**1. Introduction **

Partition calculus is the area of set theory that deals with Ramsey theory; it is devoted to Ramsey’s theorem and its infinite and infinitary generalizations. This means both strengthenings of Ramsey’s theorem for sets of natural numbers (like the Carlson-Simpson or the Galvin-Prikry theorems characterizing the completely Ramsey sets in terms of the Baire property) and for larger cardinalities (like the -Rado theorem), as well as variations in which the homogeneous sets are required to possess additional structure (like the Baumgartner-Hajnal theorem).

Ramsey theory is a vast area and by necessity we won’t be able to cover (even summarily) all of it. There are many excellent references, depending on your particular interests. Here are but a few:

- Paul András Hajnal, Attila Máté, Richard Rado,
**Combinatorial set theory: partition relations for cardinals**, North-Holland, (1984). - Ronald Graham, Bruce Rothschild, Joel Spencer,
**Ramsey theory**, John Wiley & Sons, second edn., (1990). - Neil Hindman, Dona Strauss,
**Algebra in the Stone- compactification**, De Gruyter, (1998). - Stevo
*High-dimensional Ramsey theory and Banach space geometry*, in**Ramsey methods in Analysis**, Spiros Argyros, Stevo Birkhäuser (2005), 121–257. - András Hajnal, Jean Larson,
*Partition relations*, in**Handbook of set theory**, Matthew Foreman, Akihiro Kanamori, eds., to appear.

I taught a course on Ramsey theory at Caltech a couple of years ago, and expect to post notes from it at some point. Here we will concentrate on infinitary combinatorics, but I will briefly mention a few finitary results.

In the early 1950s, Rado introduced the partition calculus notation, several variants of which we will be using through these lectures.

Definition 1Let be a set and be a cardinal. Then

Let be a cardinal. A function is referred to as a-partition or -coloring of. Sometimes one writes the partition “explicitly” as where

Let be cardinals, and let be a sequence of cardinals. Thearrow notationis defined as follows:

iff for any set of size and any coloring there is an and a set such that Here, denotes the pointwise image of the set under the function We say that is

monochromatic,orhomogeneousor-homogeneousfor the coloring and also say thatadmitsan -homogeneous subset of of the prescribed size

One reads asarrows. The negation of the relation above is denoted by replacing with

Although we are mostly interested in the version above, we will also consider the following:

Definition 2Anorder typeis an equivalence class of the collection of partial orders under the relation of order isomorphism; we denote the order type of a poset by

When talking about order types of partially ordered sets, by we mean the order type of and by we mean the order type of The order type of an ordinal is denoted also by so the notation is consistent with the standard use for well-ordered sets.

If is a partially ordered set and is an order type, then denotes

There is risk of ambiguity here, so when it matters we use for the set of subsets of of

sizewhile is used for the collection of subsets of (the partially ordered set) oforder typeIf are order types, means that whenever and there is an order preserving injection (an

embedding) of intoIf denote order types, is a cardinal, and is a sequence of order types, then

iff for any poset of order type and any coloring there is an and a set such that As before, such an is called homogeneous for

The notation is very flexible; if all the are equal to one simply writes and if then one simply says For example, and have the same meaning, and are true whether denotes size or order type.

However daunting it appears at first sight, this notation is a very useful shorthand, and it compresses a large amount of information. We assume choice throughout unless otherwise stated; the results vary wildly in the absence of choice.

Note that if then for any and such that for all

**2. The classical Ramsey’s theorem **

Ramsey theory is to a large extent the study of the “arrow” relation, in a sense, a series of (vast) generalizations of the **pigeonhole** (or **Schubfach-**, or **Dirichlet’s**) principle.

The result that gave birth to the field is Frank Ramsey’s theorem from 1928, published in 1930, and obtained in order to show a decidability result.

*Proof:* The result is clear if by the pigeonhole principle.

We proceed by induction on Assume the result for and let’s show it for Let be given. For let be the function By induction define a sequence as follows:

- By induction, there is a number and an infinite set that is -homogeneous for
- Inductively, let There is then a and an infinite that is -homogeneous for

Now consider the function given by Let be infinite and homogeneous for Then the set is infinite and homogeneous for

This proof has the advantage that it is easily finitizable. By this I mean, from the proof one can easily extract bounds that allow us to prove the following:

*Proof:* Argue by induction on We denote by the least as in the statement, and our task is to show that in fact More generally, we let be the smallest number such that so

To begin with, by the pigeonhole principle.

