Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 130 (Ramsey Theory)

Let $[m]$ be an alphabet. A combinatorial line in $[m{]}^N$ is obtained by choosing a nonempty set of active coordinates, fixing every other coordinate, and putting the same variable letter in all active coordinates. The Hales-Jewett theorem states that, for every pair of positive integers $m,k$, some $N$ makes every $k$-coloring of $[m{]}^N$ contain a monochromatic combinatorial line.

We prove it by mathematical induction on $m$. The case $m=1$ is immediate. Suppose the result is known for the alphabet $[m-1]$, with any finite number of colors. Order each line by its variable letter, and call its last point its focus. Say that $r$ lines are color-focused when they share a focus, each line with its focus removed is monochromatic, and those $r$ punctured lines have different colors.

We claim that for every $1\le r\le k$ there is $N_r$ such that every $k$-coloring of $[m{]}^{N_r}$ contains either a monochromatic line or $r$ color-focused lines. For $r=1$, restrict a coloring to $[m-1{]}^{N_1}$ with $N_1=HJ(m-1,k)$. A monochromatic line there becomes a punctured line over $[m]$ after adjoining the point obtained by putting $m$ in every active coordinate. It is either already monochromatic or is one color-focused line.

Assume $n=N_r$ works for $r$, and putView $[m{]}^{n+N}$ as $[m{]}^n\times [m{]}^N$. For $b\in [m-1{]}^N$, record the entire color patternThis is a coloring with at most $k^{m^n}$ colors, so the induction hypothesis on the alphabet supplies a line on which this pattern is constant. Adjoin its focus $b^{+}$, whose active coordinates contain the letter $m$, and write $L$ for the resulting line over $[m]$. Thus $c(a,b)$ is independent of $b\in L\setminus \{b^{+}\}$ for every $a$. It defines a $k$-coloring $c^{\prime }(a)$ of $[m{]}^n$.

If $c^{\prime }$ has a monochromatic line, fixing any $b\in L\setminus \{b^{+}\}$ gives one for $c$. Otherwise there are $r$ color-focused lines for $c^{\prime }$, with common focus $f$. Couple the active coordinates of each of those lines with the active coordinates of $L$; this produces $r$ punctured lines focused at $(f,b^{+})$. The line obtained by fixing the first component at $f$ and varying along $L$ supplies one more. Its punctured color differs from the preceding $r$ colors, since equality with one of them would make the corresponding $c^{\prime }$-line monochromatic. Hence there are $r+1$ color-focused lines. The claim follows by induction on $r$.

Take $r=k$. The common focus has one of the $k$ colors, so it completes the punctured line of that color to a monochromatic line. This proves the Hales-Jewett theorem.

A $d$-dimensional combinatorial subspace has $d$ disjoint nonempty active coordinate sets, one for each independent variable. The Extended Hales-Jewett theorem says that, given $m,k,d$, every $k$-coloring of $[m{]}^n$ contains such a monochromatic subspace when $n$ is large enough. To deduce it, chooseand identifyA monochromatic combinatorial line over the alphabet $[m{]}^d$ becomes a monochromatic $d$-parameter set over $[m]$: within every active block, its first coordinates form the first variable set, its second coordinates form the second, and so on. Thus $n=dN$ suffices.

Let the given coloring use $k$ colors, and choose $N$ from the Extended Hales-Jewett theorem for alphabet $\{0,1\}$ and dimension $n$. Color a word $w\in \{0,1{\}}^N$ by the color of the positive integerA monochromatic $n$-parameter set has disjoint active coordinate sets $I_1,\dots ,I_n$ and fixed letters outside their union. PutEvery $x_s$ is positive. As the $n$ independent variable letters range over zero and one, their images under $\Phi$ are preciselyAll these integers have one color, so they form the required monochromatic Hilbert cube.

