# 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 a
positively expansive homeomorphism^{} $f:X\to X$, 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 $f:X\to X$ 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 $$ implies $$ for all $n\in \mathbb{N}$. Then there exist $\u03f5>0$, a point $y$ with $$, and an increasing sequence $\{{n}_{k}\}$ such that $d({f}^{{n}_{k}}(y),{f}^{{n}_{k}}(x))>\u03f5$ for each $k$ By uniform expansivity, there is $N>0$ such that for every $u$ and $v$ such that $d(u,v)>\u03f5$ there is $n\in \mathbb{Z}$ with $$ 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 $\u03f5>0$ and a sequence^{} $\{{x}_{k}\}$ of points of ${K}_{x}$ such that $d({f}^{{n}_{k}}(x),{f}^{{n}_{k}}({x}_{k}))>\u03f5$, 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 $r=\mathrm{min}\{d(y,{x}_{i}):1\le i\le m\}>0$. 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 ${K}_{\u03f5}=\{(x,y)\in X\times X:d(x,y)\ge \u03f5\}$ for $\u03f5>0$ 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 $\u03f5>0$ there is ${N}_{\u03f5}$ such that for all $(x,y)\in {K}_{\u03f5}$, there is $n\in \{1,\mathrm{\dots},{N}_{\u03f5}\}$ such that ${F}^{n}(x,y)\in U$.

We will prove that for each $\u03f5>0$, there is $\delta >0$ such that
${F}^{n}({K}_{\u03f5})\subset {K}_{\delta}$ 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 $K={\bigcup}_{n=0}^{{N}_{\u03f5}}{F}^{n}({K}_{\u03f5})$, and let ${\delta}_{0}=\mathrm{min}\{d(x,y):(x,y)\in K\}$.
Since $K$ is compact, the minimum distance ${\delta}_{0}$ is reached at some point of $K$; i.e. there exist $(x,y)\in {K}_{\u03f5}$ and
$0\le n\le {N}_{\u03f5}$ 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 $\alpha \in K-{K}_{\u03f5}$, there is $\beta \in {K}_{\u03f5}$ and some $$ such that $\alpha ={F}^{m}(\beta )$, and ${F}^{k}(\beta )\notin {K}_{\u03f5}$ for $$. Also, there is $n$ with $$ such that ${F}^{n}(\beta )\in U\subset {K}_{\u03f5}$. Hence $$, and $F(\beta )={F}^{m+1}(\alpha )\in {F}^{m+1}({K}_{\u03f5})\subset K$; On the other hand, $F({K}_{\u03f5})\subset K$. Therefore $F(K)\subset K$, and inductively ${F}^{n}(K)\subset K$ for any $n\in \mathbb{N}$. It follows that ${F}^{n}({K}_{\u03f5})\subset {F}^{n}(K)\subset K\subset {K}_{\delta}$ for each $n\in \mathbb{N}$ as required.

Title | the only compact metric spaces that admit a positively expansive homeomorphism are discrete spaces |
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Canonical name | TheOnlyCompactMetricSpacesThatAdmitAPositivelyExpansiveHomeomorphismAreDiscreteSpaces |

Date of creation | 2013-03-22 13:55:11 |

Last modified on | 2013-03-22 13:55:11 |

Owner | Koro (127) |

Last modified by | Koro (127) |

Numerical id | 10 |

Author | Koro (127) |

Entry type | Theorem |

Classification | msc 37B99 |