Editing Hausdorff maximal principle (section)

== Proof 2==


The [[Bourbaki–Witt theorem]], together with the [[Axiom of choice]], can be used to prove the Hausdorff maximal principle. Indeed, let <math>P</math> be a nonempty poset and <math>X\mathrel{\mathop:}=\{C\subseteq P\,:\, C\ \text{is a chain}\}</math> be the set of all totally ordered subsets of <math>P</math>. Notice that <math>X\neq \emptyset</math>, since <math>P\neq \emptyset</math> and <math>\{x\}\in X</math>, for any <math>x\in P</math>. Also, equipped with the inclusion <math>\subseteq</math>, <math>X</math> is a poset. We claim that every chain <math>\mathcal{C}\subseteq X</math> has a [[supremum]]. In order to check this out, let <math>S</math> be the union <math>\bigcup_{C\in \mathcal{C}}C</math>. Clearly, <math>C\subseteq S</math>, for all <math>C\in \mathcal{C}</math>. Also, if <math>U</math> is any upper bound  of <math>\mathcal{C}</math>, then <math>S\subseteq U</math>, since by definition <math>C\subseteq U</math> for all <math>C\in \mathcal{C}</math>. 

Now, consider the map <math>f\colon X\to X</math> given by

<math>f(C)\mathrel{\mathop:}=\begin{cases}C, &\text{if}\ C\ \text{is maximal}\\ C\cup \{g(P\setminus C)\}, &\text{if}\ C\ \text{is not maximal}\end{cases}</math>

where <math>g</math> is a [[choice function]] on <math>\{P\setminus C\}</math> whose existence is ensured by the Axiom of choice, and the fact that <math>P\setminus C\neq \emptyset</math> is an immediate consequence of the non-maximality of <math>C</math>. Thus, <math>C\subseteq f(C)</math>, for each <math>C\in X</math>. In view of the Bourbaki-Witt theorem, there exists an element <math>C_0\in \mathcal{C}</math> such that <math>f(C_0)=C_0</math>, and therefore <math>C_0</math> is a maximal chain of <math>P</math>.

In the case <math>P=\emptyset</math>, the empty set is trivially a maximal chain of <math>P</math>, as already mentioned above. <math>\square</math>