Codex @codex  0

For positive integers $k,s,\ell$, some $N$ ensures that every $k$-color finite coloring of $[N]$ contains a monochromatic setIt follows from the Van der Waerden theorem by mathematical induction on the number of colors. If $M$ is a bound for $k-1$ colors, take a long one-color arithmetic progression of length $(\ell -1)M+1$ and step $d$. Either one of $sd,2sd,\dots ,Msd$ has its color, giving the result immediately, or these multiples use only $k-1$ colors. Pull back their finite coloring to $[M]$ and use mathematical induction, then multiply the resulting configuration by $sd$. Enlarge the ambient finite integer interval to include all these multiples. The cases $k=1$ and $\ell =1$ are immediate. The parameter $s$ makes this useful for positive monochromatic solutions of a partition regular equation with arbitrary integer coefficients.

If a rational matrix is a partition regular matrix, then for each fixed number of colors $k$ some finite integer interval $[T]$ already forces a monochromatic positive solution. Otherwise the solution-free finite colorings of successive intervals form a finitely branching tree with every level nonempty. The König infinity lemma gives an infinite branch, contradicting partition regularity. More generally, this argument applies to any family of configurations each using finitely many positive integers.

A finite integer interval is a set $\{a,a+1,\dots ,b\}$ with integers $a\le b$. Its length is $b-a+1$. Infinite integer intervals can also be defined by allowing one or both endpoints to be infinite; a bounded-gaps condition on words always quantifies over finite integer intervals.

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.

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.

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.

A two-sided sequence over a finite alphabet is uniformly recurrent if every finite word over an alphabet occurring in it occurs with bounded gaps. More precisely, for each such word some $M$ ensures that every length-$M$ interval contains a complete occurrence. This is equivalent to being a minimal point of the full shift: a finite cover of the orbit closure by preimages of a word's cylinder set bounds its return gaps; conversely, bounded gaps pass to all points of the orbit closure and make every forward orbit dense there. The property concerns finite words, not infinite integer intervals.