Ramsey theory. Andrés Eduardo Caicedo. Undergraduate Math Seminar, March 22, Department of Mathematics Boise State University

Transcription

1 Andrés Eduardo Department of Mathematics Boise State University Undergraduate Math Seminar, March 22, 2012

2 Thanks to the NSF for partial support through grant DMS My work is mostly in set theory, but I am also interested in finite combinatorics. can be seen as part of both these fields.

3 Shameful self-promotion: Next term, I ll be teaching a topics course on set theory, centered on. Please contact me if you think you d be interested in taking the course.

4 This is the news I woke up to yesterday: professor wins prestig.html

5

6 Szemerédi s most famous theorem is at the heart of. It is a strengthening of the following result of van der Warden: Theorem (van der Waerden, 1927) Suppose the natural numbers are split into finitely many sets, N = A 1 A 2 A k. Then at least one of the sets A i contains arbitrarily large arithmetic progressions.

7 Recall that an arithmetic progression is a sequence of numbers of the form a, a + b, a + 2b,..., a + kb. For certain partitions, it is easy to verify the theorem. For example, N = Even Odd, and both sets contain arithmetic progressions of any length (in fact, both contain an infinite progression).

8 For another example, N = Squares Non-squares. In this case, the non-squares contain arbitrarily large progressions, because the gaps between squares get larger and larger.

9 Aside: What about the squares? Do they also contain arbitrarily large progressions? Here is an example of length 3: 1 2 = 1, 5 2 = 25 = , 7 2 = 49 =

10 Aside: What about the squares? Do they also contain arbitrarily large progressions? Here is an example of length 3: 1 2 = 1, 5 2 = 25 = , 7 2 = 49 =

11 Aside: What about the squares? Do they also contain arbitrarily large progressions? Here is an example of length 3: 1 2 = 1, 5 2 = 25 = , 7 2 = 49 = Fermat asked in 1640 whether we could find an example of length 4. Euler proved that this is not the case.

12 Aside: What about the squares? Do they also contain arbitrarily large progressions? Here is an example of length 3: 1 2 = 1, 5 2 = 25 = , 7 2 = 49 = Fermat asked in 1640 whether we could find an example of length 4. Euler proved that this is not the case.

13 Of course, the strength of van der Waerden s theorem comes from the fact that it applies to all possible partitions, even into sets for which no nice structure is apparent. It is then natural to ask whether there is a way, given a partition N = A 1... A k, to identify a piece A i with arbitrarily large progressions.

14 Of course, the strength of van der Waerden s theorem comes from the fact that it applies to all possible partitions, even into sets for which no nice structure is apparent. It is then natural to ask whether there is a way, given a partition N = A 1... A k, to identify a piece A i with arbitrarily large progressions.

15 First attempt: Maybe one of the A i actually contains an infinite progression. Counterexample: N = A B, where A = {1, 3, 4, 7, 8, 9, 13, 14, 15, 16,... } and B = {2, 5, 6, 10, 11, 12, 17, 18, 19, 20,... }.

16 First attempt: Maybe one of the A i actually contains an infinite progression. Counterexample: N = A B, where A = {1, 3, 4, 7, 8, 9, 13, 14, 15, 16,... } and B = {2, 5, 6, 10, 11, 12, 17, 18, 19, 20,... }.

17 First attempt: Maybe one of the A i actually contains an infinite progression. Counterexample: N = A B, where A = {1, 3, 4, 7, 8, 9, 13, 14, 15, 16,... } and B = {2, 5, 6, 10, 11, 12, 17, 18, 19, 20,... }.

18 Paul Erdős and Paul Turán proposed a different attempt.

19 A set A N has density δ iff A {1,..., n} lim = δ. n n This notion is too restrictive; many sets do not have density. More generally, we can say that A has positive upper density iff there is a sequence k 1 < k 2 < k 3 <... such that A {1,..., k n } lim > 0. n k n The usefulness of the notion comes from the fact that if N = A 1 A k, then at least one of the A i has positive upper density. Would this suffice to ensure the presence of progressions?

