Rainbow Colourings - Portal · → λ. Y ⊂ X is an f-rainbowiff f ↾ Y n is 1–1 . Problem (Erdos, Hajnal, 1967, Problem 68)˝ Assume that f : ω1 2 → 3 establishes ω1 →

Rainbow Colourings

Lajos Soukup

Alfréd Rényi Institute of MathematicsHungarian Academy of Sciences

http://www.renyi.hu/ ∼soukup

Logic Colloquium 2008

Soukup, L (Rényi Institute) Rainbow Colourings LC2008 1 / 36

The beginnings

Ramsey theory : find large homogeneous sets

monochromatic sets

anti Ramsey theory : find large inhomogeneous sets

polychromatic sets, rainbow sets

first anti Ramsey theorems : Rado, 1973.

polychromatic Ramsey, rainbow Ramsey


The beginnings


monochromatic sets






The beginnings


monochromatic sets






The beginnings


monochromatic sets






The beginnings


monochromatic sets






The beginnings


monochromatic sets






The beginnings


monochromatic sets






The beginnings

Definition

Let f :[

X]n

→ λ.Y ⊂ X is an f -rainbow iff f ↾

[

Y]n is 1–1 .


The beginnings

Definition

Let f :[

X]n


[

Y]n is 1–1 .

X


The beginnings

Definition

Let f :[

X]n


[

Y]n is 1–1 .

Y

X


The beginnings

Definition

Let f :[

X]n


[

Y]n is 1–1 .

Y

X


The beginnings

Let f :[

X]n

→ λ. Y ⊂ X is an f -rainbow iff f ↾[

Y]n

is 1–1 .

Problem (Erdos, Hajnal, 1967, Problem 68)

Assume that f :[

ω1]2

→ 3 establishes ω1 6→[

ω1]2

3.Does there exist an f -rainbow triangle ?


The beginnings

Let f :[

X]n


Y]n

is 1–1 .


Assume that f :[

ω1]2


ω1]2



The beginnings

Let f :[

X]n


Y]n

is 1–1 .


Assume that f :[

ω1]2


ω1]2


ω1X


The beginnings

Let f :[

X]n


Y]n

is 1–1 .


Assume that f :[

ω1]2


ω1]2


ω1X


The beginnings

Let f :[

X]n


Y]n

is 1–1 .


Assume that f :[

ω1]2


ω1]2


ω1X

Notation

f establishes that ω1 6→[

ω1]2

3:


The beginnings

Let f :[

X]n


Y]n

is 1–1 .


Assume that f :[

ω1]2


ω1]2


ω1X

Notation

f establishes that ω1 6→[

ω1]2

3: f ⊢ ω1 6→[

ω1]2

3


Rainbow triangle

Assume f ⊢ ω1 6→[

ω1]2

3. Does there exist an f -rainbow triangle ?

Fact

If f ⊢ ω1 6→[

(ω, ω1)]2

3 then there is a rainbow triangle .

Definition

Let d :[

X]2

→ λ, f :[

Y]2

→ λ. f realizes d (d =⇒ f )

iff ∃Φ : X 1–1−→ Y s.t. ∀{ζ, ξ} ∈

[

X]2 d(ζ, ξ) = f (Φ(ζ),Φ(ξ)).


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


Definition

Let d :[

X]2

→ λ, f :[

Y]2


iff ∃Φ : X 1–1−→ Y s.t. ∀{ζ, ξ} ∈

[

X]2 d(ζ, ξ) = f (Φ(ζ),Φ(ξ)).


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


f ⊢ ω1 6→[

(ω, ω1)]2

3


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


f ⊢ ω1 6→[

(ω, ω1)]2

3 iff ∀A ∈[

ω1]ω

∀B ∈[

ω1]ω1

ω1A B


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


f ⊢ ω1 6→[

(ω, ω1)]2

3 iff ∀A ∈[

ω1]ω

∀B ∈[

ω1]ω1 f ′′[A, B] = 3.

ω1A B


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


Definition

Let d :[

X]2

→ λ, f :[

Y]2


iff ∃Φ : X 1–1−→ Y s.t. ∀{ζ, ξ} ∈

[

X]2 d(ζ, ξ) = f (Φ(ζ),Φ(ξ)).


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


Definition

Let d :[

X]2

→ λ, f :[

Y]2


iff ∃Φ : X 1–1−→ Y s.t. ∀{ζ, ξ} ∈

[

X]2 d(ζ, ξ) = f (Φ(ζ),Φ(ξ)).

X Y


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


Definition

Let d :[

X]2

→ λ, f :[

Y]2


iff ∃Φ : X 1–1−→ Y s.t. ∀{ζ, ξ} ∈

[

X]2 d(ζ, ξ) = f (Φ(ζ),Φ(ξ)).

X Y


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


Definition

Let d :[

X]2

→ λ, f :[

Y]2


iff ∃Φ : X 1–1−→ Y s.t. ∀{ζ, ξ} ∈

[

X]2 d(ζ, ξ) = f (Φ(ζ),Φ(ξ)).

X YΦ


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


Definition

Let d :[

X]2

→ λ, f :[

Y]2


iff ∃Φ : X 1–1−→ Y s.t. ∀{ζ, ξ} ∈

[

X]2 d(ζ, ξ) = f (Φ(ζ),Φ(ξ)).

X YΦ


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


Definition

Let d :[

X]2

→ λ, f :[

Y]2


iff ∃Φ : X 1–1−→ Y s.t. ∀{ζ, ξ} ∈

[

X]2 d(ζ, ξ) = f (Φ(ζ),Φ(ξ)).

X YΦ


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


Definition

Let d :[

X]2

→ λ, f :[

Y]2


iff ∃Φ : X 1–1−→ Y s.t. ∀{ζ, ξ} ∈

[

X]2 d(ζ, ξ) = f (Φ(ζ),Φ(ξ)).


Rainbow triangle


ω1]2


Fact

If f ⊢ ω1 6→[

(ω, ω1)]2


Definition

Let d :[

X]2

→ λ, f :[

Y]2


iff ∃Φ : X 1–1−→ Y s.t. ∀{ζ, ξ} ∈

[

X]2 d(ζ, ξ) = f (Φ(ζ),Φ(ξ)).

