504 APPENDIX B. THE HAUSDORFF MAXIMAL THEOREM

and so xC0 ∈ ∪C̃ ⊆ ∪D . Therefore, D ∈B∩Y0 = Y1. 4.) holds. Two cases remain, toshow that Y1 satisfies 3.).

case 1: D ⊋ C0. Then by definition of B, xC0 ∈ ∪D and so xC0 ∈ ∪θD so θD ∈ Y1.case 2: D ⊆ C0. θD ∈ Y0 so θD is comparable to C0. First suppose θD ⊋ C0. Thus

D ⊆ C0 ⊊ D ∪{xD} . If x ∈ C0 and x is not in D then D ∪{x} ⊆ C0 ⊊ D ∪{xD}. Thisis impossible. Consider x. Thus in this case that θD ⊋ C0, D = C0. It follows thatxD = xC0 ∈ ∪θC0 = ∪θD and so θD ∈ Y1. The other case is that θD ⊆ C0 so θD ∈Bby definition. This shows 3.) so Y1 is a tower and must equal Y0.

Claim 2: Any two chains in Y0 are comparable.Proof of Claim 2: Let Y1 consist of all chains of Y0 which are comparable to every

chain of Y0. {x0} is in Y1 by definition. All chains of Y0 have x0 in their union. IfS ⊆Y1, is ∪S ∈Y1? Given D ∈Y0 either every chain of S is contained in D or at leastone contains D . Either way D is comparable to ∪S so ∪S ∈ Y1. It remains to show 3.).Let C ∈ Y1 and D ∈ Y0. Since C is comparable to all chains in Y0, it follows from Claim1 either C ⊊ D when xC ∈ ∪D and θC ⊆ D or C ⊇ D when θC ⊇ D . Hence Y1 = Y0because Y0 is as small as possible.

Since every pair of chains in Y0 are comparable and Y0 is a tower, it follows that∪Y0 ∈ Y0 so ∪Y0 is a chain. However, θ ∪Y0 is a chain which properly contains ∪Y0and since Y0 is a tower, θ ∪Y0 ∈ Y0. Thus ∪(θ ∪Y0) ⊋ ∪(∪Y0) ⊇ ∪(θ ∪Y0) which isa contradiction. Therefore, for some chain C it is impossible to obtain the xC describedabove and so, this C is a maximal chain. ■

If X is a nonempty set,≤ is an order on X if

x≤ x,

and if x, y ∈ X , theneither x≤ y or y≤ x

andif x≤ y and y≤ z then x≤ z.

≤ is a well order and say that (X ,≤) is a well-ordered set if every nonempty subset of Xhas a smallest element. More precisely, if S ̸= /0 and S ⊆ X then there exists an x ∈ S suchthat x≤ y for all y ∈ S. A familiar example of a well-ordered set is the natural numbers.

Lemma B.0.3 The Hausdorff maximal principle implies every nonempty set can be well-ordered.

Proof: Let X be a nonempty set and let a ∈ X . Then {a} is a well-ordered subset of X .Let

F = {S⊆ X : there exists a well order for S}.Thus F ̸= /0. For S1, S2 ∈F , define S1 ≺ S2 if S1 ⊆ S2 and there exists a well order for S2,≤2 such that

(S2,≤2) is well-ordered

and ify ∈ S2 \S1 then x≤2 y for all x ∈ S1,

and if ≤1is the well order of S1 then the two orders are consistent on S1. Then observe that≺ is a partial order on F . By the Hausdorff maximal principle, let C be a maximal chainin F and let

X∞ ≡ ∪C .