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Martin Gould Ramsey Theory

Ramsey Theory

Martin Gould

Contents

1 Introduction 3

2 Preliminaries 3

3 The Birth of Ramsey Theory 3

4 Ramsey’s Theorem: Infinite Version 4

5 Ramsey’s Theorem: Finite Version 6

6 Ramsey Numbers 7

7 The Compactness Theorem 11

8 Discussion and The Future 12

References 15

1


2


1 Introduction

With its roots in an old decision problem and its branches reaching areas as varied as algebra, combinatorics,set theory, logic, analysis, and geometry, Ramsey theory has played an important role in a plethora ofmathematical developments throughout the last century. Ramsey theory is concerned with the preservationof structure under partitions – it is the study of unavoidable regularity in large structures. In this essay, Iexplore some of the core ideas underpinning Ramsey theory and present a variety of problems to which itcan provide interesting and elegant solutions.

Ramsey theory remains an active area of research today. In the last decade, an enormous amount ofprogress has been made on a wealth of open problems in Ramsey theory (e.g., [12, 11, 17, 20]) yet entireareas still remain poorly understood today ([7, 8]).

2 Preliminaries

In this section, I state some key definitions and formalities that are used freely throughout the remainder ofthe essay.

Definition. A graph G = (V,E) is an ordered pair such that V is a set of elements, called vertices or nodes,and E ⊆ {{v1, v2} | v1, v2 ∈ V, v1 6= v2} is the set of edges.

Definition. An edge e = {e1, e2} ∈ E is adjacent to the vertex v ∈ V if e1 = v or e2 = v.

Definition. Edges e = {e1, e2} , f = {f1, f2} ∈ E are adjacent to each other if either of e1 or e2 is equal toeither of f1 or f2.

Definition. Nodes v1, v2 ∈ V are adjacent to each other if ∃ e ∈ E such that e = {v1, v2}.

Definition. A complete graph Kn is a graph such that |V | = n and E = {{v1, v2} |v1, v2 ∈ V, v1 6= v2}. Inother words, Kn is a graph with n nodes, each of which is adjacent to every other.

Definition. An r−colouring of the vertices of a graph G = (V,E) is a function χV : V → {1, 2, . . . , r}. Anr−colouring of the edges of a graph G = (V,E) is a function χE : E → {1, 2, . . . , r}. More generally, anr−colouring of a set S is a function χ : S → {1, 2, . . . , r}.

Definition. A proper r−colouring of the vertices (edges) of a graph G = (V,E) is a colouring of the vertices(edges) of the graph such that no two adjacent vertices (edges) share the same colour.

Throughout the essay, I take N to be the set {1, 2, . . .} – that is, I do not consider zero to be a naturalnumber.

3 The Birth of Ramsey Theory

Although the ideas central to Ramsey theory have interested mathematicians for many years, most modernexpositions on the subject cite van der Waerden’s 1927 paper [21] – which predates Ramsey’s own publication[19] by three years – as being the foundation of its rigorous study. Van der Waerden’s theorem states thatfor any r, p ∈ N, there exists N ∈ N such that whenever the first N natural numbers are partitioned into rsets, one of these sets will contain an arithmetic progression1 of p terms.

Three years later, as an intermediate step towards presenting a solution to a decision problem, FrankRamsey published two other “existence” proofs [19] that I discuss in Sections 4 and 5. Rather than solelysearching for structure on N, as van der Waerden had done, Ramsey’s work was concerned with findingstructure in both the countably infinite and the finite, an example of which I explore now.

1An arithmetic progression is a subsequence of N such that the difference between subsequent terms is constant.

3


A Classic Result for the Google Generation

The following is a claim which follows from a classic result in Ramsey theory:

Given any six members of the social networking site www.facebook.com, there willalways exist either a trio whom are all friends, or a trio of which none are friends.

The classical statement of this problem involves friendships between six attendees at a dinner party, butrecasting the problem to use facebook removes any ambiguities (and potential asymmetries) stemming fromthe definition of “friendship” that might occur otherwise. This result will be derived as a consequence ofbounds on Ramsey numbers in Section 5, but for now it is possible to prove the assertion directly. Label thesix members in question A,B,C,D,E, and F . There must exist a subset S ⊂ {B,C,D,E, F} of size threesuch that A is either friends with all of S or is friends with none of S, because there are 5 elements in theset {B,C,D,E, F} and only two possible friendship states (“friend” or “not friend”). Assuming, withoutloss of generality,2 that A is friends with the three members of S, simply look at the friendship states of themembers of S. If any two members S1, S2 ∈ S are friends, then the trio A,S1, S2 are the required trio offriends. If no pair of members of S are friends, then S itself is a trio none of whom are friends. In eithercase, the required trio exists.

