Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 130 (Ramsey Theory)

Write $\mathbb{N}=\{1,2,\dots \}$ and let $\chi$ be the given finite coloring of the edges of the complete graph on the positive integers. Construct an increasing sequence $(v_i)$ and nested infinite sets $A_0\supset A_1\supset \cdots$ as follows. Set $A_0=\mathbb{N}$. Having chosen $A_{i-1}$, take $v_i=\min A_{i-1}$, partition $A_{i-1}\setminus \{v_i\}$ by the color of $\{v_i,w\}$, and use the infinite pigeonhole principle to choose an infinite set $A_i$ on which this color is constant, say $r_i$. Every element of $A_i$ exceeds $v_i$.

The infinite pigeonhole principle applied again to $(r_i)$ gives an infinite set of indices $I$ and a color $r$ such that $r_i=r$ for all $i\in I$. For $i<j$ in $I$, nesting gives $v_j\in A_i$, so $\chi (\{v_i,v_j\})=r$. Thus the infinite set $X=\{v_i:i\in I\}$ is a monochromatic complete graph:This proves the required infinite case of Ramsey's theorem; finiteness of the color set is used at both applications of the infinite pigeonhole principle.

Pull the given finite coloring $\chi :\mathbb{N}\to [k]$ back to a finite coloring of unordered pairs by puttingThe ordering $a<b$ makes this a well-defined function on the edges of a complete graph. By Ramsey's theorem, some infinite set $X=\{x_1<x_2<\cdots \}$ has all its pairs of one color $r$ under $\tilde{\chi }$. ConsequentlyPossible coincidences between these products cause no difficulty, since the original finite coloring assigns each positive integer a single color.

Use the cyclic logarithmic coloringHere the floor function places $x$ in the unique half-open real interval $[(3/2{)}^j,(3/2{)}^{j+1})$. If $y/x\in [1.9,2]$, thenbecause $3/2<1.9\le 2<(3/2{)}^2$. For any real $t$ and $1<u<2$, $\lfloor t+u\rfloor -\lfloor t\rfloor$ is either $1$ or $2$. Applying this to the logarithms of $x$ and $y$ shows that their bin indices differ by $1$ or $2$, and hence have different residues in modular arithmetic modulo $3$. This proves the required separation, including both endpoints of the prescribed ratio interval. The half-open bins remove any ambiguity at their boundaries.

Let $\rho$ be the cyclic logarithmic coloring from part (iii). Define a finite coloring of the positive integers byThe inner logarithm is at least $1$ whenever $n\ge 2$, so this is defined everywhere. Suppose an increasing infinite sequence $(x_i)$ made every product with distinct indices monochromatic. Fix an index $i$ with $x_i\ge 2$ and write $a={\log }_2{x}_i>0$. The sequence is unbounded, so choose $j>i$ with $b={\log }_2{x}_j\ge 28a$. Both ordered products are among the supposedly monochromatic values, but their logarithms satisfyIndeed, the first inequality is equivalent to $b\ge 28a$, and the second uses $a>0$. Part (iii) therefore gives $\chi (x_i{x}_j^2)\ne \chi (x_j{x}_i^2)$, a contradiction. Thus the assertion with all distinct ordered indices is false. The obstruction needs an infinite unbounded sequence; it does not assert that every pair of different positive integers gives different colors.

We prove the Hales-Jewett theorem by mathematical induction on the alphabet size. A combinatorial line has a nonempty active coordinate set, with one common letter in all those coordinates. More generally, a $d$-dimensional combinatorial subspace has $d$ disjoint nonempty active coordinate sets, each with an independently variable letter.

