Show that two open balls with respect to different metrics (and different radii) on $[0,1)$ coincide

Question

Let $e$ be the Euclidian metric on $\mathbb{R}$. That is, $$e(x,y):=|x-y|.$$ We define another metric $d$ on $[0,1)$ by $$d(x,y):=\inf_{k\in\mathbb{Z}}|x-y+k|.$$ For $N\in\mathbb{N}$ we can define a map $f\colon[0,1)\to[0,1)$ by $$f_{N}(x):=Nx-\lfloor Nx\rfloor.$$ I want to compare orbits of points in $[0,1)$ with respect to $d$. For $x,y\in[0,1)$ we can compare their orbit segments $$\{x,f_{N}(x),f_{N}^{2}(x),\ldots,f_{N}^{n}(x)\}\qquad\text{and}\qquad\{y,f_{N}(y),f_{N}^{2}(y),\ldots,f_{N}^{n}(y)\}$$ of length $n\in\mathbb{N}\cup\{0\}$ with another metric $d_{n}$ on $[0,1)$ defined by $$d_{n}(x,y):=\max_{0\leq i\leq n}d(f_{N}^{i}(x),f_{N}^{i}(y)).$$

Suppose that $0<\delta\leq1$. I want to prove that for all $x\in [0,1)$, $$B_{d_{n}}(x,\delta)=B_{e}(x,\delta/2N^{n})\mod1.$$ Here $B_{\text{metric}}(\text{point},\text{radius})$ denotes an open ball with respect to the metric in the subscript.

I was hoping to prove "$\supset$" with one of my previous posts, but I didn't really succeed. I'm really struggling with the infimum in the definition of $d$. Any help would be greatly appreciated. Thanks in advance!

Answer 1

We can simplify your problem a little bit. First, we can interpret $d$ as the shortest path in $\Bbb R/\Bbb Z$ (I don't think this has a formal definition, but it helps to see what is happening).

Next, note that for any integer $k$, $$d(x+k,y)=d(x,y).$$ Therefore $$d_n(x,y)=\max_{0\leq i\leq n}d(N^ix,N^iy).$$ Fix $x\in[0,1)$. Then, for $y\in B_e(x,\frac\delta{N^n})$ and $i\in\{0,\ldots,n\}$, $$d(N^ix,N^iy)\leq|N^ix-N^iy|<N^i\frac\delta{N^n}\leq\delta,$$ so $y\in B_{d_n}(x,\delta)$. This proves the $\supseteq$ inclusion.

Answer 2

As observed by @IsaacRen, $[0, 1)$ may be identified with a circle of circumference $1$. Then, $d$ is the legnth of the shorter of the two paths connecting the two points while $e$ is always the length of the path not crossing $0$. It follows that $d(x, y) ≤ e(x, y) ∈ \{d(x, y), 1 - d(x, y)\}$. We also have that $B_d(0, δ) ≠ B_e(0, δ)$ unless both balls are the whole space.

Also note that when $f_N$ corresponds to the map $z ↦ z^N$ when we consider the circle to be the unit circle in the complex plane (while normalizing its length to be $1$). That means that $f_N$ stretches all $d$-distances by the factor of $N$ (locally – if no winding overlapping occurs). We have $B_{d_n}(x, δ) = B_{d}(x, δ/N^n)$ if $δ ≤ 1/2$. For this to be equal to $B_e(x, δ/N^n)$ we additionally need that the $d$-ball does not contain $0$, i.e. $x ∈ (δ/N^n, 1 - δ/N^n)$.

Regarding the inclusions, we have $d ≤ e$ and if the winding overlapping occurs, $f_N$ makes the distance smaller than $N$ times the original one. Hence, we always have $B_{d_n}(x, δ) ⊇ B_{d}(x, δ/N^n) ⊇ B_{e}(x, δ/N^n)$.