This result highlights the underlying flavour of Ramsey theory – the idea that inheriting some sort ofstructure is, in many scenarios, unavoidable. Regardless of which six out of the hundreds of millions offacebook users are chosen, the result always holds.

4 Ramsey’s Theorem: Infinite Version

Before examining the first of Ramsey’s theorems, I introduce some notation:

Notation. Given a set X and a natural number k, let[Xk

]denote the set of subsets of X that have k elements.

In other words,

A ∈[X

k

]⇔ A ⊆ X, |A| = k.

Notice that it follows immediately that∣∣∣[Xk ]∣∣∣ =

(Xk

)(but I will not use this fact).

Theorem 1. Let X be a countably infinite set.3 Then for all k, r ∈ N and for every partition of the set[Xk

]into r classes, one of the classes contains every element of

[Yk

]for some infinite set Y ⊆ X.

The proof of this theorem will make use of an induction, but before proceeding with the formal details itis illuminating to consider some specific values of k.

First consider k = 1. It is clear from the definition that[X1

]= X. The result of the theorem then follows

from the infinite version of the pigeonhole principle [1] – if an infinite number of pigeons are placed into afinite number of pigeonholes, there must exist a pigeonhole that contains an infinite number of pigeons.

Next consider k = 2. This is clearly much harder than the k = 1 case, but it is still possible to think ofthe problem in a familiar way by making use of graph theory. If X is considered to be the set of verticesin an infinite graph, then elements of

[X2

]can be treated as edges on that graph. In this way, the graph

G =(X,[X2

])is an infinite complete graph. Furthermore, the act of partitioning these edges into r classes

can be thought of as applying a colouring χE to the edges of the graph. In this context, the Theorem 1 saysthat there exists some infinite subset Y ⊆ X of vertices such that every edge between every vertex in Y hasthe same colour under the given colouring.

2If instead A is not friends with the members of S, the same logic holds with the roles of “friend” and “not friend” reversed.3In order to keep the focus of the essay on Ramsey theory and not on axiomatic set theory, I will work only on a countably

infinite set. However, if the reader is prepared to admit the axiom of choice [14] and thus the Well-Ordering Theorem [4], theresult can be generalised to any infinite set X by using the well-ordering given by the theorem.

4


To see that this is true, first define a well-ordering ≺ on X such that no two elements of X are consideredto be equal. Then choose an element θ ∈ X such that there are infinitely many elements x′ ∈ X satisfyingθ ≺ x′. For example, if X consists of the negative integers, then choosing θ = −1 with the ordering −1 ≺−2 ≺ −3 ≺ . . . suffices. It is always possible to find such an ordering and such a θ because X is countable (or,via the Well-Ordering Theorem, if X is not countable but the axiom of choice is assumed). After applyingthe r−colouring χE to the edges of graph, examine the colour of every edge of the form {θ, x}, x ∈ X forwhich θ ≺ x. There are an infinite number of such edges but only r < ∞ different colours, so (by the samelogic as in the k = 1 case) there must exist some colour c0 ∈ {1, 2, . . . , r} such that infinitely many edgesadjacent to θ have the colour c0 under χE .

Define U := {x ∈ X| χE ({θ, x}) = c0}, and notice that U is an infinite subset of X. By the Well-OrderingPrinciple [2], there exists a smallest element of U . Call this smallest element u0, and notice that θ ≺ u0.

Next, consider all edges of the form {u0, u}, u ∈ U . Again, there are infinitely many such edges, so (bythe same logic as above) there exists a colour c1 ∈ {1, 2, . . . , r} such that infinitely many edges adjacent tou0 have the colour c1 under χE . Notice that it is not necessarily the case that c0 = c1.

Define V := {u ∈ U | χE ({u0, u}) = c1}, and notice that V is an infinite subset of X. Again, by theWell-Ordering Principle, there exists a smallest element of V . Call this smallest element v0, and notice thatθ ≺ u0 ≺ v0.

Continuing in this manner indefinitely yields the infinite set of vertices T = {θ, u0, v0, . . .} with θ ≺ u0 ≺v0 ≺ . . .. Notice that for any edge connecting a pair of vertices t, t′ ∈ T , the colour χE ({t, t′}) depends onlyon min (t, t′) . Hence, it is possible to define a new r−colouring χV on the vertices in T by χV (v) = χE ({v, v′})for any v′ � v, with v, v′ ∈ T . However, there are infinitely many vertices in T , so there exists some colour cand some infinite set Y ⊆ T such that every member of Y has the colour c under χV . This Y is an infinitesubset of X for which every edge

[Y2

]has the same colour under χE , as required.