20 A set A N has density δ iff A {1,..., n} lim = δ. n n This notion is too restrictive; many sets do not have density. More generally, we can say that A has positive upper density iff there is a sequence k 1 < k 2 < k 3 <... such that A {1,..., k n } lim > 0. n k n The usefulness of the notion comes from the fact that if N = A 1 A k, then at least one of the A i has positive upper density. Would this suffice to ensure the presence of progressions?

21 A set A N has density δ iff A {1,..., n} lim = δ. n n This notion is too restrictive; many sets do not have density. More generally, we can say that A has positive upper density iff there is a sequence k 1 < k 2 < k 3 <... such that A {1,..., k n } lim > 0. n k n The usefulness of the notion comes from the fact that if N = A 1 A k, then at least one of the A i has positive upper density. Would this suffice to ensure the presence of progressions?

22 Theorem (Szemerédi, 1975) If A N has positive upper density, then it contains arbitrarily large arithmetic progressions.

23

24 As a silly example, if A is the set of squares, then A {1,..., n} n 1 n 0, which means that N \ A contains arbitrarily large arithmetic progressions. (This example is silly, because one can prove this by much easier methods.)

25 The theorem admits a refinement: Theorem Suppose that 0 < δ < 1 and n are given. Then there is an N large enough such that if A {1,..., N} N δ, then A contains an arithmetic progression of length n.

26 During the Abel prize awarding ceremony, Timothy Gowers presented Szemerédi s work. Here is how he explained this result: Imagine the following one-player game. You are told a small number, such as 5, and a large number, such as 10,000. Your job is to choose as many integers between 1 and 10,000 as you can, and the one rule that you must obey is that from the integers you choose it should not be possible to create a five-term arithmetic progression. For example, if you were accidentally to choose the numbers 101, 1103, 2105, 3107 and 4109 (amongst others), then you would have lost, because these numbers form a five-term progression with step size Obviously you are destined to lose this game eventually. But Szemerédi s theorem tells us something far more interesting: even if you play with the best possible progression-avoiding strategy, you will lose long before you get anywhere near choosing all the numbers.

27 Paul Erdős was not a rich man, but he used to assign a monetary reward to his questions. As Wikipedia tells us: These ranged from $25 for problems that he felt were just out of the reach of current mathematical thinking, to several thousand dollars for problems that were both difficult to attack and mathematically significant. There are thought to be at least a thousand such outstanding prizes, though there is no official or comprehensive list. The prizes remain active despite Erdős s death; Ronald Graham is the (informal) administrator of solutions. Winners can get either a check signed by Erdős (for framing only) or a cashable check from Graham.

28 Szemerédi s theorem earned Szemerédi an Erdős prize of $1,000. This is not the end of the story. Because some sets A N contain arbitrarily large progressions, even though they have density 0. For an easy example, consider A = { 1, 10 10, , , , , , , , ,... }.

29 Szemerédi s theorem earned Szemerédi an Erdős prize of $1,000. This is not the end of the story. Because some sets A N contain arbitrarily large progressions, even though they have density 0. For an easy example, consider A = { 1, 10 10, , , , , , , , ,... }.

30 Szemerédi s theorem earned Szemerédi an Erdős prize of $1,000. This is not the end of the story. Because some sets A N contain arbitrarily large progressions, even though they have density 0. For an easy example, consider A = { 1, 10 10, , , , , , , , ,... }.

31 The prime number theorem tells us that if A is the set of primes, then A {1,..., n} n 1 log n, so the primes have density 0. Are there arbitrarily large arithmetic progressions of prime numbers?

32 The prime number theorem tells us that if A is the set of primes, then A {1,..., n} n 1 log n, so the primes have density 0. Are there arbitrarily large arithmetic progressions of prime numbers?

33 This question was recently solved by Terence Tao and Benjamin Green.

34 Theorem (Green-Tao, 2004) The primes contain arbitrarily large arithmetic progressions.

35 Not even this is the end of the story. There is a natural notion of largeness, different from positive upper density, under which the prime numbers are actually large: Recall that the harmonic series diverges, that is, = +. It is a famous theorem of Euler that the same is true of the series of reciprocals of primes: = +.

