Indian Institute of Information Technology Design and Manufacturing, Kancheepuram Instructor Chennai 600 127, India N.Sadagopan An Autonomous Institute under MHRD, Govt of India Scribe: An Institute of National Importance S.Dhanalakshmi www.iiitdm.ac.in COM205T Discrete Structures for Computing-Lecture Notes Functions and Infinite Sets Objective: We shall introduce functions, special functions and related counting problems. We shall see the importance of functions in the context of innite sets and discuss innite sets in detail. Further, we introduce the notion of counting in the context of innite sets. Denition 1 (Function) Let A and B be two non-empty sets. A function or a mapping f from A to B, written as f : A ! B, where every element a 2 A is mapped to a unique element b 2 B. Note: 1. The element b 2 B is called the image of 0a0 under f and is written as f(a). 2. If f(a) = b then 0a0 is called the pre-image of b under f. 3. A is called the domain of f and ff(a) j a 2 Ag is called the range of f. B is called the co-domain of f. 4. Examples for function: a) Domain = Set of apples and Range = weight of apples. b) Domain = Set of students and Range = CGPA of students. 5. Every function is a relation but the converse is not true. i.e. Every relation is not a function. 6. For every element a 2 A, there exist an image f(a). i.e., f(a) is well dened. 7. Let j A j= n and j B j= m. The number of dierent functions f : A ! B is mn, where jAj represents the cardinality of A, the number of elements of A. 8. Let j A j= n and j B j= m. If X = fR j R is a relation dened w.r.t (A; B)g and Y = fR j R is a function w.r.t (A; B)g then, j X |≥| Y j (Since, j X j= 2mn and j Y j= mn). Things to know: 1. Is there a function f : A ! B, if A = ; and B 6= ; ? Yes, an empty function. 2. Is there a function f : A ! B, if A 6= ; and B = ; ? No. 3. Is there a function f : A ! B, if A = ; and B = ; ? Yes, void function. Denition 2 (One-one function/Injective) A function is said to be injective if for every element in the range there exists a unique pre-image. i.e., no two elements in the domain map to same element in the co-domain. Denition 3 (Onto function/Surjective) A function is said to be surjective if for every el- ement in the co-domain there exists a pre-image. • A function is said to be bijective if it is both one-one and onto function. A B A B 1 a 1 a 2 b 2 b 3 c 3 c 4 d 4 d e (a) (b) A B A B 1 a 1 a 2 b 2 b 3 3 c 4 c 4 d (d) (c) Fig. 1. (a) Not one-one and not onto (b) one-one but not onto (c) onto but not one-one (d) one-one and onto • Let A = B = N(set of natural numbers). - Give an example of a function f : N ! N, that is one-one and not onto ? f(x) = x + 1 - Give an example of a function , that is not one-one but onto x f : N ! N ? f(x) = d 2 e - Give an example of a function f : N ! N, that is one-one and onto ? f(x) = x • Let A = B = I(set of integers). - Give an example of a function f : I ! I, that is one-one and not onto ? f(x) = 2x - Give an example of a function , that is not one-one but onto x , if f : I ! I ? f(x) = d 2 e x ≥ 1 and f(x) = x, if x ≤ 0. - Give an example of a function f : I ! I, that is one-one and onto ? f(x) = x • Give a bijective function f : (0; 1) ! (4; 5) in real numbers ? f(x) = x + 4. Let and . The number of dierent one-one functions are . • j A j= n j B j= m mCn ×n! = mPn Food for Thought 1. Let j A j= n and j B j= m. The number of dierent onto functions ? 2. Let j A j= n and j B j= m. The number of dierent bijective functions ? 3. Give a bijective function f : (0; 1) ! (a; b) in real numbers ? 