The logic employed in the k = 2 case is instructive of the general strategy for the formal induction. Froma colouring of

[X2

], it was possible to “induce” a colouring of

[X1

]that had a sufficiently regular structure on

an infinite subset of X. When considering the cases k ≥ 3, the framework of graph colouring needs to begeneralised slightly by using hypergraphs:

Definition. A hypergraph H = (V,E) is an ordered pair such that V is a set of elements, called vertices ornodes, and E ⊆ {{v1, v2, . . .} | v1, v2, . . . ∈ V, v1 6= v2 6= . . .} is the set of hyperedges – that is, a hypergraphis a graph in which edges can connect two or more vertices.

For the cases k ≥ 3,[Xk

]can be thought of as being the set of hyperedges of a k−uniform hypergraph:

Definition. A k−uniform hypergraph is a hypergraph such that every hyperedge has size k — that is, everyhyperedge connects exactly k vertices.

I now present the formal proof.

Proof. I use induction on k.The theorem holds for k = 1, by the infinite version of the pigeonhole principle (see discussion above).Let θ be the smallest element of X under some (strict) well-ordering ≺. For the inductive step, consider

any r−colouring χ of[Xk+1

](i.e., χ :

[Xk+1

]→ {1, 2, . . . , r}).

Define a new colouring χ0 :[X\{θ}k

]→ {1, 2, . . . , r} by χ0(A) := χ (A ∪ {θ}) .

Using the inductive hypothesis, there exists an infinite set X0 ⊆ X such that every element of[X0k

]has

the same colour under the χ0-colouring. Let c0 ∈ {1, 2, . . . , r} denote this colour.Set X0

′ := {x ∈ X0| θ ≺ x0}, let x1 := min (X0), and define the colouring χ1 :[X0′

k

]→ {1, 2, . . . , r} by

χ1(A) = χ (A ∪ {x1}). Then there exists an infinite X1 ⊆ X0′ such that every element of

[X1k

]has the same

colour under the χ1 colouring. Let c1 ∈ {1, 2, . . . , r} denote this colour.

5


Continue deconstructing the elements θ ≺ x1 ≺ x2 ≺ . . . and noting the colours c0, c1, c2, . . . . Finally,define the colouring χ′ : {x1, x2, x3, . . .} → {1, 2, . . . , r} by χ′(xj) = cj , j = 1, 2, . . ..

Then, because there are infinitely many xjs and only r < ∞ colours, there must exist an infinite Y ⊆{x1, x2, . . .} ⊆ X such that χ′ assigns the same colour to every element of Y . This Y is the required infiniteset.

5 Ramsey’s Theorem: Finite Version

While undoubtedly an important result in its own right, the infinite version of Ramsey’s theorem as discussedin Section 4 wasn’t quite enough for Ramsey’s original decision problem. To solve his problem, Ramsey neededa result of a similar flavour, but dealing with finite sets.

Theorem 2. For all r, k, n ∈ N, there exists an N ∈ N such that if X is any set with at least N elements, forevery partition of

[Xk

]into r classes, one of the classes contains every element of

[Yk

]from some set Y ⊆ X

with |Y | ≥ n.

I present a full proof of Theorem 2 in Section 7, but before doing so it is illuminating to calculate someexplicit values of N to see how it grows with r, k, and n. A crucial thing to notice about Theorem 2 is thatit provides no indication of how large N might be – it only guarantees its existence. Indeed, it becomesapparent that performing such calculations becomes extremely difficult, extremely quickly.

The Ramsey Function

The Ramsey function provides a useful framework within which calculations of the value of N can be per-formed.

Definition. The Ramsey function is the function R : (k, l1, l2, . . . , lr)→ N, Rk (l1, l2, . . . , lr) = N , where Nis the smallest natural number such that for any set X with |X| ≥ N , there exists an i ∈ {1, 2, . . . , r} anda set Y ⊆ X with |Y | = li, such that under any partition of

[Xk

]into r classes, one of the classes contains

every element of the set[Yk

].4

Once again, it is illuminating to re-express this definition in the language of hypergraph colouring. Onlyone new concept is needed:

Definition. A complete k−uniform hypergraph H = (V,E) is a k−uniform hypergraph for which E =[Vk

].

In other words, a complete k−uniform hypergraph is a k−uniform hypergraph in which every possible hyperedge(which necessarily has size k) is present.

Given the values l1, l2, . . . , lr, and k the Ramsey function outputs an integer. This integer is the smallestvalue of N such that on any complete k−uniform hypergraph X =

(V,[Vk

])with at least N vertices, for any

r−colouring χ of its hyperedges[Xk

], there will always exist a colour i ∈ {1, 2, . . . , r} and a subset Y ⊆ X of

vertices, with |Y | = li, such that every hyperedge[Yk

]has the same colour under χ.

As indicated above, the Ramsey function provides more flexibility than is required to find the N inTheorem 2 – indeed, such a calculation is precisely the special case l1 = l2 = . . . = lr = n. Despite this,calculating so-called “Ramsey numbers” of the form N = Rk (l1, l2, . . . , lr) has proven to be an exciting areaof research in its own right.

