## Path decompositions

April 2, 2016

On Thursday March 10, Peter Cholak gave a beautiful talk at the Logic Seminar at the University of Michigan, on Rado’s path decomposition theorem and its effective content. I want to review here some of the results covered by Peter. Slides for another version of the talk can be found in Peter’s page. This is joint work by Peter, Greg Igusa, Ludovic Patey, and Mariya Soskova.

As usual, given a set $X$, let ${}[X]^2$ denote the collection of 2-sized subsets of $X$. If $r$ is a positive integer, an $r$coloring of ${}[X]^2$ (or simply, a coloring, if $r$ is understood) is a map $c:[X]^2\to r$ (where we use ordinal notation, so $r=\{0,1,\dots,r-1\}$).  We can think of this as a coloring using $r$ colors of the edges of the complete graph whose underlying set of vertices is $X$. When $r=2$, we have an even simpler interpretation: A 2-coloring is just a graph on $X$.

Given an $r$-coloring  $c$ of ${}[X]^2$, a path of color $i\in r$ is a sequence $a_0,a_1,\dots$ of distinct elements of $X$ (which may be finite or infinite, or even empty, or of length 1) such that for all $k$, if $a_{k+1}$ is defined, then $c(a_k,a_{k+1})=i$. Note that this is a much weaker requirement than asking that $\{a_0,a_1,\dots\}$ be monochromatic (which would mean that $c(a_k,a_j)=i$ for all $k\ne j$).  Also, in what follows $X$ is either a finite number or $\mathbb N$. However, we do not require that the elements in the sequence be listed in their natural order: We may very well have that $a_k>a_{k+1}$ for some $k$.

The starting point is the following observation:

Fact (Erdős).  If $m$ is finite and $c$ is a 2-coloring of ${}[m]^2$, then there are paths $P_0$ of color 0 and $P_1$ of color 1 such that every (vertex) $n\in m$ appears in exactly one of the $P_i.$

In general, if $c$ is an $r$-coloring of ${}[X]^2$, we say that $P_i$, $i, is a path decomposition of $c$ iff each $P_i$ is a path of color $i$ and every vertex $x\in X$ appears in exactly one of the $P_i$. Using this notion, what the fact states is that for any finite $m$, any 2-coloring of ${}[m]^2$ admits a path decomposition.

Proof. Suppose the result holds for $m$ and $c$ is a 2-coloring of ${}[m+1]^2$. We can then find paths $P_0$ and $P_1$ of color 0 and 1 respectively such that each $n appears in exactly one of the $P_i$. We want to show that the same holds for the full coloring (which includes edges one of whose vertices is $m$) at the possible expense of having to modify the partial paths we currently have. If one of the $P_i$ is empty, this is clear. Assume then that $P_0=(a_1,\dots,a_k,a_{k+1})$ and $P_1=(b_1,\dots,b_l,b_{l+1})$. The result is also clear if $c(a_{k+1},m)=0$ or $c(b_{l+1},m)=1$. Finally, if $c(a_{k+1},m)=1$ and $c(b_{l+1},m)=0$, consider $c(a_{k+1},b_{l+1})$. If this color is 0, we can let the paths be $P_0'=P_0{}^\frown(b_{l+1},m)$ and $P_1'=(b_1,\dots,b_l)$. Similarly, if $c(a_{k+1},b_{l+1})=1$, we can let the paths be $P_0'=(a_1,\dots,a_k)$ and $P_1'=P_1{}^\frown(a_{k+1},m)$. (This is perhaps most obvious if a picture is drawn.) $\Box$

Rado’s paper is a generalization of this result and its countable version. The reference is

MR0485504 (58 #5334)
Advances in graph theory (Cambridge Combinatorial Conf., Trinity College, Cambridge, 1977).
Ann. Discrete Math. 3 (1978), 191–194.

The paper opens indicating that Erdős sketched his proof to Rado; there does not seem to be an actual reference for Erdős’s proof. Rado proceeds to prove a more general version. I will only discuss here a particular case.

First, it should be noted that, unlike typical results in Ramsey theory where, once the case of two colors is handled, the argument easily generalizes to any number of colors, the proof above does not lift to more than two. The usual way of doing this lifting is by identifying all but one of the colors. This would result in two paths $P_0$ and $P_1$, where along $P_0$ we only see color 0 and along $P_1$ we only see the other colors, but not 0. Let $c$ be the given coloring and $V$ be the set of vertices appearing in $P_1$. If the restriction of $c$ to ${}[V]^2$ does not use color 0 we could indeed proceed inductively. But there is nothing to prevent 0 from being present as well, so the “easy” lifting argument actually breaks down.