Given let and define for by for The proof of Theorem 3 above easily shows that

One can see that, in fact, for all there is a constant such that for all

where the tower of 2s has height

The case of Ramsey’s theorem has received special attention, the statement being particularly well-known. Write for for This is the smallest number of vertices required to guarantee that whenever each edge of the complete graph is colored either blue or red, there is either going to be a blue copy of or a red copy of

*Proof:* The proof is by induction on Note that Now argue that from which the statement follows immediately.

To see the claimed inequality, consider a graph on many vertices, a coloring of its edges in colors blue and red, and let be one of the vertices. Let be the set of vertices that are joined to by a blue edge and let be the set of vertices joined to by a red edge, so It follows that either or

Without loss, assume that It follows that either there is a red copy of with vertices from and we are done, or else, there is a blue copy of with vertices from In this case, all these vertices are joined to by blue edges so, by adding to them, we obtain a blue copy of

In fact, this was shown by Rödl. Very recently (in a paper to appear in the Annals of Mathematics), David Conlon has shown that for all there is a constant such that whenever then

In joint work by David Conlon, Jacob Fox, and Benny Sudakov, the known bounds on have also been recently improved.

** 2.1. The happy end problem **

The -Szekeres Theorem 5 above is responsible for popularizing Ramsey’s theorem. and Szekeres reproved Ramsey’s result in 1934 when studying a question in elementary geometry, sometimes referred to as the “*happy end problem*.” (Another proof of Ramsey’s result had been found in 1933 by Skolem.)

Definition 6For all natural numbers let denote the least if it exists, such that given any points in general position in the plane (i.e., no four of them lying on the same circle, no three of them being collinear, no two of the lines they determine being parallel), there are that form the vertices of a convex -gon.

For example, and as observed by Esther Klein, who also asked the general problem of showing that for all is due to Makai, but the first published proof is due to J.D. Kalbfleisch, J.G. Kalbfleisch, and R. Stanton, in 1970.

To see that notice that 4 points do not suffice by considering the three vertices of a triangle and a fourth point inside. Now, given 5 points, consider their convex hull. We are done if it is not a triangle. If it is a triangle, there are two points inside, and the line they determine does not cross one of the sides of the triangle. The quadrilateral with vertices these two points and the two vertices of the segment not crossed by is convex.

Theorem 7 (-Szekeres)For all exists.

Shortly after their paper was published, Szekeres married Klein, which gave the name to the problem.

*Proof:* In fact, This is an argument of Michael Tarsy. As an undergraduate student at the Technion University, Tarsy was taking a combinatorics course under Lewin, and missed the lecture where the usual proof of the existence of was presented so, when asked to in an exam, he produced the following instead:

Let and let be points in general position in the plane. For color blue if one encounters them in this order when going clockwise around the triangle they determine, and color them red otherwise.

One easily checks that four points determine a convex quadrilateral iff the set they form is homogeneous under the coloring just described, i.e., iff all its triples receive the same color. An easy induction shows that a set of points determines a convex -gon iff any 4 of them determine a convex quadrilateral. The result follows.

The best known upper bound is shown by Tóth and Valtr in 1998. Shortly before, Bárány and Valtr showed that for any there is a positive constant such that any sufficiently large finite set of points in the plane contains subsets each of size at least such that every set with is in convex position.

It is conjectured that for all That the right hand side is a lower bound is also due to and Szekeres.

** 2.2. A decidability result **

Ramsey proved his result to obtain an instance of the *Entscheidungsproblem*, a decidability theorem. Fix a first order language with no function symbols, and consider the formulas in the language, of the form

where is quantifier free. These are the so-called *Bernays-Schönfinkel formulas.*

Ramsey proved that there is an algorithm that decides whether these formulas have (finite) models. In contrast, it is well known (from results of Church and Trakhtenbrot) that the problem is undecidable for general first order formulas, and even for those of the form

with quantifier free.

