6 CARDINALITY OF SETS
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1 6 CARDINALITY OF SETS MATH Foundations of Pure Mathematics We all have an idea of what it means to count a finite collection of objects, but we must be careful to define rigorously what it means for a set to have a given number of elements, not least because we must be able to apply the same ideas to infinite sets, where counting can get less intuitive. In this section we will encounter some familiar combinatorial results, but here we approach them in a rigorous way, giving proofs using our definition of the cardinality of a set. Let n N. Define N n = {1, 2, 3,..., n} = { N : 1 n}. Let A be a set. We say A has cardinality n if there exists a bijection f : N n A. In this case write A = n. Define = 0. Example: Let A = {3, π, 2, 7}. Define f : N 4 A by f is a bijection, so A = 4. f(1 = 3 f(2 = 7 f(3 = π f(4 = 2. Let A be a set. If A = n for some n N {0}, then we say that A is finite. Otherwise we say A is infinite. The definition of A involved choice of f : N n A. It s not trivial to show that A does not depend on the choice of f. In proving this, we will derive results such as the pigeonhole principle. Theorem 6.1 Let m, n N. If there is a 1-1 function f : N m N n, then m n. Proof. We use induction on m. For m N, let p(m be the statement n N, if there is a 1-1 function f : N m N n, then m n. For m = 1, we have m n for all n N, so p(1 is trivially true. Let N, and suppose that p( is true. Let n N and suppose that we have a 1-1 function f : N +1 N n. Write b = f( + 1. Define a new function f 1 : N N n \ {b} by f 1 (x = f(x. It is easy to chec that f 1 is
2 Define g : N n \ {b} N n 1 by g(x = { x if x b 1 x 1 if x b + 1. It is easy to chec that g is 1-1. Since f 1 and g are both 1-1, it follows that g f 1 : N N n 1 is 1-1. Hence by the inductive hypothesis n 1. Hence + 1 n. So p( + 1 is true. Hence by induction p(m is true for all m N. Corollary 6.2 Let A be a set. Suppose that m, n N and that there are bijections f : N m A and g : N n A. Then m = n. Proof. Let A be a set. Suppose that m, n N and that there are bijections f : N m A and g : N n A. Now g 1 f : N m N n is a bijection, and so 1-1. Hence by the theorem above m n. On the other hand, f 1 g : N n N m is also 1-1, so by the theorem n m. So m = n. We have proved that the cardinality of a finite set is well-defined, i.e., it does not depend on choice. Corollary 6.3 Let A and B be finite sets and let f : A B be a 1-1 function. Then A B. If further f is a bijection, then A = B. Proof. If A =, then the result is immediate. So we may assume that A (and B. Write A = m and B = n, where m, n N. So there are bijections α : N m A and β : N n B. Then β 1 f α : N m N n is 1-1. Hence by the theorem m n. The contrapositive of this is the pigeonhole principle: Theorem 6.4 (Pigeonhole principle Let A and B be non-empty finite sets and let f : A B. If A > B, then there exist x 1, x 2 A with x 1 x 2 and f(x 1 = f(x 2. Example: In a group of 13 people, 2 will have a birthday in the same month. Definition 6.5 Let A be a set. We say that A is countable if it is finite or if there exists a bijection f : N A. In the latter case we say that A is countable infinite. Remar: An example of an uncountable set is R. We ll discuss uncountably infinite sets later. 2
