The Ohio State University October 10, 2016 Combinatorics | 1 Functions and sets Enumerative combinatorics (or counting), at its heart, is all about finding functions between different sets in such a way that reveals their size. A little combinatorics goes a long way: you'll probably run into it in every discipline that involves numbers. Beyond numbers and sets, there is not much additional formal theory needed to get started. For sets A and B; a function f : A ! B is any assignment of elements of B defined for every element of A: All f needs to do to be a function from A to B is that there is a rule defined for obtaining f(a) 2 B for every element of a 2 A: In some situations, it can be non-obvious that a rule in unambiguous (we will see examples), in which case it may be necessary to prove that f is well-defined. We will be interested in counting, which motivates the following three definitions. Definition 1: A function f : A ! B is injective if and only if for all x; y 2 A; f(x) = f(y) ) x = y: These are sometimes called one-to-one functions. Think of injective functions as rules for placing A inside of B in such a way that we can undo our steps. Lemma 1: A function f : A ! B is injective if and only if there is a function g : B ! A so that 8x 2 A g(f(x)) = x: This function g is called a left-inverse for f Proof. ) : We need to define g on all B: Let S ⊂ B be all those elements of B that can be reached using f (also called the range of f). Formally: S = fb 2 B : 9 a 2 A; f(a) = bg : Then for each b 2 S; define g(b) to be an arbitrary a so that f(a) = b: For each b 2 B nS (the elements of B not in S), just define g to be any arbitrary assignment. Then by definition, for any a 2 A; a0 = g(f(a)) is an element of A so that f(a0) = f(a); which by assumption implies a = a0 = g(f(a)); as desired. (: Suppose x; y 2 A have f(x) = f(y): Then x = g(f(x)) = g(f(y)) = y: There is a complementary definition for surjective functions. Math 3345 Combinatorics1 Fall 2016 Definition 2: A function f : A ! B is surjective if and only if for all y 2 B there exists x 2 A so that f(x) = y: These are sometimes called onto functions. Think of surjective functions as rules for surely (but possibly inefficiently) covering every B by elements of A. Lemma 2: A function f : A ! B is surjective if and only if there is a function g : B ! A so that 8y 2 B f(g(y)) = y: This function is called a right-inverse for f: Proof. ) : We need to define g on all B: By assumption, for any b 2 B; there is an a 2 A so that f(a) = b: Hence, let g(b) be any arbitrary choice of such an a: Then by definition, for any b 2 B; f(g(b)) = b: (: Suppose y 2 B is arbitrary. Let a = g(y): Then f(a) = f(g(y)) = y; and so since y was arbitrary, f is surjective. Combining these two lemmas, we see that if g is the left-inverse of f (so that f is injective) then we also have that f is a right-inverse of g: This automatically implies that the left-inverse of an injective function f is surjective. By the same argument, the right-inverse of a surjective function f is injective. Finally, we define the main object of combinatorics, the function that is both injective and surjective. Definition 3: A function f : A ! B is bijective if it is both injective and bijective. Just like with injective and surjective functions, we can characterize bijective functions according to what type of inverse it has. Lemma 3: A function f : A ! B is bijective if and only if there is a function g : B ! A so that 1. 8a 2 A; g(f(a)) = a: 2. 8b 2 B; f(g(b)) = b: This function g is called a two-sided-inverse for f: Proof. ) : Using Lemmas 1 and 2, we know there are functions g; h : B ! A so that 1. 8a 2 A; g(f(a)) = a: 2. 