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74-hw6s

# 74-hw6s - MATH 74 HOMEWORK 6 SOLUTIONS 1(a By#5 from last...

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MATH 74 HOMEWORK 6 SOLUTIONS 1(a). By #5 from last week’s homework, A n has 2 n subsets. 1(b). Let S n denote the set of all subsets of A n with an even number of elements, and T n the set of all subsets of A n with an odd number of elements. We claim that #( T n ) = #( S n ). Do do this, we will define a function f : T n S n and prove it is a bijection. Define f by f ( X ) = X ∪ { 1 } 1 X X \ { 1 } 1 X Note that no matter what, as f ( X ) has either one more or one less element than X does, f really does send elements of T n to elements of S n . Let g : S n T n be given by the same rule, g ( Y ) = Y ∪ { 1 } 1 Y Y \ { 1 } 1 Y For the same reason, g really does send elements of S n to T n . Remark. This is not the only way to do this exercise. There are a ton of other bijections between S n and T n you could write down. [If the f and g I chose above seem mysterious, write out what it they are doing in the case n = 3 or 4 . The idea should be clearer then.] Alternatively you could prove the result by induction (to prove the inductive step, you’d need to establish some relationship between # S n +1 and # S n ). There are tons of ways to do this, I only chose one of them. The same remark applies to all of the other exercises. For any X T n , if 1 X we have g ( f ( X )) = g ( X \ { 1 } ) = ( X \ { 1 } ) ∪ { 1 } = X and if 1 X we have g ( f ( X )) = g ( X ∪ { 1 } ) = ( X ∪ { 1 } ) \ { 1 } = X so that g ( f ( X )) = X for all X T n . Similarly, for any Y S n f (

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