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Unformatted text preview: MATH 109 Assignment 5 Due: 5/2/11 #10.1 Proof. We are given that X = { x 1 ,...,x n } is a finite set of cardinality n . Let Y be a set, and first suppose there is a bijection f : X → Y . Since f is injective, we have f ( x 1 ) ,...,f ( x n ) are all distinct. Hence  Y  ≥ n . On the other hand, since f is surjective, we have Y ⊂ { f ( x 1 ) ,...,f ( x n ) } ; hence  Y  ≤ n . This implies that  Y  = n . Now suppose that Y = { y 1 ,...,y n } has cardinality n . It is easy to see that the function f : X → Y defined by f ( x i ) = y i is a bijection. Hence, we have shown that if X is a finite set of cardinality n , then a set Y has  Y  = n if and only if there exists a bijection X → Y . #11.5(ii) Proof. We are given that X and Y are finite sets with X ⊆ Y . First suppose that  X  <  Y  . We must have some y ∈ Y is not contained in X (otherwise  X  ≥  Y  ) so X 6 = Y ; this implies that X ⊂ Y ....
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 Spring '06
 Knutson
 Math, Logic, Natural number, base case, induction hypothesis, Xn

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