3 Example: Z is countable. To see this, tae the bijection f : N Z defined as follows: f(1 = 0 f(2 = 1 f(3 = 1 f(4 = 2 f(5 = 2 f(6 = 3 f(7 = 3 f(8 = 4 f(9 = 4 etc. We will wor towards the inclusion-exclusion principle for finding the size of a union of sets. We will do this in a number of stages. Theorem 6.6 Let X and Y be finite sets such that X Y = (we say that X and Y are disjoint. Then X Y = X + Y. Proof. If X = 0, then X = and X Y = Y, and the result holds. Similarly if Y = 0. Hence we may suppose that X = m > 0 and Y = n > 0. So there exist bijections f : N m X and g : N n Y. Define h : N m+n X Y by { h(i = f(i if 1 i m g(i m if m + 1 i m + n. We claim that h is a bijection. h is onto: Let z X Y. If z X, then j N m such that f(j = z (since f is onto. So h(j = f(j = z. If z Y, then j N n such that g(j = z (since g is onto. So h(m+j = g(j = z. In either case we have found an element mapping to z, so h is onto. h is 1-1 : Let h(i = h(j for some i, j N m+n. Write z = h(i = h(j. Since z X Y and X Y =, either z X or z Y, but not both. If z X (i.e., i, j m, then h(i = f(i and h(j = f(j. Since f is 1-1, we have i = j. If z Y (i.e., i, j, > m, then h(i = g(i m and h(j = g(j m. Since g is 1-1, we have i m = j m, so i = j. Hence h is 1-1. We have shown that h is a bijection. So X Y = m + n = X + Y. Corollary 6.7 Suppose that X 1,..., X n are pairwise disjoint finite sets. Then X 1 X n = n X i is a finite set and n n X i = X i. 3
4 Proof. By induction on n, using Theorem 6.6. What if X and Y in Theorem 6.6 are not disjoint? We have the inclusion-exclusion principle, which turns out to be an application of Corollary 6.7. Theorem 6.8 (Inclusion-exclusion principle Let X and Y be finite sets. Then X Y = X + Y X Y. Proof. Note that X Y = (X \ Y (Y \ X (X Y, a union of pairwise disjoint sets. By Corollary 6.7 X Y = X \ Y + Y \ X + X Y. Now X = (X \ Y (X Y, a union of disjoint sets, so by Theorem 6.6 X = X \ Y + X Y. Similarly Y = Y \ X + X Y. Substituting, we get X Y = ( X X Y + ( Y X Y + X Y = X + Y X Y. This can be generalized to more than two sets, as we will see later. We now turn our attention to Cartesian products. Theorem 6.9 Let X and Y be finite sets, with X = m and Y = n. Then X Y is a finite set and X Y = mn. Proof. Consider first the case X = or Y =. Then X Y = and the result holds in this case. Suppose that X and Y. Write X = {x 1,..., x m }. Then X Y = m ({x i } Y. Now i, {x i } Y = Y, since we may define a bijection f i : Y {x i } Y by f i (y = (x i, y and by Corollary 6.3 this implies the sets have the same cardinality. Note that the {x i } Y are pairwise disjoint (for distinct x i, so by Corollary 6.7 m X Y = ({x i } Y = m {x i } Y = m Y = m Y = mn. 4
5 Corollary 6.10 Let X 1,..., X m be finite sets, where X i = n i for each i. Then X 1 X m = n 1 n 2 n m. Proof. Use Theorem 6.9 and induction on m. We can now use these results to count functions. Corollary 6.11 Let X and Y be non-empty finite sets, where X = m and Y = n. Then the number of functions X Y is n m. Proof. Write X = {x 1,..., x m }. A function f : X Y determines an element of Y m = Y } {{ Y } by m times (f(x 1, f(x 2,..., f(x m. Conversely, each element (y 1,..., y m Y m determines a function f : X Y by i, f(x i = y i. So (given our labelling x 1,..., x m we have constructed a bijection g : {f : X Y is a function} Y m. [To see that g is 1-1, suppose f 1, f 2 are functions X Y and that g(f 1 = g(f 2. Then (f 1 (x 1,..., f 1 (x m = (f 2 (x 1,..., f 2 (x m. So f 1 (x = f 2 (x for all x X. So f 1 = f 2. To see that g is onto, let (y 1,..., y m Y m. Define f : X Y as above by f(x i = y i. Then g(f = (y 1,..., y m.] The result follows since Y m = Y m = n m by Corollary Notation: For n N, define n! = n(n Define 0! = 1 Theorem 6.12 Let A and B be finite sets with A = B = n. Then there are precisely n! bijections A B. In particular, there are precisely n! permutations of A. Proof. We use induction on n. The result is clear for n = 1. Let N. Suppose that the result is true for n =. Suppose that A = B = + 1. Fix a A. For each b B, count bijections f : A B for which f(a = b. By our assumption, there are! bijections A \ {a} B \ {b}. So there are! bijections f : A B satisfying f(a = b. There are +1 choices for b, so the total number of bijections is ( +1! = ( +1!. So by induction the result is true for all n. 5
