1.4. THE HAUSDORFF MAXIMAL THEOREM 11

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 .

Define an order, ≤, on X∞ as follows. If x, y are elements of X∞, pick S ∈ C such that x, yare both in S. Then if ≤S is the order on S, let x≤ y if and only if x≤S y. This definition iswell defined because of the definition of the order,≺. Now let U be any nonempty subset ofX∞. Then S∩U ̸= /0 for some S ∈ C . Because of the definition of ≤, if y ∈ S2 \S1, Si ∈ C ,then x ≤ y for all x ∈ S1. Thus, if y ∈ X∞ \ S then x ≤ y for all x ∈ S and so the smallestelement of S∩U exists and is the smallest element in U . Therefore X∞ is well-ordered.Now suppose there exists z ∈ X \X∞. Define the following order, ≤1, on X∞∪{z}.

x≤1 y if and only if x≤ y whenever x,y ∈ X∞

x≤1 z whenever x ∈ X∞.

Then letC̃ = {S ∈ C or X∞∪{z}}.

Then C̃ is a strictly larger chain than C contradicting maximality of C . Thus X \X∞ = /0and this shows X is well-ordered by ≤. This proves the lemma.

With these two lemmas the main result follows.

Theorem 1.4.4 The following are equivalent.

The axiom of choice

The Hausdorff maximal principle

The well-ordering principle.

Proof: It only remains to prove that the well-ordering principle implies the axiom ofchoice. Let I be a nonempty set and let Xi be a nonempty set for each i ∈ I. Let X = ∪{Xi :i ∈ I} and well order X . Let f (i) be the smallest element of Xi. Then f ∈∏i∈I Xi. ■

There are some other equivalences to the axiom of choice proved in the book by Hewittand Stromberg [22].