Andr es Eduardo CaicedoCaicedo Ramsey theory of ordinals. Countable ordinals The ordinals are the order types of well-ordered sets. Recall that a linear order(X;

Ramsey theory of very small countable ordinals

Andres Eduardo Caicedo

Department of MathematicsBoise State University

English version of the talk given at the Conference50 anos de la Carrera de Matematicas

Universidad de los Andes, Bogota, September 17–19, 2014

Caicedo Ramsey theory of ordinals

Introduction

My interest in Ramsey theory started as an undergrad, in LosAndes, reading a set of notes by Ronald Graham, and the book byGraham, Rothschild, and Spencer.


My favorite anecdote to motivate Ramsey’s theorem is a story atthe beginning of Extremal and Probabilistic Combinatorics, byNoga Alon and Michael Krivelevich (The Princeton companionto mathematics, Timothy Gowers, June Barrow-Green, ImreLeader, eds., PUP, 2008. Chapter IV.19, pp. 562–575).


In the course of an examination of friendship betweenchildren some fifty years ago, the Hungarian sociologistSandor Szalai observed that among any group of abouttwenty children he checked, he could always find fourchildren any two of whom were friends, or else fourchildren no two of whom were friends.Despite the temptation to try to draw sociologicalconclusions, Szalai realized that this might well be amathematical phenomenon rather than a sociologicalone. Indeed, a brief discussion with the mathematiciansErdos, Turan, and Sos convinced him this was the case.

What Szalai is observing, in the languaje of Ramsey theory, is thatr(4, 4) ≤ 20, that is, any graph on 20 vertices either contains acopy of K4, the complete graph on 4 vertices, or a copy of K4, theindependent set of size 4. (In fact, r(4, 4) = 18).








We denote byn→ (m, l)2

the statement that any graph on n vertices has a copy of Km orKl.This is equivalent to saying that whenever the edges of Kn arecolored red and blue, either we have a red copy of Km, or a bluecopy of Kl. We say that the relevant subgraph is monochromatic.


Theorem (Ramsey)

For all m, l there is an n such that any graph on n verticescontains a copy of Km or Kl, that is, n→ (m, l)2.

Define the Ramsey number r(m, l) as the smallest possible valueof n.

Clearly, r(n,m) = r(m,n), r(1,m) = 1, and r(2,m) = m.


Frank Plumpton Ramsey (22 Feb., 1903 – 19 Jan., 1930).Philosopher, economist, mathematician.


For example:

r(3, 3) = 6.

To show that it is at least 6, it suffices to consider the followingcoloring of K5:


r(3, 3) = 6.

To see that it is at most 6, note that in any coloring of K6 theremust be three edges with the same color and sharing a commonvertex:

(The mathematical coloring book, Soifer, p.242.)


r(4, 3) = 9.

To see that it is at least 9, it suffices to consider the followinggraph:

(http://plus.maths.org/content/friends-and-strangers)


http://plus.maths.org/content/friends-and-strangers

r(4, 3) = 9.

To see that it is at most 10, argue as with r(3, 3) ≤ 6, consideringthe following diagram:

(Ramsey theory, Graham-Rothschild-Spencer, p.4.)

To see that it is at most 9 requires an additional argument.


r(4, 4) = 18.

To see that it is at least 18, it suffices to consider the followinggraph:

(The mathematical coloring book, Soifer, p.246.)


r(3, 5) = 14, r(3, 6) = 18, r(3, 7) = 23, r(3, 8) = 28,r(3, 9) = 36.

40 ≤ r(3, 10) ≤ 42.

r(4, 5) = 25, 36 ≤ r(4, 6) ≤ 41.

These results require extensive computations. For instance, thatr(4, 5) ≥ 25 was shown by Kalbfleisch in 1965. That r(4, 5) ≤ 25was shown by McKay and Radziszowski in 1993.


r(3, 5) = 14, r(3, 6) = 18, r(3, 7) = 23, r(3, 8) = 28,r(3, 9) = 36.

40 ≤ r(3, 10) ≤ 42.

r(4, 5) = 25, 36 ≤ r(4, 6) ≤ 41.



r(3, 5) = 14, r(3, 6) = 18, r(3, 7) = 23, r(3, 8) = 28,r(3, 9) = 36.

