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Unformatted text preview: Honors Introduction to Analysis I Homework II Solution February 9, 2009 Problem 1 If A , X , Y are sets then A X denotes the set of all functions from X to A and X × Y denotes the Cartesian product of X and Y . Verbally describe the sets ( A X ) Y and A X × Y and construct a bijection between them. It follows that these sets have the same cardinality, in analogy with the basic identity ( a x ) y = a xy for ordinary exponentials. Solution. ( A X ) Y = g : Y → A X = { g : Y → { f : X → A }} It is the set of all functions g that assign to each element of Y a function f : X → A . So for a fixed y , g ( y ) is a function of x , so we can write g ( y )( x ), which would translate as for a fixed y evaluate g ( y ) at x . A X × Y = { h : X × Y → A } It is the set of all functions h that assign to each couple ( x,y ) ∈ X × Y an element in A . Define the following bijection F : ( A X ) Y → A X × Y by F ( g ) = h , where h is defined as h ( x,y ) = g ( y )( x ). This translates as: for a given (fixed) pair ( x,y ) ∈ A X × Y evaluate g ( y ) at the point x . • (injectivity) Suppose F ( g 1 ) = F ( g 2 ). Then this means that for each pair ( x,y ) (in particular for each x ): g 1 ( y )( x ) = g 2 ( y )( x ), i.e. g 1 ( y ) = g 2 ( y ) , ∀ y , hence g 1 = g 2 , so F is injective. • (surjectivity) Given h ∈ A X × Y , define g ∈ ( A X ) Y in the following way: for each (fixed) y , h ( y,x ) becomes a function in one variable, x . Let this function be g ( y ) (so g evaluated at the fixed y ). Doing this for each y ∈ Y , we define a function g in the whole domain Y . By our construction we see that F ( g ) = h , so F is surjective. Hence F is bijective, so the two sets have the same cardinality. Problem 2 Prove that addition and multiplication of fractions are well defined : if one of the fractions is replaced by an equivalent fraction (representing the same rational number), the values of the sum and the product do not change....
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This note was uploaded on 09/22/2010 for the course MATH 413 at Cornell University (Engineering School).
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