Definition 8Let be a linearly ordered set. Let Two sequences and of elements ofhave the same ordering,

iff for all iff

Equivalently, if and then iff

The somewhat awkward definition allows for the possibility that two or more of the (and consequently, of the ) are the same.

Definition 9A -ary relation on a linearly ordered set iscanonical oniff whenever then

For example, the canonical unary relations on a set are “true” (i.e., ) and “false” (i.e., ). The canonical binary relations are “false,” “” “” “” “” “” “,” and “true.”

Lemma 10For all there is such that whenever is -colored for all with then there is an such that is monochromatic for all such

*Proof:* Let and for let Then we can take

Theorem 11For all there is such that whenever (whether finite or infinite), and is a set of many -ary relations on for all with then there is on which all the relations in all the are canonical.

*Proof:* This is a consequence of Lemma 10. Consider a cardinal and sets of -ary relations on for as in the statement of the theorem.

For define an equivalence relation on by saying that whenever , if and are the increasing enumerations of and respectively, then iff the following holds: For all with all surjections and all

For example, the equivalence class of a singleton is completely determined by the tuple of truth values of the relations for

Clearly, there is an absolute (and easily computable) bound on the number of equivalence classes of these relations; “absolute” means here that only depends on and but is independent of and of the specific relations in

Now use Lemma 10 with the just described and to find the required noting that if is monochromatic for the partition of into equivalence classes for all with then all the relations in are canonical on

Theorem 12 (Ramsey)Consider a language with only relation symbols, and let be a finite theory all of whose axioms are universal, i.e., of the form

where is quantifier free. There is a finite number that only depends on the number of variables, and the number and arity of the relations used in such that for any (whether finite or infinite), there is a model of of size iff there is acanonicalmodel of with universe Here, we say that a model is canonical iff all the relations in its language are canonical.

*Proof:* Letting be the number of -ary relations mentioned in for (for an appropriate ), we can take as in Theorem 11.

First, if there is a model of with universe there is certainly a canonical substructure of of size since the language of only has relation symbols. The universality of the axioms guarantees that this substructure is indeed a model of

Second, if there is a canonical model of with universe then there is one with universe any larger cardinal, obtained simply by extending each relation in the same canonical fashion. (For example, if a binary relation symbol is interpreted in by then we also let be its interpretation in any larger model.) That this extended structure is indeed a model of follows easily from canonicity.

It is now relatively easy to extend the above result to give the decidability of the question of whether a Bernays-Schönfinkel sentence in a language with no function symbols is consistent. Essentially, one first replaces such a formula with a finite disjunction of universal formulas as above (in a possibly larger language).

Now, for any such formula take and as in the statement of Theorem 12, and note that whether or not has a model of size reduces to whether it has a canonical model with universe There are only finitely many such models, so one can easily determine whether this is the case. If not, then one only has to examine the finitely many structures in the language of of size less than to see whether at least one of them models

Finally, whether the original formula is consistent or not is equivalent to whether our search was successful for at least one of the formulas in the disjunction

** 2.3. Tree arguments **

It is convenient to examine other proofs of Ramsey’s Theorem 3, as they may illustrate different features than the argument above. Here is a proof that generalizes to the context of large cardinals; moreover, the technique it uses (trees and end-homogeneous sets) is widely applied in partition calculus. For the purpose of proving the argument is certainly more involved than the one we have given; on the other hand, it provides us with a fairly concrete (*constructive*) description of how to find homogeneous sets for given colorings:

*Proof:* The proof is by induction on For the result is obvious.

The key to the inductive step is a mixture of König’s lemma and the construction of what now are called *end-homogeneous* sets. We first present the case

Definition 13Atreeis a poset such that the -predecessors of any element of the tree are well-ordered by

To any tree we can associate a rank function as usual,

For each the set of with is called the **-th level** of the tree. The first such that the -th level of is empty is called the **height** of We are primarily interested in -trees, i.e., trees of height

Definition 14Abranchthrough a tree is a maximal linearly ordered subset of

Lemma 15 (König)Let be an -tree such that all its levels are finite (we say that is infinite andfinite branching). Then there is an infinite branch through

*Proof:* Define a branch by induction, so each is in the -th level of and is infinite.