40 ≤ r(3, 10) ≤ 42.

r(4, 5) = 25, 36 ≤ r(4, 6) ≤ 41.



r(3, 5) = 14, r(3, 6) = 18, r(3, 7) = 23, r(3, 8) = 28,r(3, 9) = 36.

40 ≤ r(3, 10) ≤ 42.

r(4, 5) = 25, 36 ≤ r(4, 6) ≤ 41.



The two implementations required 3.2 years and 6 yearsof cpu time on Sun Microsystems computers (mostlySparcstation SLC). This was achieved without unduedelay by employing a large number of computers (up to110 at once).

McKay-Radziszowski (1993)


43 ≤ r(5, 5) ≤ 49.

102 ≤ r(6, 6) ≤ 165.

Suppose aliens invade the Earth and threaten to destroyit in a year if human beings do not find r(5, 5). It is(probably) possible to save the Earth by putting togetherthe world’s best minds and computers. If, however, theinvaders were to demand r(6, 6), the human beingsmight as well attempt a preemptive strike without eventrying to ponder the problem.

P. Erdos (1993)


43 ≤ r(5, 5) ≤ 49.

102 ≤ r(6, 6) ≤ 165.


P. Erdos (1993)


43 ≤ r(5, 5) ≤ 49.

102 ≤ r(6, 6) ≤ 165.


P. Erdos (1993)



Ramsey’s theorem also admits infinitary versions, and in thiscontext we talk of the partition calculus.


The notation generalizes as expected: If κ, λ, µ are cardinals(“sizes” of sets), κ→ (λ, µ)2 is defined just as in the finite case.Recall that ℵ0 = |N|.

Theorem (Ramsey)

ℵ0 → (ℵ0,ℵ0)2, that is, any graph on infinitely many verticescontains either a complete infinite graph, Kℵ0 , or an infiniteindependent graph, Kℵ0 .


Erdos and Rado generalized this to all cardinals:

Theorem (Erdos-Rado)

For all cardinals κ, λ there is a cardinal µ such that µ→ (κ, λ)2.

The number µ tends to be much larger than κ and λ. For instance,if κ and λ are uncountable, then µ must be larger than c = |R|.We are interested in a variant where, rather than the size of the setof vertices, what matters is its order type. In this talk, we restrictourselves to considering ordinals.


Countable ordinals

The ordinals are the order types of well-ordered sets. Recall that alinear order(X,<) is a well-order if and only if every non-emptysubset of X has a first element. The usual order of the naturalnumbers is an example.Any two ordinals can be compared (one is an initial segment of theother). What we obtain with the ordinals is a transfinite way ofcontinuing the sequence of the naturals.


We have two operations at our disposal to continue the sequence:

Given an ordinal α, we can “add 1” and obtain the successorordinal α+ 1.

Or we can consider a non-empty collection of ordinals,without maximum, and add its supremum. The ordinals soobtained are the limit ordinals.

We denote by ω the first infinite ordinal: We begin with thenaturals, 0, 1, 2, . . . , and then

ω, ω+1, ω+2, . . . , ω2, ω2+1, . . . , ω3, . . . , ω4, . . . , ω2, . . . ωω, . . . , ω1, . . .






ω, ω+1, ω+2, . . . , ω2, ω2+1, . . . , ω3, . . . , ω4, . . . , ω2, . . . ωω, . . . , ω1, . . .






ω, ω+1, ω+2, . . . , ω2, ω2+1, . . . , ω3, . . . , ω4, . . . , ω2, . . . ωω, . . . , ω1, . . .






ω, ω+1, ω+2, . . . , ω2, ω2+1, . . . , ω3, . . . , ω4, . . . , ω2, . . . ωω, . . . , ω1, . . .






ω, ω+1, ω+2, . . . , ω2, ω2+1, . . . , ω3, . . . , ω4, . . . , ω2, . . . ωω, . . . , ω1, . . .


In this talk I will concentrate on ordinals smaller than ω2, but ω1

(the first uncountable ordinal) makes a brief cameo. As usual, weidentify each ordinal with its set of predecessors, so that α = [0, α).

Definition

If α, β, γ are ordinals, we say that α→ (β, γ)2 if and only if forany red-blue coloring of the edges of Kα we can find a subsetH ⊆ α with KH monochromatic and such that, if its color is red,then H has order type at least β while, if it is blue, then H hasorder type at least γ.As before, r(β, γ) is the least α such that α→ (β, γ)2.


