)</math> of <math>P</math>.

For each <math>C \in F</math>, let <math>C^*</math> be the set of all <math>x \in P - C</math> such that <math>C \cup \{ x \}</math> is in <math>F</math>. If <math>C^* = \emptyset</math>, then let <math>\widetilde{C} = C</math>. Otherwise, let
:<math>\widetilde{C} = C \cup \{ f(C^*) \}.</math>
Note <math>C</math> is a maximal element if and only if <math>\widetilde{C} = C</math>. Thus, we are done if we can find a <math>C</math> such that <math>\widetilde{C} = C</math>.

Fix a <math>C_0</math> in <math>F</math>. We call a subset <math>T \subset F</math> a ''tower (over <math>C_0</math>)'' if

# <math>C_0</math> is in <math>T</math>.
# The union of each totally ordered subset <math>T' \subset T</math> is in <math>T</math>, where "totally ordered" is with respect to set inclusion.
# For each <math>C</math> in <math>T</math>, <math>\widetilde{C}</math> is in <math>T</math>.

There exists at least one tower; indeed, the set of all sets in <math>F</math> containing <math>C_0</math> is a tower. Let <math>T_0</math> be the intersection of all towers, which is again a tower.

Now, we shall show <math>T_0</math> is totally ordered. We say a set <math>C</math> is ''comparable in <math>T_0</math>'' if for each <math>A</math> in <math>T_0</math>, either <math>A \subset C</math> or <math>C \subset A</math>. Let <math>\Gamma</math> be the set of all sets in <math>T_0</math> that are comparable in <math>T_0</math>. We claim <math>\Gamma</math> is a tower. The conditions 1. and 2. are straightforward to check. For 3., let <math>C</math> in <math>\Gamma</math> be given and then let <math>U</math> be the set of all <math>A</math> in <math>T_0</math> such that either <math>A \subset C</math> or <math>\widetilde{C} \subset A</math>.

We claim <math>U</math> is a tower. The conditions 1. and 2. are again straightforward to check. For 3., let <math>A</math> be in <math>U</math>. If <math>A \subset C</math>, then since <math>C</math> is comparable in <math>T_0</math>, either <math>\widetilde{A} \subset C</math> or <math>C \subset \widetilde{A} </math>. In the first case, <math>\widetilde{A}</math> is in <math>U</math>. In the second case, we have <math>A \subset C \subset \widetilde{A}</math>, which implies either <math>A = C</math> or <math>C = \widetilde{A}</math>. (This is the moment we needed to collapse a set to an element by the axiom of choice to define <math>\widetilde{A}</math>.) Either way, we have <math>\widetilde{A}</math> is in <math>U</math>. Similarly, if <math>C \subset A</math>, we see <math>\widetilde{A}</math> is in <math>U</math>. Hence, <math>U</math> is a tower. Now, since <math>U \subset T_0</math> and <math>T_0</math> is the intersection of all towers, <math>U = T_0</math>, which implies <math>\widetilde{C}</math> is comparable in <math>T_0</math>; i.e., is in <math>\Gamma</math>. This completes the proof of the claim that <math>\Gamma</math> is a tower.

Finally, since <math>\Gamma</math> is a tower contained in <math>T_0</math>, we have <math>T_0 = \Gamma</math>, which means <math>T_0</math> is totally ordered.

Let <math>C</math> be the union of <math>T_0</math>. By 2., <math>C</math> is in <math>T_0</math> and then by 3., <math>\widetilde C</math> is in <math>T_0</math>. Since <math>C</math> is the union of <math>T_0</math>, <math>\widetilde C \subset C</math> and thus <math>\widetilde C = C</math>. <math>\square</math>