First prove the alphabet insensitivity lemma, rather than assume it. Fix distinct letters $a,b\in [m]$, $k$ colors, and dimension $d$. Partition the coordinates into consecutive blocks of lengths $L_1,\dots ,L_d$, chosen from left to right. Put $S_0=0$ and chooseConstruct the variable blocks from right to left. When working on block $i$, the later blocks have already become $(d-i)$-parameter words, while all earlier $S_{i-1}$ coordinates remain unrestricted. For each $t\in \{0,\dots ,L_i\}$, take the color vector of the word having $a^t{b}^{L_i-t}$ in block $i$, indexed by every earlier word and every assignment to the later parameters. There are $m^{S_{i-1}+d-i}$ entries and hence at most $L_i$ different vectors. The pigeonhole principle gives $t<s$ with identical vectors. Fix the first $t$ positions of this block to $a$, fix its last $L_i-s$ positions to $b$, and make the intervening $s-t$ positions a new variable block. Substituting $a$ or $b$ there gives exactly the two equal color vectors.

This equality holds for every earlier word and every later parameter assignment. Restricting the earlier coordinates in subsequent steps therefore preserves it. Likewise, the insensitivity already proved in a later block survives every restriction of earlier coordinates. At the end, we obtain a $d$-dimensional combinatorial subspace whose induced finite coloring is unchanged whenever any parameter switches between $a$ and $b$, with all other parameters fixed. Multiple switches preserve the color by successive single switches. This is the alphabet insensitivity lemma in full.

For alphabet size $1$, one coordinate is already a monochromatic combinatorial line. Suppose the Hales-Jewett theorem has been proved for alphabet size $m-1$, and let $d=n(m-1,k)$. Apply the proved alphabet insensitivity lemma with letters $m-1,m$ to get a $d$-dimensional combinatorial subspace in $[m{]}^N$. Restrict its parameter words to $[m-1{]}^d$. The induction hypothesis gives a monochromatic combinatorial line in these parameters. The union of its active parameter blocks is nonempty, so its image is a combinatorial line in the original coordinates. Its additional point with active letter $m$ has the same color as the point with active letter $m-1$, by insensitivity. Therefore all $m$ points have one color, provingThe construction supplies a finite recursive bound; optimal bounds are not needed.

Let the required arithmetic progression have length $m\ge 2$, let $k$ be the number of colors, and take $n=n(m,k)$ from the Hales-Jewett theorem. Given any finite coloring $\chi$ of $[mn]$, color a word over an alphabet $w\in [m{]}^n$ by $\chi (w_1+\cdots +w_n)$. For a monochromatic combinatorial line, let $A\ne \varnothing$ be its active coordinate set and let $C$ be the sum of its fixed coordinates. The sums of its $m$ words areThey all lie in $[mn]$, have one color, and form an arithmetic progression with strictly positive common difference $|A|$. Thus, writing $W(k,m)$ for the finite Van der Waerden theorem bound,The length-one case is immediate. Restricting a finite coloring of the positive integers to this finite integer interval gives the infinite-domain formulation of the Van der Waerden theorem as well.

A run of a word is a maximal consecutive block of equal letters. For every $N\ge 1$, use the following finite coloring of $[3{]}^N$, with at most nine colors:We show that this run-count obstruction to interval-active lines excludes every monochromatic combinatorial line with an interval active set $[a,b]$. If $a=1$, its three words already have different first coordinates, so their colors differ. Suppose $a>1$ and denote the letter immediately to the left by $p$. All contributions to $R(w)$ away from the two possible boundaries of $[a,b]$ are independent of its active letter $x$; no internal active boundary contributes a change.

If $b=N$, the only variable contribution is $1_{\{p\ne x\}}$, which is $0$ at $x=p$ and $1$ at another letter. If $b<N$, write $q$ for the letter immediately to the right. The variable contribution isWhen $p=q$, its values are $0$ and $2$. When $p\ne q$, it equals $1$ at $x=p$ or $x=q$, and equals $2$ at the third letter. In every case these values are distinct modulo $3$. Hence no dimension works for alphabet size three and nine colors, which disproves the proposed universal strengthening. For the PDF's illustrative line, choosing either adjacent letter merges one active run with a neighboring run, while choosing any other letter merges neither; this is exactly the boundary effect measured above. Recording the first letter also handles active intervals meeting the beginning, including the entire coordinate set.