Por example, r(ω, ω) = ω (Ramsey), but already r(ω + 1, ω) = ω1

(Erdos-Rado, Specker).In fact, ω1 → (ω1, ω + 1) (Erdos-Rado). But this means that if wewant to restrict ourselves to countable resources, then we mustnecessarily accept that one of the ordinals under consideration isactually finite.On the other hand, Erdos and Milner proved that if α is countable,and n is finite, then r(α, n) is also countable.








Can we be more precise? Consider an example:

r(ω + 1, 3) = ω2 + 1.

ω2 is not enough: It suffices to color red all edges in either copy ofω, and color blue all edges between both copies. As the diagramsuggests, we are basically considering a graph on two vertices.

[0, ω) [ω, ω2)



r(ω + 1, 3) = ω2 + 1.


[0, ω) [ω, ω2)



r(ω + 1, 3) = ω2 + 1.


[0, ω) [ω, ω2)



r(ω + 1, 3) = ω2 + 1.


[0, ω) [ω, ω2)


r(ω + 1, 3) = ω2 + 1.

To see that ω2 + 1 is an upper bound, consider a coloring ofKω2+1. Since ω → (ω, ω)2, we may assume that all edges in K[0,ω)

and K[ω,ω2) are red.If some α > ω is connected to infinitely many vertices in [0, ω)with a red edge, we are done. Essentially, this means we havereproduced the previous diagram. The blue edge means that everypoint in [ω, ω2) is connected to almost every point in [0, ω) with ablue edge.

[0, ω) [ω, ω2)

But we have not considered yet the last point.


r(ω + 1, 3) = ω2 + 1.



[0, ω) [ω, ω2)



r(ω + 1, 3) = ω2 + 1.



[0, ω) [ω, ω2)



r(ω + 1, 3) = ω2 + 1.



[0, ω) [ω, ω2)



r(ω + 1, 3) = ω2 + 1.

[0, ω) [ω, ω2)

If the last point is connected to one of the big dots with a rededge, we are done (we have a red copy of Kω+1). If the edges areblue we are done as well (now we have a blue copy of K3):

[0, ω) [ω, ω2)

The point is that we have reduced the infinitary problem ofidentifying r(ω+ 1, 3) to a problem about colorings of finite graphs.


r(ω + 1, 3) = ω2 + 1.

[0, ω) [ω, ω2)


[0, ω) [ω, ω2)



r(ω + 1, 3) = ω2 + 1.

[0, ω) [ω, ω2)


[0, ω) [ω, ω2)



The same ideas allow us to show, for example:

r(ω + 2, 3) = ω2 + 4.

r(ω + 2, 4) = ω3 + 7.

r(ω + 3, 4) = ω3 + 16.

In each case, lifting and projecting operations allow us to identifycolorings of infinite graphs with colorings of finite graphs, reducingthe computation of the ordinals r(ω + n,m) to similar problems tothe computation of finite Ramsey numbers.



r(ω + 2, 3) = ω2 + 4.

r(ω + 2, 4) = ω3 + 7.

r(ω + 3, 4) = ω3 + 16.




r(ω + 2, 3) = ω2 + 4.

r(ω + 2, 4) = ω3 + 7.

r(ω + 3, 4) = ω3 + 16.




r(ω + 2, 3) = ω2 + 4.

r(ω + 2, 4) = ω3 + 7.

r(ω + 3, 4) = ω3 + 16.




r(ω + 2, 3) = ω2 + 4.

r(ω + 2, 4) = ω3 + 7.

r(ω + 3, 4) = ω3 + 16.



For example, the following finite graph verifies thatω3 + 15 6→ (ω + 3, 4). For the sake of clarity, only the blue edgesare depicted.

v0 : ω v1 : ω v2 : ω

v3 v4 v5 v6 v7 v8 v9 v10 v11 v12 v13 v14 v15 v16 v17


For example, the following finite graph verifies thatω3 + 15 6→ (ω + 3, 4). For the sake of clarity, only the blue edgesare depicted.

v0 : ω v1 : ω v2 : ω



The previous graph is one of only five possible. They have all beenfound by Kyle Beserra, an undergraduate student at BSU.


