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Unformatted text preview: HOMEWORK 1  SOLUTIONS 1. By induction on n : Base Case ( n = 1): f n = f 1 = f which is surjective by assumption. Inductive Step : Suppose f n is surjective for all 1 n k . We want to show that f k +1 is surjective. Let x X be arbitrary. We want to find y X such that f k +1 ( y ) = x . Well f k +1 ( y ) = f ( f k ( y )) and we know f is surjective, so first pick some z X such that f ( z ) = x . Next, by our inductive hypothesis, we know that f k is surjective so pick some y such that f k ( y ) = z . This choice of y does it, since: f k +1 ( y ) = f ( f k ( y )) = f ( z ) = x 2. (GRADED  EASY). By induction on  X  : Base Case ( n = 0): If  X  = 0, then its the empty set and has only one subset, namely itself, the empty set. And indeed: P ( X )  = {} = 1 = 2 = 2  X  Inductive Step : Assume that for all n k and every set X such that  X  = n , P ( X )  = 2  X  . Now suppose X is a set with  X  = k + 1. Lets write X = { x 1 ,...,x k +1 } . Now every subset of X either contains x k +1 or it doesnt, and its easy to see that there are an equal number of subsets of X which contain x k +1 as there are which dont. Thus the number of subsets of X is double the number of subsets of { x 1 ,...,x k } . This latter set has cardinality k , so we can apply the inductive hypothesis to obtain:...
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This note was uploaded on 03/28/2012 for the course MATH 110 taught by Professor Gurevitch during the Fall '08 term at University of California, Berkeley.
 Fall '08
 GUREVITCH
 Math

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