Use the standard definition of a partition regular matrix: for every finite coloring of the positive integers, there is a positive solution whose coordinates all have one color. Coordinates may repeat. Clearing denominators lets us assume $a_i\in \mathbb{Z}\setminus \{0\}$; multiplication of the row by a nonzero rational number changes neither its zero-sum subsets nor its solutions. We prove both directions of the Rado theorem for one equation directly, without invoking any form of Rado's theorem.

For necessity, choose a prime number $p>{\sum }_i|a_i|$ and use the last nonzero digit coloring: write $x=p^{v_p(x)}u$, with $p\nmid u$, and assign color $u\mathrm{mod}p$. Given a monochromatic solution, let $v={\min }_i{v}_p(x_i)$, let $I=\{i:v_p(x_i)=v\}$, and let $r\ne 0$ be the common color. Divide ${\sum }_i{a}_i{x}_i=0$ by $p^v$ and reduce modulo $p$. Terms outside $I$ vanish; the others giveThe nonzero residue $r$ is invertible modulo the prime number $p$, so $p$ divides ${\sum }_{i\in I}{a}_i$. This sum has absolute value less than $p$, forcing it to be zero. The set $I$ is nonempty by its definition.

For sufficiency we first derive the needed Brauer progression theorem from the permitted Van der Waerden theorem. Fix positive integers $s,\ell$. We claim that for every number of colors $r$, a finite integer interval $[B(r,s,\ell )]$ forces a monochromatic setThe case $\ell =1$ is immediate by taking $a=s$ and $d=1$. For $r=1$ and $\ell \ge 2$, take $a=d=1$ in an interval of length at least $\max (s,\ell )$. Suppose the claim holds for $r-1$ and write $M=B(r-1,s,\ell )$. By the Van der Waerden theorem, some $R$ forces an arithmetic progression of length $(\ell -1)M+1$ in $[R]$ with one color $q$. Work in $[\max (R,sMR)]$ so that all needed multiples also lie in the coloring's domain. If a subprogression of length $\ell$ and step $td$ has $std$ of color $q$, we are done. Otherwise, for every $1\le t\le M$, the initial long arithmetic progression contains such a subprogression, and $std$ avoids color $q$. The induced finite coloring $t\mapsto \chi (sdt)$ of $[M]$ therefore uses at most $r-1$ colors. The induction hypothesis gives a monochromatic set $\{se,b,b+e,\dots ,b+(\ell -1)e\}$ for this induced finite coloring. Multiplying by $sd$ gives the desired configuration in the original coloring, with initial term $sdb$ and step $sde$. This proves the claim for $\ell \ge 2$; the only use below has $\ell \ge 3$.

Now suppose ${\sum }_{i\in I}{a}_i=0$ for a nonempty $I$. Choose $i_0\in I$, put $B={\sum }_{i\notin I}{a}_i$, and put $s=|a_{i_0}|$. If $B=0$, then ${\sum }_i{a}_i=0$ and any constant positive vector is already a monochromatic solution. Otherwise take $L=|B|$ and apply the proved Brauer progression theorem with $\ell =2L+1$. All of $sd$ and $a+jd$ for $0\le j\le 2L$ have one color. DefineEvery coordinate is a positive integer in that monochromatic set, since $L-\mathrm{sgn}(a_{i_0})B$ is either $0$ or $2L$. Using the zero sum on $I$ givesThus both directions are proved:For $n=1$ the right side is impossible, agreeing with the absence of positive solutions for a single nonzero coefficient.

First justify the finite bound used in the hint. Fix $k$. If no finite $T$ forced a monochromatic positive solution of ${\sum }_i{a}_i{x}_i=0$, consider the rooted tree whose level $t$ consists of the solution-free finite colorings of $[t]$, with restriction as the predecessor map. Every level is nonempty and every vertex has at most $k$ children. The König infinity lemma gives an infinite branch, hence a finite coloring of all positive integers with no such solution, contradicting the partition regular matrix hypothesis. This is the compactness bound for partition regularity.