Haddad-Sabbagh

To describe the general result, we need a lemma.

Lemma

For all positive integers n,m there exists a positive integerk ≥ n,m such that any red-blue coloring of the edges of K[0,k),and such that K[0,m) is blue, admits a subset H ⊂ [0, k) with KH

monochromatic, and

1 Either KH is blue, and |H| = m+ 1,

2 Or KH is red, |H| = n+ 1, and H contains a number in[0,m).

(If we remove the requirements that K[0,m) is blue andH ∩ [0,m) 6= ∅, this is Ramsey’s theorem.)Denote by rHS(n+ 1,m+ 1) the least number k witnessing thelemma.(HS is for Haddad and Sabbagh.)


Haddad-Sabbagh


Lemma


monochromatic, and





Haddad-Sabbagh


Lemma


monochromatic, and





Haddad-Sabbagh


Lemma


monochromatic, and





Haddad-Sabbagh


Lemma


monochromatic, and





Haddad-Sabbagh


Lemma


monochromatic, and





For example:

rHS(2,m) = m = (m− 1) + 1.

rHS(3,m) =m(m+ 1)

2= (m− 1) +

(m(m− 1)

2+ 1

).

rHS(n+ 1, 3) = r(n, 3) + n+ 1 = 2 + (r(n, 3) + n− 1).

rHS(4, 3) = 10 = 2 + 8.

rHS(4, 4) = 19 = 3 + 16.


For example:

rHS(2,m) = m = (m− 1) + 1.

rHS(3,m) =m(m+ 1)

2= (m− 1) +

(m(m− 1)

2+ 1

).

rHS(n+ 1, 3) = r(n, 3) + n+ 1 = 2 + (r(n, 3) + n− 1).

rHS(4, 3) = 10 = 2 + 8.

rHS(4, 4) = 19 = 3 + 16.


For example:

rHS(2,m) = m = (m− 1) + 1.

rHS(3,m) =m(m+ 1)

2= (m− 1) +

(m(m− 1)

2+ 1

).

rHS(n+ 1, 3) = r(n, 3) + n+ 1 = 2 + (r(n, 3) + n− 1).

rHS(4, 3) = 10 = 2 + 8.

rHS(4, 4) = 19 = 3 + 16.


For example:

rHS(2,m) = m = (m− 1) + 1.

rHS(3,m) =m(m+ 1)

2= (m− 1) +

(m(m− 1)

2+ 1

).

rHS(n+ 1, 3) = r(n, 3) + n+ 1 = 2 + (r(n, 3) + n− 1).

rHS(4, 3) = 10 = 2 + 8.

rHS(4, 4) = 19 = 3 + 16.


For example:

rHS(2,m) = m = (m− 1) + 1.

rHS(3,m) =m(m+ 1)

2= (m− 1) +

(m(m− 1)

2+ 1

).

rHS(n+ 1, 3) = r(n, 3) + n+ 1 = 2 + (r(n, 3) + n− 1).

rHS(4, 3) = 10 = 2 + 8.

rHS(4, 4) = 19 = 3 + 16.


For example:

rHS(2,m) = m = (m− 1) + 1.

rHS(3,m) =m(m+ 1)

2= (m− 1) +

(m(m− 1)

2+ 1

).

rHS(n+ 1, 3) = r(n, 3) + n+ 1 = 2 + (r(n, 3) + n− 1).

rHS(4, 3) = 10 = 2 + 8.

rHS(4, 4) = 19 = 3 + 16.


rHS(4, 4) = 19 = 3 + 16.

The graph verifying that rHS(4, 4) > 18 is precisely

v0 : ω v1 : ω v2 : ω



rHS(4, 4) = 19 = 3 + 16.


v0 : ω v1 : ω v2 : ω



rHS(4, 4) = 19 = 3 + 16.


v0 : ω v1 : ω v2 : ω



Theorem (Haddad-Sabbagh)

Given positive integers n and m, we have that

r(ω + n,m) = ω(m− 1) + t,

where m− 1 + t = rHS(n+ 1,m).


Haddad and Sabbagh announced this result in Comptes Rendus in1969, but never published the proof. This announcementconstituted the first of three notes in Comptes Rendus theypublished that year.

In the second, they presented an elementary proof of

ω2 → (ω2, n)2 ∀n > 0,

a result due to Specker.

