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Unformatted text preview: Math 245A Homework 1 Brett Hemenway October 19, 2005 3. a. Let M be an infinite algebra on a set X Let B x be the intersection of all E M with x E . Suppose y B x , then B y B x because B x is a set in M which contains y . But B x B y because if y E for any E M , then x E , otherwise y 6 B x = B x E c . So B x = B y . This means that for all x, y , either B x = B y or B x B y = . The collection B x x X must be infinite since by hypothesis M was infinite, and every element E of M either contains B x or is disjoint from it for each x . So the B x form an infinite collection of disjoint sets in M . b. If there are uncountably many different B x M we are done. Otherwise there are countably many, i.e. we have a sequence { B x n } n =1 where B x i B x j = when i 6 = j . The set { , 1 } = { , 1 }{ , 1 } , . . . is uncountable, and we have an injection { , 1 } , M { a i } i =1 7 [ i =1 B x i if a i = 1 if a i = 0 So M is uncountable. 4. Let A an algebra, suppose A is closed under countable increasing unions. then we wish to show that A is closed under arbitrary count able unions. Let { E n } n =1 be a sequence of sets in A . Let F n = n m =1 E n . Then { F n } n =1 is an increasing sequence of functions and N n =1 E n = N n =1 F n . By our hypothesis, n =1 F n A , so n =1 E n A . The converse, that if A is a algebra it is closed under countable increasing unions, follows from the definition of a algebra....
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.
 Spring '10
 tao/analysis
 Algebra

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