Introduction to Mathematical Analysis I - Second Edition

13 (b) f ( f − 1 ( B )) ⊂ B . Proof: We prove (a) and leave (b) as an exercise. (a) Let x ∈ A . By the definition of image, f ( x ) ∈ f ( A ) . Now, by the definition of preimage, x ∈ f − 1 ( f ( A )) . Theorem 1.2.4 Let f : X → Y be a function, let A , B ⊂ X , and let C , D ⊂ Y . The following hold: (a) If C ⊂ D , then f − 1 ( C ) ⊂ f − 1 ( D ) ; (b) f − 1 ( D \ C ) = f − 1 ( D ) \ f − 1 ( C ) ; (c) If A ⊂ B , then f ( A ) ⊂ f ( B ) ; (d) f ( A \ B ) ⊃ f ( A ) \ f ( B ) . Proof: We prove (b) and leave the other parts as an exercise. (b) We prove first that f − 1 ( D \ C ) ⊂ f − 1 ( D ) \ f − 1 ( C ) . Let x ∈ f − 1 ( D \ C ) . Then, from the definition of inverse image, we get f ( x ) ∈ D \ C . Thus, f ( x ) ∈ D and f ( x ) 6∈ C . Hence x ∈ f − 1 ( D ) and x 6∈ f − 1 ( C ) . We conclude that x ∈ f − 1 ( D ) \ f − 1 ( C ) . Next we prove f − 1 ( D ) \ f − 1 ( C ) ⊂ f − 1 ( D \ C ) . Let x ∈ f − 1 ( D ) \ f − 1 ( C ) . Thus, x ∈ f − 1 ( D ) and x 6∈ f − 1 ( C ) . Therefore, f ( x ) ∈ D and f ( x ) 6∈ C . This means f ( x ) ∈ D \ C and, so, x ∈ f − 1 ( D \ C ) . Theorem 1.2.5 Let f : X → Y be a function, let { A α } α ∈ I be an indexed family of subsets of X , and let { B β } β ∈ J be an indexed family of subsets of Y . The following hold: (a) f ( S α ∈ I A α ) = S α ∈ I f ( A α ) ; (b) f ( T α ∈ I A α ) ⊂ T α ∈ I f ( A α ) ; (c) f − 1 ( S β ∈ J B β ) = S β ∈ J f − 1 ( B β ) ; (d) f − 1 ( T β ∈ J B β ) = T β ∈ J f − 1 ( B β ) . Proof: We prove (a) and leave the other parts as an exercise. (a) Let y ∈ f ( S α ∈ I A α ) . From the definition of image of a set, there is x ∈ S α ∈ I A α such that y = f ( x ) . From the definition of union of a family of sets, there is α 0 ∈ I such that x ∈ A α 0 . Therefore, y = f ( x ) ∈ f ( A α 0 ) and, so, y ∈ S α ∈ I f ( A α ) . . Given functions f : X → Y and g : Y → Z , we define the composition function g ◦ f of f and g as the function g ◦ f : X → Z given by ( g ◦ f )( x ) = g ( f ( x )) for all x ∈ X . Theorem 1.2.6 Let f : X → Y and g : Y → Z be two functions and let B ⊂ Z . The following hold: (a) ( g ◦ f ) − 1 ( B ) = f − 1 ( g − 1 ( B )) ; (b) If f and g are injective, then g ◦ f is injective; (c) If f and g are surjective, then g ◦ f is surjective; (d) If g ◦ f is injective, then f is injective; (e) If g ◦ f is surjective, then g is surjective.