120 Solutions to Exercises has a smallest element, say y, by the induction hypothesis. If y > x, then x is the smallest element in 8, while, if x > y, then y is the smallest element in 8. In either case, 8 has a smallest element, as required. By the Principle of Mathematical Induction, the result is true for all n ;:::: 1. (b) The real numbers is an infinite set without a smallest element: For any r E R, r -1 is always smaller than r. On the other hand, the natural numbers does have a smallest element. (c) The Well-Ordering Principle! 18. [BB] For n = 1, n~=1 Bi = Blo and so A U (n~=1 Bi) = AU BI = n~=1 (A UBi) and the result holds for n = 1. Now assume that A U (n~=1 = n~=1 for k ;:::: 1. Then, given a set A and k + 1 sets Bl, B2, ... , Bk+1. we have A U (n~~l = AU [( n~=1 n Bk+I] = [A U (n~=1 ] n (A U Bk+1 ) since A U (B n C) = U B) n U C) = [n~=1 (A UBi)] n U Bk+I) (by the induction hypothesis) = n~~11 ( A U ) , giving the result for k + 1. By the Principle of Mathematical Induction, we conclude that A U ( n~=1 = n~=1 ( AU Bi ), for all n ;:::: 1. 19. For n = U~=I = BI
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.