Introduction to Mathematical Analysis I - Second Edition

14 1.2 FUNCTIONS Proof: We prove (d) and leave the other parts as an exercise. (d) Suppose g ◦ f is injective and let x , x 0 ∈ X be such that f ( x ) = f ( x 0 ) . Then ( g ◦ f )( x ) = g ( f ( x )) = g ( f ( x 0 )) = ( g ◦ f )( x 0 ) . Since g ◦ f is injective, it follows that x = x 0 . We conclude that f is injective. Definition 1.2.4 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 . Definition 1.2.5 We say that set A is finite 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 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 N is infinite. The following result will be useful when studying sequences and accumulation points. Theorem 1.2.7 Suppose A is an infinite set. Then there exists a one-to-one function f : N → A . Proof: Let A be an infinite set. We define f as follows. Choose any element a 1 ∈ A and set f ( 1 ) = a 1 . Now the set A \ { a 1 } is again infinite (otherwise A = { a } ∪ ( A \ { a 1 } ) would be the union of two finite sets). So we may choose a 2 ∈ A with a 2 6 = a 1 and we define f ( 2 ) = a 2 2 . Having defined f ( 1 ) , . . . , f ( k ) , we choose a k + 1 ∈ A such that a k + 1 ∈ A \ { a 1 , . . . , a k } and define f ( k + 1 ) = a k + 1 (such an a k + 1 exists because A \{ a 1 , . . . , a k } is infinite and, so, nonempty). The function f so defined clearly has the desired properties. To paraphrase, the previous theorem says that in every infinite set we can find a sequence made up of all distinct points. Exercises 1.2.1 I 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 ) = x 2 − 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 . (b) f ( A \ B ) = f ( A ) \ f ( B ) for A , B ⊂ X . 1.2.5 Prove part (2) of Theorem 1.2.3 . 2 This fact relies on a basic axiom of set theory called the Axiom of Choice. See [ Lay13 ] for more details.