2.8. THE HAUSDORFF MAXIMAL THEOREM 61

This is done by considering a set B ⊆ Y0 consisting of D which acts like C in theabove and showing that it actually equals Y0 because it is a tower.

Proof of Claim 1: Consider B ≡{D ∈ Y0 : D ⊆ C0 or xC0 ∈ ∪D

}. Let Y1 ≡Y0∩B.

I want to argue that Y1 is a tower. By definition all chains of Y1 contain x0 in their unions.If D ∈ Y1, is θD ∈ Y1? If S ⊆ Y , is ∪S ∈ Y1? Is {x0} ∈B?{x0} cannot properly contain C0 since x0 ∈ ∪C0. Therefore, C0 ⊇ {x0} so {x0} ∈B.If S ⊆ Y1, and D ≡ ∪S , is D ∈ Y1? Since Y0 is a tower, D is comparable to C0.

If D ⊆ C0, then D is in B. Otherwise D ⊋ C0 and in this case, why is D in B? Why isxC0 ∈ ∪D? The chains of S are in B so one of them, called C̃ must properly contain C0

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,either x≤ y or y≤ x

if x≤ y and y≤ z then x≤ z.

and≤ is a well order if (X ,≤) if every nonempty subset of X has a smallest element. Moreprecisely, if S ̸= /0 and S ⊆ X then there exists an x ∈ S such that x ≤ y for all y ∈ S. Afamiliar example of a well-ordered set is the natural numbers.

Lemma 2.8.3 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 , defineS1 ≺ S2 if S1 ⊆ S2 and there exists a well order for S2,≤2 such that (S2,≤2) is well-orderedand if y ∈ S2 \ S1 then x ≤2 y for all x ∈ S1, and if ≤1is the well order of S1 then the twoorders are consistent on S1. Then observe that ≺ is a partial order on F . By the Hausdorffmaximal principle, let C be a maximal chain in F and let X∞ ≡ ∪C . Define an order, ≤,