= Solution
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 $\{\mathcal L^{-j}V:j\geq0\}$ 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 $\mathcal L^j$, $0\leq j\leq J$, some $\delta>0$ ensures
$$
d(z,z')<\delta\ \Longrightarrow
d(\mathcal L^jz,\mathcal L^jz')<\eta\quad(0\leq j\leq J).
$$
<Proximality> 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 $t\geq0$, 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 $j\leq J$ has $\mathcal L^{t+j}y\in V$. Then $\mathcal L^{t+j}x\in U$ also. Hence arbitrarily large positive $n$ satisfy
$$
\mathcal L^n x\in U,\qquad \mathcal L^n y\in U.
$$
This 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 $F_r=\{0\}\cup\operatorname{FS}(a_1,\ldots,a_r)$, where $\operatorname{FS}$ denotes the <finite-sums set>, and maintain
$$
y(s)=q\quad(s\in F_r),\qquad x(s)=q\quad(s\in F_r\setminus\{0\}).
$$
The 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. Therefore
$$
\boxed{\chi(s)=q\quad\text{for every }s\in\operatorname{FS}(a_1,a_2,\ldots).}
$$
This proves the <Hindman theorem> using only the permitted proximal-minimal existence result and the <compactness> and return arguments supplied above.
Back to article page
