13 ■ Example 1.2.2 Let f : R → R be given by f (x)= 3x− 1. Let A =[0,2) and B = {1,−4,5}. Then f (A)=[−1,5) and f −1(B)= {2 3 ,−1,2}. ■ Example 1.2.3 Let f : R → R be 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 → R be 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 A be a subset of X, and let B be 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 defnition of image, f (x) ∈ f (A). Now, by the defnition 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 frst f −1(D\C) ⊂ f −1(D) \ f −1(C). Let x ∈ f −1(D \C). Then, from the defnition of inverse image, we get f (x) ∈ D\C. Thus, f (x) ∈ D and f (x) ̸ ∈ C. Hence x ∈ f −1(D) and x ̸ ∈ 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 ̸ ∈ f −1(C). Therefore, f (x) ∈ D and f (x) ̸ ∈ C. This means f (x) ∈ D\C and, 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: S S (i) f ( α ∈I Aα )= α ∈I f (Aα ). T T (ii) f ( α ∈I Aα ) ⊂ α ∈I f (Aα ). (iii) f −1( S β ∈J Bβ )= S β ∈J f −1(Bβ ). (iv) f −1( T β ∈J Bβ )= T β ∈J f −1(Bβ ). Proof: We prove (i) and leave the other parts as an exercise. S S S First we show f ( α ∈I Aα ) ⊂ α ∈I f (Aα ). Let y ∈ f ( α ∈I Aα ). From the defnition of image of S a set, there is x ∈ α ∈I Aα such that y = f (x). From the defnition of union of a family of sets, there S is α0 ∈ I such that x ∈ Aα0. Therefore, y = f (x) ∈ f (Aα0 ) and, so, y ∈ α ∈I f (Aα ).
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