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Unformatted text preview: 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)...
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This note was uploaded on 10/04/2010 for the course ECON 204 taught by Professor Anderson during the Summer '08 term at University of California, Berkeley.
 Summer '08
 ANDERSON

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