We prove the Hilbert cube theorem directly by mathematical induction on its dimension. Dimension zero is immediate. Suppose every finite coloring contains a monochromatic Hilbert $n$-cube, and let $c:\mathbb{N}\to [k]$ be a $k$-coloring. Refine it to the finite coloringBy the induction hypothesis, some Hilbert $n$-cubeis monochromatic for $C$. Consequently, for every $j\in \{0,\dots ,k\}$ the translate $H+j$ is monochromatic for $c$; denote its color by ${\gamma }_j$. By the pigeonhole principle, two of the $k+1$ colors ${\gamma }_0,\dots ,{\gamma }_k$ agree, say ${\gamma }_s={\gamma }_t$ with $s<t$. Thenis a monochromatic Hilbert $(n+1)$-cube. This proves the result without using the Finite sums theorem or the Van der Waerden theorem.

Let $A$ be a rational matrix with columns $a_1,\dots ,a_n$. It is a partition regular matrix when every finite coloring of the positive integers has a monochromatic vector $x$ with $Ax=0$. It has the columns property when the column indices have an ordered partition $B_1,\dots ,B_r$ such thatand, for every $s>1$,Rado's theorem states that $A$ is partition regular if and only if it has the columns property.

First suppose $A$ is partition regular, and clear its denominators. For a prime number $p$, apply the last nonzero digit coloring in base $p$: if $x=p^{v_p(x)}u$ with $p\nmid u$, its color is $u\mathrm{mod}p$. Choose a monochromatic solution and group its coordinates into blocks $B_1,\dots ,B_r$ of equal $p$-adic valuation, in increasing order of valuation. Only finitely many ordered partitions are possible, so one partition occurs for infinitely many primes.

For any such prime, divide ${\sum }_i{x}_i{a}_i=0$ by the lowest power of $p$ and reduce modulo $p$. All coordinates in $B_1$ have the same nonzero leading digit $d$, while later blocks vanish, soThe integer vector ${\sum }_{i\in B_1}{a}_i$ is divisible by infinitely many primes and therefore equals zero.

Now fix $s>1$, and let $t$ be the common valuation on $B_s$. Reduction modulo $p^{t+1}$ givesIf the first sum of columns were outside the rational linear span of the earlier columns, an integer linear functional would vanish on every earlier column but not on that sum. Applying it to the congruence would say that infinitely many primes divide one fixed nonzero integer, a contradiction. Thus the displayed partition has the columns property.

Conversely, suppose $A$ has the columns property. For $s>1$, choose rational numbers $q_{is}$ such thatLet $d_{is}$ equal $1$ for $i\in B_s$, equal $-q_{is}$ in the earlier blocks, and equal zero in the later blocks. For $s=1$, let $d_{i1}$ be the indicator of $B_1$. Thenfor every $s$. Choose a positive integer $c$ clearing all denominators and a positive integer $p$ with $|c{d}_{is}|\le p$ whenever $s$ is later than the block containing $i$.

By the monochromatic m-p-c set theorem, the coloring contains a monochromatic $(r,p,c)$-set with generators $z_1,\dots ,z_r$. DefineIf $i\in B_t$, thenso every $y_i$ lies in that one monochromatic M-p-c set. Moreover,Thus $A$ is partition regular, proving Rado's theorem.

For the equation $4x+3y-6z=0$, a monochromatic solution in the last nonzero base-$p$ digit coloring would, at the least $p$-adic valuation among $x,y,z$, force a nonempty subset sum of $4,3,-6$ to vanish modulo $p$. The seven possible sums areNone is divisible by $5$, so base $5$ gives no monochromatic solution. The smaller primes do admit solutions: $(x,y,z)=(3,2,3)$ works for base $2$, whose coloring has one color, and $(3,4,4)$ is monochromatic in base $3$. Hence the smallest prime is

For $x+2y-7z=0$, use the last nonzero base-$11$ digit itself. The nonempty subset sums of $1,2,-7$ arenone zero modulo $11$. This is the required $10$-coloring.

