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)

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

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

History 1924, S. Banach and A. Tarski 1939, Kurt G ödel Early 1960s, Paul Cohen

1924, S. Banach and A. Tarski

1939, Kurt G ödel

Early 1960s, Paul Cohen

When do we need it? When we have a finite number of sets? let X 1 ={a,b} and X 2 ={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?

When we have a finite number of sets?

let X 1 ={a,b} and X 2 ={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 

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)

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.

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.

Equivalents to 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.

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.

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.

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-ordering Corollary of the Axiom of Choice . Let X be a set. There exists a function f : 2 X -> 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: 2 X -> X be a function, as in corollary above, whose existence is guaranteed by the Axiom of Choice. Set x 1 = f (X) and x n = f (X – (union of x j for j=1 to j=1-n for n ≥2)) The sequence of {x n } 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.

Corollary of the Axiom of Choice . Let X be a set. There exists a function f : 2 X -> 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: 2 X -> X be a function, as in corollary above, whose existence is guaranteed by the Axiom of Choice. Set x 1 = f (X) and x n = f (X – (union of x j for j=1 to j=1-n for n ≥2))

The sequence of {x n } 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 {x n } relative to the ordering induced by the natural numbers are the sets of the form {x 1 , x 2 , … , x m } 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 = {x n } is in F .

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 {x n } relative to the ordering induced by the natural numbers are the sets of the form {x 1 , x 2 , … , x m } 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 = {x n } 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 D 1 and D 2 . Then one of the two domains, say, for example, D 1 , is a segment for the other, say, for example, D 2 , with respect to the corresponding ordering ~ 2 . Moreover, ~ 1 and ~ 2 coincide on such a segment.

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 D 1 and D 2 . Then one of the two domains, say, for example, D 1 , is a segment for the other, say, for example, D 2 , with respect to the corresponding ordering ~ 2 . Moreover, ~ 1 and ~ 2 coincide on such a segment.

Let D 0 be the union of the domains of the elements of F . Also let ~ 0 be that ordering on D 0 that coincides with the ordering ~ in F on its domain D. By Lemma 2, this is a linear ordering on D 0 and satisfies requirement (*) of the class F . Therefore, by Lemma 1, it is a well-ordering on D 0 . It remains to show that D 0 =X. Consider the set D′ 0 = D 0 union { f (X - D 0 )} and the ordering ~′ 0 that coincides with ~ 0 on D 0 and by which (X - D 0 ) follows any element of D 0 . Therefore, D′ 0 = D 0 . However, this is a contradiction unless (X - D 0 ) is the empty set. □

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

D′ 0 = D 0 union { f (X - D 0 )}

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

Consequences of the Axiom of Choice Set theory Algebra General topology

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

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