In the third, they announced the existence of an algorithm tocompute the Ramsey numbers

r(ω2a+ ωb+ c, d) (a, b, c, d ∈ N).


As it was, the proof was written down, bare, with nocomments or hints. Our note got very little attention, infact it got almost none! So no further details were everpublished.

L. Haddad (2006)


In 1971, Milner published an algorithm deciding the validity ofany statement of the form ωn → (ωk,m)2 with n, k,m finite.

In 2012, Weinert presented a result giving us an algorithm tocompute the numbers r(ω2n,m).

45 years later, no proof of the theorems of Haddad and Sabbaghhas been published.As part of a more general proyect to study the partition calculus ofcountable ordinals, I am currently completing a survey presentingthe details.


A topological extension

There are several ways of extending the results above.

Definition (Baumgartner)

A set H of ordinals is order-homeomorphic to an ordinal α if andonly if α is the order type of H, and the unique order isomorphismbetween H and α is a homeomorphism (here, α has the ordertopology, and H has the subspace topology, considering H as asubset of an ordinal).

This is stronger than requiring that H be homeomorphic to α. Forinstance, ω + 1 and ω + 2 are homeomorphic.


Theorem (Baumgartner)

For all finite n larger than 0 there is a γ < ε0 = ωω...

such that forall finite l larger than 1, we have that

γ →cl (ωn + 1)2l,2n.

The notation here means that if the edges of Kγ are colored with lcolors, there is a subset H ⊆ γ order-homeomorphic to ωn + 1 andsuch that in KH at most 2n colors are present.(cl is for closed.)


Recently, Claribet Pina, a student of Todorcevic, has improvedsome of the bounds obtained by Baumgartner. For example,

For all l, k > 1, ωωk →cl (ω2 + 1)2l,5.

ω3 →cl (ω2 + 1)2l,11 for all l > 1, and this fails with 10 in placeof 11.


Another possible direction is the study of rcl(ωn+m, k), for finiten,m, k. This is the least α such that

α→cl (ωn+m, k)2.

This function is significantly subtler than r(ωn+m, k).


Theorem (C.)

For all finite n,m, k, rcl(ωn+m, k) < ωω.

The proof gives more information, but it is harder to describe thanin the case of the Haddad-Sabbagh results.Erdos and Rado proved that

ω1 →cl (α, n)2

for all countable α and finite n. The difficulty is to determinewhether, for each α and n, ω1 can be replaced with a countableordinal.


Theorem (C.)

For all finite n,m, k, rcl(ωn+m, k) < ωω.

The proof gives more information, but it is harder to describe thanin the case of the Haddad-Sabbagh results.Erdos and Rado proved that

ω1 →cl (α, n)2

for all countable α and finite n. The difficulty is to determinewhether, for each α and n, ω1 can be replaced with a countableordinal.


Open problem

If α ≥ ω2 is countable and n is finite, does it follow thatrcl(α, n) < ω1?

Recall that Erdos and Milner proved that r(α, n) is countable.Their argument does not seem to adapt to the case of rcl(α, n).The results of Baumgartner do not seem to solve the problemeither.


Open problem

If α ≥ ω2 is countable and n is finite, does it follow thatrcl(α, n) < ω1?

Recall that Erdos and Milner proved that r(α, n) is countable.Their argument does not seem to adapt to the case of rcl(α, n).The results of Baumgartner do not seem to solve the problemeither.


For example:

r(ω + 1, 3) = ω2 + 1, but

rcl(ω + 1, 3) = ω2 + 1.

r(ω + 2, 3) = ω2 + 4, but

rcl(ω + 2, 3) = ω22 + ω + 2.

An essential ingredient of the argument is a closed version of thepigeonhole principle.


For example:

r(ω + 1, 3) = ω2 + 1, but

rcl(ω + 1, 3) = ω2 + 1.

r(ω + 2, 3) = ω2 + 4, but

rcl(ω + 2, 3) = ω22 + ω + 2.



For example:

r(ω + 1, 3) = ω2 + 1, but

rcl(ω + 1, 3) = ω2 + 1.

r(ω + 2, 3) = ω2 + 4, but

rcl(ω + 2, 3) = ω22 + ω + 2.



