13 ■ Example 1.2.2 Let f : R→Rbe given by f (x) =3x−1. Let A= [0,2) andB={1,−4,5}. Then f (A) = [−1,5) and f −1(B) ={2 3,−1,2}. ■ Example 1.2.3 Let f : R→Rbe given by f (x) =−x+7. Let A= [0,2) and B= (−∞,3]. Then f (A) = (5,7] and f −1(B) = [4,∞). ■ Example 1.2.4 Let f : R→Rbe given by f (x) =x 2. Let A= (−1,2) and B= [1,4). Then f (A) = [0,4) and f −1(B) = (−2,−1] ∪[1,2). Theorem 1.2.2 Let f : X →Y be a function, let Abe a subset of X, and let Bbe a subset of Y. The following hold: (i) A⊂f −1(f (A)). (ii) f (f −1(B)) ⊂B. Proof: We prove (i) and leave (ii) as an exercise. Let x ∈A. By the definition of image, f (x) ∈ f (A). Now, by the definition of preimage, x ∈ f −1(f (A)). □ Theorem 1.2.3 Let f : X →Y be a function, let A,B⊂X, and let C,D⊂Y. The following hold: (i) If C⊂D, then f −1(C) ⊂f −1(D). (ii) f −1(D\C) =f −1(D)\ f −1(C). (iii) If A⊂B, then f (A) ⊂f (B). (iv) f (A\B) ⊃f (A)\ f (B). Proof: We prove (ii) and leave the other parts as an exercise. We show first 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)∈Dand f (x)̸∈C. Hencex∈f −1(D) andx̸∈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 x̸∈ f −1(C). Therefore, f (x) ∈Dand f (x)̸ ∈C. This means f (x) ∈D\Cand, so, x∈ f −1(D\C). □ Theorem 1.2.4 Let f : X →Y be a function, let {Aα}α∈I be an indexed family of subsets of X, and let {Bβ}β∈J be an indexed family of subsets of Y. The following hold: (i) f (Sα∈I Aα) =Sα∈I f (Aα). (ii) f (Tα∈I Aα) ⊂Tα∈I f (Aα). (iii) f −1(Sβ∈J Bβ) =Sβ∈J f − 1(Bβ). (iv) f −1(Tβ∈J Bβ) =Tβ∈J f − 1(Bβ). Proof: We prove (i) and leave the other parts as an exercise. First we show f (Sα∈I Aα) ⊂Sα∈I f (Aα). Let y ∈ f (Sα∈I Aα). From the definition of image of a set, there is x ∈Sα∈I Aα such that y =f (x). From the definition of union of a family of sets, there is α0 ∈I such that x ∈Aα0. Therefore, y =f (x) ∈ f (Aα0) and, so, y ∈Sα ∈I f (Aα).
RkJQdWJsaXNoZXIy NTc4NTAz