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Why if one have an $\varepsilon$-expansive homeomorphism $T:X \rightarrow X$ ($X$ a compact metric space) and a given partition $D$ of $X$ which has diameter smaller than $\varepsilon$ the sequence of refined partitions $D_n = \bigvee_{i = -n}^n T^{-i} D$ has diameter converging to zero ?

Recall that a $\varepsilon$-expansive homeomorphism $T$ is such that given any two distinct points $x$ and $y$ there exist $n \in \mathbf{Z}$ such that $d(T^nx, T^ny) > \varepsilon$

I can see intituively why this is true, somehow the refined partitions have less an less points in its members precisely because they have diameter less than epsilon but $T$ keeps separating points (and i fact open sets of points) at distance greater than $\varepsilon$.

Thanks in advance!

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2 Answers 2

up vote 3 down vote accepted

Fix $\delta>0$ and let $$ A_n=\{(x,y): d(x,y)\ge\delta, d(T^ix,T^iy)\le\varepsilon,|i|\le n\} $$ Sets $A_n$ are closed, nested and, by expansiveness, $\cap A_n=\varnothing$. Hence $A_N=\varnothing$ for some $N=N(\delta)$. Clearly, $diam D_N\le\delta$.

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Thank you very much! –  Felipe Olmos Feb 22 '10 at 14:01
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It is not clear to me that the sequence $D_n$ necessarily has diameter converging to zero. In fact, consider the much more concrete subexample of an Anosov diffeomorphism $T$ on a compact manifold, and let $D$ be a Markov partition. The partition $D_n$ is then also a Markov partition. Yet the proofs (there are different proofs by Sinai and Bowen) that there exist Markov partitions of arbitrarily small diameter are nontrivial. For a sketch of such a proof, see Invariant measures for hyperbolic dynamical systems by Chernov, available at #45 here.

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BTW, the proof is on page 9. –  Steve Huntsman Feb 16 '10 at 14:37
    
I am in fact reading the book "Equilibirium States and the Ergodic Theory of Anosov Diffeomorphisms" by Bowen, and this statement appears as a comment in Proposition 2.5. (A proof you mention is also in the book but later). I think the fact that there exists Markov partitions of arbitrarily small diameter for an Anosov Diffeomorphism is independent from what I am asking since the homeomorphism has to be $epsilon$-expansive but i'm still not familiarized with Anosov Diffeomorphisms so I can be very wrong. Thanks for you response! –  Felipe Olmos Feb 18 '10 at 15:07
    
I'll have to check on this. You're probably ahead of me, I'm a newcomer to this area as well. –  Steve Huntsman Feb 19 '10 at 3:12
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