Given an $m\in\mathbb{N}_{\geq2}$, define the map $f\colon[0,1)\to[0,1)$ by $f(x):=mx\mod1$. Using this map, for $n\in\mathbb{N}$ we define the metric $$d_n(x,y):=\max_{0\leq i\leq n-1}\min\{|f^{i}(x)-f^{i}(y)|,1-|f^{i}(x)-f^{i}(y)|\}$$ on $[0,1)$. Given $a\in[0,1)$ and $\varepsilon>0$, what is the (Euclidian) diameter of the ball (or actually the interval) $B_{d_n}(a,\varepsilon)$? For simplicity, it is okay to make assumptions on $\varepsilon$ (e.g. $\varepsilon<1/m$, if needed).
Note that $d_{n}$ measures the maximum distance between the first $n$ $f$-iterates of $x$ and $y$.
EDIT: I read somewhere that $$B_{d_n}(x,\varepsilon)=B_d(x,\varepsilon/m^{n-1}),$$ where $d(x,y)=\min\{|x-y|,1-|x-y|\}$ and $0<\varepsilon<1/2$, but I don't see why this is true. This would imply that the Euclidian diameter of $B_{d_n}(x,\varepsilon)$ is $2\varepsilon/m^{n-1}$.