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

Lemma 1. If $(X,d)$ is a compact metric space and there is an expansive homeomorphism $f\colon X\to X$ 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 such that $d(x,y)<\delta$ implies $d(f^n(x),f^n(y))<c$ for all $n\in \N$ . Then there exist $\epsilon>0$ , a point $y$ with $d(x,y)<\delta$ , and an increasing sequence $\{n_k\}$ such that $d(f^{n_k}(y),f^{n_k}(x))>\epsilon$ for each $k$ By uniform expansivity, there is $N>0$ such that for every $u$ and $v$ such that $d(u,v)>\epsilon$ there is $n\in \Z$ with $|n|<N$ such that $d(f^n(x),f^n(y))>c$ . Choose $k$ so large that $n_k>N$ . Then there is $n$ with $|n|<N$ 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 $f\colon X\to X$ 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 $\lim_{n\to\infty} d(f^n(x),f^n(y)) = 0$ . We assert that $\lim_{n\to\infty}\diam(f^n(K_x))=0$ . In fact, if that is not the case, then there is an increasing sequence of positive integers $\{n_k\}$ , some $\epsilon>0$ and a sequence $\{x_k\}$ of points of $K_x$ such that $d(f^{n_k}(x),f^{n_k}(x_k))>\epsilon$ , 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 $d(y,z)<\delta$ , then $d(f^n(y),f^n(z))<\epsilon/2$ for all $n>0$ . In particular $d(f^{n_{k_i}}(x_{k_i}),f^{n_{k_i}}(y))<\epsilon/2$ if $i$ is large enough. But also $d(f^{n_{k_i}}(y), f^{n_{k_i}}(x))<\epsilon/2$ if $i$ is large enough, because $y\in K_x$ . Thus, for large $i$ , we have $d(f^{n_{k_i}}(x_{k_i}),f^{n_{k_i}}(x))<\epsilon$ . That is a contradiction from our previous claim.

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

Proof of the theorem. Consider the sets $K_\epsilon = \{(x,y)\in X\times X : d(x,y)\geq \epsilon\}$ for $\epsilon>0$ and $U=\{(x,y)\in X\times X : d(x,y)>c\}$ , where $2c$ is the expansivity constant of $f$ , and let $\F\colon X\times X\to X\times X$ 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 $\epsilon>0$ there is $N_\epsilon$ such that for all $(x,y)\in K_\epsilon$ , there is $n\in\{1,\dots,N_\epsilon\}$ such that $\F^n(x,y)\in U$ .

We will prove that for each $\epsilon>0$ , there is $\delta>0$ such that $F^n(K_\epsilon)\subset K_\delta$ for all $n\in \N$ . 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 $K=\bigcup_{n=0}^{N_\epsilon} \F^n(K_\epsilon)$ , 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 $(x,y)\in K_\epsilon$ and $0\leq n\leq N_\epsilon$ 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 $\alpha\in K-K_\epsilon$ , there is $\beta\in K_\epsilon$ and some $0<m\leq N_\epsilon$ such that $\alpha=\F^m(\beta)$ , and $\F^k(\beta)\notin K_\epsilon$ for $0<k\leq m$ . Also, there is $n$ with $0<m<n\leq N_\epsilon$ such that $\F^n(\beta)\in U\subset K_\epsilon$ . Hence $m<N_\epsilon$ , and $\F(\beta)=\F^{m+1}(\alpha)\in \F^{m+1}(K_\epsilon)\subset K$ ; On the other hand, $\F(K_\epsilon)\subset K$ . Therefore $F(K)\subset K$ , and inductively $\F^n(K)\subset K$ for any $n\in \N$ . It follows that $\F^n(K_\epsilon)\subset F^n(K)\subset K \subset K_\delta$ for each $n\in \N$ as required.