Fact

If f ⊢ ω1 6→[

(ω, ω1)]2ω1

then f realizes each function d :[

ω]2

→ ω1.


Rainbow triangle

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.

Theorem (Shelah, 1975)

CH =⇒ ∃f ⊢ ω1 6→[

ω1]2ω, no f -rainbow triangle.

♦ =⇒ ∃f ⊢ ω1 6→[

ω1]2ω1

, no f -rainbow triangle.


Rainbow triangle

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


CH =⇒ ∃f ⊢ ω1 6→[


♦ =⇒ ∃f ⊢ ω1 6→[

ω1]2ω1



Rainbow triangle

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


CH =⇒ ∃f ⊢ ω1 6→[


♦ =⇒ ∃f ⊢ ω1 6→[

ω1]2ω1



Rainbow triangle

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


CH =⇒ ∃f ⊢ ω1 6→[


♦ =⇒ ∃f ⊢ ω1 6→[

ω1]2ω1



Rainbow triangle

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


CH =⇒ ∃f ⊢ ω1 6→[


♦ =⇒ ∃f ⊢ ω1 6→[

ω1]2ω1



Proof: CH =⇒ ∃f ⊢ ω1 6→[

ω1]2ω, no f -rainbow triangle

First try:

By transfinite induction on α < ω1 define f (ξ, α) for ξ < α.

First challenge : ω1 6→[

ω1]2ω

fix a |• -sequence {Aζ : ζ < ω1} ⊂

[

ω1]ω

if ζ < α and Aζ ⊂ α then f ′′[Aζ , {α}] = ω

Second challenge : no f -rainbow triangle ???


Proof: CH =⇒ ∃f ⊢ ω1 6→[


First try:



ω1]2ω


[

ω1]ω




Proof: CH =⇒ ∃f ⊢ ω1 6→[


First try:



ω1]2ω


[

ω1]ω




Proof: CH =⇒ ∃f ⊢ ω1 6→[


First try:



ω1]2ω


[

ω1]ω




Proof: CH =⇒ ∃f ⊢ ω1 6→[


First try:



ω1]2ω


[

ω1]ω




Proof: CH =⇒ ∃f ⊢ ω1 6→[


First try:



ω1]2ω


[

ω1]ω




Proof: CH =⇒ ∃f ⊢ ω1 6→[


First try:



ω1]2ω


[

ω1]ω




Proof: CH =⇒ ∃f ⊢ ω1 6→[


Second try: Fix the colouring first! Colour[

2ω]2 as follows:

∆(x , y) = min{n : x(n) 6= y(n)}. No ∆-rainbow triangle!

h : ω → ω s.t. h−1{k} is infinite for each k ∈ ω.

f (x , y) = h(∆(x , y)). No f -rainbow triangle!


Proof: CH =⇒ ∃f ⊢ ω1 6→[



2ω]2 as follows:





Proof: CH =⇒ ∃f ⊢ ω1 6→[



2ω]2 as follows:




2ω


Proof: CH =⇒ ∃f ⊢ ω1 6→[



2ω]2 as follows:




2ω

∆

x y ∆(x , y)ω


Proof: CH =⇒ ∃f ⊢ ω1 6→[



2ω]2 as follows:




2ω

∆

x y ∆(x , y)ω


Proof: CH =⇒ ∃f ⊢ ω1 6→[



2ω]2 as follows:




2ω

∆

x y ∆(x , y)ω ω

h

k


Proof: CH =⇒ ∃f ⊢ ω1 6→[



2ω]2 as follows:




2ω

∆

x y ∆(x , y)ω ω

h

k


Proof: CH =⇒ ∃f ⊢ ω1 6→[



2ω]2 as follows:




2ω

∆

x y ∆(x , y)ω ω

h

k


Proof: CH =⇒ ∃f ⊢ ω1 6→[


f (x , y) = h(∆(x , y)). No f -rainbow triangle.

Choose S = {sα : α < ω1} ⊂ 2ω s .t. f ↾[

S]2 works.

Enumerate[

ω1]ω as {Aξ : ξ < ω1}. Write Sξ = {sν : ν ∈ Aξ}.

if ξ < α and sα ∈ Sξ then f ′′[{α}, Sξ ] = ω.


Proof: CH =⇒ ∃f ⊢ ω1 6→[




S]2 works.

Enumerate[




Proof: CH =⇒ ∃f ⊢ ω1 6→[




S]2 works.

Enumerate[




Proof: CH =⇒ ∃f ⊢ ω1 6→[




S]2 works.

Enumerate[


if ξ < α and sα ∈ Sξ then f ′′[{α}, Sξ ] = ω.Sξ sα


Proof: CH =⇒ ∃f ⊢ ω1 6→[




S]2 works.

Enumerate[



2ω

Sξ sα ω ω

h

k

Summary


A weaker partition relation

Fact: If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.

f ⊢ ω1 6→[

(ω, ω1)]2ω1

iff ∀A ∈[

ω1]ω

∀B ∈[

ω1]ω1 f ′′[A, B] = ω1.

Theorem (Erdos, Hajnal, 1978)

If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

∀A, B ∈[

ω1]ω1 ∀ν < ω1 ∃α ∈ A ∃β ∈ B f (α, β) = ν and α < β.




(ω, ω1)]2ω1


ω]2

→ ω1.

f ⊢ ω1 6→[

(ω, ω1)]2ω1

iff ∀A ∈[

ω1]ω

∀B ∈[

ω1]ω1 f ′′[A, B] = ω1.


If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

∀A, B ∈[





(ω, ω1)]2ω1


ω]2

→ ω1.

f ⊢ ω1 6→[

(ω, ω1)]2ω1

iff ∀A ∈[

ω1]ω

∀B ∈[

ω1]ω1 f ′′[A, B] = ω1.


If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

∀A, B ∈[


A

B

ν

ω1




(ω, ω1)]2ω1


ω]2

→ ω1.

f ⊢ ω1 6→[

(ω, ω1)]2ω1

iff ∀A ∈[

ω1]ω

∀B ∈[

ω1]ω1 f ′′[A, B] = ω1.


