axiom of choice final version
DESCRIPTION
Axiom of Choice Final Version Math 101 Fall 2008TRANSCRIPT
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Axiom of Choice
by Catherine Janes
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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)
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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)
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History
1924, S. Banach and A. Tarski
1939, Kurt Gödel
Early 1960s, Paul Cohen
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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?
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Shoes and socks
Shoes are well-ordered!
Socks are not well-ordered
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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)
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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. □
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Consequences of the Axiom of Choice
Set theory
Algebra
General topology
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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