Choose such a $T$ and let $S=\mathrm{lcm}(1,\dots ,T)$, the least common multiple. For a given finite coloring $\chi$ of the positive integers, pull it back to $[T]$ byEach $S/t$ is a positive integer in $[S]$. The defining property of $T$ gives $x_1,\dots ,x_n\in [T]$ of one color under $\tilde{\chi }$, with ${\sum }_i{a}_i{x}_i=0$. Set $y_i=S/x_i$. Their colors under $\chi$ agree, andThus reciprocal partition regularity follows. This construction is an involution on the divisors of $S$; it permits repeated coordinates and never requires a reciprocal of a positive integer to itself be integral without the common scaling factor $S$.

Use the product topology on $\mathcal{C}=[k{]}^{\mathbb{Z}}$, with $[k]$ discrete, and the left shift $(\mathcal{L}c)(j)=c(j+1)$. A basic cylinder set specifies finitely many coordinates. This is a compact metric space, and $\mathcal{L}$ is a homeomorphism. Write $Y=\overline{\{{\mathcal{L}}^{n}c:n\ge 0\}}$ for the forward orbit closure. A minimal point means that $Y$ is a minimal dynamical system, equivalently that every point of $Y$ has a dense forward orbit in $Y$; it need not be a fixed point.

There is a convention needed in the source: the bounded-gaps property concerns every finite integer interval $I$, hence every finite word over an alphabet. Literally allowing $I=\mathbb{Z}$ would make the displayed condition impossible for finite $U$, although a constant coloring is a minimal point. Under the standard finite-word interpretation, the property is precisely uniform recurrence.

Suppose first that $Y$ is a minimal dynamical system. Its nonempty compact forward-invariant subset $\mathcal{L}(Y)$ must equal $Y$. As the ambient left shift is invertible, all integer translates of $c$ belong to $Y$. Let $I=[a,b]\cap \mathbb{Z}$ have length $\ell =b-a+1$, and let $V=\{z\in Y:z{|}_I=c{|}_I\}$. This is a nonempty clopen set. Each forward orbit in $Y$ meets $V$, so $\{{\mathcal{L}}^{-n}V:n\ge 0\}$ covers $Y$. By compactness, finitely many suffice; let $D$ be the largest index in this finite cover. For any integer $u$, the point ${\mathcal{L}}^{u-a}c$ lies in $Y$, so some $0\le n\le D$ satisfies ${\mathcal{L}}^{u-a+n}c\in V$. The prescribed word therefore occurs at positions $[u+n,u+n+\ell -1]$, entirely inside $[u,u+D+\ell -1]$. Taking $M=D+\ell$ proves the bounded-gaps property for every interval of length at least $M$.

Conversely suppose $c$ is uniformly recurrent. Every finite word from any integer translate of $c$ occurs arbitrarily far to the right, so every such translate belongs to $Y$. Moreover, the property that every length-$M$ block contains a specified word of $c$ passes to every $z\in Y$: a finite block of $z$ is a limit of blocks of forward translates of $c$, and the finite discrete alphabet forces eventual exact agreement on that block. Given $I=[a,b]\cap \mathbb{Z}$, look for its word inside $z([a,a+M-1])$. An occurrence starts at $a+n$ for some $n\ge 0$, so ${\mathcal{L}}^{n}z$ agrees with $c$ on $I$. Taking $I=[-r,r]\cap \mathbb{Z}$ for arbitrarily large $r$ proves that $c$ belongs to the forward orbit closure of every $z\in Y$. That closure is a closed forward-invariant subset of $Y$, so it also contains every forward translate of $c$ and hence all of $Y$. Every forward orbit is therefore dense in $Y$, provingThe same conclusion holds if orbit closure is defined using all integer iterates: the bounded-gaps condition makes the forward and two-sided closures equal.