For example:

r(ω + 1, 3) = ω2 + 1, but

rcl(ω + 1, 3) = ω2 + 1.

r(ω + 2, 3) = ω2 + 4, but

rcl(ω + 2, 3) = ω22 + ω + 2.



For example:

r(ω + 1, 3) = ω2 + 1, but

rcl(ω + 1, 3) = ω2 + 1.

r(ω + 2, 3) = ω2 + 4, but

rcl(ω + 2, 3) = ω22 + ω + 2.



For example:

r(ω + 1, 3) = ω2 + 1, but

rcl(ω + 1, 3) = ω2 + 1.

r(ω + 2, 3) = ω2 + 4, but

rcl(ω + 2, 3) = ω22 + ω + 2.



Closed version of the pigeonhole principle

Recall that any nonzero ordinal admits a Cantor normal form:

α = ωδ0n0 + ωδ1n1 + · · ·+ ωδknk,

with k finite, δ0 > δ1 > · · · > δk, and n0, n1, . . . , nk finite.

Definition (Hessenberg)

Given ordinals α, β with normal forms

α = ωδ0n0 + ωδ1n1 + · · ·+ ωδknk

andβ = ωδ0m0 + ωδ1m1 + · · ·+ ωδkmk,

their natural sum is

α#β = ωδ0(n0 +m0) + ωδ1(n1 +m1) + · · ·+ ωδk(nk +mk).


Closed version of the pigeonhole principle

Recall that any nonzero ordinal admits a Cantor normal form:

α = ωδ0n0 + ωδ1n1 + · · ·+ ωδknk,

with k finite, δ0 > δ1 > · · · > δk, and n0, n1, . . . , nk finite.

Definition (Hessenberg)

Given ordinals α, β with normal forms

α = ωδ0n0 + ωδ1n1 + · · ·+ ωδknk


their natural sum is

α#β = ωδ0(n0 +m0) + ωδ1(n1 +m1) + · · ·+ ωδk(nk +mk).


For example:

(ω22 + ω + 3) + (ω3 + ω2 + ω2 + 5) = ω3 + ω2 + ω2 + 5,

(ω3 + ω2 + ω2 + 5) + (ω22 + ω + 3) = ω3 + ω23 + ω + 3, and

(ω22 + ω + 3)#(ω3 + ω2 + ω2 + 5) =(ω3 + ω2 + ω2 + 5)#(ω22 + ω + 3) = ω3 + ω23 + ω3 + 8.


For example:

(ω22 + ω + 3) + (ω3 + ω2 + ω2 + 5) = ω3 + ω2 + ω2 + 5,

(ω3 + ω2 + ω2 + 5) + (ω22 + ω + 3) = ω3 + ω23 + ω + 3, and

(ω22 + ω + 3)#(ω3 + ω2 + ω2 + 5) =(ω3 + ω2 + ω2 + 5)#(ω22 + ω + 3) = ω3 + ω23 + ω3 + 8.


For example:

(ω22 + ω + 3) + (ω3 + ω2 + ω2 + 5) = ω3 + ω2 + ω2 + 5,

(ω3 + ω2 + ω2 + 5) + (ω22 + ω + 3) = ω3 + ω23 + ω + 3, and

(ω22 + ω + 3)#(ω3 + ω2 + ω2 + 5) =(ω3 + ω2 + ω2 + 5)#(ω22 + ω + 3) = ω3 + ω23 + ω3 + 8.


For example:

(ω22 + ω + 3) + (ω3 + ω2 + ω2 + 5) = ω3 + ω2 + ω2 + 5,

(ω3 + ω2 + ω2 + 5) + (ω22 + ω + 3) = ω3 + ω23 + ω + 3, and

(ω22 + ω + 3)#(ω3 + ω2 + ω2 + 5) =(ω3 + ω2 + ω2 + 5)#(ω22 + ω + 3) = ω3 + ω23 + ω3 + 8.


Definition

α→ (β, γ)1 if and only if for all partitions α = A ∪B, either Ahas order type at least β, or B has order type at least γ.

Denote by r(β, γ; 1) the least value of α such that α→ (β, γ)1.


Theorem (Milner-Rado)

Ifα = ωδ0n0 + ωδ1n1 + · · ·+ ωδknk


with nk +mk > 0, then

r(β, γ; 1) + ωδk = β#γ.