6 6.1 Counting subsets and the binomial theorem Let A be a set and N {0}. A -subset of A is a subset X A with X =. Write P (A = {X A : X = }. Consequently, if A = n, then n P(A = P (A. =0 ( n We define to be the cardinality P (N n of P (N n. ( n is called a binomial coefficient, or n choose. WARNING: This is the definition used in this course - we will see later that it is equivalent to a more familiar one. ( n Remar: If A is a set with A = n, then P (A = P (N n. Hence number of -element subsets of A. Example: Let A = {a, b, c}. ( 3 P 0 (A = { } (so = 1. 0 ( 3 P 1 (A = {{a}, {b}, {c}} (so = 3. 1 ( 3 P 2 (A = {{a, b}, {b, c}, {a, c}} (so = 3. ( 2 3 P 3 (A = {A} (so = 1. 3 We collect some facts about ( n : Lemma ( 6.13 Let n, N {0}. Then n (i = 0 if > n, ( ( ( n n n (ii = = 1 and = n, ( 0 ( n 1 n n (iii =, n (iv if 0 < n, then ( ( n n 1 = 6 + ( n 1 1. is the
7 Proof. (i and (ii are clear. (iii f : P (A P n (A, f(x = A \ X defines a bijection. (iv See exercises. Theorem 6.14 Let n, N {0} with n. Then ( n n! =!(n!. Proof. Uses Lemma 6.13 and induction on n. See [IMR, ]. If n = 0, then = 0 and ( 0 = 1 = 0! 0 0!0!, so the result is true for n = 0. If n = 1, then = 0 or = 1. In either case ( 1 = 1 = 1! 1!0!. Let m N. Suppose that the result is true for n = m (and all. We will show that it is true for n = m + 1. Let N {0} with 0 m + 1. If = 0 or = m + 1, then ( m + 1 so we may assume that 1 m. By ( Lemma 6.13 ( m + 1 m = + ( m 1 = 1 = = m!!(m! + m! ( 1!(m ( 1! (since true for m (m + 1! 0!(m + 1!, = m!(m+1 + m!!(m+1! (!(m+1! = m!(m+1 +!(m+1! (m+1! =!(m+1! So the result is true for m + 1. Hence by induction the result is true for all n. An application of this is: Theorem 6.15 (Binomial theorem Let n N and let a, b R. Then (a + b n = n =0 7 ( n a b n.
8 Setch proof. (a + b n = (a + b... (a + b. }{{} n times Label the factors on the right hand side 1,..., n. (a + b n is obtained as a sum of 2 n products, each of which is a product of n a s and b s. Each product corresponds to a subset X of N n = {1,..., n} defined by i X if and only if the product involves a from factor i. Let Z with 0 n. Then X P (N n if and only if the corresponding product is a b n. So the number of times that a b n occurs in (a + b n is P (N n = ( n. 6.2 Bonus nonexaminable material Recall that a set A is countable if either A is finite or there is a bijection N A. (i There is a bijection N N N. The bijection f : N N N is given by f(1 = (1, 1, f(2 = (2, 1, f(3 = (1, 2, f(4 = (1, 3, f(5 = (2, 2, f(6 = (3, 1, f(7 = (4, 1, f(8 = (3, 2, f(9 = (2, 3, f(10 = (1, 4, etc. By induction, for all n N, there is a bijection N N n, i.e., N n is countable. More generally, a finite Cartesian product of countable sets is countable. (ii Existence of non-countable sets Theorem 6.16 (Cantor Let A be a set. Then there is no onto map A P(A. Proof. Suppose that f : A P(A is onto. Let x A. Note that f(x A, and so either x f(x or x f(x. This allows us to define a subset S = {x A : x f(x} A. Now since f is onto, there is a A such that f(a = S. Suppose that a f(a. Then a S since f(a = S, so by definition of S we have a f(a, a contradiction. Hence a f(a. But then a S by definition of S, and so a f(a since S = f(a, a contradiction. Corollary 6.17 P(N is not countable. Proof. Suppose that P(N is countable. Then there is a bijection f : N P(N. Then f is onto, a contradiction by the Theorem. 8
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