Given define a tree as follows: The **nodes** of the tree are the elements of We define by induction; in what follows, refers always to the ordering of rather than the usual relation among integers. Denote by the -th level of We define by inductively defining the levels Let ; the **potential successors** of are the elements of Given and let be the set of potential successors of Divide into the sets

for The **immediate successors** of (all of which belong to ) are the least elements of these sets (there may be less than many of them, if one of the is empty). The potential successors of are the elements of if any. This completes the construction of the tree. Notice that if is infinite, at least one of the is infinite as well. It follows that the tree is an -tree and that all its levels are finite; in fact, for each

By König’s lemma, there is an infinite branch through By construction, this branch is **end-homogeneous**, i.e., it has the property that for any and any (larger) numbers

If this value is say that is an *-node*. Now apply the case of the theorem: For some the set of -nodes is infinite, and we have ensured it is homogeneous for This completes the proof of Ramsey’s theorem for

Assume Ramsey’s theorem for and let By a further trivial induction, we may also assume that We define an -tree all of whose levels are finite, and explain how to extract from any branch through an infinite homogeneous set for

The first levels of are simply given by To choose the successors of divide into two classes,

and

and define as the set of least elements of each class (so has size at most 2; it may be a singleton if one of the is empty). Say Then the elements of are to be chosen among the remaining elements of (if any).

Suppose we have already defined that and that we have selected a subset of of potential successors of Let so Define an equivalence relation on by

The immediate successors of are the least elements of the -equivalence classes, and the successors of any such element are to be chosen among the remaining elements of its class.

It follows that each node has finitely many immediate successors (since for each the equivalence relation on admits only finitely many classes). If is infinite, at least one of the -classes is infinite as well, and it follows that is an -tree. By König’s lemma, it has an infinite branch Again, is *end-homogeneous*: By construction, has the property that given any is constant for all

Say that is *of type * if this value is 0 and *of type * otherwise. This gives a coloring By induction, there is an infinite -homogeneous set But then is also -homogeneous, and we are done.

** 2.4. Compactness **

As it is well-known, an easy compactness argument shows that Ramsey’s theorem implies Theorem 4. The disadvantage of using compactness is that no discernible finite bounds seem to follow from such an argument:

*Proof:* We want to show that for all natural numbers there is an such that Suppose otherwise and fix counterexamples so for all there is a function that does not admit homogeneous sets of size Given two counterexamples say that iff i.e., there must be integers such that and

This gives a tree structure to the set of all such functions. Notice that this tree is infinite, since there is a counterexample with domain for all it is also finite branching, since there are only finitely many counterexamples with domain for any given because, in fact, there are only finitely many functions

By König’s lemma there is a branch through this tree. The functions along this branch cohere, and we can then define a function by taking their union. But, by construction, does not admit a homogeneous set of size which of course contradicts Ramsey’s theorem that claims that there must in fact be an *infinite* homogeneous set.

This is a typical example of a *compactness argument*, called this way because it can be recast as a corollary of the compactness of the Tychonoff product of certain finite (discrete) sets It is also a consequence of the compactness of first-order logic. There is in fact a very general compactness result that applies even in situations where no tree argument as above is available.

Theorem 16 (Rado’s selection principle)Let be a finite nonempty set for all and let For set and let be given. Then there exists such that for all there is such that and

*Proof:* For let

and

We need to show that

This is an immediate consequence of Tychonoff’s theorem: Note that each is a closed subset of the Tychonoff product of the discrete sets because each is finite, being a subset of the finite set and we can write

as a finite union of closed sets. Let

We claim that has the finite intersection property. If this is the case, it follows from the compactness of that and we are done.

To prove the claim, let and let for be finite subsets of Then is finite, and has an extension By definition,

for all so for all and we are done.

Actually, we can quickly provide a self-contained proof of the remainder of the argument: Since has the finite intersection property, it can be extended to an ultrafilter over For and let

so for any fixed This is a finite union, so it follows that for each there is some such that Let be the function Then as required: Let Then

since is finite, all the are in and In particular, this intersection is nonempty, and we can fix one of its elements, say Then and Hence, and we are done.

One can also prove Rado’s selection principle as a consequence of the compactness of first order logic; for example, it is easy to give a proof using the ultrapower construction as in the proof of *2 implies 6* of Theorem 8 in lecture II.11.

