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204PS1Solution2009

# 204PS1Solution2009 - Econ 204 Summer 2009 Problem Set 1...

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Econ 204 Summer 2009 Problem Set 1 Solutions 1. Cardinality For each pair of set A and set B, show that A and B are numerically equivalent. (Hint: Show that there exists a bijection f : A ! B; i.e. f is one to one and onto.) (a) A = ( ° 1 ; 1) B = ( °1 ; + 1 ) (b) A = [0 ; 1] B = (0 ; 1) (c) A is an in°nite uncountable set, B = A [ C where C is an in°nite countable set. Solution: (a) f ( x ) = tan ° 2 x , x 2 ( ° 1 ; 1) (b) f ( x ) = 8 > < > : 1 2 if x = 0 1 n +2 if x = 1 n ; n = 1 ; 2 ; :::; x otherwise , x 2 [0 ; 1] (c) Since A is an in°nite set, we can obtain an in°ite sequence f a 1 ; a 2 ; : : : g from A: Let A 1 = f a 1 ; a 2 ; : : : g : A 1 ± A and A n A 1 6 = ; as A is uncountabl• e. There are three cases: Case 1: C \ A 1 = ; Since C is an in°nite countable set, let C = f c 1 ; c 2 ; : : : g . f ( x ) = 8 > < > : a i if x = a 2 i ; i = 1 ; 2 ; :::; c i if x = a 2 i ° 1 ; i = 1 ; 2 ; :::; x if x 2 A n A 1 , x 2 A Case 2: C \ A 1 6 = ; and C n A 1 is a °nite set. Let C n A 1 = f k 1 ; k 2 ; : : : ; k m g where m is a natural number. f ( x ) = 8 > < > : a i if x = a i + m ; i = 1 ; 2 ; :::; k i if x = a i ; i = 1 ; 2 ; :::; m x if x 2 A n A 1 , x 2 A Case 3: C \ A 1 6 = ; and C n A 1 is a in°nite countable set. Let C n A 1 = f s 1 ; s 2 ; : : : g f ( x ) = 8 > < > : a i if x = a 2 i ; i = 1 ; 2 ; ::: s i if x = a 2 i ° 1 ; i = 1 ; 2 ; ::: x if x 2 A n A 1 , x 2 A 2. Induction Using mathematical induction, show the following: n = 1 ; 2 ; 3 ; : : : (a) P n i =1 k ° i = 1 ° 1 k n k ° 1 , k 6 = 1 : (b) P 1 i = n ( k ° 1) k ° i = k 1 ° n , k > 1 : (c) P n i =1 1 p i ² p n Solution: (a) For n = 1 , 1 k = 1 ° 1 k k ° 1 .

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