The situation is indeed worse:

Theorem (Pokrovskiy). For any $r>2$ and any $m$ there is an $M>m$ and an $r$-coloring of ${}[M]^2$ that does not admit a path decomposition.

The proof can be found in:

MR3194196
Pokrovskiy, Alexey.
J. Combin. Theory Ser. B 106 (2014), 70–97.

On the other hand, we have:

Theorem (Rado). For any finite $r$, any $r$-coloring of ${}[\mathbb N]^2$ admits a path decomposition.

As already mentioned, Rado’s result is more general, in particular allowing the use of countably many colors. However, the arguments that follow only apply directly to the stated version.

Before sketching the proof, note that even for $r=2$, the result does not follow as usual from the finite version: Given a 2-coloring $c$ of $\mathbb N$, the standard approach would consist of letting $P_0,P_1$ be the paths resulting from successively applying Erdős’s theorem to the restrictions of $c$ to $m=2,3,\dots$. But the inductive argument we presented allows the paths to be modified from one value of $m$ to the next, which means that we cannot ensure that the process will successfully identify (via initial segments) paths for the full coloring (the partial paths do not “stabilize”). Together with Pokrovskiy’s negative result just indicated, this leaves us with a curious Ramsey-theoretic statement to which the usual compactness arguments do not apply. (Its finite counterpart, Erdős’s result, is weaker in the sense that it only applies to two colors, and requires a different argument.)

Proof. Consider a nonprincipal ultrafilter $\mathcal U$ on $\mathbb N$. The ultrafilter provides us with a notion of largeness. Given $r\in\mathbb N$ and a coloring $c:[\mathbb N]^2\to r$, define for $x\in\mathbb N$ and $i\in r$ the set of neighbors of $x$ in color $i$ as

$N(x,i)=\{y\mid c(x,y)=i\}.$

Note that for any $x$ the $N(x,i)$ partition $\mathbb N\setminus\{x\}$ and therefore there is exactly one $i$ such that $N(x,i)$ is large (that is, it is in $\mathcal U$). For $i\in r$, define

$A_i=\{x\mid N(x,i)\in\mathcal U\}$,

and note that the $A_i$ partition $\mathbb N$.

We proceed by stages to define the paths $P_0,\dots,P_{r-1}$ as required. We set $P^0_i=\emptyset$ for all $i$. In general, at the beginning of any given stage $n$ we have defined (finite) partial approximations $P^n_i$ to each path $P_i$, say $P^n_i$ has length $n_i$, with $P^n_i=(a_{i,1},\dots,a_{i,n_i})$, using the convention that $n_i=0$ indicates that $P^n_i=\emptyset$. For each $i$, we will ensure that $P^{n+1}_i$ end extends $P^n_i$ (for all $n$), and simply set $P_i$ as the resulting path. Inductively, we require that each $P^n_i$ is a path of color $i$, and that if $n_i>0$, then $a_{i,n_i}\in A_i$.

Now, at stage $n$, we simply consider the least $y$ not yet in any of the $P^n_i$. There is a unique $i$ with $y\in A_i$. We set $P^{n+1}_j=P^n_j$ for all $j\ne i$. If $P^n_i=\emptyset$, then set $P^{n+1}_i=(y)$. Finally, if $n_i>0$, the point is that since $N(y,i)$ and $N(a_{i,n_i},i)$ are both large, then so is their intersection (all we really need is that the intersection of sets in $\mathcal U$ is nonempty). Let $x$ be a point in their intersection, and set $P^{n+1}_i=P^n_i{}^\frown(x,y)$. The induction hypothesis is preserved, and this completes stage $n$ of the construction.

It should be immediate that the $P_i$ so constructed indeed provide a path decomposition of $c$, and this completes the proof. $\Box$

It is interesting to note that the notation just developed allows us to give a quick proof of Ramsey’s theorem for pairs: Given a coloring $c:[\mathbb N]^2\to r$, use notation as above, and note that for exactly one $i\in r$, the set $A_i$ is in $\mathcal U$. We argue that there is an infinite subset $H\subseteq A_i$ that is homogeneous for $c$ with color $i$, that is, $c''[H]^2=\{i\}$. Indeed, we can simply set $H=\{h_n\mid n\in\mathbb N\}$, where the $h_i$ are defined recursively so that $h_0\in A_i$ and $h_{j+1}\in A_i\cap\bigcap_{k\le j}N(h_k,i)$ for all $j$.