8b 2 B; f(h(b)) = b: The only trouble is that we don't know g = h just from invoking the lemmas, but it turns out that must be true. Let b 2 B be arbitrary. Then h(b) = g(f(h(b))) = g(b); Math 3345 Combinatorics2 Fall 2016 where in the first equality, we apply f followed by g to h(b); and in the second equality, we use that f(h(b)) = b: This shows g = h: (: Because g exists, by Lemmas 1 and 2, we have f is both injective and surjective, and hence bijective. Example 1: Let f : Z ! Z be given by f(x) = 2x: Then f is injective because if f(x) = f(y); then 2x = 2y; which implies x = y: However, f is not surjective because there is no x 2 Z so that f(x) = 1: Example 2: Let f : R ! R be given by f(x) = x3: Then f is bijective. The function g : R ! R given by x 7! x1=3 is a two-sided inverse for f: Example 3: Let f : R ! R≥0; where R≥0 = fx 2 R : x ≥ 0g ; 2 be given by f(x) =px : Then f is surjective but not injective. The function g : R≥0 ! R defined by g(x) = x which gives the non-negative square-root of a non-negative number x is a right-inverse for f: This is to say that 8x 2 R≥0; f(g(x)) = x: Therefore f is surjective. However, f(1) = 1 = f(−1); and so f is not injective. Also, 8x 2 R≥0; g(f(x)) = jxj; so g is not a left-inverse for f: Exercise 1: Fix a 2 R with 0 ≤ a ≤ 2π: Define the function fa(x) = sin(a + x) mapping [0; π] to [−1; 1]: For what values of a is fa injective? Surjective? Exercise 2: Fix b 2 N: By the division algorithm, for every z 2 Z; there are unique q; r 2 Z with f0; 1; 2; : : : ; b − 1g so that z = bq + r: Let fb : Z ! f0; 1; 2; : : : ; b − 1g be the function that sends z to r: (a) Prove that the function fb is surjective. (b) Prove that the function fb is not injective. (c) Fix a 2 Z: Define the function g : f0; 1; 2; : : : ; b − 1g ! f0; 1; 2; : : : ; b − 1g by g(z) = fb(a + z): Prove that the function g is bijective. (d) Fix a; b 2 N: Define the function m : f0; 1; 2; : : : ; b − 1g ! f0; 1; 2; : : : ; b − 1g by m(z) = fb(a · z): Prove that the function m is bijective if gcd(a; b) = 1: Comment: this is the basis of modular arithmetic, which allows one to define an addition and multiplication rule on f0; 1; 2; ··· ; b − 1g that satisfy the normal axioms of addition and multiplication. Even better, one can show this has multiplicative inverses when b is prime. Math 3345 Combinatorics3 Fall 2016 Composition We also need to be able to talk about compositions of functions. If f : A ! B and g : B ! C are functions, then g ◦ f : A ! C (said \g compose f") is the function that for all x 2 A; (g ◦ f)(x) = g(f(x)): Injectivity, surjectivity, and bijectivity all behave nicely with respect to composition Lemma 4: Let f : A ! B and g : B ! C be two functions. 1. If both f and g are injective, then g ◦ f is injective. 2. If both f and g are surjective, then g ◦ f is surjective. 3. If both f and g are bijective, then g ◦ f is bijective. Exercise 3: Prove this lemma. Set operations Finally, we need to define a few common set operations. For any two sets, A and B we define the union, intersection, and set difference by the following rules. x 2 A [ B () (x 2 A) _ (x 2 B) x 2 A \ B () (x 2 A) ^ (x 2 B) x 2 A n B () (x 2 A) ^ (x2 = B) Like with the logical operations _ and ^; the set operations have distributivity rules, but for the moment we mostly need these for notation. When we do set theory, we will cover these rules further. Example 4: Let A = f1; 2g and B = f2; 3g : Then A [ B = f1; 2; 3g A \ B = f2g A n B = f1g 2 Cardinality One way to think of two sets A and B for which there exists a bijection f : A ! B is that B is somehow a relabeling of A: The function f tells us how to do the relabeling, and its properties ensure that for every element of B there is one and only one element of A in correspondence with it. The only information left over is the size, or cardinality, of the set. Math 3345 Combinatorics4 Fall 2016 Definition 4: Two sets A and B have the same cardinality if and only if there exists a bijection f : A ! B: Cardinality makes sense for finite sets and infinite sets.