Let $E_m:S^1 \rightarrow S^1$ be the linear expanding map $E_m(x) = mx$ mod 1 (under the identification $[0,1] \sim S^1$). A sensitivity constant $\Delta$ is a positive real if for all $x, \in S^1$ and all $\epsilon > 0$ there exists some $y \in S^1$ such that $d(x,y) < \epsilon$ then; $$d(E^N(x), E^N(y)) \ge \Delta$$ for some $N \in \mathbb{N}$. Where $d(.,.)$ is the euclidean distance on $[0,1]$.
I'm trying to prove that for all $m$, $\frac{1}{2|m|}$ is a sensitivity constant. However I cant quite see why there is the multiple of a half there. This is what I have so far...
Pick any $x \in S^1$ represented by the sequence $x=x_1x_2...$ with $x_i \in \{0,1,...,m-1\}$ (a code). Then a neighbourhood of diameter $3^{-n}$ is characterised by the set of all points sharing the first $n$ digits of $x$, this is the so-called "cylinder". Then choosethe $(n+1)$st digit, $y_{n+1}$ to be different than $x_{n+1}$. Then I want to argue that there exists a point in this subset, the one characterised by $x_1x_2...x_ny_{n+1}$ that is at least $\frac{1}{2|m|}$ apart. But I can't see why this point must exist.
I'd appreciate any help! Thanks :)
Write $x=\sum_{k=1} \frac{x_{k}}{m^{k}} $ , where $x_{k}\in${1,...,m-1}. Choose $y=\sum_{k=1} \frac{y_{k}}{m^{k}}$, where $y_{1}=x_{1},...,y_{s-1}=x_{s-1}, y_{s}\neq x_{s},y_{k}=x_{k} $ for all $k\geq s+1$, and s satisfies $\frac{2}{m^{s-1}}<\epsilon$. We have $d(x,y)=d(\frac{x_{s}}{m^{s}},\frac{y_{s}}{m^{s}})\leq\frac{2}{m^{s-1}}$. Note that $E_{m}^{s-1}(x)=\frac{x_{s}}{m}+\frac{x_{s+1}}{m^{2}}+... $ and $E_{m}^{s-1}(y)=\frac{y_{s}}{m}+\frac{y_{s+1}}{m^{2}}+...$. Therefore $d(E_{m}^{s-1}(x),E_{m}^{s-1}(y))= \frac{d(x_{s},y_{s})}{m}\geq \frac{1}{m}$