4Notice the symmetry involved in the definition: for any permutation σ, it follows that Rk (l1, l2, . . . , lr) =Rk (σ(l1), σ(l2), . . . , σ(lr)).

6


6 Ramsey Numbers

I now consider some scenarios in which Ramsey numbers can be calculated exactly. Such exact calculationsare exceptionally difficult to perform in all but the simplest of cases, and indeed exact Ramsey numbers arecurrently only known for a tiny fraction of choices of r, k and (l1, l2, . . . , lr).

When k = 1, the pigeonhole principle provides all of the answers. In a graph colouring sense, k = 1corresponds to simply colouring the vertices, because

[X1

]= X. Consider a collection of vertices V , with

|V | =∑ri=1 li− r, of which l1− 1 vertices have colour 1, and l2− 1 vertices have colour 2, and . . . , and lr − 1

vertices have colour r. Then there is no set of vertices of size li that all have colour i for any i ∈ {1, 2, . . . , r},but adding a single vertex of any colour i ∈ {1, 2, . . . , r} would yield a set of vertices of size li that all havecolour i. Hence R1 (l1, l2, . . . , lr) =

∑ri=1 li − r + 1.

Red-Blue Graph Colouring

The most commonly studied Ramsey numbers are for the case r = k = 2. In the hypergraph colouring sense,r = k = 2 relates to colouring the edges of a standard complete graph G =

(V,[V2

]), and it is common to

name the two colours assigned by the colouring as “red” and “blue”. Under such a setup, the r = k = 2 casecan be approached in a very visual sense, by physically drawing the graphs in question using red and blueedges. Figure 1 shows an example of a red-blue colouring of K4.

Figure 1: A red-blue colouring of K4 (colour online)

Given any s, t ∈ N, R2(s, t) is the smallest N ∈ N such that every red-blue colouring of the edges of KN

contains either a red Ks or a blue Kt. Theorem 2 says that N must be finite for any choice of s and t, andindeed I will show directly in this section that this must be true in the r = k = 2 case.

Recall the example about any six members of the social networking site www.facebook.com. In thenotation just introduced, the example gives the result R2(3, 3) ≤ 6. In other words, by assigning red edgesto pairs that are friends and blue edges to those that aren’t (or vice-versa), the example showed that it isimpossible to colour the complete5 graph K6 in such a way that avoids having either a red K3 or a blue K3.As it stands, there is not sufficient reasoning to be able to assert that R2(3, 3) = 6, because 6 has not yetbeen proven to be the minimal such N ∈ N for which any red-blue colouring of KN must contain either a redK3 or a blue K3: for example, perhaps N = 5 is sufficient? If, however, it were possible to exhibit a red-bluecolouring of K5 that contains neither a red K3 nor a blue K3, then it would follow that R(3, 3) > 5.6 Thisis indeed the case, as Figure 2 demonstrates.

Hence R2(3, 3) ≤ 6 and R2(3, 3) > 5, so R2(3, 3) = 6.

5The graph is complete because everybody is either friends or not friends with each other member of the group.6Recall that R2(s, t) ≤ n if and only if every red-blue colouring of KN contains either a red Ks or a blue Kt, so exhibiting

any specific colouring of K5 containing neither a red Ks nor a blue Kt is strong enough to disprove the statement.

7


Figure 2: A red-blue colouring of K5 with neither a red K3 nor a blue K3 (colour online)

Exact Ramsey Numbers and Bounds

In order to assert that R2(3, 3) = 6, so far I have made logical deductions specific to the choices s = 3 andt = 3. In this section, I explore some results that can be applied more generally and use these to rederivethe R2(3, 3) example. Moreover, one such derivation will provide a proof for Theorem 2 in the case wherer = k = 2.

The first class of Ramsey numbers that I shall consider are those of the form R2(s, 2) (or, by symmetry,R2(2, t)).

Lemma 1. For every s ∈ N, R2(s, 2) = s.

Proof. In every red-blue colouring of Ks, either there exists a blue edge (and hence a blue K2) or every edgeis red, in which case there exists a red Ks. This implies that R2(s, 2) ≤ s. However, an entirely red colouringof Ks−1 contains neither a blue K2 nor a red Ks. Hence, R2(s, 2) > s− 1, so R(s, 2) = s.

In general, it is much harder to evaluate R2(s, t) when s > 2 and t > 2, but there are some bounds thatcan come in handy:

Theorem 3.

If s ≥ 2 and t ≥ 2, then R2(s, t) ≤(s+ t− 2

.

)s− 1 (1)

If also s > 2 and t > 2, then R2(s, t) ≤ R2(s− 1, t) +R2(s, t− 1). (2)

Proof. I prove (1) and (2) simultaneously. Notice that if s = t = 2, then (1) holds with equality, by Lemma1. So assume that s, t > 2.