As Peter indicated in his talk, these pretty arguments are somewhat dissatisfying in that invoking a nonprincipal ultrafilter is too strong a tool for the task at hand. He then proceeded to indicate how we can in fact do better, computationally speaking. For instance, if the coloring $c$ is computable, then we can find a path decomposition below $0''$. The key to this improvement comes from two observations.

First, we do not really need an ultrafilter to carry out the argument. It suffices to consider a set $C\subseteq\mathbb N$ that is cohesive with respect to all the $N(x,i)$, meaning that $C$ is infinite and, for any $x,i$, either $C\subseteq^* N(x,i)$ or $C\subseteq^* \mathbb N\setminus N(x,i)$, where $\subseteq^*$ is the eventual containment relation: $A\subseteq^* B$ iff there is a finite subset $s$ of $A$ such that $A\setminus s\subseteq B$, in which case we say that $A$ is almost contained in $B$.

The point is that we can replace all instances where we required that a set $N$ is in $\mathcal U$ by the new largeness condition stating that $C$ is almost contained in $N$. For instance, note that if $N_1,N_2$ are large, then so is their intersection. As before, for any $x$ there is a unique $i$ with $N(x,i)$ large, and we can redefine $A_i$ as the set of $x$ such that

$\exists k\,\forall y>k\,(y\in C\to c(x,y)=i),\qquad (a)$

or, equivalently,

$\forall k\,\exists y>k\,(y\in C\land c(x,y)=i).\qquad (b)$

With these modifications, it is straightforward to verify that the proof above goes through. This shows that a path decomposition of $c$ is $\Delta^0_2(C)$.

In more detail: Note first that these two conditions are indeed equivalent, and second, clearly the $A_i$ are pairwise disjoint since $C$ is infinite and, moreover, for all $x$ there is a unique $i$ such that $x\in A_i$:

Suppose that $(a)$ holds, and let $k_0$ be such that $\forall y>k_0\,(y\in C\to(c(x,y)=i)$. Since $C$ is infinite, we can indeed find elements $y$ of $C$ larger than $k_0$, and any such $y$ witnesses $(b)$.

Conversely, if $(b)$ holds, then $C\subseteq^* N(x,i)$, because $C$ is cohesive and has infinite intersection with $N(x,i)$. But then $(a)$ holds, as wanted.

To see that any $x$ is in a unique $A_i$, fix $x$ and use that $x$ is cohesive to conclude that if $C\subseteq^*\mathbb N\setminus N(x,i)$ for all $i$, then $C\subseteq^*\{x\}$, which contradicts the infinitude of $C$. It follows that $C\subseteq^* N(x,i)$ for some $i$ and, since the $N(x,j)$ are pairwise disjoint, this $i$ is unique. This proves that $(a)$ holds and therefore $x\in A_i$. Uniqueness follows from this same observation: If $x\in A_i$, then (as shown above) $C\subseteq^* N(x,i)$. But there is only one $i$ for which this is true.

The second observation is that there is an easy recursive construction of a set that is cohesive with respect to all the $N(x,i)$: Consider first $N(0,0),\dots,N(0,r-1)$. One of these sets is infinite (since their union is $\mathbb N\setminus\{0\}$), say $N(0,j_0)$, and let $a_0$ be its first element. Consider now

$N(0,j_0)\cap N(1,0),\dots,N(0,j_0)\cap N(1,r-1)$.

Their union is $N(0,j_0)\setminus\{1\}$, so one of these sets is infinite, say $N(0,j_0)\cap N(1,j_1)$. Let $a_1$ be its first element above $a_0$. Etc. The set $C=\{a_0,a_1,\dots\}$ so constructed is as wanted. Note that this construction explicitly obtains an infinite set $C$ that, for each $x$, is almost contained in one the $N(x,i)$, which is superficially stronger than being cohesive. However, as verified above, any set cohesive for all the $N(x,i)$ must actually have this property.

Computationally, the advantage of this construction is that it makes explicit that all we need to access a cohesive set is an oracle deciding of any $N(x,i)$ whether it is infinite. For computable $c$, these are all $\Pi^0_2$ questions.

