14 1.2 Functions S S S We now prove that α ∈I f (Aα ) ⊂ f ( α ∈I Aα ). Let y ∈ α ∈I f (Aα ). From the defnition of union of a family of sets, there is α0 ∈ I such that y ∈ f (Aα0 ). From the defnition of image of a set, there S S is x ∈ Aα0 such that y = f (x). We have x ∈ Aα0 ⊂ α ∈I Aα . Therefore, y = f (x) ∈ f ( α ∈I Aα ). By Theorem 1.1.1 the result follows. □ Defnition 1.2.5 Let f : X → Y and g: Y → Z be two functions. Then the composition function g ◦ f of f andg is the function g ◦ f : X → Z given by (g ◦ f )(x)= g( f (x)) for all x ∈ X. Theorem 1.2.5 Let f : X → Y and g: 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 and g are injective, then g ◦ f is injective. (iii) If f and g are surjective, then g ◦ f is surjective. (iv) If g ◦ f is injective, then f is injective. (v) If g ◦ f is surjective, then g 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. □ Defnition 1.2.6 A sequence of elements of a set A is a function with domain N and codomain A. We discuss sequences in detail in Chapter 2. Defnition 1.2.7 We say that set A is fnite if it is empty or if there exists a natural number n and a one-to-one correspondence f : A →{1,2,...,n}. A set is infnite if it is not fnite. We leave it as an exercise to prove that the union of two fnite sets is fnite. It is also easy to show, by contradiction, that N is infnite. Exercises 1.2.1 ▶ Let f : X → Y be a function. Prove that: (a) If f is one-to-one, then A = f −1( f (A)) for every subset A of X. (b) If f is onto, then f ( f −1(B)) = B for every subset B of Y. 1.2.2 Let f : R → R be given by f (x)= x2 − 3 and let A =[−2,1) and B =(−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 by g(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.
RkJQdWJsaXNoZXIy NTc4NTAz