Suppose that either (1) or (2) fails for some choice of s, t > 2. Pick such a pair, with s+ t minimal. Thenif u := s− 1 and v := t− 1, R2(s, v), R2(u, t), and R2(u, v) must each satisfy (1), by minimality of the choiceof s+ t. This implies that there exists a finite n ∈ N such that n = R2(s, v) +R2(u, t), because (1) holds forR2(s, v) and R2(u, t), and the “choose” function always outputs a finite answer for any finite inputs.

Consider a red-blue colouring χ of the edges of Kn =(V,[V2

]), and choose any vertex v ∈ V . Define

NR := {w ∈ V | w 6= v, χ ({v, w}) = red} ,

NB := {w ∈ V | w 6= v, χ ({v, w}) = blue} .

Clearly NR ∪ NB ∪ v = V , but |V | = n and so |NR ∪ NB | = |NR| + |NB | = n − 1. Therefore, either|NR| ≥ R2(u, t) or |NB | ≥ R2(s, v), because if |NR| < R2(u, t) and |NB | < R2(s, v), it would follow that|NR| ≤ R2(u, t)− 1 and |NB | ≤ R2(s, v)− 1 and thus that

|NR|+ |NB | ≤ R2(u, t) +R2(s, v)− 2 = n− 2

8


which is a contradiction, because |NR|+ |NB | = n− 1.If |NR| ≥ R2(u, t), then consider the complete graph only on the vertices of NR. Because |NR| ≥ R2(u, t),

any colouring of such a graph must contain either a red Ku or a blue Kt. However, if vertex v is also includedin this consideration, there must always be a red Ku+1 = Ks or a blue Kt.

Similarly, if |NB | ≥ R2(s, v), then any colouring of the complete graph on the vertices of NB must containeither a red Ks or a blue Kv. Then if vertex v is also included in this consideration, there must always be ared Ks or a blue Kv+1 = Kt.

Hence, every red-blue colouring of the edges of Kn contains either a red Ks or a blue Kt, so R2(s, t) ≤ n.It then follows that:

R2(s, t) ≤ n = R2(s, v) +R2(u, t) (so (2) holds)

≤(s+ v − 2s− 1

)+(u+ t− 2u− 1

)(by minimality of the choice of s and t, so (1) holds for R2(s, v) and R2(u, t))

=(s+ t− 3s− 1

)+(s+ t− 3s− 2

)=

(s+ t− 3)!(s− 1)!(t− 2)!

+(s+ t− 3)!

(s− 2)!(t− 1)!

=((t− 1) + (s− 1)) (s+ t− 3)!

(s− 1)!(t− 1)!=

(s+ t− 2)!(s− 1)!(t− 1)!

=(s+ t− 2s− 1

).

Hence, (1) also holds, contradicting the assumption that either (1) or (2) fails for the given choice of s, t >2.

Using (1), it is now possible to quickly derive the bound used in the facebook example without makingany logical deductions specific to the s = t = 3 setup:

R2(3, 3) ≤(

3 + 3− 23− 1

)=(

42

)= 6.

Exhibiting the graph colouring shown in Figure 2 completes the argument that R2(3, 3) = 6.As mentioned earlier, Theorem 3 also provides a proof of Theorem 2 in the r = k = 2 case:

Corollary 1. R2(s, t) <∞ for all s, t ∈ N.

Proof. By Theorem 3, R2(s, t) ≤(s+t−2s−1

)<∞ because the “choose” function always outputs a finite answer

for any s, t ∈ N.

Aside from the cases already discussed, only a handful of exact Ramsey numbers are known to date forr = k = 2. These are shown in Table 1, which is taken from Radziszowski’s excellent review [18] of recentprogress in the area.7

Given the bound provided by Theorem 3, why is it so difficult to compute exact Ramsey numbers?Exploring R2(3, 4) reveals the problem. Table 1 indicates that R2(3, 4) = 9. Figure 3 shows a red-bluecolouring of K8 that contains neither a red K3 nor a blue K4, which implies that R2(3, 4) > 8.

The problem then comes from the fact that the bound given by either of (1) or (2) in Theorem 3 isR2(3, 4) ≤ 10, which isn’t strong enough to deduce directly that R2(3, 4) = 9. Additional reasoning specificto this choice of s and t is required to find the exact solution – there is no known “one size fits all” approachthat can be implemented.

7Indeed, Radziszowski himself was involved in the original proof that R2(4, 5) = 25 back in 1995.

