the only compact metric spaces that admit a positively expansive homeomorphism are discrete spaces

Theorem. Let $(X,d)$ be a compact metric space. If there exists apositively expansive homeomorphism $f:X\to X$, then $X$ consists only of isolatedpoints, i.e. $X$ is finite.

Lemma 1. If $(X,d)$ is a compact metric space and there is an expansive homeomorphism $f:X\to X$ such that every point is Lyapunov stable, then every point is asymptotically stable.

Proof. Let $2c$ be the expansivity constant of $f$.Suppose some point $x$ is not asymptotically stable, and let $\delta$ be suchthat implies for all $n\in \mathbb{N}$. Then there exist $ϵ>0$, a point $y$ with , and an increasing sequence $\{n_k\}$ such that $d(f^{n_k}(y),f^{n_k}(x))>ϵ$ for each $k$By uniform expansivity, there is $N>0$ such that for every $u$ and $v$ such that $d(u,v)>ϵ$ there is $n\in \mathbb{Z}$ with such that $d(f^n(x),f^n(y))>c$.Choose $k$ so large that $n_k>N$. Then there is $n$ with such that $d(f^{n+n_k}(x),f^{n+n_k}(y))=d(f^n(f^{n_k}(x)),f^n(f^{n_k}(y)))>c$. But since $n+n_k>0$, this contradicts the choce of $\delta$. Hence every point is asymptotically stable.

Lemma 2 If $(X,d)$ is a compact metric space and $f:X\to X$ is a homeomorphism such that every point is asymptotically stable and Lyapunov stable, then $X$ is finite.

Proof.For each $x\in X$ let $K_x$ be a closed neighborhood of $x$ such that for all $y\in K_x$ we have ${lim}_{n\to \mathrm{\infty }}d(f^n(x),f^n(y))=0$. We assert that ${lim}_{n\to \mathrm{\infty }}\mathrm{diam}(f^n(K_x))=0$. In fact,if that is not the case, then there is an increasing sequence of positive integers $\{n_k\}$, some $ϵ>0$ and a sequence $\{x_k\}$ of points of $K_x$ such that $d(f^{n_k}(x),f^{n_k}(x_k))>ϵ$, and there is a subsequence $\{x_{k_i}\}$ converging to some point $y\in K_x$.

From the Lyapunov stability of $y$, we can find $\delta >0$ such that if , then for all $n>0$. In particular if $i$ is large enough. But also if $i$ is large enough, because $y\in K_x$. Thus, for large $i$, we have . That is a contradiction from our previous claim.

Now since $X$ is compact, there are finitely many points $x_1,\mathrm{\dots },x_m$ such that $X={\bigcup }_{i=1}^m{K}_{x_i}$,so that $X=f^n(X)={\bigcup }_{i=1}^m{f}^n(K_{x_i})$. To show that $X=\{x_1,\mathrm{\dots },x_m\}$, suppose there is $y\in X$ such that $r=\min \{d(y,x_i):1\le i\le m\}>0$. Then there is $n$ such that for $1\le i\le m$but since $y\in f^n(K_{x_i})$ for some $i$, we have a contradiction.

Proof of the theorem.Consider the sets $K_{ϵ}=\{(x,y)\in X\times X:d(x,y)\ge ϵ\}$ for $ϵ>0$ and $U=\{(x,y)\in X\times X:d(x,y)>c\}$, where $2c$ is the expansivity constant of $f$, and let $F:X\times X\to X\times X$ be the mapping given by $F(x,y)=(f(x),f(y))$. It is clear that $F$ is a homeomorphism.By uniform expansivity, we know that for each $ϵ>0$ there is $N_{ϵ}$ such that for all $(x,y)\in K_{ϵ}$, there is $n\in \{1,\mathrm{\dots },N_{ϵ}\}$ such that $F^n(x,y)\in U$.

We will prove that for each $ϵ>0$, there is $\delta >0$ such that$F^n(K_{ϵ})\subset K_{\delta }$ for all $n\in \mathbb{N}$. This is equivalent to say that every point of $X$ is Lyapunov stable for $f^{-1}$, and by the previous lemmas the proof will be completed.

Let $K={\bigcup }_{n=0}^{N_{ϵ}}{F}^n(K_{ϵ})$, and let ${\delta }_0=\min \{d(x,y):(x,y)\in K\}$.Since $K$ is compact, the minimum distance ${\delta }_0$ is reached at some point of $K$; i.e. there exist $(x,y)\in K_{ϵ}$ and$0\le n\le N_{ϵ}$ such that $d(f^n(x),f^n(y))={\delta }_0$.Since $f$ is injective, it follows that ${\delta }_0>0$ and letting$\delta ={\delta }_0/2$ we have $K\subset K_{\delta }$.

Given $\alpha \in K-K_{ϵ}$, there is $\beta \in K_{ϵ}$ and some such that $\alpha =F^m(\beta )$, and $F^k(\beta )\notin K_{ϵ}$ for .Also, there is $n$ with such that $F^n(\beta )\in U\subset K_{ϵ}$. Hence ,and $F(\beta )=F^{m+1}(\alpha )\in F^{m+1}(K_{ϵ})\subset K$; On the other hand, $F(K_{ϵ})\subset K$. Therefore $F(K)\subset K$, and inductively $F^n(K)\subset K$ for any $n\in \mathbb{N}$. It followsthat $F^n(K_{ϵ})\subset F^n(K)\subset K\subset K_{\delta }$ for each $n\in \mathbb{N}$ as required.