Axiom of Choice Final Version

Axiom of Choice

by Catherine Janes

Set Theory Axiom 1 (the axiom of extension) Axiom 2 (the axiom of the null set)

Axiom 3 (the axiom of pairing) Axiom 4 (the axiom of union) Axiom 5 (the axiom of the power set) Axiom 6 (the axiom of separation) Axiom 7 (the axiom of replacement) Axiom 8 (the axiom of infinity)

Axiom 9 (the axiom of regularity)

The Axiom of Choice

Given any nonempty set Y whose members are pairwise disjoint sets, there exists a set X consisting of exactly one element taken from each set belonging to Y. (Lay 94)

Let {Xα} be a family of nonempty sets. Then there is a set X which contains, from each set Xα, exactly one element. (Garrity 207)

History

1924, S. Banach and A. Tarski

1939, Kurt Gödel

Early 1960s, Paul Cohen

When do we need it?

When we have a finite number of sets? let X1={a,b} and X2={c,d}. let X={a,c}.

When we have an infinite number of sets whose elements are well-ordered? well-ordering of the natural numbers

When we have an infinite number of sets whose elements are not well-ordered?

Shoes and socks

Shoes are well-ordered!

Socks are not well-ordered

Infinite Number of Sets

We can also say that all sets can be well-ordered.

“The Axiom of Choice gives no method for finding the set X; it just mandates the existence of X.” (Garrity 208)

Some Terms

A total order ≤ on a set E is said to be a well-order on E provided that, for each A a subset of E, there exists an m an element of A such that m ≥b for each b an element of A. The element m is said to be the maximal element of A (on E with respect to ≤).

Given S a subset of K, we say that q an element of K is a ≤-upper bound of S provided that s≤ q for each s in S.

A relation ~ on a set X is a partial ordering of X if it is transitive (if x~y and y~z implies x~z) and antisymmetric (x~y and y~x implies x=y) For example, ≤ is a partial ordering of the real numbers.

A partial ordering ~ on a set X is a linear ordering on X if for any two elements x, y in X, either x~y or y~x. Again, the relation ≤ is a linear ordering on the real numbers.

A linearly ordered subset E of X is maximal if any linearly ordered subset of X is contained in E.

Equivalentsto the Axiom of Choice

The well-ordering principle

Given any set A, there exists a well-order in A.

Recall:

A total order ≤ on a set E is said to be a well-order on E provided that, for each A a subset of E, there exists an m an element of A such that m ≥b for each b an element of A.

Zorn’s Lemma

Let X be a partially ordered set such that every linearly ordered subset has an upper bound. Then X has a maximal element.

Zorn’s Lemma

Hausdorff Maximal Principle –Every partially ordered set contains a maximal linearly ordered subset. Partial and linear order are meant with respect to the same ordering ~.

A relation ~ on a set X is a partial ordering of X if it is transitive (if x~y and y~z implies x~z) and antisymmetric (x~y and y~x implies x=y)

A partial ordering ~ on a set X is a linear ordering on X if for any two elements x, y in X, either x~y or y~x.

A linearly ordered subset E of X is maximal if any linearly ordered subset of X is contained in E.

Zorn’s Lemma

Zorn’s Lemma. Let X be a partially ordered set such that every linearly ordered subset has an upper bound. Then X has a maximal element.

Proof. Let M be the maximal linearly ordered set claimed by the maximal principle, which states that every partially ordered set contains a maximal linearly ordered subset. An upper bound for M is a maximal element of X. □

Definition: Let X be a set partially ordered by the relation ~ and let E be a subset of X. An upper bound of a subset E of X is an element x of X such that y~x for all y in E. If x is an element of E, then x is a maximal element of E.

Well-orderingCorollary of the Axiom of Choice. Let X be a set. There exists a

function f: 2X → X such that f (E) is an element of E for every E a subset of X. That is, one may choose an element out of every subset of X.

Proof:

Let f: 2X → X be a function, as in corollary above, whose existence is guaranteed by the Axiom of Choice. Set x1 = f (X) and xn= f (X – (union of xj for j=1 to j=1-n for n ≥2))

The sequence of {xn} can be given the ordering of the natural numbers and, as such, is well-ordered. A well-ordering for X is constructed by rendering transfinite such a process.

Let D be a subset of X and let ~ be a linear ordering defined on D. A subset E of D is a segment relative to ~ if for any x an element of E, all y elements of D such that y ~ x belong to E.

The segments of {xn} relative to the ordering induced by the natural numbers are the sets of the form {x1, x2, … , xm } for some m in the natural numbers. The union and intersection of two segments is a segment. The empty set is a segment relative to any linear ordering ~.

Denote by F the family of linear orderings ~ defined on subsets D of X and satisfying the following:

If E as subset of D is a segment, then the first element of (D – E) is f (X – E). (*)

Such a family is not empty since the ordering of the natural numbers on the domain D = {xn} is in F.

Lemma 1. Every element of F is a well-ordering on its domain.

Lemma 2. Let ~1 and ~2 be two elements in F with domains D1 and D2. Then one of the two domains, say, for example, D1, is a segment for the other, say, for example, D2, with respect to the corresponding ordering ~2. Moreover, ~1 and ~2 coincide on such a segment.

Let D0 be the union of the domains of the elements of F. Also let ~0 be that ordering on D0 that coincides with the ordering ~ in F on its domain D. By Lemma 2, this is a linear ordering on D0 and satisfies requirement (*) of the class F. Therefore, by Lemma 1, it is a well-ordering on D0. It remains to show that D0 =X. Consider the set

D′0 = D0 union {f (X - D0)}

and the ordering ~′0 that coincides with ~0 on D0 and by which (X - D0) follows any element of D0. Therefore, D′0 = D0. However, this is a contradiction unless (X - D0) is the empty set. □

Consequences of the Axiom of Choice

Set theory

Algebra

General topology

The Banach-Tarski Paradox

The Banach Tarski Paradox: Let S and T be solid three-dimensional spheres of possibly different radii. Then S and T are equivalent by decomposition.

S= T=

= and