## There is no positively expansive homeomorphism

Let $f$ be a homeomorphism on a compact metric space $(X,d)$. Then $f$ is said to be $\mathbb{Z}$-expansive, if there exists $\delta>0$ such that for any two points $x,y\in X$, if $d(f^nx,f^ny)<\delta$ for all $n\in\mathbb{Z}$, then $x=y$. The constant $\delta$ is called the expansive constant of $f$.

Similarly one can define $\mathbb{N}$-expansiveness if $f$ is not invertible. An interesting phenomenon observed by Schwartzman states that

Theorem. A homeomorphism $f$ cannot be $\mathbb{N}$-expansive (unless $X$ is finite).

This result was reported in Gottschalk–Hedlund’s book Topological Dynamics (1955), and a proof was given in King’s paper A map with topological minimal self-joinings in the sense of del Junco (1990). Below we copied the proof from King’s paper.

Proof. Suppose on the contrary that there is a homeo $f$ on $(X,d)$ that is $\mathbb{N}$-expansive. Let $\delta>0$ be the $\mathbb{N}$-expansive constant of $f$, and $d_n(x,y)=\max\{d(f^k x, f^k y): 1\le k\le n\}$.

It follows from the $\mathbb{N}$-expansiveness that $N:=\sup\{n\ge 1: d_n(x,y)\le\delta \text{ for some } d(x,y)\ge\delta\}$ is a finite number. Pick $\epsilon\in(0,\delta)$ such that $d_N(x,y)<\delta$ whenever $d(x,y)<\epsilon$.

Claim. If $d(x,y)<\epsilon$, then $d(f^{-n} x, f^{-n}y)<\delta$ for any $n\ge 1$.

Proof of Claim. If not, we can prolong the $N$-string since $f^{k}=f^{k+n}\circ f^{-n}$.

Recall that a pair $(x,y)$ is said to be $\epsilon$-proximal, if $d(f^{n_i}x, f^{n_i}y)<\epsilon$ for some $n_i\to\infty$. The upshot for the above claim is that any $\epsilon$-proximal pair is $\delta$-indistinguishable: $d(f^{n}x, f^{n}y)<\delta$ for all $n$.

Cover $X$ by open sets of radius $< \epsilon$, and pick a finite subcover, say $\{B_i:1\le i\le I\}$. Let $E=\{x_j:1\le j\le I+1\}$ be a subset consisting of $I+1$ distinct points. Then for each $n\ge 0$, there are two points in $f^n E$ share the room $B_{i(n)}$, say $f^nx_{a(n)}$, and $f^nx_{b(n)}$. Pick a subsequence $n_i$ such that $a(n_i)\equiv a$ and $b(n_i)\equiv b$. Clearly $x_a\neq x_b$, and $d(f^{n_i}x_a,f^{n_i}x_b)<\epsilon$. Hence the pair $(x_a,x_b)$ is $\epsilon$-proximal and $\delta$-indistinguishable. This contradicts the $\mathbb{N}$-expansiveness assumption on $f$. QED.