Partitions and Expansiveness

Question

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!

Answer 1

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, $diam D_N\le\delta$.

Answer 2

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.