9


s t R2(s, t)3 4 93 5 143 6 183 7 233 8 283 9 364 4 184 5 25

Table 1: The only known values of R2(s, t)

Figure 3: A red-blue colouring of K8 with neither a red K3 nor a blue K4 (colour online)

Other Exact Ramsey Numbers

Almost no exact Ramsey numbers are known for situations in which r or k are larger than 2. Indeed, not asingle nontrivial exact Ramsey number is known for k ≥ 3, and only one nontrivial Ramsey number is knownfor r ≥ 3. This is R2(3, 3, 3) = 17 [10].

Bounds on Ramsey Numbers

In the absence of any practical algorithm for computing exact values of Ramsey numbers, a great deal ofresearch effort has been concentrated on obtaining bounds instead. The two bounds that I explore deal withdiagonal Ramsey numbers, i.e. Ramsey numbers of the form R2(s, s). Indeed, diagonal Ramsey numbershave received by far the most attention in the literature thus far, perhaps due to the fact that Ramsey’soriginal theorem (Theorem 2, above) deals with diagonal Ramsey numbers.

It is easy to find an upper bound on diagonal Ramsey numbers simply by using equation (1) of Theorem3:

R2(s, s) ≤(s+ s− 2s− 1

)=(

2(s− 1)s− 1

).

Using binomial expansion, it can easily8 be shown that(2kk

)≤ 22k. Hence (by setting s − 1 = k) it follows

immediately that:R2(s, s) ≤ 22s−2 ≤ 4s.

Hence, R2(s, s) ≤ 4s is an upper bound on diagonal Ramsey numbers.In 1947, Erdos published the following proof of a lower bound for diagonal Ramsey numbers [5]:

Theorem 4. Let k, n ∈ N be such that(nk

)21−(k

2) < 1. Then R2(k, k) > n.

8Consider, for example, the binomial expansion (1+1)2k =`2k

0

´·12k ·10+

`2k1

´·12k−1 ·11+. . .+

`2kk

´1k ·1k +. . .+

`2k2k

´·10 ·12k =`2k

0

´+

`2k1

´+ . . .+

`2kk

´+ . . .+

`2k2k

´. Clearly

`2kk

´≤ 22k because

`2kk

´is just one of the terms in the binomial expansion of 22k.

10


Proof. In order to show that R2(k, k) > n, it is sufficient to show that there exists a colouring of the edgesof Kn that contains no monochromatic Kk. Consider an edge colouring of Kn in which colours are assignedrandomly. Let each edge be coloured independently, and such that for all e ∈ E:

P (Edge e is coloured red) = P (Edge e is coloured blue) =12.

There are(nk

)copies of Kk in Kn. Let Ai be the event that the ith Kk is monochromatic. Then:

P (Ai) = 2 ·(

12

)(k2)

= 21−(k2)

(where the leading 2 is because there are two colours from which to choose). Then:

P (∃ a monochromatic Kk) = P (∪iAi) ≤∑i

P (Ai) =(n

k

)21−(k

2).

However,(nk

)21−(k

2) < 1 by the assumption of the theorem, so

P (∃ a colouring with no monochromatic Kk) > 0.

Hence, there exists a colouring with no monochromatic Kk.

At first glance, the theorem appears to offer only a very weak lower bound on diagonal Ramsey numbers.In fact, the bound is very useful, as the following corollary reveals:

Corollary 2. For k ≥ 3, R2(k, k) > 2k2 .

Proof. Given k ≥ 3, define n := b2 k2 c (where bxc denotes the integer part of x). Then

(n

k

)21−(k

2) ≤ nk

k!21− k(k−1)

2 ≤

(2

k2

)kk!

· 21− k22 + k

2 =21+ k

2

k!.

However, 21+ k2

k! < 1 if k ≥ 3, so Theorem 4 applies.

Corollary 2 is particularly interesting because it provides an insight into how diagonal Ramsey numbersgrow. Specifically, it shows that they grow exponentially in k.

7 The Compactness Theorem

Although there are numerous known proofs of Theorem 2, most rely on particularly messy inductions in orderto define a specific colouring. However, with Theorem 1 proven in Section 4 above, a result from the 1950s –commonly known as Erdos and de Bruijn’s Compactness Theorem [3] – provides a wonderfully elegant proofof Theorem 2.

The version of Theorem 5 that I present is a modern one (again making use of the intuitive notions ofhypergraphs and colouring), but the content of the theorem is the same as that originally published by Erdosand de Bruijn. First, however, some new definitions are needed. The concept of adjacency in a hypergraph isa natural extension of adjacency in a graph – i.e., a hyperedge e = {e1, e2, . . . , en} is adjacent to the verticese1, e2, . . . , en. Then:

Definition. Given a hypergraph H = (V,E), a proper r−colouring of H is a colouring of the vertices,χ : V → {1, . . . , r}, such that @ e ∈ E for which every vertex adjacent to e has the same colour under χ.

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Definition. The chromatic number of a hypergraph H = (V,E), denoted by χ(H), is the smallest r ∈ N forwhich a proper r−colouring of H exists.