36 Not even this is the end of the story. There is a natural notion of largeness, different from positive upper density, under which the prime numbers are actually large: Recall that the harmonic series diverges, that is, = +. It is a famous theorem of Euler that the same is true of the series of reciprocals of primes: = +.

37 Not even this is the end of the story. There is a natural notion of largeness, different from positive upper density, under which the prime numbers are actually large: Recall that the harmonic series diverges, that is, = +. It is a famous theorem of Euler that the same is true of the series of reciprocals of primes: = +.

38 The highest Erdős prize is $5,000. It will be awarded for a proof (or disproof) of the following conjecture: Suppose A N and n A 1 n = +. Then A contains arbitrarily large arithmetic progressions.

39 Currently, we do not even know if the condition implies that A contains a progression of length 3. Naively, one would think that if A is built greedily to avoid 3-term arithmetic progressions, then n A 1 n 1 n n B for any other B without 3-term arithmetic progressions, and so it would suffice to check that the first sum converges.

40 This greedy set A starts 1,2,4,5,10,11,13,... In fact, A consists precisely of those n such that n 1 only has 1s and 0s in its base 3 representation. One has < 1 n < n A Unfortunately (?), Wróblewski built in 1983 a set B with no 3-term progressions, with n B 1 n > Currently, we do not know how much higher one can reach.

41 This greedy set A starts 1,2,4,5,10,11,13,... In fact, A consists precisely of those n such that n 1 only has 1s and 0s in its base 3 representation. One has < 1 n < n A Unfortunately (?), Wróblewski built in 1983 a set B with no 3-term progressions, with n B 1 n > Currently, we do not know how much higher one can reach.

42 This greedy set A starts 1,2,4,5,10,11,13,... In fact, A consists precisely of those n such that n 1 only has 1s and 0s in its base 3 representation. One has < 1 n < n A Unfortunately (?), Wróblewski built in 1983 a set B with no 3-term progressions, with n B 1 n > Currently, we do not know how much higher one can reach.

43 Ok. So... What is?

44 Whereas the entropy theorems of probability theory and mathematical physics imply that, in a large universe, disorder is probable, certain combinatorial theorems show that complete disorder is impossible. T.S. Motzkin (1967)

45 Indeed one can view as the study of generalizations and repeated applications of the pigeonhole principle. Tao-Vu (2006)

46 The pigeonhole principle If sufficiently many objects are distributed over not too many classes, then at least one class contains many of these objects. The principle was first stated by Dirichlet in 1834, who called it the Schubfachprinzip.

47 Dirichlet used the principle to prove a result on Diophantine approximation. Given a real α, let α denote the closest distance from α to an integer. Theorem (Dirichlet) For any real α and k N there is a positive integer m k such that mα 1 k + 1. If α is irrational, this implies that there are infinitely many fractions n/m such that α n < 1 m m 2. These fractions are closely related to the continued fraction of α.

48 For example: π = Its convergents are 3, , , = 3, 22 7, , = 15 3, , ,....

49 Note that π 3 1 < 1, π 22 7 = < 1 49 = , π = < = ,...

50 Thus for any irrational α, we can find infinitely many fractions n m with α n < 1 m m 2. On the other hand, a deep number theoretic result, the Thue-Siegel-Roth theorem, states that if α is irrational and algebraic, then the equation α n < 1 m m 2+ɛ has only finitely many solutions for any ɛ > 0. One of the key ingredients of the proof is the construction of what is called an auxiliary function. Curiously, one of the main steps of the construction is an application of the pigeonhole principle.

51 The typical framework of a problem in is as follows: Given a certain size, a large structure is partitioned (colored) into a small number of pieces. Must it be the case that there is a monochromatic substructure of the given size? For example, if the structures are arithmetic progressions, van der Waerden s theorem is precisely an affirmative answer.

52 Ramsey s theorem, the result from which the field takes its name, is a generalization of the following statement: Given a number n, let K n denote the complete graph on n-vertices, and let K n denote the graph on n-vertices with no edges. Theorem (Ramsey) For any n, m there is an M so large that any graph on M vertices either contains a copy of K n, or a copy of K m.

