An Excursion in Discrete Mathematics

An Excursion in Discrete Mathematics by Harold Reiter One of the simplest sounding yet far-reaching concepts in discrete mathematics in the pigeonhole principle (PHP). Its use permeates all levels of expertise, enabling beginning students to feel comfortable and completely convinced of the truth of propositions whose proofs would otherwise reside in the twilight zone of students’ consciousness–that of being easily believed, but only vaguely understood. It also provides advanced problem posers and problems solvers a tool for limiting the size of an optimal set of objects in a combinatorial setting. This talk will take us on a journey through a small area of discrete mathematics, starting with a rather simple problem and modifying it bit by bit, either weakening the hypothesis, strengthening the conclusion, or reformulating the problem in a more general setting. First we state the pigeonhole principle: Suppose n is a positive integer and n + 1 pigeons are required to be distributed among n pigeonholes. No matter how that is done, the process must result in at least two pigeons occupying the same hole. Proposition 1. Let S be a subset of {1, 2, 3, . . . , 100} with at least 51 members. Then S contains two consecutive integers. Proof. If S has more than 51 members, consider just the subset S 0 consisting of the smallest 51 members of S. If the result is true of S 0 , then it is surely true of S. So we assume S has exactly 51 elements. Apply PHP with these 51 numbers as the pigeons, and the 50 pairs {1, 2}, {3, 4}, {5, 6}, . . . , {99, 100} as the holes. A pigeon is assigned to a hole in the obvious way: x goes into the hole which has x as one of its two members. The principle then asserts that two on the members of S belong to the same pigeonhole. This can only happen if S contains a pair of the form 2n − 1, 2n, for some n = 1, 2, 3, . . . , 50. The next proof, which does not explicitly reference PHP, sheds some light on the problem and its generalizations. Proof 2. Using the set S, construct a new set S 0 as follows: S 0 = {t + 1 | t ∈ S and t + 1 ≤ 100}. Note that S 0 has 50 elements if 100 ∈ S and 51 elements if 100 6∈ S. We use the notation |T | to mean the number of elements of the finite set T . The following formula follows by noting that each element of S ∪ S 0 is counted once on each side of the equation: |S ∪ S 0 | = |S| + |S 0 | − |S ∩ S 0 |. 1

Since |S| + |S 0 | ≥ 101 and |S ∪ S 0 | ≤ 100, it follows that |S ∩ S 0 | = |S| + |S 0 | − |S ∪ S 0 | ≥ 101 − 100 = 1. We may conclude that S ∩ S 0 is not empty. Suppose x belongs to both S and S 0 . Then both x and x − 1 belong to S. We may regard the proposition above a providing sufficient conditions to guarentee that the number 1 is expressible as a difference of two memebrs of S. In what follows, we use PHP to establish conditions which guarentee that other positive integers can be expressed as a difference of two members of S. Suppose k is a positive integer less than 100 which we wish to express as a difference of two members of S. Let f (k) denote the smallest number such that if any subset S of {1, 2, 3, . . . , 100} with more than f (k) members has the property that k can be expressed as the difference of some two members of S. That f is well-defined is clear: For any k ≤ 100, is S contains 100 members then k is so expressible. The proof above shows that f (1) ≤ 50. To see that f (1) = 50, note that {1, 3, 5, 7, . . . , 99} is a 50 element subset which does not contain any two consecutive integers. In what follows we use the notation bxc to mean the largest integer not exceeding the real number x. Proposition 2.Let k be a positive integer less than 100 and let S be a subset of {1, 2, 3, . . . , 100} with more than b 100 ck + min{k, 100 − 2kb 100 c} members. Then S 2k 2k contains a pair of numbers x, y such that x − y = k. Proof. We show that {1, 2, 3, . . . , 100} can be partitioned into f (k) subsets, each either a singleton or a doubleton such that each doubleton {u, v} satisfies |u−v| = k. We can then apply PHP with members of S as the pigeons and sets of the partition as the holes to complete the proof. We split the partitioning into two cases, k ≤ 50 and k ≥ 51. Case 1. In this case, partition the set {1, 2, . . . , b 100 c2k} into b 100 ck doubletons 2k 2k {1, k+1}, {2, k+2}, {3, k+3}, . . . , {k, 2k}, . . . , {2k+1, 3k+1}, {2k+2, 3k+2}, {2k+ 3, 3k + 3}, . . . , {3k, 4k}, . . . , {(b 100 c − 1)2k + 1, (b 100 c − 1)2k + k + 1}, . . . , {(b 100 c− 2k 2k 2k 100 1)2k + k, (b 2k c)2k}. The other sets of the partition are singletons. The example k = 15 should shed some light on the idea. The number b 100 c2k = b 100 c30 = 90 2k 30 100 is the largest even multiple of 15 less than 100. Each of the b 30 c = 3 groups of 30 is partitioned into doubletons, leaving 100 − 3 · 30 = 10 singletons. Thus f (15) = b 100 c15 + min{15, 100 − 30b 100 c} = 3 · 15 + 10 = 55. The example 30 30 k = 30 is also instructive. The number b 100 c2k = b 100 c60 = 60 is the largest 2k 60 100 even multiple of 30 less than 100. This b 60 c = 1 group of 60 is partitioned into doubletons. The set of numbers 61, 62, 63, . . . , 100 remaining contains 10 doubletons {61, 91}, {62, 92}, . . . , {70, 100}, each of which have absolute difference 30, and 20 100 other numbers 71, 72, . . . , 90. Thus f (30) = 30 + min{30, 100 − 2 · 30b 2·30 c} = 60. 100 100 Case 2. For k ≥ 51, f (k) = b 2k ck + min{k, 100 − 2kb 2k c} = 0 · k + min{k, 100 − 2

