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Notes on Counting Finite Sets

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Notes on Counting Finite Sets Notes on counting finite sets Murray Eisenberg February 26, 2009 Contents 0 Introduction2 1 What is a finite set?2 2 Counting unions and cartesian products4 2.1 Sum rules......................................5 2.2 Product rules: ordered selection with independent choices...........6 2.3 Generalized product rules: ordered selection with dependent choices.....8 2.4 Inclusion-Exclusion Principle........................... 10 3 The Pigeonhole Principle 11 3.1 The basic Pigeonhole Principle.......................... 11 3.2 The Generalized Pigeonhole Principle...................... 13 4 Counting subsets: unordered selection with no repetitions 14 5 Sets of functions between finite sets 17 5.1 Permutations.................................... 18 5.2 Injections...................................... 21 5.3 Derangements.................................... 22 6 Indistinguishable objects 24 6.1 MISSISSIPPI models............................... 24 6.2 \Stars-and-bars" models.............................. 25 6.3 \Xs-and-wedges" models.............................. 26 7 Partitions 27 7.1 Surjections..................................... 28 Copyright c 2003{2009 by Murray Eisenberg. All rights reserved. 0 Introduction The purposes of these notes are: • to construct mathematical objects|\models"|for various kinds of things that may be counted systematically; • to formulate basic counting methods as precise mathematical theorems about such mathematical objects; and • to prove those mathematical theorems. In short, the aim here is to construct the mathematical infrastructure for the subject of combinatorics. The fundamental objects considered are sets and functions between sets. See the Math- ematica notebook SetsAndFunctions.nb for information about sets, subsets, unions, inter- sections, etc., and about injective (one-to-one) functions, surjective (\onto") functions, and bijective functions (one-to-one correspondences). Only a few motivating applications are included in this draft of these notes. Consult your textbook for many more examples. For numerical calculations, you may want to use my notebook Combinatorics.nb, which references some of the functions from the Mathematica Add-On package Combinatorica8. 1 What is a finite set? The empty set|the set f g having no elements whatsoever|is said to be finite. The idea that a nonempty set A be finite is that it has exactly n elements for some positive integer n. And this means that, for some positive integer n, the set A can be expressed in the form A = fa1; a2; : : : ; ang (*) subject to the restriction that distinct subscripts label distinct elements of A, that is, ai 6= aj whenever i 6= j: (**) Thus the elements of the \standard" finite set f1; 2; : : : ; ng with n elements can be used to count the elements a1; a2; : : : ; an of A. Subscripting elements of A with the integers 1; 2; : : : ; n amounts to having a function f : f1; 2; : : : ; ng ! A; with f(k) = ak (k = 1; 2; : : : ; n): Condition (*) means that f : f1; 2; : : : ; ng ! A is surjective (that is, \onto"); condition (**) means that the function f : f1; 2; : : : ; ng ! A is injective (that is, one-to-one). Thus the idea that a set be finite may be defined as follows. Definition 1. A set A is said to be finite when either A is empty or else there is some positive integer n and some bijection f : f1; 2; : : : ; ng ! A. In short, a set A is finite if and only if it is empty or else can be put into a one-to-one correspondence with f1; 2; : : : ; ng for some positive integer n. We want to call such n the \number of elements" of A, but before doing that we must know that there is only one such n. This is a consequence of the following proposition. Proposition 2. If n and m are positive integers with n 6= m, then there does not exist a bijection h: f1; 2; : : : ; ng ! f1; 2; : : : ; mg. 2 The preceding proposition can be proved, but the proof is somewhat complicated and rests upon some very fundamental properties of the natural numbers. We shall simply accept the truth of the proposition. From it we can deduce the result we want: Corollary 3. Let A be a nonempty finite set. Suppose n and m are positive integers and suppose f : f1; 2; : : : ; ng ! A and g : f1; 2; : : : ; mg ! A are bijections. Then m = n. Proof. Let n and m, f and g be in the statement. Then the inverse function g−1 : A ! f1; 2; : : : ; mg is also bijective. Hence