If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

∀A, B ∈[


A

B

α

β

ν

ω1




(ω, ω1)]2ω1


ω]2

→ ω1.

f ⊢ ω1 6→[

(ω, ω1)]2ω1

iff ∀A ∈[

ω1]ω

∀B ∈[

ω1]ω1 f ′′[A, B] = ω1.


If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

∀A, B ∈[


A

B

α

β

ν

ω1




If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

Proof: If A, B0, B1, · · · ∈[

ω1]ω1 and ν0, ν1, · · · ∈ ω1 then ∃α ∈ A,

∃B′i ∈

[

Bi]ω1 such that f ′′[B′

i , {α}] = {νi}.




If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

Proof: If A, B0, B1, · · · ∈[

ω1]ω1 and ν0, ν1, · · · ∈ ω1 then ∃α ∈ A,

∃B′i ∈

[


i , {α}] = {νi}.




If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

Proof: If A, B0, B1, · · · ∈[

ω1]ω1 and ν0, ν1, · · · ∈ ω1 then ∃α ∈ A,

∃B′i ∈

[


i , {α}] = {νi}.

A B0 B1 Bn

ν0 ν1 νn




If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

Proof: If A, B0, B1, · · · ∈[

ω1]ω1 and ν0, ν1, · · · ∈ ω1 then ∃α ∈ A,

∃B′i ∈

[


i , {α}] = {νi}.

A B0 B1 Bn

B′0

B′1 B′

nαν0 ν1 νn




If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

Proof: If A, B0, B1, · · · ∈[

ω1]ω1 and ν0, ν1, · · · ∈ ω1 then ∃α ∈ A,

∃B′i ∈

[


i , {α}] = {νi}.

A B0 B1 Bn

B′0

B′1 B′

nαν0 ν1 νn

Summary


One more partition relation

If f ⊢ ω1 6→[

(ω1;ω1)]2

ω1then f realizes each function d :

[

ω]2

→ ω1.

∀A, B ∈[


Problem


(ω1,ω1)]2ω1

. Does f realize each function

d :[

ω]2

→ ω1?

∀A, B ∈[

ω1]ω1 ∀ν < ω1 ∃α ∈ A ∃β ∈ B f (α, β) = ν



If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

∀A, B ∈[


Problem


(ω1,ω1)]2ω1


d :[

ω]2

→ ω1?

∀A, B ∈[

ω1]ω1 ∀ν < ω1 ∃α ∈ A ∃β ∈ B f (α, β) = ν



If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

∀A, B ∈[


Problem


(ω1,ω1)]2ω1


d :[

ω]2

→ ω1?

∀A, B ∈[

ω1]ω1 ∀ν < ω1 ∃α ∈ A ∃β ∈ B f (α, β) = ν

A

B

ν

ω1



If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

∀A, B ∈[


Problem


(ω1,ω1)]2ω1


d :[

ω]2

→ ω1?

∀A, B ∈[

ω1]ω1 ∀ν < ω1 ∃α ∈ A ∃β ∈ B f (α, β) = ν

A

B

α

β

ν

ω1



If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

∀A, B ∈[


Problem


(ω1,ω1)]2ω1


d :[

ω]2

→ ω1?

∀A, B ∈[

ω1]ω1 ∀ν < ω1 ∃α ∈ A ∃β ∈ B f (α, β) = ν

A

B

α

β

ν

ω1



If f ⊢ ω1 6→[

(ω1;ω1)]2ω1


ω]2

→ ω1.

∀A, B ∈[


Problem


(ω1,ω1)]2ω1


d :[

ω]2

→ ω1?

∀A, B ∈[

ω1]ω1 ∀ν < ω1 ∃α ∈ A ∃β ∈ B f (α, β) = ν

A

B

α

β

ν

ω1


A new method


(ω1,ω1)]2

ω1. Does f realize each function d :

[

ω]2

→ ω1?

Fact

If f ⊢ ω1 6→[

(ω1, ω1)]2ω1

then d =⇒ f for each d :[

3]2

→ ω1.

Theorem (Todorcevic)

ω1 6→[

ω1]2ω1

.

Theorem (Shelah)

ω2 6→[

(ω2;ω2)]2ω2

.

Theorem (Moore)

ω1 6→[

(ω1;ω1)]2ω1

.


A new method


(ω1,ω1)]2


[

ω]2

→ ω1?

Fact

If f ⊢ ω1 6→[

(ω1, ω1)]2ω1


3]2

→ ω1.


ω1 6→[

ω1]2ω1

.

Theorem (Shelah)

ω2 6→[

(ω2;ω2)]2ω2

.

Theorem (Moore)

ω1 6→[

(ω1;ω1)]2ω1

.


A new method


(ω1,ω1)]2


[

ω]2

→ ω1?

Fact

If f ⊢ ω1 6→[

(ω1, ω1)]2ω1


3]2

→ ω1.


ω1 6→[

ω1]2ω1

.

Theorem (Shelah)

ω2 6→[

(ω2;ω2)]2ω2

.

Theorem (Moore)

ω1 6→[

(ω1;ω1)]2ω1

.


A new method


(ω1,ω1)]2


[

ω]2

→ ω1?

Fact

If f ⊢ ω1 6→[

(ω1, ω1)]2ω1


3]2

→ ω1.


ω1 6→[

ω1]2ω1

.

Theorem (Shelah)

ω2 6→[

(ω2;ω2)]2ω2

.

Theorem (Moore)

ω1 6→[

(ω1;ω1)]2ω1

.


A new method


(ω1,ω1)]2


[

ω]2

→ ω1?

Fact

If f ⊢ ω1 6→[

(ω1, ω1)]2ω1


3]2

→ ω1.


ω1 6→[

ω1]2ω1

.

Theorem (Shelah)

ω2 6→[

(ω2;ω2)]2ω2

.

Theorem (Moore)

ω1 6→[

(ω1;ω1)]2ω1

.


Recent results

If f ⊢ ω1 6→[

(ω1, ω1)]2ω1


3]2

→ ω1.

Theorem (Hajnal)

∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 s.t. d 6=⇒ f for some rainbow d :[

5]2

→ 10.


Recent results

If f ⊢ ω1 6→[

(ω1, ω1)]2ω1


3]2

→ ω1.