For a $5$-coloring, identify each nonzero residue $d$ modulo $11$ with $-d$, giving the five colorsIf a monochromatic solution existed, reduction at the least $11$-adic valuation would give a signed nonempty subset sum of $1,2,-7$ equal to zero modulo $11$. Singles have absolute residues $1,2,7$; pairs have absolute residues $1,3,5,6,8,9$; and triples have absolute residues $4,6,8,10$. None is zero modulo $11$, which proves that this $5$-coloring works.

Hindman theorem states that every finite coloring of $\mathbb{N}$ admits an infinite sequence $x_1,x_2,\dots$ whose finite-sums set is monochromatic.

Identify the Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ with the compact Hausdorff space of ultrafilters on $\mathbb{N}$, equipped with addition on the Stone-Čech compactification of the natural numbers. This makes $\beta \mathbb{N}$ a compact Hausdorff left-topological semigroup. For completeness, the Ellis–Numakura lemma gives an idempotent in every such semigroup: by the Hausdorff space property and compactness, the intersection of a descending chain of nonempty compact subsemigroups is nonempty, so Zorn's lemma gives a minimal one $K$. For $a\in K$, the compact subsemigroup $K+a$ equals $K$. Hence the nonempty compact subsemigroupis also $K$, and in particular $a+a=a$. Choose the resulting idempotent ultrafilter on the natural numbers $\mathcal{U}$.

One color class $A$ belongs to $\mathcal{U}$. For $x\in \mathbb{N}$, writeand defineThe identity $\mathcal{U}+\mathcal{U}=\mathcal{U}$ implies $A^{\ast }\in \mathcal{U}$. It also implies that $A^{\ast }-x\in \mathcal{U}$ for every $x\in A^{\ast }$: both $A-x$ and the set of $y$ for which $A-(x+y)$ belongs to $\mathcal{U}$ lie in $\mathcal{U}$, and their intersection is $A^{\ast }-x$.

Choose $x_1\in A^{\ast }$. Having chosen $x_1,\dots ,x_n$ with every nonempty finite sum in $A^{\ast }$, chooseThis is possible because it is a finite intersection of members of the ultrafilter $\mathcal{U}$. Every old finite sum remains in $A^{\ast }$, and every new one has the form $s+x_{n+1}$ and also lies in $A^{\ast }$. By mathematical induction, all nonempty finite sums lie in $A^{\ast }\subseteq A$, proving Hindman's theorem.

Now putfor each $A\in \mathcal{U}$. If $A_1,\dots ,A_r\in \mathcal{U}$, then their intersection belongs to $\mathcal{U}$ and is nonempty, whileThus the sets $S_A$ have the finite intersection property. Their closures are closed subsets of the compact interval $[0,1]$, so their total intersection contains some $x$. Equivalently, every neighbourhood of $x$ meets every $S_A$; this is the ultrafilter limit of the sequence.

The point is unique. If distinct points $x,y$ both had this property, choose disjoint neighborhoods $U,V$. The index set $I_U=\{n:x_n\in U\}$ must belong to $\mathcal{U}$, because otherwise its complement would belong to $\mathcal{U}$ and the associated $S_A$ would miss $U$. Similarly $I_V\in \mathcal{U}$. Their intersection is empty, contradicting the definition of an ultrafilter.

Let $X=\{v_1,\dots ,v_m\}$ be the vertices of a regular polygon, indexed cyclically so that $v_i\mapsto v_{i+1}$ is an isometry. We first prove the product lemma needed for the cyclic-symmetry argument.

Suppose $A\subseteq X$ has the property that, for every number of colors, a finite witness forces a copy of $X$ on which the vertices corresponding to $A$ are monochromatic. We claim that for every $N,k$ there is a finite witness forcing an $A$-invariant copy of $X^N$. Proceed by mathematical induction on $N$. The case $N=1$ is the assumption. For the step, choose a finite set $S$ that forces an $A$-invariant $X^{N-1}$ for a coloring with $k^{|X|-|A|+1}$ colors, and choose $T$ that forces an $A$-monochromatic $X$ for a coloring with $k^{|S|}$ colors.

