14 1.2 Functions We now prove that Sα ∈I f (Aα)⊂f (Sα ∈I Aα). Let y∈Sα ∈I f (Aα). From the definition of union of a family of sets, there is α0 ∈I such that y∈ f (Aα0). From the definition of image of a set, there is x ∈Aα0 such that y =f (x). We have x ∈Aα0 ⊂Sα ∈I Aα. Therefore, y =f (x) ∈ f (Sα ∈I Aα). By Theorem 1.1.1 the result follows. □ Definition 1.2.5 Let f : X →Y and g: Y →Z be two functions. Then the composition function g◦ f of f andgis the functiong◦ f : X →Z given by (g◦ f )(x)=g(f (x)) for all x ∈X. Theorem 1.2.5 Let f : X →Y andg: Y →Z be two functions and let B⊂Z. The following hold: (i) (g◦ f )−1(B)=f −1(g−1(B)). (ii) If f andg are injective, theng◦ f is injective. (iii) If f andg are surjective, theng◦ f is surjective. (iv) If g◦ f is injective, then f is injective. (v) If g◦ f is surjective, theng is surjective. Proof: We prove (iv) and leave the other parts as an exercise. Suppose g◦ f is injective and let x1,x2 ∈X be such that f (x1)= f (x2). Then (g◦ f )(x1)= g(f (x1))=g(f (x2))=(g◦ f )(x2). Since g◦ f is injective, it follows that x1 =x2. We conclude that f is injective. □ Definition 1.2.6 Asequence of elements of a set Ais a function with domainNand codomainA. We discuss sequences in detail in Chapter 2. Definition 1.2.7 We say that set Ais finite if it is empty or if there exists a natural number nand a one-to-one correspondence f : A→{1,2,...,n}. Aset is infinite if it is not finite. We leave it as an exercise to prove that the union of two finite sets is finite. It is also easy to show, by contradiction, that Nis infinite. Exercises 1.2.1 ▶Let f : X →Y be a function. Prove that: (a) If f is one-to-one, thenA=f −1(f (A)) for every subset Aof X. (b) If f is onto, then f (f −1(B))=Bfor every subset Bof Y. 1.2.2 Let f : R→Rbe givenby f (x)=x2−3 and let A=[−2,1) andB=(−1,6). Find f (A) and f −1(B). 1.2.3 Prove that each of the following functions is bijective. (a) f : (−∞,3] →[−2,∞) given by f (x)=|x−3|−2. (b) g: (1,2) →(3,∞) given byg(x)=3/(x−1). 1.2.4 Prove that if f : X →Y is injective, then the following hold: (a) f (A∩B)=f (A)∩f (B) for A,B⊂X.
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