Theorem (Hajnal)

∃f ⊢ ω1 6→[

(ω1, ω1)]2


5]2

→ 10.


Recent results

If f ⊢ ω1 6→[

(ω1, ω1)]2ω1


3]2

→ ω1.

Theorem (Hajnal)

∃f ⊢ ω1 6→[

(ω1, ω1)]2


5]2

→ 10.


Recent results

If f ⊢ ω1 6→[

(ω1, ω1)]2ω1


3]2

→ ω1.

Theorem (Hajnal)

∃f ⊢ ω1 6→[

(ω1, ω1)]2


5]2

→ 10.


∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2

→ 2 be a Sierpinski colouring: {rα : α < ω1} ⊂ R; forα < β < ω1 let g(α, β) = 1 iff rα < rβ .Let e :

[

5]2

→ 2 be the “pentagon without diagonals”: e(α, β) = 1iff α ≡ β + 1 mod 5

(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2

g′′[A′; B′] = {i} and g′′[B′; A′] = {1 − i}


∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2

g′′[A′; B′] = {i} and g′′[B′; A′] = {1 − i}


∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2

g′′[A′; B′] = {i} and g′′[B′; A′] = {1 − i}


∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2

g′′[A′; B′] = {i} and g′′[B′; A′] = {1 − i}


∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2

g′′[A′; B′] = {i} and g′′[B′; A′] = {1 − i}


∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2

g′′[A′; B′] = {i} and g′′[B′; A′] = {1 − i}

A

Bω1


∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2

g′′[A′; B′] = {i} and g′′[B′; A′] = {1 − i}

A

B

A′

B′

ω1


∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2

g′′[A′; B′] = {i} and g′′[B′; A′] = {1 − i}

A

B

A′

B′

ω1


∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2

g′′[A′; B′] = {i} and g′′[B′; A′] = {1 − i}Let m ⊢ ω1 6→

[

(ω1;ω1)]2

5Put f (α, β) = 2m(α, β) + g(α, β).Assume that A, B ∈

[

ω1]ω1. Let ℓ = 2n + i < 10.

∃A′ ∈[

A]ω1 ∃B′ ∈

[

B]ω1 g′′[A′; B′] = {i} (or g′′[B′; A′] = {i})

∃α ∈ A′ ∃β ∈ B′ α < β and m(α, β) = n.f (α, β) = 2n + i = ℓ.∃m′ :

[

5]2

→ 5 s.t. d(α, β) = 2m′(α, β) + e(α, β) is 1–1.if d =⇒ f then d mod 2 =⇒ f mod 2, i.e. e =⇒ g.


∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2


[

(ω1;ω1)]2


[

ω1]ω1. Let ℓ = 2n + i < 10.

∃A′ ∈[

A]ω1 ∃B′ ∈

[

B]ω1 g′′[A′; B′] = {i} (or g′′[B′; A′] = {i})


[

5]2



∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2


[

(ω1;ω1)]2


[

ω1]ω1. Let ℓ = 2n + i < 10.

∃A′ ∈[

A]ω1 ∃B′ ∈

[

B]ω1 g′′[A′; B′] = {i} (or g′′[B′; A′] = {i})


[

5]2



∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2


[

(ω1;ω1)]2


[

ω1]ω1. Let ℓ = 2n + i < 10.

∃A′ ∈[

A]ω1 ∃B′ ∈

[

B]ω1 g′′[A′; B′] = {i} (or g′′[B′; A′] = {i})


[

5]2



∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2


[

(ω1;ω1)]2


[

ω1]ω1. Let ℓ = 2n + i < 10.

∃A′ ∈[

A]ω1 ∃B′ ∈

[

B]ω1 g′′[A′; B′] = {i} (or g′′[B′; A′] = {i})


[

5]2



∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2


[

(ω1;ω1)]2


[

ω1]ω1. Let ℓ = 2n + i < 10.

∃A′ ∈[

A]ω1 ∃B′ ∈

[

B]ω1 g′′[A′; B′] = {i} (or g′′[B′; A′] = {i})


[

5]2



∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2


[

(ω1;ω1)]2


[

ω1]ω1. Let ℓ = 2n + i < 10.

∃A′ ∈[

A]ω1 ∃B′ ∈

[

B]ω1 g′′[A′; B′] = {i} (or g′′[B′; A′] = {i})


[

5]2



∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2


[

(ω1;ω1)]2


[

ω1]ω1. Let ℓ = 2n + i < 10.

∃A′ ∈[

A]ω1 ∃B′ ∈

[

B]ω1 g′′[A′; B′] = {i} (or g′′[B′; A′] = {i})


[

5]2



∃f ⊢ ω1 6→[

(ω1, ω1)]2

10 ∃d :[

5]2 1–1

−→ 10 d 6=⇒ f .

Let g :[

ω1]2


[

5]2


(⋆) e 6=⇒ g.(†) ∀A, B ∈

[

ω1]ω1 ∃A′ ∈

[

A]ω1 ∃B′ ∈

[

B]ω1 ∃i < 2


[

(ω1;ω1)]2


[

ω1]ω1. Let ℓ = 2n + i < 10.

∃A′ ∈[

A]ω1 ∃B′ ∈

[

B]ω1 g′′[A′; B′] = {i} (or g′′[B′; A′] = {i})


[

5]2



Recent results

[Hajnal] ∃f ⊢ ω1 6→[

(ω1, ω1)]2


5]2

→ 10.

Theorem (Hajnal)

If f ⊢ ω1 6→[

(ω1, ω1)]2ω

then there exists an infinite f -rainbow set .

Summary


Recent results

[Hajnal] ∃f ⊢ ω1 6→[

(ω1, ω1)]2


5]2

→ 10.

Theorem (Hajnal)

If f ⊢ ω1 6→[

(ω1, ω1)]2ω

then there exists an infinite f -rainbow set .

Summary


Recent results

Shelah:CH =⇒ ∃f ⊢ ω1 6→

[

ω1]2ω


♦ =⇒ ∃g ⊢ ω1 6→[

ω1]2ω1

, no g-rainbow triangle.

Theorem (Hajnal)

∃f ⊢ ω1 6→[

ω1]2

6 and ∃d :[

4]2

→ 6 rainbow s.t. d 6=⇒ f .

