# a1sol - MATH 242 Assignment 1 Solutions 1 TRUE We have y...

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Unformatted text preview: MATH 242 Assignment 1 Solutions 1. TRUE. We have y ∈ f ( A ∩ f- 1 ( B )) ⇔ ∃ x ∈ A ∩ f- 1 ( B ) such that f ( x ) = y ⇔ ∃ x ∈ A such that f ( x ) ∈ B and f ( x ) = y ⇔ ( ∃ x ∈ A such that f ( x ) = y ) and y ∈ B ⇔ y ∈ f ( A ) and y ∈ B ⇔ y ∈ f ( A ) ∩ B 2. FALSE. You can start out by arguing as in question 1. y ∈ f ( A ∪ f- 1 ( B )) ⇔ ∃ x ∈ A ∪ f- 1 ( B ) such that f ( x ) = y ⇔ ∃ x ∈ A such that f ( x ) ∈ B or f ( x ) = y ⇔ ( ∃ x ∈ A such that f ( x ) = y ) or ( ∃ x ∈ f- 1 ( B ) such that f ( x ) = y ) ⇔ y ∈ f ( A ) or y ∈ f ( f- 1 ( B )) The argument shows that f ( A ∪ f- 1 ( B )) = f ( A ) ∪ f ( f- 1 ( B )) . It is clear that f ( f- 1 ( B )) ⊆ B , but f- 1 ( B ) can be empty when B isn’t. For a specific example take X = Y = { 1 , 2 } , f (1) = 1 , f (2) = 1 , A = ∅ , B = { 2 } . Then f- 1 ( B ) and A ∪ f- 1 ( B ) are both empty, and hence, so is f ( A ∪ f- 1 ( B )) , but f ( A ) ∪ B = B = { 2 } . So f ( A ∪ f- 1 ( B )) 6 = f ( A ) ∪ B in this case. 3. TRUE. Probably best to show containment in both directions. We will show that ( g ◦ f )( T ) ⊆ g ( f ( T )) . Let z ∈ ( g ◦ f )( T ) , then there exists x ∈ T such that z = ( g ◦ f )( x ) = g ( f ( x )) . But f ( x ) ∈ f ( T ) and z = g ( f ( x )) ∈ g ( f ( T )) . To show that ( g ◦ f )( T ) ⊇ g ( f ( T )) , let z ∈ g ( f ( T )) . Then there exists y ∈ f ( T ) such that z = g ( y ) . Further, there exists x ∈ T such that y = f ( x ) . Hence z = g ( f ( x )) = ( g ◦ f )( x ) and x ∈ T so that z ∈ ( g ◦ f )( T ) ....
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a1sol - MATH 242 Assignment 1 Solutions 1 TRUE We have y...

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