Theorem 4 is an easy consequence of Theorem 16: Given we want to show that for some we have Assume otherwise. It follows that for each finite there is some function without a homogeneous set of size Let and for each let For each finite let be as above. For an arbitrary pick some finite such that and let be Rado’s selection principle now gives us an -coloring of without homogeneous sets of size as any such set would be contained in some finite, such that and we have a contradiction.

Another well-known consequence of Rado’s principle is the de Bruijn- theorem on the chromatic number of infinite graphs:

Definition 17If is a graph, thechromatic numberof is the smallest cardinal such that there is a map where is the set of vertices of such that whenever then there is no edge in between and

Theorem 18 (de Bruijn-)Let be a graph. If and everyfinitesubgraph of has chromatic number at most then the same holds of

*Proof:* This is also an easy consequence of Theorem 16. Now take to be the set of vertices of and each to be and take for to be a coloring witnessing that the chromatic number of the subgraph of spanned by is at most

Unlike the finite version of Ramsey’s theorem, the de Bruijn- result is not a consequence of König’s lemma, since may well be uncountable. It follows immediately from Theorem 18 and the well-known 4-color theorem that every planar graph is 4-colorable.

A well-known open problem in geometric graph theory, the **Hadwiger-Nelson problem**, asks for the minimum number of colors required to color the plane such that no two points at distance one from each other have the same color. Abusing notation a bit, let’s call this number It is known that the drawing below (taken from the wikipedia entry on the problem) shows both a coloring of the plane with 7 colors witnessing and a finite graph showing that

Whatever the true value of Theorem 18 shows that any lower bound can be verified by examining some finite configuration. A recent remark of Shelah and Soifer suggests that the value of may depend on whether the axiom of choice is adopted or not; Falconer has shown, for example, that if we require the monochromatic sets to be Lebesgue measurable. In particular, this must be the case in Solovay’s model, or under determinacy, even if every finite subset of the plane has chromatic number at most 4.

Truszczynski-Tuza and Rav, among others, have studied consequences of Rado’s principle in algebra and combinatorics.

** 2.5. Ultrafilters **

There is yet another proof of Ramsey’s theorem that is worth discussing. It explains that infinite homogeneous sets can be found because one of the colors is “large” in a precise sense. For this, we need a construction closely related to the product of ultrafilters discussed in lecture II.11.

It is convenient to think of ultrafilters (over in this discussion, but the argument holds in general) as generalized quantifiers: Let be an ultrafilter over . Given a formula we interpret as saying that “almost all” have property

Clearly, we have:

- Similarly with any of the standard binary connectives in place of

Definition 19Given a nonprincipal ultrafilter over and , define as the set of all such that

Note that since is nonprincipal. It then follows that for any iff Similarly, is closed under finite intersections and supersets, so it is an ultrafilter. It is nonprincipal, since for any finite from which one easily gets that any is infinite.

We now fix a nonprincipal and use to give a proof that In fact, given we will obtain an infinite homogeneous set such that is contained in a color in Without loss, we may assume that

- Note that there is a color More precisely, for some

Given let

We then have:

- whenever

where “” means “”

Let Then:

- If then

Hence we can build an increasing sequence such that for all and all we have But then, letting we have that and we are done.

(Notice the similarities between this and the tree argument.)

** Bibliography **

Here are some additional references (besides those mentioned at the beginning) consulted while preparing this note:

- Paul , George Szekeres,
*A combinatorial problem in geometry*, Compositio Math.**2**, (1935), 463–470. - András Hajnal,
*Infinite combinatorics*, in**Handbook of combinatorics, Vol. 2**, Elsevier (1995), 2085–2116. - Walter Morris, Valeriu Soltan,
*The -Szekeres problem on points in convex position—a survey*, Bull. Amer. Math. Soc. (N.S.),**37 (4)**, (2000), 437–458. - Frank Ramsey,
*On a problem of formal logic*, Proc. London Math. Soc.,**30**, (1930), 264–286. - Miroslaw Truszczynski, Zsolt Tuza,
*Rado’s Selection Principle: applications to binary relations, graph and hypergraph colorings and partially ordered sets*, Discrete Mathematics,**103**, (1992), 301–312.

*Typeset using LaTeX2WP. Here is a printable version of this post.*

Thank you for this introduction to partition calculus!

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