Summary


Recent results


[

ω1]2

ω, no f -rainbow triangle.

♦ =⇒ ∃g ⊢ ω1 6→[

ω1]2

ω1, no g-rainbow triangle.

Theorem (Hajnal)

∃f ⊢ ω1 6→[

ω1]2

6 and ∃d :[

4]2


Summary


Recent results


[

ω1]2


♦ =⇒ ∃g ⊢ ω1 6→[

ω1]2

ω1, no g-rainbow triangle.

Theorem (Hajnal)

∃f ⊢ ω1 6→[

ω1]2

6 and ∃d :[

4]2


Summary


Stepping up:colourings of ω2

Negative theorems

Shelah: CH =⇒ ∃f ⊢ ω1 6→[

ω1]2ω


Theorem (Shelah)

CH + 2ω1 = ω2 =⇒ ∃f ⊢ ω2 6→[

ω2]2ω1




Negative theorems


ω1]2


Theorem (Shelah)

CH + 2ω1 = ω2 =⇒ ∃f ⊢ ω2 6→[

ω2]2ω1




Negative theorems


ω1]2


Theorem (Shelah)

CH + 2ω1 = ω2 =⇒ ∃f ⊢ ω2 6→[

ω2]2ω1




Negative theorems


ω1]2


Theorem (Shelah)

CH + 2ω1 = ω2 =⇒ ∃f ⊢ ω2 6→[

ω2]2ω1




Hajnal: ∃f ⊢ ω1 6→[

(ω1, ω1)]2

10s.t. d 6=⇒ f for some rainbow d :

[

5]2

→ 10.

Theorem

CH =⇒ ∃f ⊢ ω2 6→[

(ω2, ω2)]2

10 s.t. d 6=⇒ f for some rainbow

d :[

5]2

→ 10.


ω1]2

6 and ∃d :[

4]2


Theorem

CH =⇒ ∃f ⊢ ω2 6→[

ω2]2

6 and ∃d :[

4]2





(ω1, ω1)]2


[

5]2

→ 10.

Theorem

CH =⇒ ∃f ⊢ ω2 6→[

(ω2, ω2)]2


d :[

5]2

→ 10.


ω1]2

6 and ∃d :[

4]2


Theorem

CH =⇒ ∃f ⊢ ω2 6→[

ω2]2

6 and ∃d :[

4]2





(ω1, ω1)]2


[

5]2

→ 10.

Theorem

CH =⇒ ∃f ⊢ ω2 6→[

(ω2, ω2)]2


d :[

5]2

→ 10.


ω1]2

6 and ∃d :[

4]2


Theorem

CH =⇒ ∃f ⊢ ω2 6→[

ω2]2

6 and ∃d :[

4]2





(ω1, ω1)]2


[

5]2

→ 10.

Theorem

CH =⇒ ∃f ⊢ ω2 6→[

(ω2, ω2)]2


d :[

5]2

→ 10.


ω1]2

6 and ∃d :[

4]2


Theorem

CH =⇒ ∃f ⊢ ω2 6→[

ω2]2

6 and ∃d :[

4]2




Positive theorems

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


(ω1, ω2)]2

2. Does f realize each function

d :[

ω1]2

→ 2?


(a) If f ⊢ ω2 6→[

(ω1;ω2)]2

2 then V Fn(ω,2) |= f ⊢ ω2 6→[

(ω1;ω2)]2

2.

(b) In V Fn(ω,2) there is a colouring c :[

ω1]2

→ 2 such that c 6=⇒ f forany f ∈ V.

(c) It is consistent that ∃f ⊢ ω2 6→[

(ω1;ω2)]2

2 such that c 6=⇒ f for some

c :[

ω1]2

→ 2.



Positive theorems

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


(ω1, ω2)]2


d :[

ω1]2

→ 2?


(a) If f ⊢ ω2 6→[

(ω1;ω2)]2

2 then V Fn(ω,2) |= f ⊢ ω2 6→[

(ω1;ω2)]2

2.


ω1]2



(ω1;ω2)]2


c :[

ω1]2

→ 2.



Positive theorems

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


(ω1, ω2)]2


d :[

ω1]2

→ 2?


(a) If f ⊢ ω2 6→[

(ω1;ω2)]2

2 then V Fn(ω,2) |= f ⊢ ω2 6→[

(ω1;ω2)]2

2.


ω1]2



(ω1;ω2)]2


c :[

ω1]2

→ 2.



Positive theorems

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


(ω1, ω2)]2


d :[

ω1]2

→ 2?


(a) If f ⊢ ω2 6→[

(ω1;ω2)]2

2 then V Fn(ω,2) |= f ⊢ ω2 6→[

(ω1;ω2)]2

2.


ω1]2



(ω1;ω2)]2


c :[

ω1]2

→ 2.



Positive theorems

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


(ω1, ω2)]2


d :[

ω1]2

→ 2?


(a) If f ⊢ ω2 6→[

(ω1;ω2)]2

2 then V Fn(ω,2) |= f ⊢ ω2 6→[

(ω1;ω2)]2

2.


ω1]2



(ω1;ω2)]2


c :[

ω1]2

→ 2.



Positive theorems

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


(ω1, ω2)]2


d :[

ω1]2

→ 2?


(a) If f ⊢ ω2 6→[

(ω1;ω2)]2

2 then V Fn(ω,2) |= f ⊢ ω2 6→[

(ω1;ω2)]2

2.


ω1]2



(ω1;ω2)]2


c :[

ω1]2

→ 2.



Positive theorems

If f ⊢ ω1 6→[

(ω, ω1)]2ω1


ω]2

→ ω1.


(ω1, ω2)]2


d :[

ω1]2

→ 2?


(a) If f ⊢ ω2 6→[

(ω1;ω2)]2

2 then V Fn(ω,2) |= f ⊢ ω2 6→[

(ω1;ω2)]2

2.


ω1]2



(ω1;ω2)]2


c :[

ω1]2

→ 2.


Colourings of ω2

Problem (Erdos, Hajnal, 1978)

Assume GCH and f ⊢ ω2 6→(ω1.+ ω)2


f :[

ω1]2

→ 2?

Theorem (Hajnal)

Con ( GCH + ∃f ⊢ ω2 6→[(ω1;ω)]2ω1).