Peter further refined this analysis in his talk via the notion of a set being $\mathsf{PA}$ over $0'$: This is any set $X$ such that for any uniformly computable sequence of pairs of $\Pi^0_2$ sentences $\phi_{i,0},\phi_{i,1}$ for $i\in\mathbb N$ such that at least one is true, there is an $f\le_T X$ that predicts the true sentence of each pair in the sense that for all $i$, if $f(i)=j$, then $\phi_{i,j}$ is true. In symbols, say that $X>>0'$. The point of the notion is that a result of Jockusch and Stephen gives us that if $X>>0'$ then there is a cohesive set $C$ such that $C'\le_1 X$. The relevant paper is:

MR1270396 (95d:03078)
Jockusch, Carl; Stephan, Frank
A cohesive set which is not high.
Math. Logic Quart. 39 (1993), no. 4, 515–530.

MR1477624 (99a:03044)
Correction.
Math. Logic Quart. 43 (1997), no. 4, 569.

This shows that a path decomposition for a computable coloring $c$ can actually be found below $0''$  (and more).

Peter concluded his talk by indicating how for special colorings the complexity can be further improved. For instance, say that a coloring $c:[\mathbb N]^2\to r$ is stable iff $\lim_y c(x,y)$ exists for all $x$. One can check that for stable $c$, we can use cofinite as a notion of largeness in the preceding arguments, and that a path decomposition can accordingly be found when $c$ is computable below $0'$. On the other hand, this is optimal, in that one can find a stable computable $c$ such that any path decomposition computes $0'$.

## Summer Kisner – Schur’s theorem

May 19, 2013

My student Summer Kisner completed her M.S. this term, and graduated on Saturday.

2013–5-18 Summer

Here is a copy of the slides she used on her defense. (The slides display incorrectly on my computer, but it seems to be a problem on my end. If it is not, please let me know, and I’ll see what I can do.)

Her thesis, Schur’s theorem and related topics in Ramsey theory, discusses Schur’s theorem, one of the first result in what we now call Ramsey theory. The result states that if the positive integers $\mathbb Z^+$ are partitioned into finitely many sets, $\mathbb Z^+=A_1\cup\dots\cup A_n$, then for some $i$, $1\le i\le n$, there are integers $x,y,z$ (not necessarily different), all of them in $A_i$, such that $x+y=z$. One usually describes this in terms of colors: We color the positive integers with finitely many colors, and there is a monochromatic triple $x,y,z$ with $x+y=z$.

This result is a cornerstone of Ramsey theory. It was significantly generalized by Rado (using the notion of partition regularity), and is connected to van der Waerden’s and Szemerédi’s famous results.

Nowadays, Schur’s theorem is typically proved as a corollary of Ramsey’s theorem. This is usually stated in terms of graphs, but I will use the notation from the partition calculus. Let ${}[X]^k$ denote the collection of $k$-sized subsets of the set $X$. Suppose that $X$ is infinite, and consider a coloring $c:[X]^k\to C$, where the set $C$ of colors is finite. Ramsey’s theorem asserts that under these assumptions, there is an infinite subset $H$ of $X$ that is homogeneous or monochromatic for $c$, in the sense that $c$ assigns the same color to all $k$-sized subsets of $H$. In fact, we have finitary versions of this result: For any $n$ and any $l=|C|$, if we only require that $H$ has size at least $n$, then there is an $m$ such that it suffices to assume that $X$ has size at least $m$. Even for $k=l=2$, the study of Ramsey numbers, the least $m$ seen as a function of $n$, proves to be incredibly difficult and computationally unfeasible. For example, if $n=5$, then we know that $43\le m\le 49$, but its exact value is not known.

To deduce Schur’s theorem from Ramsey’s, let $r$ be such that, for any for coloring of ${}[\{1,\dots,r\}]^2$ using $n$ colors, there is a monochromatic set of size $3$. The least such $r$ we denote $R_n(3)$. We want to show that if $\mathbb Z^+=A_1\cup\dots\cup A_n$, then there is a monochromatic solution to the equation $x+y=z$. In fact, we claim that it suffices to consider $\{1,\dots,r-1\}$ rather than the whole set of positive integers. Indeed, given a partition $\{1,\dots,r-1\}=A_1\cup\dots\cup A_n$, consider the coloring of ${}[\{1,\dots,r\}]^2$ where if $a, then the set $\{a,b\}$ has color $i$, where $b-a\in A_i$. By definition of $r$, we can find $a such that $\{a,b\},\{a,c\},\{b,c\}$ all have the same color. Now notice that $c-a=(c-b)+(b-a)$, that is, $b-a,c-b,c-a$ are monochromatic for the original coloring of $\{1,\dots,r-1\}$.

