Ramsey Theory: Proofs and Theorems of Ramsey's Graph Theory, Study notes of Mathematical Analysis

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RAMSEY THEORY

Ramsey’s Theorem

Suppose we 2-colour the edges of K 6 of Red and Blue. There must be either a Red triangle or a Blue triangle.

This is not true for K 5.

Ramsey’s Theorem

For all positive integers k, there exists R(k,) such that if N ≥ R(k, ) and the edges of KN are coloured Red or Blue then then either there is a “Red k-clique” or there is a “Blue-clique. A clique is a complete subgraph and it is Red if all of its edges are coloured red etc.

R( 1 , k) = R(k, 1 ) = 1 R( 2 , k) = R(k, 2 ) = k

Theorem

R(k, ) ≤ R(k, − 1 ) + R(k − 1 , `).

Proof Let N = R(k, − 1 ) + R(k − 1 ,).

1

V

V

Red edges

Blue edges

R

B

VR = {(x : ( 1 , x) is coloured Red} and VB = {(x : ( 1 , x) is coloured Blue}.

Theorem

R(k, `) ≤

k + ` − 2 k − 1

Proof Induction on k + . True for k + ≤ 5 say. Then

R(k, ) ≤ R(k, − 1 ) + R(k − 1 , `)

≤

k + ` − 3 k − 1

k + ` − 3 k − 2

k + ` − 2 k − 1

So, for example,

R(k, k) ≤

2 k − 2 k − 1

≤ 4 k

Theorem

R(k, k) > 2 k/^2

Proof We must prove that if n ≤ 2 k/^2 then there exists a Red-Blue colouring of the edges of Kn which contains no Red k-clique and no Blue k-clique. We can assume k ≥ 4 since we know R( 3 , 3 ) = 6. We show that this is true with positive probability in a random Red-Blue colouring. So let Ω be the set of all Red-Blue edge colourings of Kn with uniform distribution. Equivalently we independently colour each edge Red with probability 1/2 and Blue with probability 1/2.

Pr (ER ∪ EB ) ≤ Pr (ER ) + Pr (EB ) = 2 Pr (ER )

= 2 Pr

⋃^ N

j= 1

ER,j

∑^ N

j= 1

Pr (ER,j )

∑^ N

j= 1

)(k 2 ) = 2

n k

)(k 2 )

nk k!

)(k 2 )

2 k

(^2) / 2

k!

)(k 2 )

21 +k/^2 k! < 1.

Very few of the Ramsey numbers are known exactly. Here are a few known values.

R( 3 , 3 ) = 6

R( 3 , 4 ) = 9

R( 4 , 4 ) = 18

R( 4 , 5 ) = 25

43 ≤ R( 5 , 5 ) ≤ 49

Ramsey’s Theorem in general

Note that

(a) N(p, q; 1 ) = p + q − 1 (b) N(p, r ; r ) = p(≥ r ) N(r , q; r ) = q(≥ r )

We proceed by induction on r. It is true for r = 1 and so assume r ≥ 2 and it is true for r − 1 and arbitrary p, q. Now we further proceed by induction on p + q. It is true for p + q = 2 r and so assume it is true for r and all p′, q′^ with p′^ + q′^ < p + q. Let

p 1 = N(p − 1 , q; r ) p 2 = N(p, q − 1 ; r )

These exist by induction.

Ramsey’s Theorem in general

Now we prove that

N(p, q; r ) ≤ 1 + N(p 1 , q 1 ; r − 1 )

where the RHS exists by induction.

Suppose that n ≥ 1 + N(p 1 , q 1 ; r − 1 ) and we color

([n] r

with 2 colors. Call this coloring σ.

From this we define a coloring τ of

([n− 1 ] r − 1

as follows: If X ∈

([n− 1 ] r − 1

then give it the color of X ∪ {n} under σ.

Now either (i) there exists A ⊆ [n − 1 ], |A| = p 1 such that (under τ ) all members of

( A

r − 1

are Red or (ii) there exists B ⊆ [n − 1 ], |A| = q 1 such that (under τ ) all members of

( B

r − 1

are Blue.

Ramsey’s Theorem in general

Now consider the case of s colors. We show that

N(q 1 , q 2 ,... , qs; r ) ≤ N(Q 1 , Q 2 ; r )

where

Q 1 = N(q 1 , q 2 ,... , qbs/ 2 c; r ) Q 2 = N(qbs/ 2 c+ 1 , qbs/ 2 c+ 2 ,... , qs; r )

(^ Let^ n^ =^ N(Q^1 ,^ Q^2 ;^ r^ )^ and assume we are given an^ s-coloring of [n] r

First temporarily re-color Red, any r -set colored with i ≤ bs/ 2 c and re-color Blue any r -set colored with i > bs/ 2 c.

Ramsey’s Theorem in general

Then either (a) there exists a Q 1 -subset A of [n] with

(A

r

colored Red or (b) there exists a Q 2 -subset B of [n] with

(A

r

colored Blue.

W.l.o.g. assume the first case. Now replace the colors of the r( -sets of A by there original colors. We have a bs/ 2 c-coloring of A r

. Since |A| = N(q 1 , q 2 ,... , qbs/ 2 c; r ) there must exist some i ≤ bs/ 2 c and a qi -subset S of A such that all of

(S

r

has color i. 

Hales-Jewett Theorem

Theorem For all positive integers r , m there exists a least integer HJ(r , m) such that if n ≥ HJ(r , m) and [m]n^ is colored with r colors, then there is a monochromatic combinatorial line.

Proof: Note that every r -coloring of [m]n^ contains a monochromatic line, then so does every r -coloring of [m]n+^1. (If we fix xn+ 1 = 1 then we are esentially dealing with [m]n.)

Given a line L we define L−, L+^ to be its first and last points, as the active indices increase from 1 to m.

Hales-Jewett Theorem

Lines L 1 , L 2 ,... , Ls are focused at a point f if L+ i = f for i = 1 , 2 ,... , s.

Lines L 1 , L 2 ,... , Ls are color focused at a point f if they are focused at f and the truncated lines Li \ {L+ i } are monochromatic with different colors.

For the proof we use induction on m. The case m = 1 is trivial.

We will show by induction on s that for each 1 ≤ s ≤ r there exists N = FHJ(r , s, t) such that any r -coloring of [m]n^ contains either

(^1) A monochromatic line, or (^2) s color-focused lines.