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Unformatted text preview: Math 350 Advanced Calculus I Spring 2008 Homework Solutions 2 12: Textbook § 8 1. (8.3) Show that each of the following pairs of sets S and T are equinumerous by finding a specific bijection between them. a S = [0 , 1] and T = [1 , 3] Solution We need a function that maps the interval S = [0 , 1] onto the interval T = [1 , 3]. Define f : S → T by f ( x ) = 2 x + 1 for each x ∈ S = [0 , 1]. We need to show that f is both injective and surjective. Injective. Suppose x 1 and x 2 are arbitrary elements in S = [0 , 1]. Suppose f ( x 1 ) = f ( x 2 ) . We need to show x 1 = x 2 . Since f ( x 1 ) = f ( x 2 ), then 2 x 1 + 1 = 2 x 2 + 1 which implies x 1 = x 2 , as required. Therefore f is injective. Surjective. Suppose y is an arbitrary element in T = [1 , 3]. We need to show there exists an element x ∈ S = [0 , 1] such that f ( x ) = y . From y = 2 x + 1 we solve for x to obtain x = y 1 2 . Let x = y 1 2 . Since y ∈ [1 , 3], then 1 ≤ y ≤ 3 which implies 0 ≤ x ≤ 1 so x ∈ [0 , 1]. Also, f ( x ) = 2 x + 1 = 2[ y 1 2 ] + 1 = y, as required. Therefore f is surjective. Since f is a bijection from S onto T , then S = [0 , 1] and T = [1 , 3] are equinumerous. d S = (0 , 1) and T = (0 , ∞ ) Solution We need a function that maps the interval S = (0 , 1) onto the interval T = (0 , ∞ ). We can choose a function that has a vertical asymptote at x = 0 or at x = 1. Define f : S → T by f ( x ) = 1 x 1 for each x ∈ S = (0 , 1). Injective. Suppose x 1 and x 2 are arbitrary elements in S = (0 , 1). Suppose f ( x 1 ) = f ( x 2 ) . We need to show x 1 = x 2 . Since f ( x 1 ) = f ( x 2 ), then 1 x 1 1 = 1 x 2 1 1 x 1 = 1 x 2 x 1 = x 2 So x 1 = x 2 , as required. Therefore f is injective. 1 Surjective. Suppose y is an arbitrary element of T = (0 , ∞ ). We need to show there exists an element x ∈ S = (0 , 1) such that f ( x ) = y . From y = 1 x 1 we solve for x to obtain x = 1 y 1 . Let x = 1 y 1 . Since y ∈ (0 , ∞ ), then y > 0 so y + 1 > 1 which implies 0 < 1 y +1 < 1. Therefore x = 1 y +1 ∈ S = (0 , 1). Also, f ( x ) = f 1 y + 1 ¶ = ( y + 1) 1 = y, as required. Therefore f is surjective. Since f is a bijection from S onto T , then S = (0 , 1) and T = (0 , ∞ ) are equinumerous. 2. (8.10) Prove: For all sets S , if S is denumerable, then S is equinumerous with a proper subset of itself. Solution Since S is denumerable, then S is equinumerous with N so we can list the elements in S S = { s 1 ,s 2 ,s 3 ,... } , where all elements in the listing are distinct. Let T be the set of all elements of S with an even subscript in this listing T = { s 2 ,s 4 ,s 6 ,... } ....
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This homework help was uploaded on 04/16/2008 for the course MATH 350 taught by Professor Qian during the Spring '08 term at CSU Fullerton.
 Spring '08
 QIAN
 Calculus, Sets

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