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

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)) for all $n\in\mathbb{N}$. 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 $n\in\mathbb{Z}$ with $|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| 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 $\lim_{n\to\infty}\operatorname{diam}(f^{n}(K_{x}))=0$. 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 $\operatorname{diam}(f^{n}(K_{x_{i}})) 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 $n\in\mathbb{N}$. 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 $K=\bigcup_{n=0}^{N_{\epsilon}}F^{n}(K_{\epsilon})$, 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 such that $\alpha=F^{m}(\beta)$, and $F^{k}(\beta)\notin K_{\epsilon}$ for $0. Also, there is $n$ with $0 such that $F^{n}(\beta)\in U\subset K_{\epsilon}$. Hence $m, 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 $n\in\mathbb{N}$. It follows that $F^{n}(K_{\epsilon})\subset F^{n}(K)\subset K\subset K_{\delta}$ for each $n\in\mathbb{N}$ as required.

Title the only compact metric spaces that admit a positively expansive homeomorphism are discrete spaces TheOnlyCompactMetricSpacesThatAdmitAPositivelyExpansiveHomeomorphismAreDiscreteSpaces 2013-03-22 13:55:11 2013-03-22 13:55:11 Koro (127) Koro (127) 10 Koro (127) Theorem msc 37B99