Editing Hausdorff maximal principle (section)

== Application ==
By the Hausdorff maximal principle, we can show every [[Hilbert space]] <math>H</math> contains a maximal orthonormal subset <math>A</math> as follows.<ref>{{harvnb|Rudin|1986|loc=Theorem 4.22.}}</ref> (This fact can be stated as saying that <math>H \simeq \ell^2(A)</math> as Hilbert spaces.)

Let <math>P</math> be the set of all orthonormal subsets of the given Hilbert space <math>H</math>, which is partially ordered by set inclusion. It is nonempty as it contains the empty set and thus by the maximal principle, it contains a maximal chain <math>Q</math>. Let <math>A</math> be the union of <math>Q</math>. We shall show it is a maximal orthonormal subset. First, if <math>S, T</math> are in <math>Q</math>, then either <math>S \subset T</math> or <math>T \subset S</math>. That is, any given two distinct elements in <math>A</math> are contained in some <math>S</math> in <math>Q</math> and so they are orthogonal to each other (and of course, <math>A</math> is a subset of the unit sphere in <math>H</math>). Second, if <math>B \supsetneq A</math> for some <math>B</math> in <math>P</math>, then <math>B</math> cannot be in <math>Q</math> and so <math>Q \cup \{ B \}</math> is a chain strictly larger than <math>Q</math>, a contradiction. <math>\square</math>

For the purpose of comparison, here is a proof of the same fact by Zorn's lemma. As above, let <math>P</math> be the set of all orthonormal subsets of <math>H</math>. If <math>Q</math> is a chain in <math>P</math>, then the union of <math>Q</math> is also orthonormal by the same argument as above and so is an upper bound of <math>Q</math>. Thus, by Zorn's lemma, <math>P</math> contains a maximal element <math>A</math>. (So, the difference is that the maximal principle gives a maximal chain while Zorn's lemma gives a maximal element directly.)