Notice that the common notion of proper colouring of standard graphs (as laid out in Section 2) is thensimply the specific case of the above definition that deals with 2-regular hypergraphs.

Definition. Given a hypergraph H = (V,E) and a subset W ⊆ V , the restriction of H to W , denoted byH|W , is the hypergraph H|W = (W, E|W ), where E|W = {e ∈ E|e ⊆W}.

Theorem 5. Let H = (V,E) be a hypergraph for which |e| <∞for all e ∈ E, but where V need not be finite.Suppose that for all finite W ⊆ V , χ (H|W ) ≤ r. Then χ(H) ≤ r.

The proof that I present below makes the assumption that V is countable. For a proof that doesn’t makethis assumption (but instead makes use of the axiom of choice), see [10].

Proof. If |V | < ∞, simply take W = V and the proof is complete. If not, then V is countably infinite.To ease exposition, set V ≡ N.9 By the assumption of the theorem, for all n ∈ N there exists a colouringχn : {1, . . . , n} → {1, . . . r} such that there is no e ∈ E for which every vertex adjacent to e is the same colourunder the given colouring. In other words, for all n ∈ N there exists a proper r−colouring χn of N|{1,...,n}.

I define a colouring χ∗ : N → {1, . . . , r} by induction. First, define χ∗(1) = 1. For the inductive step, ifχ∗(1), . . . , χ∗(j − 1) have been defined such that

Sj−1 := {n| n ≥ j − 1 and χ∗(i) = χn(i) for every i ∈ {1, . . . , j − 1}}

is infinite, then for some colour c ∈ {1, . . . , r} the set

T := {n ∈ Sj−1| χn(j) = c}

is also infinite. Set χ∗(j) = c and Sj = T . Then χ∗ is indeed a proper r−colouring of H (consider anye = {v1, v2, . . . , vm} ∈ E, with v1 < v2 < . . . < vm, where the ordering is inherited from the fact thatV = N). Then Svm

6= ∅, so there exists n ≥ vm such that χn(i) = χ∗(i) for all i ≤ n. In particular,χn(vj) = χ∗(vj) for all vj ≤ n, i.e., for all vj ∈ e.

Because χn is a proper r−colouring of {1, . . . , n}, e is not monochromatic under χn. Thus, e is notmonochromatic under χ∗.

Theorem 2 then immediately follows as a corollary of Theorem 5:

Corollary 3. Fix k ∈ N and let K be a family of finite subsets of N. Suppose that for any r−colouringχ :

[Nk

]→ {1, . . . , r}, there exists some A ∈ K such that

[Ak

]is monochromatic. Then for all r ∈ N, there

exists n0 ∈ N such that for all n ≥ n0, if[nk

]is r−coloured then there exists A ∈ K, A ⊆ {1, . . . , n} such that[

Ak

]is monochromatic.

Proof. Use Theorem 5 with V :=[Nk

].

8 Discussion and The Future

The first important comment to make about the main ideas in this essay is that the order in which I havepresented them is far from chronological. There has been a healthy atmosphere of conjecture and explorationdriving research in the field forwards throughout the last century, with some researchers (first Erdos [5] then,

9Any countably infinite set is in bijection with N, so there exists a permutation σ such that for all v ∈ V, there existsn ∈ N with v = σ(n). By forcing V ≡ N, I am simply ignoring the permutation element of the above expression in order tomake the arguments clearer.

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more recently, Graham [8]) personally offering monetary rewards for answers to some of the most prominentoutstanding problems. Most results in the field that are today stated as theorems were originally posed asconjectures many years before a proof was found, and there are many outstanding conjectures still awaitingproof or disproof [8] .

It is also important to highlight the fact that I have chosen to focus this essay upon the theorems ofRamsey and the work most directly related to them due to the appealing possibility of setting the problemsin the framework of graph theory. There are, however, plenty of alternative “flavours” of Ramsey theory,giving rise to what are known as “Ramsey-type” theorems. Numerous such alternatives have been presented[9], including:

Theorem. (Hales-Jewett) Given any finite set A and any r ∈ N, there exists some N(A, r) ∈ N such thatfor all n ≥ N(A, r), any r−colouring of An always contains a monochromatic combinatorial line.

Theorem. (Schur) For every r ∈ N, there exists some N(t) ∈ N such that for every partition of the set{1, 2, . . . , N} into r classes, one of the classes contains the numbers x and y and their sum x+ y.

It is perhaps unsurprising that proving many of these “Ramsey-type” theorems amounts to little morethan considering special cases of Ramsey’s theorems. For example, Schur’s theorem can be considered tobe a direct corollary of Theorem 5 [6]. On the other hand, there are many “Ramsey-type” theorems whoseproofs are unknown, or whose only known proofs stem from a different approach entirely.