2 An invitation to innite sets Motivation: - Many interesting sets such as (i) Set of prime numbers (ii) Set of C-programs (iii) Set of C-programs with exactly three statements are innite in nature. - Between I and N, which set is bigger ? - Between I and N × N, which set is bigger ? - Is [0; 1] is bigger than R ? - Can we list all C-programs ? Denition 4 (Finite) A set 0A0 is nite if there exists n 2 N, n-represents the cardinality of A such that there is a bijection f : f0; 1; : : : ; n − 1g ! A. A is innite if A is not nite. Denition 5 (Innite) Let A be a set. If there exists a function f : A ! A such that f is an injection and f(A) ⊂ A then A is innite. • Let B be a nite set, B = f1; 2;:::; 10g. Can you establish a 1-1 function f : B ! B such that f(B) ⊂ B ? No. • Can you establish a 1-1 function f : N ! N such that f(N) ⊂ N ? Yes, f(x) = x + 1. • Every 1-1 function from f : B ! B is also a bijection from B ! B if B is nite. Problem 1: Show that N is innite. Proof by contradiction: Suppose N is nite. By denition, there exists n, n represents the cardinality of N such that f(n) = an. Let K = MAXff(0); f(1); : : : ; f(n − 1)g + 1. There does not exist a x 2 f0; 1; : : : ; n − 1g such that f(x) = k. Therefore, f is not onto and thus, f is not a bijection. Hence, our assumption is wrong and N is innite. Problem 2: Show that R is innite. Proof: To prove R is innite, establish a function f : R ! R such that f is 1-1 and f(R) ⊂ R. Consider f : R ! R such that f(x) = x + 1, if x ≥ 0 and f(x) = x, if x < 0. This function f is 1-1 but not onto (Since, 0 does not have a pre-image) and f(R) ⊂ R. Hence, R is innite. Problem 3: Let P = fa; bg and f : P∗ ! P∗. Show that P∗ is innite. Proof: To prove P∗ is innite, establish a function f : P∗ ! P∗ such that f is 1-1 and f(P∗) ⊂ P∗. Consider f : P∗ ! P∗ such that f(x) = ax. The elements , b; ba; bb; : : : will not have a pre- image. Therefore f is not onto but 1-1 and f(P∗) ⊂ P∗. Hence, P∗ is innite. 3 Problem 4: Show that [0; 1] is innite. Proof: Consider such that x . This function is 1-1 but not onto f : [0; 1] ! [0; 1] f(x) = 2 f (Since, 1 does not have a pre-image) and . Hence, is innite. ( 2 ; 1] f([0; 1]) ⊂ [0; 1] [0; 1] Claim: 1 Let A0 be a subset of A. If A0 is innite then A is innite. Proof: Given A0 is innite. Therefore, there exist a function g : A0 ! A0 such that g is 1-1 and g(A0) ⊂ A0. Consider a function f : A ! A such that f(x) = x, if x 2 AnA0 and f(x) = g(x), if x 2 A0. The function f is also 1-1 and f(A) ⊂ A (Since, g is 1-1 and g(A0) ⊂ A0). Thus, A is innite. Corollary: Every subset of a nite set is a nite set. Claim 2: Let f : A ! B be an injection. If A is innite then B is innite. Proof: Since f is 1-1 and A is innite, f(A) is innite. By previous claim, B is innite (since, f(A) ⊆ B). Problem 5: A is innite. Show that (i) P (A); power set of A is innite. (ii) A [ B is innite. (iii) A × B is innite. (iv) AB, the set of all functions from B to A, is innite. Solutions: (i) Consider a function f : A ! P (A) such that f(x) = fxg. The function f is 1-1 but not onto i.e.,A ⊆ P (A). Thus, P (A) is innite. (ii) We know that, A ⊂ A [ B. Since A is innite, A [ B is innite (by claim 1). (iii) Consider a function f : A ! A × B such that f(x) = (x; a) for some a 2 B. Clearly, the function f is 1-1 but not onto. Thus, A × B is innite. (iv) Every element in AB is a function from B ! A. Consider a function f : A ! AB such that f(x) = g, g is a function from B ! A such that g(b) = x, 8 b 2 B. This function f is 1-1 but not onto. Thus, AB is innite. Denition 6 (Countable) A set A is countable if A is nite or if A has an enumeration (Listing elements of A) or if there exists a bijection from N to A.