The dynamical proof of Hindman's theorem gives the following result. The Hindman theorem asserts that every finite coloring of the positive integers admits an infinite strictly increasing sequence $(a_i)$ for which all nonempty finite sums of distinct terms have one color. We will in fact arrange $a_{r+1}>a_1+\cdots +a_r$, so all these sums also have unique representations.

Extend the given coloring arbitrarily to a point $x\in [k{]}^{\mathbb{Z}}$, retaining the prescribed colors at every positive coordinate. Let $X$ be its forward orbit closure under the left shift. The product topology makes $X$ a compact metric space, and the left shift is a continuous map. There is a nonempty minimal subsystem $Y\subseteq X$: order the nonempty closed forward-invariant subsets by reverse inclusion, use compactness and the finite intersection property to intersect any chain, and apply the Zorn lemma. Minimality also implies $\mathcal{L}(Y)=Y$. The result permitted in the question now supplies a minimal point $y\in Y$ proximal to $x$.

We need the joint return lemma for a proximal minimal pair, which we prove here. It suffices to consider an open neighborhood $U$ of $y$ in $X$. Choose an open neighborhood $V$ of $y$ with $\overline{V}\subseteq U$. Every forward orbit in $Y$ meets $V$, and a finite subcover of $\{{\mathcal{L}}^{-j}V:j\ge 0\}$ on $Y$ supplies a bound $J$ on the needed return index. For a compatible metric $d$, choose $\eta >0$ smaller than the distance from $\overline{V}$ to $X\setminus U$ when the latter is nonempty. By uniform continuity of the finitely many maps ${\mathcal{L}}^j$, $0\le j\le J$, some $\delta >0$ ensuresProximality supplies arbitrarily large $t$ with $d({\mathcal{L}}^{t}x,{\mathcal{L}}^{t}y)<\delta$. To see that the times can be large under the definition using an infimum over $t\ge 0$, either $x=y$, in which case this is automatic, or injectivity of the left shift makes every finite collection of distances strictly positive, so a sufficiently smaller proximal distance occurs beyond that collection. Some $j\le J$ has ${\mathcal{L}}^{t+j}y\in V$. Then ${\mathcal{L}}^{t+j}x\in U$ also. Hence arbitrarily large positive $n$ satisfyThis uses the minimal point property for bounded returns and proximality for closeness; closeness alone would not guarantee a return near $y$.

Put $q=y(0)$. Inductively, let $F_r=\{0\}\cup \mathrm{FS}(a_1,\dots ,a_r)$, where $\mathrm{FS}$ denotes the finite-sums set, and maintainThe initial condition at $r=0$ is just $y(0)=q$; no condition on $x(0)$ is required. The cylinder set $U_r=\{z:z(s)=q\text{ for every }s\in F_r\}$ is a neighborhood of $y$. Use the proved joint return lemma to choose $a_{r+1}>{\sum }_{i=1}^r{a}_i$ with both ${\mathcal{L}}^{a_{r+1}}x$ and ${\mathcal{L}}^{a_{r+1}}y$ in $U_r$. All new sums belong to $a_{r+1}+F_r$, and the old sums remain in $F_r$, so both inductive conditions persist. Every nonzero sum is positive, where $x$ agrees with the original coloring. ThereforeThis proves the Hindman theorem using only the permitted proximal-minimal existence result and the compactness and return arguments supplied above.

Let $x(j)=1$ for every $j\in \mathbb{Z}$, and let $y$ differ from $x$ only at $j=0$, where $y(0)=2$. Under the left shift, the unique defect of ${\mathcal{L}}^{n}y$ lies at coordinate $-n$. Every fixed finite coordinate set eventually avoids the defect, soin the product topology. If a metric $d$ induces this topology, convergence and the triangle inequality imply $d({\mathcal{L}}^{n}x,{\mathcal{L}}^{n}y)\to 0$. But left shift invariance would givefor every $n$, a contradiction. Thus no compatible metric invariant under the left shift exists. The discrete metric $1_{\{x\ne y\}}$ is shift invariant, but it induces the discrete topology rather than the product topology; compatibility is the essential restriction.