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 fix any a∈ f −1(f (A)). Then f (a) ∈ f (A), so there exists a′ ∈Asuch that f (a) = f (a′). Since f is one-to-one, a=a′ ∈A. Therefore, f −1(f (A)) ⊂Aand 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) ∈Band, hence, f (f −1(B)) ⊂B. This inclusion does not require the surjectivity of f . Now fixb∈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 Without the injectivity of f , the equality in part (a) is no longer valid.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, andA= [−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 Bfor which the equality in part (b) does not hold true.
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