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

TheoremMathworldPlanetmath. Let (X,d) be a compactPlanetmathPlanetmath metric space. If there exists a positively expansive homeomorphismPlanetmathPlanetmath f:XX, then X consists only of isolated pointsMathworldPlanetmath, i.e. X is finite.

Lemma 1. If (X,d) is a compact metric space and there is an expansive homeomorphism f:XX 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 δ be such that d(x,y)<δ implies d(fn(x),fn(y))<c for all n. Then there exist ϵ>0, a point y with d(x,y)<δ, and an increasing sequence {nk} such that d(fnk(y),fnk(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 with |n|<N such that d(fn(x),fn(y))>c. Choose k so large that nk>N. Then there is n with |n|<N such that d(fn+nk(x),fn+nk(y))=d(fn(fnk(x)),fn(fnk(y)))>c. But since n+nk>0, this contradicts the choce of δ. Hence every point is asymptotically stable.

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

Proof. For each xX let Kx be a closed neighborhoodMathworldPlanetmathPlanetmath of x such that for all yKx we have limnd(fn(x),fn(y))=0. We assert that limndiam(fn(Kx))=0. In fact, if that is not the case, then there is an increasing sequence of positive integers {nk}, some ϵ>0 and a sequenceMathworldPlanetmath {xk} of points of Kx such that d(fnk(x),fnk(xk))>ϵ, and there is a subsequenceMathworldPlanetmath {xki} converging to some point yKx.

From the Lyapunov stability of y, we can find δ>0 such that if d(y,z)<δ, then d(fn(y),fn(z))<ϵ/2 for all n>0. In particular d(fnki(xki),fnki(y))<ϵ/2 if i is large enough. But also d(fnki(y),fnki(x))<ϵ/2 if i is large enough, because yKx. Thus, for large i, we have d(fnki(xki),fnki(x))<ϵ. That is a contradictionMathworldPlanetmathPlanetmath from our previous claim.

Now since X is compact, there are finitely many points x1,,xm such that X=i=1mKxi, so that X=fn(X)=i=1mfn(Kxi). To show that X={x1,,xm}, suppose there is yX such that r=min{d(y,xi):1im}>0. Then there is n such that diam(fn(Kxi))<r for 1im but since yfn(Kxi) for some i, we have a contradiction.

Proof of the theorem. Consider the sets Kϵ={(x,y)X×X:d(x,y)ϵ} for ϵ>0 and U={(x,y)X×X:d(x,y)>c}, where 2c is the expansivity constant of f, and let F:X×XX×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 Nϵ such that for all (x,y)Kϵ, there is n{1,,Nϵ} such that Fn(x,y)U.

We will prove that for each ϵ>0, there is δ>0 such that Fn(Kϵ)Kδ for all n. This is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath 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=n=0NϵFn(Kϵ), and let δ0=min{d(x,y):(x,y)K}. Since K is compact, the minimum distance δ0 is reached at some point of K; i.e. there exist (x,y)Kϵ and 0nNϵ such that d(fn(x),fn(y))=δ0. Since f is injectivePlanetmathPlanetmath, it follows that δ0>0 and letting δ=δ0/2 we have KKδ.

Given αK-Kϵ, there is βKϵ and some 0<mNϵ such that α=Fm(β), and Fk(β)Kϵ for 0<km. Also, there is n with 0<m<nNϵ such that Fn(β)UKϵ. Hence m<Nϵ, and F(β)=Fm+1(α)Fm+1(Kϵ)K; On the other hand, F(Kϵ)K. Therefore F(K)K, and inductively Fn(K)K for any n. It follows that Fn(Kϵ)Fn(K)KKδ for each n as required.

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