53 Theorem (Ramsey) For any n, m there is an M so large that any graph on M vertices either contains a copy of K n, or a copy of K m. This is another instance of the framework mentioned before: Now we have complete graphs as structures, and we are splitting the graph into two pieces. The smallest M as above is the Ramsey number R(n, m). Clearly, R(n, m) = R(m, n), R(1, m) = 1, and R(2, m) = m.

54 Frank Plumpton Ramsey (22 Feb., Jan., 1930). Philosopher, Economist, Mathematician.

55 Statements of the form R(n, m) M are usually stated in the following way: In any party of at least M people, either there are n that know each other, or else there are m that do not know one another. We can also think of them as follows: Color each edge of K M red or blue. Then there is either a red copy of K n, or a blue copy of K m. Let s see some examples.

56 R(3, 3) = 6. That it is at least 6 follows from considering the following graph:

57 R(3, 3) = 6. That it is at most 6 follows from noticing that in any coloring of K 6, there must be 3 edges from the same vertex with the same color:

58 R(3, 4) = 9. That it is at least 9 follows from considering the following graph:

59 R(4, 3) = 9. That it is at most 10 follows from considering the following: To see that it is at most 9 is a bit more involved.

60 R(4, 4) = 18. That it is at least 18 follows from considering the following graph:

61 R(3, 5) = 14, R(3, 6) = 18, R(3, 7) = 23, R(3, 8) = 28, R(3, 9) = R(3, 10) 43. R(4, 5) = 25, 35 R(4, 6) 41. That R(4, 5) 25 was proved by Kalbfleisch in That R(4, 5) 25 was proved by McKay and Radziszowski in Computers were used in an essential fashion.

62 R(3, 5) = 14, R(3, 6) = 18, R(3, 7) = 23, R(3, 8) = 28, R(3, 9) = R(3, 10) 43. R(4, 5) = 25, 35 R(4, 6) 41. That R(4, 5) 25 was proved by Kalbfleisch in That R(4, 5) 25 was proved by McKay and Radziszowski in Computers were used in an essential fashion.

63 R(3, 5) = 14, R(3, 6) = 18, R(3, 7) = 23, R(3, 8) = 28, R(3, 9) = R(3, 10) 43. R(4, 5) = 25, 35 R(4, 6) 41. That R(4, 5) 25 was proved by Kalbfleisch in That R(4, 5) 25 was proved by McKay and Radziszowski in Computers were used in an essential fashion.

64 R(3, 5) = 14, R(3, 6) = 18, R(3, 7) = 23, R(3, 8) = 28, R(3, 9) = R(3, 10) 43. R(4, 5) = 25, 35 R(4, 6) 41. That R(4, 5) 25 was proved by Kalbfleisch in That R(4, 5) 25 was proved by McKay and Radziszowski in Computers were used in an essential fashion.

65 The two implementations required 3.2 years and 6 years of cpu time on Sun Microsystems computers (mostly Sparcstation SLC). This was achieved without undue delay by employing a large number of computers (up to 110 at once). McKay-Radziszowski (1993)

66 43 R(5, 5) R(6, 6) 165. Suppose aliens invade the Earth and threaten to destroy it in a year if human beings do not find R(5, 5). It is (probably) possible to save the Earth by putting together the world s best minds and computers. If, however, the invaders were to demand R(6, 6), the human beings might as well attempt a preemptive strike without even trying to ponder the problem. P. Erdős (1993)

67 43 R(5, 5) R(6, 6) 165. Suppose aliens invade the Earth and threaten to destroy it in a year if human beings do not find R(5, 5). It is (probably) possible to save the Earth by putting together the world s best minds and computers. If, however, the invaders were to demand R(6, 6), the human beings might as well attempt a preemptive strike without even trying to ponder the problem. P. Erdős (1993)

