Topology induced by the Möbius distance Consider the open unit disc $\mathbb{D}$ in the complex plane $\mathbb{C}$. Consider the metric called the Möbius distance on $\mathbb{D}$ given by:
$m(a,b)=\left|\frac{a-b}{1-\bar{a}b}\right|$. Is the topology induced by this metric the same as the Euclidean topology on the unit disc $\mathbb{D}$? Let $\tau_m$ denote the topology induced by the Möbius distance and $\tau_e$ denote the topology induced by the Euclidean distance.
I could show that $\tau_e$ is finer that $\tau_m$, can anyone tell how to prove the other way round?
 A: For a fixed $z\in\Bbb D$, define $f_z:\Bbb D\to\Bbb C$ by $s\mapsto\frac{z-s}{1-\overline{z}\cdot s}$. Note $f_z$ is holomorphic and $f_z(z)=0$.
The fact that $\tau_e$ is at least as fine as $\tau_m$ is really the statement that $f_z$ is a Euclidean-continuous function. That means there is, for all $\epsilon>0$, a $\delta>0$ that $|z-s|<\delta$ shall imply $m(z,s)=|f_z(s)-f_z(z)|<\epsilon$, so that the Euclidean $\delta$-ball about $z$ is contained in the Möbius $\epsilon$-ball about $z$.
So entirely in the same vein, to show $\tau_m$ is at least as fine as $\tau_e$ is to show Euclidean-continuity of $f^{-1}$. Given $w=f_z(s)=\frac{z-s}{1-\overline{z}\cdot s}$, we can compute: $$w-z=s\cdot(\overline{z}\cdot w-1)$$And we can retrieve $s$ from this through: $$f^{-1}(w)=\frac{z-w}{1-\overline{z}\cdot w}=f(w)$$
Which is everywhere well-defined and also continuous as we’ve already seen… so $\tau_m$ is at least as fine as $\tau_e$, and $\tau_m=\tau_e$.
A: Here is a pedestrian argument. Let $B_m(a,r)$ and $B_e(a,r)$ denote the open balls of radius $r$ around $a$ in the Möbius and Euclidean metrics, respectively. Fix $a\in\mathbb D$. Since
$$1-|a|\le|1-\bar ab|\le1+|a|$$
for all $b\in\mathbb D$, we have
$$\bigl(1-|a|\bigr)m(a,b)\le|a-b|\le\bigl(1+|a|\bigr)m(a,b),$$
thus for any $r>0$,
$$B_e(a,r)\supseteq B_m\bigl(a,r/(1+|a|)\bigr),\qquad
B_m(a,r)\supseteq B_e\bigl(a,(1-|a|)r\bigr).$$
That is, the two topologies have the same neighbourhood bases at each point, and as such are the same topology.
