Introduction to Mathematical Analysis I - Second Edition

5. Solutions and Hints for Selected Exercises SECTION 1.1 Exercise 1.1.2 . Applying basic rules of operations on sets yields ( X \ Y ) ∩ Z = Y c ∩ Z = Z \ Y . and Z \ ( Y ∩ Z ) = ( Z \ Y ) ∪ ( Z \ Z ) = ( Z \ Y ) ∪ /0 = Z \ Y . Therefore, ( X \ Y ) ∩ Z = Z \ ( Y ∩ Z ) . SECTION 1.2 Exercise 1.2.1 . (a) For any a ∈ A , we have f ( a ) ∈ f ( A ) and, so, a ∈ f − 1 ( f ( A )) . This implies A ⊂ f − 1 ( f ( A )) . Note that this inclusion does not require the injectivity of f . Now fix any a ∈ f − 1 ( f ( A )) . Then f ( a ) ∈ f ( A ) , so there exists a 0 ∈ A such that f ( a ) = f ( a 0 ) . Since f is one-to-one, a = a 0 ∈ A . Therefore, f − 1 ( f ( A )) ⊂ A and the equality holds. (b) Fix any b ∈ f ( f − 1 ( B )) . Then b = f ( x ) for some x ∈ f − 1 ( B ) . Thus, b = f ( x ) ∈ B and, hence, f ( f − 1 ( B )) ⊂ B . This inclusion does not require the surjectivity of f . Now fix b ∈ B . Since f is onto, there exists x ∈ X such that f ( x ) = b ∈ B . Thus, x ∈ f − 1 ( B ) and, hence, b ∈ f ( f − 1 ( B )) . We have shown that B ⊂ f ( f − 1 ( B )) and the equality holds. Without the injectivity of f , the equality in part (a) is no longer valid. Consider f ( x ) = x 2 , x ∈ R , and A = [ − 1 , 2 ] . Then f ( A ) = [ 0 , 4 ] and, hence, f − 1 ( f ( A )) = [ − 2 , 2 ] , which strictly contains A . It is also not hard to find an example of a function f and a set B for which the equality in part (b) does not hold true.

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