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Unformatted text preview: ECE 3250 HOMEWORK I SOLUTIONS Fall 2009 1. Most of the proofs I’ve seen involve drawing pictures and defining bijective mappings between various subsets of A and various subsets of B (where A and B are the two sets you’re trying to define a bijection between). It’s kind of a cool theorem. 2. Let κ : P o ( A ) → A be a choice function. For every a ∈ A , we must have κ ( { a } ) = a since a is the only element of { a } . We also must have κ ( A ) = a o for some a o ∈ A . Thus κ ( { a o } ) = κ ( A ) = a o , so κ is not injective unless A = { a o } , i.e. unless A contains only one element. 3. (a) Suppose f and g are both injective. Given a 1 and a 2 are different elements of A , we need to show that h ( a 1 ) 6 = h ( a 2 ) Let b 1 = f ( a 1 ) and b 2 = f ( a 2 ). Since f is injective, b 1 6 = b 2 . Since g is injective, g ( b 1 ) 6 = g ( b 2 ). But g ( b 1 ) = g ( f ( a 1 )) = h ( a 1 ) and g ( b 2 ) = g ( f ( a 2 )) = h ( a 2 ), so h ( a 1 ) 6 = h ( a 2 ). (b) Suppose f and g are both surjective. We need to show that for every c ∈ C there’s some a ∈ A such that h ( a ) = c . Suppose c ∈ C . Since g is surjective, there’s some b ∈ B such that g ( b ) = c . Since f is surjective, there’s some a ∈ A such that f ( a ) = b . But then h ( a ) = g ( f ( a )) = g ( b ) = c . Suppose both f and g are bijective. Then both f and g are both injective and surjective. By (a), h is then injective. By the argument in the preceding paragraph, h is also surjective. Hence h is bijective....
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 '07
 DELCHAMPS
 Mathematical analysis, Limit of a sequence, Finite set, NK, absolutely summable sequence

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