∀A ∈[

ω2]ω1 ∀B ∈

[

ω2]ω if sup A < min B then f ′′[A, B] = ω1


Colourings of ω2




f :[

ω1]2

→ 2?

Theorem (Hajnal)

Con ( GCH + ∃f ⊢ ω2 6→[(ω1;ω)]2ω1).

∀A ∈[

ω2]ω1 ∀B ∈

[



Colourings of ω2




f :[

ω1]2

→ 2?

Theorem (Hajnal)

Con ( GCH + ∃f ⊢ ω2 6→[(ω1;ω)]2ω1).

∀A ∈[

ω2]ω1 ∀B ∈

[



Colourings of ω2




f :[

ω1]2

→ 2?

Theorem (Hajnal)

Con ( GCH + ∃f ⊢ ω2 6→[(ω1;ω)]2ω1).

∀A ∈[

ω2]ω1 ∀B ∈

[


ω2A B


Colourings of ω2




f :[

ω1]2

→ 2?

Theorem (Hajnal)

Con ( GCH + ∃f ⊢ ω2 6→[(ω1;ω)]2ω1).

∀A ∈[

ω2]ω1 ∀B ∈

[


ω2A B


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).

Assume CH. Define P = 〈P,�〉 σ-complete, ω2-c.c.Underlying set : 〈X , c,A, ξ〉

X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1

〈Y , d ,B, ζ〉 � 〈X , c,A, ξ〉 iffY ⊃ X ,d ⊃ c, B ⊃ A, ζ ≥ ξ

(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].

Y X ω2


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].

Y X ω2Aβ


Con ( GCH + ∃f ⊢ ω2 6→[(ω1; ω)]2ω1).


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1


(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].

Y X ω2Aβ


Colourings of ω2

Problem of Erdos and Hajnal:Assume GCH and f ⊢ ω2 6→[(ω1; ω)]2ω1

. Does f realize every c :[

ω1]2

→ 2?

(Shelah): If f ⊢ ω2 6→[

(ω1; ω2)]2

2then V Fn(ω,2) |= f ⊢ ω2 6→

[

(ω1; ω2)]2

2.

Theorem

(a) If f ⊢ ω2 6→[(ω1;ω)]2ω1then V Fn(κ,2) |= “ f ⊢ ω2 6→[(ω1;ω)]2ω1

” .

(b) Con (GCH +∃f ⊢ ω2 6→[

(ω1;ω)]2ω1

and ∃c :[

ω1]2

→ 2 s.t. c 6=⇒ f .)


Colourings of ω2



ω1]2

→ 2?


(ω1; ω2)]2

2then V Fn(ω,2) |= f ⊢ ω2 6→

[

(ω1; ω2)]2

2.

Theorem


” .

(b) Con (GCH +∃f ⊢ ω2 6→[

(ω1;ω)]2ω1

and ∃c :[

ω1]2

→ 2 s.t. c 6=⇒ f .)


Colourings of ω2



ω1]2

→ 2?


(ω1; ω2)]2

2then V Fn(ω,2) |= f ⊢ ω2 6→

[

(ω1; ω2)]2

2.

Theorem


” .

(b) Con (GCH +∃f ⊢ ω2 6→[

(ω1;ω)]2ω1

and ∃c :[

ω1]2

→ 2 s.t. c 6=⇒ f .)


Colourings of ω2



ω1]2

→ 2?


(ω1; ω2)]2

2then V Fn(ω,2) |= f ⊢ ω2 6→

[

(ω1; ω2)]2

2.

Theorem


” .

(b) Con (GCH +∃f ⊢ ω2 6→[

(ω1;ω)]2ω1

and ∃c :[

ω1]2

→ 2 s.t. c 6=⇒ f .)


Colourings of ω2



ω1]2

→ 2?


(ω1; ω2)]2

2then V Fn(ω,2) |= f ⊢ ω2 6→

[

(ω1; ω2)]2

2.

Theorem


” .

(b) Con (GCH +∃f ⊢ ω2 6→[

(ω1;ω)]2ω1

and ∃c :[

ω1]2

→ 2 s.t. c 6=⇒ f .)


Rainbow on ω2

Problem (Hajnal)

Assume GCH + f ⊢ ω2 6→[(ω1;ω)]2ω1. Does there exist an uncountable

f -rainbow set?

Theorem

It is consistent that GCH holds and there is a function f :[

ω2]2

→ ω1

such that

(1) f ⊢ ω2 6→[(ω1;ω)]2ω1,

(2) there is no uncountable f -rainbow.


Rainbow on ω2

Problem (Hajnal)


f -rainbow set?

Theorem


ω2]2

→ ω1

such that

(1) f ⊢ ω2 6→[(ω1;ω)]2ω1,



Rainbow on ω2

Problem (Hajnal)


f -rainbow set?

Theorem


ω2]2

→ ω1

such that

(1) f ⊢ ω2 6→[(ω1;ω)]2ω1,



Rainbow on ω2

Problem (Hajnal)


f -rainbow set?

Theorem


ω2]2

→ ω1

such that

(1) f ⊢ ω2 6→[(ω1;ω)]2ω1,



Proof: ∃f ⊢ ω2 6→[(ω1; ω)]2ω1, no uncountable rainbow

First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1

Ordering: 〈Y , d ,B, ζ〉 � 〈X , c,A, ξ〉 iffY ⊃ X ,d ⊃ c, B ⊃ A, ζ ≥ ξ

(1) ∀A ∈ A ∀β ∈ (Y \ X) ∩ min A ξ ⊂ d ′′[{β}, A].(2) Seal certain countable rainbow sets !

∀A ∈ A ∀β ∈ (Y \ X) the set A ∪ {β} is not a d-rainbow .

〈P,�〉 does not satisfy ω2-c.c.!



First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1







First try:


X ∈[

ω2]ω

, c :[

X]2

→ ω1

A ⊂ P(X), |A| ≤ ω,ξ ∈ ω1






Colouring of 2ω1

It is consistent that GCH holds and ∃f ⊢ ω2 6→[(ω1; ω)]2ω1s.t. there is no un-

countable f -rainbow.