the composite function g−1 ◦ f : f1; 2; : : : ; ng ! f1; 2; : : : ; mg is bijective. But this contradicts the preceding proposition unless n = m. In view of the preceding corollary, the following definition now makes sense. Definition 4. Let A be a finite set. If A 6= ;, then there is a unique positive integer n for which A can be put into one-to-one correspondence with f1; 2; : : : ; ng; we call n the number of elements of A and write n = #A. If A = ;, we say that 0 is the number of elements of A and write # A = 0. The number of elements of a finite set Ais also called its cardinality, denoted by card(A). Examples 5. (1) For a positive integer n, the set A = f1; 2; : : : ; ng is itself finite, and # A = n. In fact, the identity function i: f1; 2; : : : ; ng ! A is a bijection. (2) Let A be the set of all even positive integers that are less than 15; that is, A = f2; 4; 6; 8; 10; 12; 14g: Then A is finite and # A = 7 because the function f : f1; 2; 3; 4; 5; 6; 7g ! A defined by f(j) = 2j (1 ≤ j ≤ 7) is bijective. (3) The set A = f0; 1; 2; 3g is finite because the function f : f1; 2; 3; 4g ! A defined by f(j) = j − 1 (1 ≤ j ≤ 4) is bijective. (4) The set ∗ N = f1; 2; 3;::: g of all natural numbers is not finite. In fact, just suppose N∗ is finite and let n = # N∗. Since N∗ is not empty, then n > 0. Since n = # N∗, there is some bijection ∗ f : f1; 2; : : : ; ng ! N : Construct a new bijection ∗ g : f1; 2; : : : ; n; n + 1g ! N by the formula ( 1 + f(j) if 1 ≤ j ≤ n g(j) = 1 if j = n + 1. Then the function g−1 ◦ f : f1; 2; : : : ; ng ! f1; 2; : : : ; n; n + 1g is also a bijection. But this is impossible according to Proposition2. 3 Definition 6. A set is said to be infinite if it is not finite. According the preceding example, the set N∗ of all positive integers is infinite. You may show, similarly, that the set N of all natural numbers|0 together with all positive integers|is also infinite. Since the composition of two bijections is itself a bijection, any set A0 that can be put into one-to-one correspondence with a given finite set A is also finite and has the same number of elements as A: Proposition 7. Let A and A0 be sets, and suppose there is some bijection g : A ! A0. Then A is finite if and only if A0 is finite, and in this case #(A0) = #(A): We shall accept the following result without proving it. Proposition 8. Let A and B be sets A ⊂ B. If B is finite, then A is also finite; moreover, in this case # A ≤ # B. Corollary 9. Let A and B be sets A ⊂ B. If A is infinite, then B is also infinite. According to this corollary, each of the sets Z (the set of all integers), Q (the set of all rational numbers), R (the set of all real numbers), and C (the set of all complex numbers) is infinite. In fact, each has the infinite set N as a subset. Eventually, you will learn how to count the k-element sets of a finite set. Here is a start. Example 10. Let S be a finite set, with # (S) = n. (a) How many 0-element subsets does S have? Answer: Just 1, namely, the empty subset of S|the subset f g that has no elements whatsoever. (The empty set is often denoted by ;, but the notation f g is very suggestive.) (b) How many 1-element subsets does S have? Answer: n. Explanation: Let P1(S) be the set of 1-element subsets of S. Then the function f : S !P1(S); defined by f(x) = fxg (x 2 S) is bijective. For example, suppose S = fa; b; cg is a 3-element set. Then P1(S) = fag; fbg; fcg has 3 elements, as does S. However, the elements of P1(S) are not the same as the elements of S itself: no matter what x is, x 6= fxg: In fact, fxg is a set having exactly 1 element, namely, x, whereas x is just the single element of that set. (c) How many n−1-element subsets does S have. Answer: n. Can you supply the reason? 2 Counting unions and cartesian products This section concerns the finiteness of various sets|unions, cartesian products, etc.|formed from given finite sets, and the number of elements in the sets that result. 4 2.1 Sum rules Suppose you are going to order a single beverage|either a cup of coffee or else a bottle of soda, but not both. If there are 8 kinds of coffee and 5 flavors of soda from which you can choose, then you have a total of 48 + 5 = 13 possible choices of beverage. Why? This situation may be modelled by the union A [ B, where set A represents the 8 kinds of coffee and set B represents the 5 flavors of soda. Then the sets A and B are disjoint|they have no element in common (at least if the none of sodas is coffee-flavored).
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