Given a $k$-coloring $c$ of $S\times T$, color $t\in T$ by its complete vector $(c(s,t){)}_{s\in S}$. In the resulting copy of $X$ in $T$, that vector is the same for every member of $A$. Each $s\in S$ can therefore be colored by the vector consisting of $c(s,a)$ for one $a\in A$ and the colors $c(s,x)$ for $x\in X\setminus A$. Applying the choice of $S$ gives an $A$-invariant copy of $X^{N-1}$, and its product with the chosen copy of $X$ in $T$ is the desired $A$-invariant copy of $X^N$. This proves the lemma rather than assuming it.

We now prove by induction on $r$ that a finite witness forces a copy of $X$ whose first $r$ vertices have one color. The case $r=1$ is trivial. For $r=2$, use that a line segment is a Euclidean Ramsey set; each occurrence of a segment congruent to $v_1{v}_2$ can be extended to a copy of the regular polygon, and only finitely many extensions are needed for a finite witness.

Assume the assertion for $r$, and put $A=\{v_1,\dots ,v_r\}$. The product lemma lets us work inside an $A$-invariantly colored copy of $X^N$, with $N$ as large as required. For an increasing $(m-1)$-element subset $H=\{a_1<\cdots <a_{m-1}\}$ of $[N]$ and $1\le i\le r$, define a word $w_i(H)$ by putting $v_{i+j}$ in coordinate $a_j$, with subscripts read modulo $m$, and putting $v_1$ in every other coordinate. Color $H$ by the $r$-tupleBy the Finite Ramsey theorem, for sufficiently large $N$ there are coordinates $b_1<\cdots <b_m$ on which all $(m-1)$-subsets have the same tuple color.

For $1\le i\le r+1$, let $z_i$ be the word whose entries in coordinates $b_1,\dots ,b_m$ arecyclically, and whose other entries are $v_1$. For $i\le r$, the word $z_i$ differs from $w_i(\{b_2,\dots ,b_m\})$ only in coordinate $b_1$, where the two entries $v_i,v_1$ lie in $A$. Similarly $z_{i+1}$ differs from $w_i(\{b_1,\dots ,b_{m-1}\})$ only in coordinate $b_m$, again by two members of $A$. The $A$-invariance and the homogeneous choice of the $b_j$ therefore giveHence $z_1,\dots ,z_{r+1}$ have one color.

The cyclic words $z_1,\dots ,z_m$ form an isometric copy of $\sqrt{m}X$: in each of the $m$ varying coordinates the cyclic shift is an isometry, so every squared distance is multiplied by $m$. Rescaling the finite witness by $1/\sqrt{m}$ gives a copy of $X$. The induction reaches $r=m$, proving that every regular polygon is a Euclidean Ramsey set.

For three consecutive vertices $p_0,p_1,p_2$ of a regular $M$-gon scaled to have side length one,Since the polygon is Euclidean Ramsey, choosing $M$ large enough proves that $\{0,1,2\}$ is an approximately Euclidean Ramsey set.

In fact every finite $X\subseteq {\mathbb{R}}^d$ is approximately Ramsey. First perturb each coordinate of each point onto a sufficiently fine lattice $\delta \mathbb{Z}$, changing all pairwise distances by less than $\epsilon /2$. For each coordinate, map the finitely many required integers to consecutive vertices of a very large regular polygon, scaled so that one angular step has arc length $\delta$. If the step angle is $\theta$, the chord replacing a difference of $q$ lattice steps has lengthas $\theta \to 0$, uniformly over the finitely many differences involved. Taking the orthogonal Cartesian product of $d$ such polygons therefore embeds the perturbed points with a further distance error below $\epsilon /2$. Each polygon is Euclidean Ramsey, and their product is Euclidean Ramsey by the product theorem for Euclidean Ramsey sets. A monochromatic copy of that product contains the corresponding approximate copy of $X$, completing the proof.