Define rcl(β, γ; 1) similarly:

Definition

α→cl (β, γ)1 if and only if for all partitions α = A ∪B, either Acontains a subset order-homeomorphic to β, or B contains asubset order-homeomorphic to γ.rcl(β, γ; 1) is the least value of α such that α→cl (β, γ)1.


Define rcl(β, γ; 1) similarly:

Definition

α→cl (β, γ)1 if and only if for all partitions α = A ∪B, either Acontains a subset order-homeomorphic to β, or B contains asubset order-homeomorphic to γ.rcl(β, γ; 1) is the least value of α such that α→cl (β, γ)1.


It is easy to see that if β and γ are countable, then ω1 →cl (β, γ)1.It is more involved to verify that rcl(β, γ; 1) is countable.

Theorem (Baumgartner, Weiss (?))

There is an algorithm that computes rcl(β, γ; 1) from the Cantornormal forms of β, γ, and their exponents.


It is easy to see that if β and γ are countable, then ω1 →cl (β, γ)1.It is more involved to verify that rcl(β, γ; 1) is countable.

Theorem (Baumgartner, Weiss (?))

There is an algorithm that computes rcl(β, γ; 1) from the Cantornormal forms of β, γ, and their exponents.


This result does not appear in the literature. However, in the paperwhere Baumgartner studies the relation β →cl (α)2n,m, he presentsa lemma (“essentially due to W. Weiss”) from which one candeduce the existence of the algorithm, and its details.


For instance:

rcl(ω + 1, ω + 1; 1) = ω2 + 1.

rcl(ω + 2, ω + 2; 1) = ω2 + ω + 2.

rcl(ωω+23 + ω22 + ω + 2, ωω4 + ω17 + 1; 1) =

ωω2+24 + ωω+317 + ωω+23 + ω22 + ω + 2.


For instance:

rcl(ω + 1, ω + 1; 1) = ω2 + 1.

rcl(ω + 2, ω + 2; 1) = ω2 + ω + 2.

rcl(ωω+23 + ω22 + ω + 2, ωω4 + ω17 + 1; 1) =

ωω2+24 + ωω+317 + ωω+23 + ω22 + ω + 2.


For instance:

rcl(ω + 1, ω + 1; 1) = ω2 + 1.

rcl(ω + 2, ω + 2; 1) = ω2 + ω + 2.

rcl(ωω+23 + ω22 + ω + 2, ωω4 + ω17 + 1; 1) =

ωω2+24 + ωω+317 + ωω+23 + ω22 + ω + 2.


For instance:

rcl(ω + 1, ω + 1; 1) = ω2 + 1.

rcl(ω + 2, ω + 2; 1) = ω2 + ω + 2.

rcl(ωω+23 + ω22 + ω + 2, ωω4 + ω17 + 1; 1) =

ωω2+24 + ωω+317 + ωω+23 + ω22 + ω + 2.


To illustrate the connection between the functions rcl(β, γ) yrcl(β, γ; 1), I mention that, in order to show

rcl(ω + 2, 3) = ω22 + ω + 2,

what one actually proves is that

rcl(ω + 2, 3) = rcl(ω + 1, 3) + rcl(ω + 2, ω + 2; 1),

and

rcl(ω + 2, ω + 2; 1) = rcl(ω + 1, ω + 1; 1) + (ω + 2).

The computation of upper bounds for the numbers rcl(ωn+m, k)with n > 1 is more cumbersome, and requieres additional results ofErdos-Rado on the function r(ωn, k).


To illustrate the connection between the functions rcl(β, γ) yrcl(β, γ; 1), I mention that, in order to show

rcl(ω + 2, 3) = ω22 + ω + 2,

what one actually proves is that

rcl(ω + 2, 3) = rcl(ω + 1, 3) + rcl(ω + 2, ω + 2; 1),

and

rcl(ω + 2, ω + 2; 1) = rcl(ω + 1, ω + 1; 1) + (ω + 2).

The computation of upper bounds for the numbers rcl(ωn+m, k)with n > 1 is more cumbersome, and requieres additional results ofErdos-Rado on the function r(ωn, k).


The end.


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Andr es Eduardo CaicedoCaicedo Ramsey theory of ordinals. Countable ordinals The ordinals are the order types of well-ordered sets. Recall that a linear order(X;