2k · 0} = k. The partition consists of {1, k + 1}, {2, k + 2}, {3, k + 3}, . . . , {100 − k, 100}, {100 − k + 1}, . . . , {k}. It is easy to see that we can do no better. The set {1, 2, 3, . . . , k} has k members no two of which differ by k. The example k = 60 may be helpful. In this case the partition consists of the doubletons {1, 61}, {2, 62}, . . . , {40, 100} and the singletons {41}, {42}, . . . , {60}. Next, we turn our attention to problem of expressing an integer is the difference or the sum of two members of S. Since the sum of two members of S is often bigger than 100 and the difference is ofter negative, we consider instead which numbers k, 0 ≤ k ≤ 99 are congruent modulo 100 to either y − x or y + x for x and y in S. Reasoning as above, we are led to the following theorem. Theorem 1. Let S be a subset of {1, 2, 3, . . . , 100} with at least 51 members and let k be a nonnegtive integer less than 100. Then S contains a pair x, y such that x + y ≡ k(mod 100). Also, S contains a pair u, v such that u − v ≡ k(mod 100). Proof. Construct a new set S 0 from S as follows. Let S 0 = {(k − u)(mod 100) |u ∈ S}. If u1 6= u2 then (k − u1 )(mod 100) 6= (k − u2 )(mod 100). Thus S 0 has at least 51 members. Now the equation |S ∪ S 0 | = |S| + |S 0 | − |S ∩ S 0 | again yields the desired results. In fact, we have |S| + |S 0 | ≥ 102 and |S ∪ S 0 | ≤ 101 so |S ∩ S 0 | = |S| + S 0 | − |S ∪ S 0 | ≥ 102 − 101 = 1. The conclusion that S ∪ S 0 is not empty means that S contains a number x such that x = k − y(mod 100) for some y in S. Adding y to both sides gives us the first of the two conclusions in the theorem. The other result follows by considering S 0 = {k + u(mod 100)| u ∈ S} Finally, we are ready to state and prove the generalization to topological groups mentioned in the introduction. Theorem 2. Let G be a compact, connected topological group with Haar measure µ (normalized so that µ(G) = 1). Let S be a closed subset of G with µ(S) ≥ 1/2. Then G ⊂ S + S and G ⊂ S − S. If G is not connected, the hypothesis µ(S) > 1/2 is sufficient. Proof Let S be as hypothesized and let Z be an arbitrary element of G. Let Sz = {u| u = z + x for some x ∈ S}. Now Sz is a closed subset of G, and µ(Sz ) = µ(S) ≥ 1/2. We claim that Sz ∩ S 6= ∅. If S ∩ S 0 = ∅, then µ(S ∪ S 0 ) = µ(Sz ) + µ(S) − µ(Sx ∩ S) ≥ 1/2 + 1/2 − 0 = 1. Thus G−(Sx ∪S) is an open set of measure zero. This is possible only if G−(Sz ∪S) = ∅. Hence G is the union of two disjoint closed sets, contradicting the connectedness of G. We must conclude, therefore, that Sz ∩ S 6= ∅. Let u ∈ Sz ∩ S. Then there 3

exists x in S such that u = z + x. That is, z = u − x. To prove that G ⊂ S + S, let z be as above and consider the set Sz0 = {u| u = z − x, x ∈ S}. Reasoning a above, we find that Sz0 ∩ S 6= ∅, and this leads to the desired conclusion. If G is not known to be connected, the hypothesis µ(S) > 1/2is sufficient to guarantee that Sz ∩ S 6= ∅ (S 0 ∩ S 6= ∅) from which the desired conclusions follow.

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