Future Research in the Field

Despite providing an intuitive introduction to the necessary theory, many researchers have recently come tobelieve that graph theory might not be the best way to proceed with future explorations of Ramsey theory [6].The main reason for this is that such an approach to Ramsey theory often requires a lengthy enumeration ofall possible cases. Recall, for example, that proving R2(s, t) = n without using some kind of logical argumentwould require exhibiting a single red-blue colouring of Kn−1 with no red Ks and no blue Kt (in order toshow that R2(s, t) > n− 1) and also an enumeration of every possible red-blue colouring of Kn (in order todemonstrate that each one of them contains either a red Ks or a blue Kt). This approach is arduous for largevalues of n, as the number of red-blue colourings of Kn grows as 2n. It is possible to reduce this numbersomewhat by making use of the various symmetries involved, but enumeration is still a prohibitively long taskin general. Numerous alternatives have been proposed, and it is currently unclear whether or not one specificapproach will dominate future research in Ramsey theory (as the graph theoretic approach has arguablydone in the past). One such approach has made use of category theory. As interest in this area blossomedin the second half of the twentieth century, it became apparent that it could provide a rich framework forRamsey theory. A seminal work by Leeb [13] introduced the idea of using category theory in this way, andhas more recently been followed by other publications (e.g., [15] and [16]) following the same principles. Itwill be very interesting to see whether some of the problems that have remained unsolved in Ramsey theorywill eventually be attributed to future advances in category theory – or maybe even vice-versa.

Acknowledgements

I would like to thank Alex Scott for first introducing me to Ramsey theory during his lecture course “GraphTheory” given at The University of Oxford in Michaelmas Term, 2009. I would also like to thank PuckRombach, Oliver Riordan, and Sam Howison for useful discussions during the project, and Mason Porter forhis help and assistance in revising and shaping this essay.

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References

[1] N.L. Biggs, Discrete mathematics, 2nd ed., Oxford University Press, 2002.

[2] K. Ciesielski, Set theory for the working mathematician, Cambridge University Press, 1997.

[3] N.G. de Bruijn and P. Erdos, A colour problem for infinite graphs and a problem in the theory of relations,Nederl. Akad. Wetensch. Indag. Math. 13 (1951), 311–313.

[4] F.R. Drake and D. Singh, Intermediate set theory, John Wiley and Sons, 1996.

[5] P. Erdos, Some remarks on the theory of graphs, Bull. Amer. Math. Soc. 53 (1947), 292–294.

[6] R.L. Graham, Rudiments of Ramsey theory, Regional Conference Series in Mathematics 45 (1980).

[7] , Open problems in Euclidean Ramsey theory, Geombinatorics 13 (2004), 165–177.

[8] , Some of my favorite problems in Ramsey theory, Integers: Electronic J. Combinatorial NumberTheory 7(2) (2007), A15.

[9] R.L. Graham, M. Grotschel, and L. Lovasz (eds.), Handbook of combinatorics, vol. 2, Elsevier Science,1995.

[10] R.L. Graham, B.L. Rothschild, and J.H. Spencer, Ramsey theory, 2nd ed., Wiley Interscience, 1990.

[11] G. Halasz, L. Lovasz, M. Simonovits, and V.T. Sos, Paul Erdos and his mathematics, vol. 2, Springer,Budapest, 2002.

[12] B. Landman, A. Robertson, and C. Culver, Some new exact Van der Waerden numbers, Integers:Electronic J. Combinatorial Number Theory 5 (2005), no. 2, A10.

[13] K. Leeb, Vorlesungen uber Pascaltheorie, Universitat Erlangen, 1973.

[14] G.H. Moore, Zermelo’s axiom of choice: Its origins, development and influence, Springer, New York,1982.

[15] J Nesetril and V Rodl, Partition (Ramsey) theory – a survey, Colloq. Math. Soc. Janos Bolyai 18 (1978),754–792.

[16] J. Nesetril and V. Rodl, Partition theory and its applications, Surveys in Combinatorics: London Math.Soc. Lecture Note Series 38 (1979), 96–149.

[17] J. O’Rourke, Computational geometry column 46, SIGACT News 35 (2004), no. 3, 42–45.

[18] S.P. Radziszowski, Small Ramsey numbers, Electronic Journal of Combinatorics 1 (1994), 28.

[19] F.P. Ramsey, On a problem of formal logic, Proc. London Math. Soc. 2 (1930), no. 30, 264–286.

[20] S. Surahmat, E.T. Baskoro, S. Uttunggadewa, and H. Broersma, An upper bound for the Ramsey numberof a cycle of length four versus wheels, in Combinatorial Geometry and Graph Theory 3330 (2005), 181–184.

[21] B.L. van der Waerden, Beweis einer baudetschen Vermutung, Nieuw Arch. Wiskd. 15 (1927), 212–216.

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