68 43 R(5, 5) R(6, 6) 165. Suppose aliens invade the Earth and threaten to destroy it in a year if human beings do not find R(5, 5). It is (probably) possible to save the Earth by putting together the world s best minds and computers. If, however, the invaders were to demand R(6, 6), the human beings might as well attempt a preemptive strike without even trying to ponder the problem. P. Erdős (1993)

69 In general, we know that m 1 (n)2 n/2 < R(n, n) < m 2 (n)2 2n where m 1, m 2 are modest functions of n, by which I simply mean that they are small compared to the exponential term. This means that, if lim R(n, n)1/n n exists, then it is somewhere between 2 and 4. Ronald Graham offers a $100 prize for a proof that lim R(n, n)1/n n exists, and a prize of $250 for a computation of this limit.

70 In general, we know that m 1 (n)2 n/2 < R(n, n) < m 2 (n)2 2n where m 1, m 2 are modest functions of n, by which I simply mean that they are small compared to the exponential term. This means that, if lim R(n, n)1/n n exists, then it is somewhere between 2 and 4. Ronald Graham offers a $100 prize for a proof that lim R(n, n)1/n n exists, and a prize of $250 for a computation of this limit.

71 Here is an easy application of Ramsey s theorem: Suppose that X is an infinite linear order. Then it either contains an infinite increasing sequence, or an infinite decreasing sequence. That is, X contains a copy of N, or a copy of the negative integers. What if X is a nice order? Suppose that the rationals are split into two sets. Then one of the sets contains a copy of the rationals.

72 Here is an easy application of Ramsey s theorem: Suppose that X is an infinite linear order. Then it either contains an infinite increasing sequence, or an infinite decreasing sequence. That is, X contains a copy of N, or a copy of the negative integers. What if X is a nice order? Suppose that the rationals are split into two sets. Then one of the sets contains a copy of the rationals.

73 Here is an easy application of Ramsey s theorem: Suppose that X is an infinite linear order. Then it either contains an infinite increasing sequence, or an infinite decreasing sequence. That is, X contains a copy of N, or a copy of the negative integers. What if X is a nice order? Suppose that the rationals are split into two sets. Then one of the sets contains a copy of the rationals.

74 Let K Q be the complete graph with set of vertices Q. Theorem (Galvin) Color the edges of K Q using finitely many colors. Then there is a copy of Q that uses at most 2 colors. (By the way, this is one of the results I expect to discuss in detail during the course next term.)

75 has many applications in number theory, analysis, and algebra. For example, beyond the pigeonhole principle, the earliest recorded Ramsey theoretic result is due to Hilbert. He used it as a lemma towards a proof of the irreducibility of certain polynomials.

76

77 An n-dimensional affine cube is a set Q Z for which there exist a, x 1,..., x n N such that Q = {a + i F x i F {1,..., n}}. Theorem (Hilbert) For every pair of positive integers r, n, there exists an M such that if {1,..., M} is colored with at most r colors, then there is a monochromatic n-dimensional affine cube.

78 Let s go back to the first result we mentioned. Theorem (Van der Waerden) For any n, r there is an N such that if {1, 2,..., N} is split into r pieces, then one of the pieces contains an arithmetic progression of size at least n. Call W (n, r) the smallest possible N.

79 Not much is known in terms of bounds. The original proof is a double induction and gives an Ackermannian bound.

80 Ackermannian bounds are as bad as they get. For example, the diagram above shows that W (3, 3) 7( )(2 3 7( ) + 1) Actually, W (3, 3) = 27.

81 Ackermannian bounds are as bad as they get. For example, the diagram above shows that W (3, 3) 7( )(2 3 7( ) + 1) Actually, W (3, 3) = 27.

82 Van der Waerden s proof is from In 1988, Shelah found a new proof that gives much better bounds (still huge). For example, even for two colors, the bounds obtained by Shelah are given by towers of towers of powers of 2. In general, he obtains bounds that are towers of towers of towers of powers of 2.

83 The best known bounds are due to Timothy Gowers. In work from the early 2000s, Gowers established that W (n, 2) n+9.

84 Ronald Graham conjectures that W (n, 2) 2 n2. He offers $1000 for a proof of the conjecture.

85