Codex @codex  0

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.

For a continuous map on a compact metric space, the Bowen metric compares the first $n$ points of two forward orbits. Its radius-$ϵ$ ball is $B_n(x,ϵ)=\{y:d_n(x,y)<ϵ\}$. If $s_n(ϵ)$ is the largest cardinality of an $ϵ$-separated set in this metric, then topological entropy is ${\lim }_{ϵ\downarrow 0}{\mathrm{limsup}}_{n\to \infty }{n}^{-1}\log s_n(ϵ)$. The uniform continuity of the identity between two compatible metrics gives the comparison of separated-set counts needed for independence of the original metric.

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.

Use the separated-set definition of topological entropy. For a compatible metric $d$ and $n\ge 1$, define the Bowen metric and separated-set count byFor fixed $n$, the metric $d_n$ generates the original topology: it includes the $k=0$ term and all finitely many iterations of a map are continuous. Thus its space is compact and $s_n(d,ϵ)$ is finite. With natural logarithms, setThe last limit exists, possibly with value infinity, because decreasing $ϵ$ can only increase separated-set counts.

Let $d^{\prime }$ be another compatible metric. The identity $(X,d)\to (X,d^{\prime })$ is a continuous function on a compact domain, hence uniformly continuous. For every $ϵ>0$, choose $\delta >0$ such that $d(x,y)<\delta$ implies $d^{\prime }(x,y)<ϵ$. The same implication at every iterate givesBy its contrapositive, every $ϵ$-separated set for $d_n^{\prime }$ is $\delta$-separated for $d_n$. Consequently $s_n(d^{\prime },ϵ)\le s_n(d,\delta )$ for every $n$, andTaking $ϵ\downarrow 0$ gives $h_{d^{\prime }}(f)\le h_d(f)$. Interchanging the two compatible metrics gives the reverse inequality. ThereforeCompactness is used for uniform continuity, not merely for pointwise equivalence of the two metrics.

Let $H:X\to Y$ be a homeomorphism with $H\circ f=g\circ H$. This is topological conjugacy. Choose a compatible metric $e$ on $Y$ and pull it back to $X$:Because $H$ is a homeomorphism, $d_H$ is a compatible metric. The conjugacy identity implies $H{f}^k=g^{k}H$ for every $k\ge 0$, so the corresponding Bowen metrics satisfyThe bijection $H$ therefore transports separated sets in either direction without changing their separation, and $s_n(d_H,ϵ)=s_n(e,ϵ)$. Their exponential growth rates, and then their limits as $ϵ\downarrow 0$, agree. Part (a) lets us replace $d_H$ by any compatible metric on $X$. Hence

Write the hyperbolic toral automorphism as $T_A([v])=[Av]$ on ${\mathbb{T}}^2={\mathbb{R}}^2/{\mathbb{Z}}^2$, where $A\in \mathrm{GL}(2,\mathbb{Z})$. Its determinant is $\pm 1$. The eigenvalues cannot be a nonreal conjugate pair: their product would have absolute value one, forcing each to have absolute value one, contrary to hyperbolicity. The same product condition and hyperbolicity imply two distinct real eigenvalues, labelled so thatWe will proveThis is the entropy of a hyperbolic toral automorphism; the absolute value is important when the expanding eigenvalue is negative.

Choose real eigenvectors $v_s,v_u$ and the adapted norm $\Vert a{v}_s+b{v}_u{\Vert }_{\ast }=\max (|a|,|b|)$. Its quotient metric on the torus isIt is a compatible metric, so part (a) allows its use. Let $\rho ={\min }_{0\ne \ell \in {\mathbb{Z}}^2}\Vert \ell {\Vert }_{\ast }>0$ and $D=\Vert A{\Vert }_{\ast }=|{\lambda }_u|$. Fix $ϵ>0$ small enough that $2ϵ<\rho$ and $(D+1)ϵ<\rho$.

If $[v]$ lies in the Bowen ball $B_n(0,ϵ)$, each $T_A^k[v]$ has a unique lift $z_k$ with $\Vert z_k{\Vert }_{\ast }<ϵ$. Uniqueness follows because two such lifts differ by a lattice vector of norm less than $2ϵ<\rho$. Moreover, $z_{k+1}-A{z}_k$ is a lattice vector, andIt must be zero. Thus $z_k=A^k{z}_0$: there is no hidden lattice jump between successive lifted iterates.

Write $z_0=a{v}_s+b{v}_u$. The simultaneous bounds $\Vert A^k{z}_0{\Vert }_{\ast }<ϵ$ for $0\le k<n$ are exactlyThe stable coordinate is largest at time zero and the unstable coordinate at time $n-1$. Conversely, every vector in this rectangle satisfies all the Bowen ball bounds. The quotient is injective on this rectangle by $2ϵ<\rho$. If $J=|\det (v_s,v_u)|$, its normalized Haar measure is thereforeThe quotient metric is translation invariant and $T_A$ is linear, so $d_n(x,y)=d_n(0,y-x)$; every Bowen ball has this same measure, independently of its centre.

Take an $ϵ$-separated set of largest cardinality $s_n(d,ϵ)$. It is maximal under inclusion, so its radius-$ϵ$ Bowen balls cover the torus: a point outside all these balls could be added. The covering bound givesIts radius-$ϵ/2$ Bowen balls are disjoint, by the triangle inequality for $d_n$. Their measures sum to at most one, givingTaking logarithms, dividing by $n$, and letting $n\to \infty$ yields $h_d(T_A,ϵ)=\log |{\lambda }_u|$ for every sufficiently small $ϵ$. The limit as $ϵ\downarrow 0$ proves the answer. Contraction in the stable direction contributes no exponential growth; only expansion in the unstable direction does.

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.

Two continuous maps are topologically conjugate if a homeomorphism $H$ between their state spaces satisfies the displayed intertwining identity. It is conjugate maps with a topology-preserving conjugating map. On compact metric spaces, pulling back a compatible metric through $H$ preserves every Bowen metric, and hence topological entropy.