TheoremIt is consistent that CH holds, 2ω1 is arbitrarily large and∃g ⊢ 2ω1 6→[(ω1, ω2)]

2ω1

s. t. there is no uncountable g-rainbow subsetof 2ω1.


Colouring of 2ω1




2ω1



Colouring of 2ω1




2ω1



Colouring of 2ω1




2ω1



Colouring of 2ω1




2ω1



The →∗ relation

f :[

X]n

→ C is κ-bounded iff |f−1{c}| ≤ κ for each c ∈ C.

λ →∗ (α)nκ−bdd iff for every κ-bounded colouring of

[

λ]n there is a

rainbow set of order type α,

→poly

Theorem (Galvin)

λ → (α)nk implies λ →∗ (α)n

k−bdd.


The →∗ relation

f :[

X]n



[

λ]n there is a


→poly

Theorem (Galvin)


k−bdd.


The →∗ relation

f :[

X]n



[

λ]n there is a


→poly

Theorem (Galvin)


k−bdd.


The →∗ relation

f :[

X]n



[

λ]n there is a


→poly

Theorem (Galvin)


k−bdd.


The →∗ relation

f :[

X]n



[

λ]n there is a


→poly

Theorem (Galvin)


k−bdd.


The →∗ relation

f :[

X]n



[

λ]n there is a


→poly

Theorem (Galvin)


k−bdd.


The →∗ relation

f :[

X]n



[

λ]n there is a


→poly

Theorem (Galvin)


k−bdd.


λ → (α)nκ implies λ →∗ (α)n

κ−bdd

Proof:

Let f :[

λ]n

→ λ be κ-bounded

There is a function g :[

λ]n

→ κ such that: if f (A) = f (B) theng(A) 6= g(B)

If Y ⊂ λ is g-monochromatic then Y is f -rainbow



κ−bdd

Proof:

Let f :[

λ]n



λ]n





κ−bdd

Proof:

Let f :[

λ]n



λ]n





κ−bdd

Proof:

Let f :[

λ]n



λ]n





κ−bdd

Proof:

Let f :[

λ]n



λ]n




Basic theorems on →∗


k−bdd.

Corollary(Galvin)

ω1 →∗ (α)22−bdd for α < ω1.

Theorem (Galvin)

CH implies that ω1 6→∗(ω1)

22−bdd .


PFA implies that ω1 →∗ (ω1)22−bdd .




k−bdd.

Corollary(Galvin)

ω1 →∗ (α)22−bdd for α < ω1.

Theorem (Galvin)


22−bdd .






k−bdd.

Corollary(Galvin)

ω1 →∗ (α)22−bdd for α < ω1.

Theorem (Galvin)


22−bdd .






k−bdd.

Corollary(Galvin)

ω1 →∗ (α)22−bdd for α < ω1.

Theorem (Galvin)


22−bdd .




Under Martin’s Axiom

If c :[

ω1]2

→ ω1 is 2-bounded then T.F.A.E.

c ⊢ ω1 6→∗(ω1)

22−bdd

there is no uncountable c-rainbow

Theorem (Abraham, Cummings, Smyth)

It is consistent that there is a function c :[

ω1]2

→ ω1 whichc.c.c-indestructibly establishes ω1 6→

∗(ω1)22−bdd .

If CH holds and there is a Suslin-tree then there is a functionc′ :

[

ω1]2

→ 2 and there is a c.c.c poset Q such that

(a) V |= there is no uncountable c′-rainbow set,

(b) V Q |= there is an uncountable c′-rainbow set.



If c :[

ω1]2


c ⊢ ω1 6→∗(ω1)

22−bdd




ω1]2


∗(ω1)22−bdd .


[

ω1]2






If c :[

ω1]2


c ⊢ ω1 6→∗(ω1)

22−bdd




ω1]2


∗(ω1)22−bdd .


[

ω1]2






If c :[

ω1]2


c ⊢ ω1 6→∗(ω1)

22−bdd




ω1]2


∗(ω1)22−bdd .


[

ω1]2






If c :[

ω1]2


c ⊢ ω1 6→∗(ω1)

22−bdd




ω1]2


∗(ω1)22−bdd .


[

ω1]2






Con ( ∃c :[

ω1]2

→ ω1 s.t for each c.c.c P V P |= f ⊢ ω1 6→∗(ω1)

22−bdd .)

Con( ∃c′ :[

ω1]2




TheoremIf GCH holds then for each k ∈ ω there is a k-bounded colouringf :

[

ω1]2

→ ω1 and there are two c.c.c posets P and Q such thatVP |= “f c.c.c-indestructibly establishes ω1 6→

∗[(ω1;ω1)]k−bdd ”,but

VQ |= “ ω1 is the union of countably many f -rainbow sets ”.

f is k-bounded,for each A ∈

[

ω1]ω1 and B ∈

[

ω1]ω1 there is ξ such that

|{〈α, β〉 ∈ A × B : α < β ∧ f (α, β) = ξ}| = k .



Con ( ∃c :[

ω1]2


22−bdd .)

Con( ∃c′ :[

ω1]2





[

ω1]2





[

ω1]ω1 and B ∈

[


|{〈α, β〉 ∈ A × B : α < β ∧ f (α, β) = ξ}| = k .



Con ( ∃c :[

ω1]2


22−bdd .)

Con( ∃c′ :[

ω1]2





[

ω1]2





[

ω1]ω1 and B ∈

[


|{〈α, β〉 ∈ A × B : α < β ∧ f (α, β) = ξ}| = k .



Con ( ∃c :[

ω1]2


22−bdd .)

Con( ∃c′ :[

ω1]2





[

ω1]2





[

ω1]ω1 and B ∈

[


|{〈α, β〉 ∈ A × B : α < β ∧ f (α, β) = ξ}| = k .



Con ( ∃c :[

ω1]2


22−bdd .)

Con( ∃c′ :[

ω1]2





[

ω1]2





[

ω1]ω1 and B ∈

[


|{〈α, β〉 ∈ A × B : α < β ∧ f (α, β) = ξ}| = k .



Con ( ∃c :[

ω1]2


22−bdd .)

Con( ∃c′ :[

ω1]2





[

ω1]2





[

ω1]ω1 and B ∈

[


|{〈α, β〉 ∈ A × B : α < β ∧ f (α, β) = ξ}| = k .


