PMATH450_S2015.pdf

# Definition a partial ordered set s is said to be well

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Definition. A partial ordered set ( S, ) is said to be well-ordered if every nonempty subset of S has a smallest element, i.e. 8 T S, T 6 = ; ) 9 t 2 T such that 8 y 2 T : t y . Note. Every well-ordered set is totally ordered, as for each pair a, b , { a, b } has a least element, so if a is the least element a b and otherwise b a . Example. (1) Z is totally ordered, but not well-ordered, since Z has no smallest element. (2) N = { 0 , 1 , 2 , . . . } is well-ordered. (3) { r 2 Q : r 0 } is not well-ordered, since { r 2 Q : r > 0 } has no smallest element. (4) Every countable set can be well-ordered, i.e. we put an ordering on it that makes it well-ordered: If S = { x i } 1 i =1 , define x i x j if i j . Definition. Suppose ( S, ) is a poset. An upper bound for A X is an element x 2 S such that x y 8 y 2 A . Example. If S = P ( X ), ordered by inclusion, then X B 8 B 2 P ( X ) An element a 2 S is maximal if whenever b 2 S and b a , then b = a . Note. Maximal elements need not exist nor be unique. Example. Let A = { a, b, c } and define a b and a c . Then both b and c are maximal. A chain is a totally ordered subset of a partially ordered set. Example. If S = P ( X ), a chain may be ; ✓ B 1 B 2 · · · . Lemma (Zorn’s Lemma) . Suppose we have a partially ordered set in which each chain has an upper bound. Then the partially ordered set has a maximal element.

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2 INTRODUCTION TO LEBESGUE INTEGRATION 4 Theorem (Application of Zorn’s Lemma) . Every vector space has a basis (linearly independent spanning set). Proof. Let V be a vector space and let S be the set of linearly independent subsets of V . Partially order S by inclusion. Let C be a chain in S . Let Y = S C 2 C C . Certainly Y C 1 8 C 1 2 C . Now we check Y 2 S . Let y 1 , . . . , y n 2 Y , and suppose P n i =1 i y i = 0. y i 2 Y , so y i 2 C j i for some C j i in the chain C . One of C j 1 , . . . , C j n is the “biggest”, say C j k . Then all y i 2 C j k , a linearly independent set. This i = 0 for all i . Thus Y is a linearly independent set, so Y 2 S and Y is an upper bound for C . Lecture 2: May 6 Proof (Continued). Thus every chain has an upper bound, so by Zorn’s lemma, S has a maximal element A . Since A 2 S , it is linearly independent, so we must verify that span( A ) = V . Suppose not. Then there is some v 2 V \ span( A ). Then A [ { v } is linearly independent, so A [ { v } 2 S . But A [ { v } ) A and this violates the maximality of A . Definition. The well-ordering principle states that every set can be well ordered. Theorem. The following are equivalent: (1) Axiom of Choice (2) Zorn’s Lemma (3) Well-Ordering Principle Proof. (3) ) (1): Suppose every set can be well-ordered. Let { A λ : λ 2 } be a nonempty family of nonempty sets. Put A = S λ 2 A λ . Let ( A, ) be a well-ordering. A λ is a nonempty subset of A , and must have a smallest element, x λ 2 A λ . Define f : ! A by f ( λ ) = x λ . 2 Introduction to Lebesgue Integration Recall the Riemann integral: Let f : [ a, b ] ! R be bounded. Let a = x 0 < x 1 < · · · < x n = b be a partition of [ a, b ].
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