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If distinct points $x,y$ of a dynamical system on a compact metric space satisfy $d(T^{n}x,T^{n}y)\to 0$ in one compatible metric, no compatible metric can make $T$ an isometry. On a compact space, all compatible metrics give the same asymptotic-pair property by uniform continuity, whereas an isometry preserves the strictly positive distance between distinct points. In a full shift, a constant sequence and a sequence differing at just one coordinate converge to each other under forward shifts, proving this obstruction directly in the product topology.

A metric is compatible with a given topology when its open balls generate exactly that topology. Compatible metrics on a compact topological space are uniformly equivalent: the identity maps between the resulting compact metric spaces are uniformly continuous. Consequently they define the same proximality and the same asymptotic-pair relation for a fixed continuous map.

The two-sided full shift on a finite alphabet $[k]$ consists of all functions $z:\mathbb{Z}\to [k]$, with the product topology and the left shift. It is a compact metric space. A compatible metric isAgreement on increasingly large finite coordinate sets is equivalent to convergence in this topology. The metric above is compatible but is not invariant under the left shift.

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.

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.

In a compact metric space with a continuous map $T$, points $x,y$ are proximal ifThis property is independent of the compatible metric. If $T$ is injective and $x\ne y$, arbitrarily small proximal distances must occur at arbitrarily large times, since each finite collection of distances is strictly positive. Proximal points need not be equal or have dense orbits.