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Axiom of Choice (Math 101 Fall 2008)

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- 1. Axiom of Choice by Catherine Janes
- 2. Set Theory <ul><li>Axiom 1 (the axiom of extension) </li></ul><ul><li>Axiom 2 (the axiom of the null set) </li></ul><ul><li>Axiom 3 (the axiom of pairing) </li></ul><ul><li>Axiom 4 (the axiom of union) </li></ul><ul><li>Axiom 5 (the axiom of the power set) </li></ul><ul><li>Axiom 6 (the axiom of separation) </li></ul><ul><li>Axiom 7 (the axiom of replacement) </li></ul><ul><li>Axiom 8 (the axiom of infinity) </li></ul><ul><li>Axiom 9 (the axiom of regularity) </li></ul>
- 3. The Axiom of Choice <ul><li>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) </li></ul><ul><li>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) </li></ul>
- 4. History <ul><li>1924, S. Banach and A. Tarski </li></ul><ul><li>1939, Kurt G ödel </li></ul><ul><li>Early 1960s, Paul Cohen </li></ul>
- 5. When do we need it? <ul><li>When we have a finite number of sets? </li></ul><ul><ul><li>let X 1 ={a,b} and X 2 ={c,d}. let X={a,c}. </li></ul></ul><ul><li>When we have an infinite number of sets whose elements are well-ordered? </li></ul><ul><ul><li>well-ordering of the natural numbers </li></ul></ul><ul><li>When we have an infinite number of sets whose elements are not well-ordered? </li></ul>
- 6. Shoes and socks <ul><li>Shoes are well-ordered! </li></ul><ul><li>Socks are not well-ordered </li></ul>
- 7. Infinite Number of Sets <ul><li>We can also say that all sets can be well-ordered. </li></ul><ul><li>“ The Axiom of Choice gives no method for finding the set X; it just mandates the existence of X.” (Garrity 208) </li></ul>
- 8. Some Terms <ul><li>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 ≤). </li></ul><ul><li>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. </li></ul><ul><li>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) </li></ul><ul><ul><li>For example, ≤ is a partial ordering of the real numbers. </li></ul></ul><ul><li>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. </li></ul><ul><ul><li>Again, the relation ≤ is a linear ordering on the real numbers. </li></ul></ul><ul><li>A linearly ordered subset E of X is maximal if any linearly ordered subset of X is contained in E. </li></ul>
- 9. Equivalents to the Axiom of Choice <ul><li>The well-ordering principle </li></ul><ul><ul><li>Given any set A, there exists a well-order in A. </li></ul></ul><ul><ul><li>Recall: </li></ul></ul><ul><ul><li>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. </li></ul></ul><ul><li>Zorn’s Lemma </li></ul><ul><ul><li>Let X be a partially ordered set such that every linearly ordered subset has an upper bound. Then X has a maximal element. </li></ul></ul>
- 10. Zorn’s Lemma <ul><ul><li>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 ~. </li></ul></ul><ul><li>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) </li></ul><ul><li>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. </li></ul><ul><li>A linearly ordered subset E of X is maximal if any linearly ordered subset of X is contained in E. </li></ul>
- 11. Zorn’s Lemma <ul><ul><li>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. </li></ul></ul><ul><ul><li>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. □ </li></ul></ul><ul><li>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. </li></ul>
- 12. Well-ordering <ul><li>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. </li></ul><ul><li>Proof : </li></ul><ul><ul><li>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)) </li></ul></ul><ul><ul><li>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. </li></ul></ul>
- 13. <ul><li>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. </li></ul><ul><li>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 ~. </li></ul><ul><li>Denote by F the family of linear orderings ~ defined on subsets D of X and satisfying the following: </li></ul><ul><ul><li>If E as subset of D is a segment, then the first element of (D – E) is f (X – E). (*) </li></ul></ul><ul><li>Such a family is not empty since the ordering of the natural numbers on the domain D = {x n } is in F . </li></ul>
- 14. <ul><li>Lemma 1. Every element of F is a well-ordering on its domain. </li></ul><ul><li>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. </li></ul>
- 15. <ul><li>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 </li></ul><ul><li> D′ 0 = D 0 union { f (X - D 0 )} </li></ul><ul><li>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. □ </li></ul>
- 16. Consequences of the Axiom of Choice <ul><li>Set theory </li></ul><ul><li>Algebra </li></ul><ul><li>General topology </li></ul>
- 17. The Banach-Tarski Paradox <ul><li>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. </li></ul><ul><li>S= T= </li></ul><ul><li> = and </li></ul>

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