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# Class Note for MATH 300 at UMass(5)

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This 2 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Massachusetts taught by a professor in Fall. Since its upload, it has received 15 views.

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Date Created: 02/06/15

Subsets of nite sets are nite1 November 28 2007 0556pm In order to prove that N is in nite we shall need Lemma 4112 restated as Lemma 1 below The text already gives a proof of this lemma but the proof below is perhaps a bit simpler Lemma 1 For every positive integer n 12nn1MILum Proof We use induction on 71 Base step 71 1 Obviously 12 i 1 Why not Inductive step Let n E N and assume 12n 1 i 12n Just suppose 1 2n 2 m 12n 1 Then there is some bijection h12n1n212nn1 The idea now is to modify h so as to make 71 2 H n 1 and then restrict the modi ed h to a bijection 12n 1 m 12n which will give a contradiction To do that rst swap the elements 71 1 and Mn 2 of 1 2 nn 1 In other words de ne the function a 12nn 1 a 12nn 1 by an1 hn2 ahn2 n1 0jjifj hn2 andj n1 Then a is a permutation of 1 2 nn 1 that is a bijection from the set 1 2 nn 1 to itself Now de ne h a o h so that h 12n 1n 2 a 12nn1 Then 11 is also bijective why and h n2n1 Hence we may restrict the domain of h to 1 2 n 1 and the codomain of h to 12n to get abijection f 12n 1 m 12n The existence of such an f contradicts the induction hypothesis D 1Copyright 2007 Murray Eisenberg All Rights Reserved In our work with nite sets we shall concentrate only upon whether sets are nite and not upon how many elements they have when they are nite Then we shall not need the more general result of Corollary 413 that 1 2 m i 1 2 n whenever m i 71 Accordingly we give below a slightly simpler proof than the one in the teXt that a subset of a nite set is nite Lemma 2 IfA is a nite set and iflJ e A then A U la is nite Proof AssumeA is nite and b e A IfA Q thenA U 10 b m 1 so that A U la is nite Now assumeA i Q Then there is some positive integernand some bijection f A a 12n Extendf to afunctionF A U la a 12nn 1 by de nining Rx fx 1fxeA n1ifxb Then F is a bijection evidently F is surjective and F is injective because f is injective b e A andn 1 e 1 2 n Hence A U la is nite D Proposition 3 Every subsetB ofa finite setA is finite Proof Since the only subset of the empty set is empty the result is trivial when A Q IfA is a nonempty nite set and B is a subset of A there is some positive integer n and some bijection f A a 1 2 n in this case fB is a subset of 1 2 n and B will be nite if and only if fB is nite Hence it suf ces to prove that for every positive integer 71 every subset of 1 2 n is nite We use induction on 71 Base step 71 1 The only subsets of 1 are Q and 1 itself and both are nite Inductive step Let n be a positive integer and assume that every subset of 12n is nite Let B C 12nn 1 It remains to deduce that B is nite Ifn 1 e B then actually B C 1 2 n so that B is nite by the inductive assumption Assume now thatn 16 B De ne D B 71 1 ThenD C 12n By the inductive assumption D is nite Since B D U n 1 and n 1 e D from the Lemma it follows that B is nite too D

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