A black box theorem

Based on some results of Abraham and Todorcevic.

Fnm(ω1, K ) = {s : s is a function, dom(s) ∈[

ω1]m

, ran(s) ⊂ K}

〈sα : α < ω1〉 ⊂ Fnm(ω1, K ) is dom-disjoint iff dom(sα)∩ dom(sβ) = ∅


A black box theorem



ω1]m

, ran(s) ⊂ K}



A black box theorem



ω1]m

, ran(s) ⊂ K}



A black box theorem



ω1]m

, ran(s) ⊂ K}



A black box theorem

DefinitionLet G be a graph on ω1 × K , m ∈ ω. We say that G is m-solid if givenany dom-disjoint sequence 〈sα : α < ω1〉 ⊂ Fnm(ω1, K ) there areα < β < ω1 such that

[sα, sβ ] ⊂ G.

G is called strongly solid iff it is m-solid for each m ∈ ω.

K

ω1

G


A black box theorem


[sα, sβ ] ⊂ G.


K

ω1

sαsβ

sγ

dom(sα) dom(sβ) dom(sγ)

G


A black box theorem


[sα, sβ ] ⊂ G.


K

ω1

sαsβ

sγ


G


A black box theorem


[sα, sβ ] ⊂ G.


K

ω1

sαsβ

sγ


G


A blackbox theorem

Let G be a graph on ω1 × K , m ∈ ω. We say that G is m-solid if given anydom-disjoint sequence 〈sα : α < ω1〉 ⊂ Fnm(ω1, K ) there are α < β < ω1 suchthat

[sα, sβ] ⊂ G.


TheoremAssume 2ω1 = ω2. If G is a strongly solid graph on ω1 × K , where|K | ≤ 2ω1 , then for each m ∈ ω there is a c.c.c poset P of size ω2 suchthat

V P |= “G is c.c.c-indestructibly m-solid. ”


A blackbox theorem


[sα, sβ] ⊂ G.





A blackbox theorem


[sα, sβ] ⊂ G.





A blackbox theorem


[sα, sβ] ⊂ G.





A blackbox theorem


[sα, sβ] ⊂ G.





A blackbox theorem


[sα, sβ] ⊂ G.





Bibliography

Abraham, U; Cummings, J; Smyth, C; Some results in polychromatic Ramsey theory. J.Symb. Log. 72, no. 3, 865-896 (2007)

Hajnal, A; Rainbow Ramsey Theorems for Colorings Establishing Negative PartitionRelations, Fund. Math. 198 (2008), 255-262.

Shelah, S; Coloring without triangles and partition relations, Israel J. Math. 20 (1975), 1-12

Soukup, L; Indestructible colourings and rainbow Ramsey theorems, arxiv preprint.

Todorcevic, S; Forcing positive partition relations. Trans. Am. Math. Soc. 280, 703-720(1983)

http://www.renyi.hu/∼soukup


Bibliography

Abraham, U; Cummings, J; Smyth, C; Some results in polychromatic Ramsey theory. J.Symb. Log. 72, no. 3, 865-896 (2007)

Hajnal, A; Rainbow Ramsey Theorems for Colorings Establishing Negative PartitionRelations, Fund. Math. 198 (2008), 255-262.

Shelah, S; Coloring without triangles and partition relations, Israel J. Math. 20 (1975), 1-12

Soukup, L; Indestructible colourings and rainbow Ramsey theorems, arxiv preprint.

Todorcevic, S; Forcing positive partition relations. Trans. Am. Math. Soc. 280, 703-720(1983)

http://www.renyi.hu/∼soukup


Summary


Summary

ZFC ω1 6→[

(ω, ω1)]2ω1

=⇒ ∀d :[

ω]2

→ ω1


Summary

ZFC ω1 6→[

(ω, ω1)]2ω1

=⇒ ∀d :[

ω]2

→ ω1

CH ω1 6→[

ω1]2ω

∧ no rainbow K3

♦ ω1 6→[

ω1]2ω1

∧ no rainbow K3


Summary

ZFC ω1 6→[

(ω, ω1)]2ω1

=⇒ ∀d :[

ω]2

→ ω1

ZFC ω1 6→[

(ω1;ω1)]2ω1

=⇒ ∀d :[

ω]2

→ ω1

CH ω1 6→[

ω1]2ω

∧ no rainbow K3

♦ ω1 6→[

ω1]2ω1

∧ no rainbow K3


Summary

ZFC ω1 6→[

(ω, ω1)]2ω1

=⇒ ∀d :[

ω]2

→ ω1

ZFC ω1 6→[

(ω1;ω1)]2ω1

=⇒ ∀d :[

ω]2

→ ω1

ZFC ω1 6→[

(ω1, ω1)]2ω1

=⇒ ∀d :[

3]2

→ ω1

ZFC ω1 6→[

(ω1, ω1)]2

10 ∧ not univ for K5-rainbows

ZFC ω1 6→[

(ω1, ω1)]2ω1

=⇒ ∃ infinite rainbow

CH ω1 6→[

ω1]2ω

∧ no rainbow K3

♦ ω1 6→[

ω1]2ω1

∧ no rainbow K3


Summary

ZFC ω1 6→[

(ω, ω1)]2ω1

=⇒ ∀d :[

ω]2

→ ω1

ZFC ω1 6→[

(ω1;ω1)]2ω1

=⇒ ∀d :[

ω]2

→ ω1

ZFC ω1 6→[

(ω1, ω1)]2ω1

=⇒ ∀d :[

3]2

→ ω1

ZFC ω1 6→[

(ω1, ω1)]2


ZFC ω1 6→[

(ω1, ω1)]2ω1

=⇒ ∃ infinite rainbow

CH ω1 6→[

ω1]2ω

∧ no rainbow K3

♦ ω1 6→[

ω1]2ω1

∧ no rainbow K3

ZFC ω1 6→[

ω1]2



Documents

Rainbow Colourings - Portal · → λ. Y ⊂ X is an f-rainbowiff f ↾ Y n is 1–1 . Problem (Erdos, Hajnal, 1967, Problem 68)˝ Assume that f : ω1 2 → 3 establishes ω1 →