1873_solutions

B n then the set a b is finite and card a b m n 54 the

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Unformatted text preview: e set A u is disjoint from AB u , Exercise 3 guarantees that card A B  card A B u ÞA u  mn  m  m n  1 6. Given that S is a finite set and card S  n, prove that the power set p S of S is a finite set and that card p S  2 n . Solution: We use mathematical induction. For each nonnegative integer n we take p n to be the assertion that whenever S is a finite set satisfying card S  n then p S is a finite set and card p S  2 n . If S  then p S  and so card p S  1  2 0  2 card S . Therefore the assertion p 0 is true. To see that the assertion p 1 is true, suppose that card S  1. We write S in the form S  u . Since pS  , u we see that card p S  2  2 1  2 card S . Now suppose that n is any positive integer for which the assertion p n is true and suppose that card S  n  1. Choose a member u of the set S. Since card S u  n we know that card p S We now define 55 u  2n. W AÞ u A pS u and we observe that W ß p S u . (To see this, define f A  A Þ u for every A W is disjoint from p S u and card W  n we deduce from Exercise 3 that card p S  card p S u Þ W  2 n  2 n  2 n 1 . pS u .) Since 7. Prove that if S is a nonempty finite set of real numbers then S has a largest member. We use mathematical induction. For each positive integer n we take p n to be the assertion that whenever S is a finite set satisfying card S  n then S has a largest member. The assertion p 1 is obvious. Now suppose that n is any positive integer for which the assertion p n is true and suppose that S is a finite set satisfying card S  n  1. We choose a member u of the set S. Since card S u  n we know that the set S u has a largest member that we shall call v. In the event that u v we see that v is the largest member of S and in the event that v  u we see that u is the largest member of S. In either event the set S must have a largest ,member. 8. Prove that none of the sets the set Z  , Z, Q, and R are finite. All of these sets are nonempty and none of them has a largest member. 9. Prove that a subset S of Z  is finite if and only if it is possible to find an integer n such that the inequality m n holds for every member m of S. We already know that if n is any positive integer then every subset of the set 1, 2, , n is finite. On the other hand, if S is a finite subset of Z  then either S is empty, in which case m 1 for every m S, or S has a largest member n, in which case m n for every m S. 10. Prove that if m and n are positive integers then the number of possible functions from the set 1, , n into the set 1, , m is m n . Solution: We want to know that if is the set of all functions from 1, 2, , n into 1, 2, , m then the set is finite and card  m n . We suppose that m is a positive integer and for each positive integer n we define n to be the set of all functions from 1, 2, , n into 1, 2, , m . For each positive integer n we take p n to be the assertion that card n  m n . 1, 2, , m to be the To see that the assertion p 1 is true we observe that...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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