Solutions and Hints for Selected Exercises SECTION 1.1 Exercise 1.1.2. Applying basic rules of operations on sets yields (X \Y) ∩ Z = Yc ∩ 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) Let A ⊂ X. 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 fx any a ∈ f −1( f (A)). Then f (a) ∈ f (A), so there exists a ′ ∈ A such that f (a)= f (a ′ ). Since f is one-to-one, a = a ′ ∈ 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 fx 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 Without the injectivity of f , the equality in part (a) is no longer valid.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)= x2, 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 fnd 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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