An easy inductive argument gives us that $R_n(3)\le 3\cdot n!$, so this gives the upper bound $3\cdot n!-1$ for the so-called $n$-th Schur number $s(n)$. To see the upper bound $R_n(3)\le 3\cdot r!$, note that $R_1(3)=3$, and verify inductively that $R_{n+1}(3)\le (n+1)R_n(3)$: Suppose that $|X|\ge (n+1)R_n(3)$, and consider a coloring of $[X]^2$ with $n+1$. Fix an element $a\in X$, and note that for some color $i$ there are $R_n(3)$ elements $b\in X$ such that $\{a,b\}$ has color $i$. Let $Y$ be the set of all these $b$, that is, $Y=\{b\in X\mid\{a,b\}$ has color $i\}$. Note that if for some $b,c\in Y$ we have that $\{b,c\}$ has color $i$ as well, then $\{a,b,c\}$ is monochromatic with color $i$. Hence we may assume that the coloring, restricted to ${}[Y]^2$, only uses $n$ colors. We are now done, since $|Y|\ge R_n(3)$.

Schur’s original proof predated Ramsey, and gives a slightly better bound than $s(n)\le 3\cdot n!$. Indeed, from his proof, we obtain that $s(n)\le \lceil n!e\rceil$.

In terms of lower bounds, one can quickly check by induction that $s(n)\ge (3^n+1)/2$. Indeed, $s(1)=2=(3^1+1)/2$, since $1+1=2$. Given a coloring $c$ of $\{1,\dots,k\}$ using colors $\{1,\dots,n\}$ and without monochromatic triples, we describe a coloring $c'$ of $\{1,\dots,3k+1\}$ using colors $\{1,\dots,n+1\}$, again without monochromatic triples. This gives the result. To define $c'$, start by letting $c'(i)=c(i)$ for $i\le k$. Now let $c'(j)=n+1$ for $k+1\le j\le 2k+1$, and finally let $c'(j)=c(j-(2k+1))$ for $2k+2\le j\le 3k+1$.

Slightly better bounds are known. For example, Anne Penfold Street shows that $s(n)\ge (89\cdot 3^{n-4}+1)/2$ in

W.D. Wallis, Anne Penfold Street, Jennifer Seberry Wallis. Combinatorics: Room squares, sum-free sets, Hadamard matrices. Lecture Notes in Mathematics, Vol. 292. Springer-Verlag, Berlin-New York, (1972). MR0392580 (52 #13397).

Schur proved his theorem in order to establish a result related to Fermat’s last theorem, namely that it cannot be established by a naive argument involving modular arithmetic: Suppose that $x, y, z$ are integers, and $x^n+y^n=z^n$. Then, for any prime $p$, the equality holds modulo $p$ and, if $p$ is large enough, then we also have that $p\not\mid xyz$. Hence, to prove that Fermat’s equation admits no solutions $(x,y,z)$, it would suffice to show that there are arbitrarily large primes $p$ such that $p$ must divide one of $x,y,z$. What Schur proved is that this is not possible and, indeed, for any $n$ and all sufficiently large primes $p$, there are nontrivial solutions to Fermat’s equation modulo $p$. To see this, let $p>s(n)$ and let $G$ be the subgroup of $(\mathbb Z/p\mathbb Z)^*$ consisting of $n$-th powers. Then $(\mathbb Z/p\mathbb Z)^*$ is union of cosets of $G$, say $(\mathbb Z/p\mathbb Z)^*=\bigcup_{i=1}^k a_iG$ where the $k$ displayed sets are disjoint. Note that $k=n/\mathrm{gcd}(n,p-1)\le n$. Now color $(\mathbb Z/p\mathbb Z)^*$ using $k$ colors, by letting $t$ have color $i$ iff $t\in a_iG$. By choice of $p$, we have a monochromatic Schur triple, that is, there are $\alpha,\beta,\gamma\in (\mathbb Z/p\mathbb Z)^*$ such that $\alpha+\beta=\gamma$ and $\alpha,\beta,\gamma\in a_iG$ for some $i$. But then there are $x,y,z$, all nonzero modulo $p$, such that $\alpha=a_i x^n$, $\beta=a_i y^n$, and $\gamma=a_i z^n$, so $(x,y,z)$ is a nontrivial solution to Fermat’s equation modulo $p$.

It is actually an interesting problem to try and determine the optimal size $\mathcal N$ of $p$ as a function of $n$. Fourier-analytic methods give here the best known bounds. Cornacchia proved in 1909 that $\mathcal N(n)\le (n-1)^2(n-2)^2+6n-2$, at least if $n$ itself is prime.

