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! 



Partitions and Expansiveness
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, $diam D\_N\le\delta$.


