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Unformatted text preview: Cardinality Part III Original Notes adopted from January 22, 2002 (Week 16) c P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics Eg . S = { a,b,c } Sets of Subsets of S: {∅ , { a } , { b } , { c } , { a,b } , { a,c } , { b,c } , { a,b,c }} . Theorem. If S has n elements ( n natural number), then it has 2 n subsets. Proof: S = { a 1 ,a 2 ,a 3 ··· a n } 2 choices: a i is either in a set or not. 2 * 2 * 2 * 2 ··· ( n times) ⇒ 2 n subsets Theorem. If S any set and P ( S ) is the set of all subsets of S (the “power set” of S ), then  P ( S )  >  S  . Proof : Must show: 1) There exists a subset T of P ( S ) such that  T  =  S  (ie  P ( S )  >  S  ) 2)  P ( S )  6 =  S  . For 1), Let T = { all subsets of S with exactly 1 element } Pair T with S by { s } ↔ s . Therefore  T  =  S  . For 2) Suppose g : S → P ( S ). Suppose g 1:1. I’ll show that g can’t be onto all of P ( S )....
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This note was uploaded on 04/26/2011 for the course MATH 246 taught by Professor Applebaugh during the Spring '10 term at University of Toronto.
 Spring '10
 Applebaugh
 Math, Sets

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