Let me close with an open problem: We could consider a multiplicative (rather than additive) version of Schur’s theorem: For any $n$ there is an $s'(n)$ such that if $\{1,2,\dots,s'(n)\}$ is colored using $n$ colors, then there is a monochromatic set $\{x,y,z\}$ with $xy=z$. Indeed, this follows as a simple corollary of Schur’s result: Just note that $s'(n)\le 2^{s(n)}$, since we could just color the powers of two and apply the additive version. what if we combine the two? It is still open whether in any finite coloring of $\mathbb Z^+$ there are integers $x,y$ such that $x,y,x+y,xy$ all receive the same color. This was originally asked by Hindman.

## 580 -Partition calculus (4)

April 9, 2009

1. Colorings of pairs. I

There are several possible ways in which one can try to generalize Ramsey’s theorem to larger cardinalities. We will discuss some of these generalizations in upcoming lectures. For now, let’s highlight some obstacles.

Theorem 1 (${\mbox{\bf Erd\H os}}$-Kakutani) ${\omega_1\not\rightarrow(3)^2_\omega.}$ In fact, ${2^\kappa\not\rightarrow(3)^2_\kappa.}$

Proof: Let ${S={}^\kappa\{0,1\}.}$ Let ${F:[S]^2\rightarrow\kappa}$ be given by

$\displaystyle F(\{f,g\})=\mbox{least }\alpha<\kappa\mbox{ such that }f(\alpha)\ne g(\alpha).$

Then, if ${f,g,h}$ are distinct, it is impossible that ${F(\{f,g\})=F(\{f,h\})=F(\{g,h\}).}$ $\Box$

Theorem 2 (Sierpiński) ${\omega_1\not\rightarrow(\omega_1)^2.}$ In fact, ${2^\kappa\not\rightarrow(\kappa^+)^2.}$

Proof: With ${S}$ as above, let ${F:[S]^2\rightarrow2}$ be given as follows: Let ${<}$ be a well-ordering of ${S}$ in order type ${2^\kappa.}$ Let ${<_{lex}}$ be the lexicographic ordering on ${S.}$ Set

$\displaystyle F(\{f,g\})=1\mbox{ iff }<_{lex}\mbox{ and }<\mbox{ coincide on }\{f,g\}.$

Lemma 3 There is no ${<_{lex}}$-increasing or decreasing ${\kappa^+}$-sequence of elements of ${S.}$

Proof: Let ${W=\{f_\alpha\colon\alpha<\kappa^+\}}$ be a counterexample. Let ${\gamma\le\kappa}$ be least such that ${\{f_\alpha\upharpoonright\gamma\colon\alpha<\kappa^+\}}$ has size ${\kappa^+,}$ and let ${Z\in[W]^{\kappa^+}}$ be such that if ${f,g\in Z}$ then ${f\upharpoonright\gamma\ne g\upharpoonright\gamma.}$ To simplify notation, we will identify ${Z}$ and ${W.}$ For ${\alpha<\kappa^+}$ let ${\xi_\alpha<\gamma}$ be such that ${f_\alpha\upharpoonright\xi_\alpha=f_{\alpha+1} \upharpoonright\xi_\alpha}$ but ${f_\alpha(\xi_\alpha)=1-f_{\alpha+1}(\xi_\alpha).}$ By regularity of ${\kappa^+,}$ there is ${\xi<\gamma}$ such that ${\xi=\xi_\alpha}$ for ${\kappa^+}$ many ${\alpha.}$

But if ${\xi=\xi_\alpha=\xi_\beta}$ and ${f_\alpha\upharpoonright\xi=f_\beta\upharpoonright\xi,}$ then ${f_\beta<_{lex} f_{\alpha+1}}$ iff ${f_\alpha<_{lex} f_{\beta+1},}$ so ${f_\alpha=f_\beta.}$ It follows that ${\{f_\alpha\upharpoonright\xi\colon\alpha<\kappa^+\}}$ has size ${\kappa^+,}$ contradicting the minimality of ${\gamma.}$ $\Box$

The lemma implies the result: If ${H\subseteq S}$ has size ${\kappa^+}$ and is ${F}$-homogeneous, then ${H}$ contradicts Lemma 3. $\Box$

Now I want to present some significant strengthenings of the results above. The results from last lecture exploit the fact that a great deal of coding can be carried out with infinitely many coordinates. Perhaps surprisingly, strong anti-Ramsey results are possible, even if we restrict ourselves to colorings of pairs.