5

I am trying to prove that the Poincaré disk $D=\left\{ z \in \mathbb{C} : |z|<1 \right\}$ equipped with the hyperbolic metric given by $d_{D}(z_1,z_2)=\inf \{ L_{D}(\gamma) \mid \gamma \text { is a continuously differentiable path with endpoints } z_1 \text{ and } z_2 \}$, where $L_{D}(\gamma)=\int_\gamma \frac{2}{(1-\lvert z \rvert^2)} \lvert dz \rvert$ is a metric space. The fact that $d_{D}(z_1,z_2) \geqslant 0$ and symmetry are obvious and I have managed to prove the triangle inequality as well. But how does $d_{D}(z_1,z_2)=0$ imply that $z_1=z_2$?

Micah
  • 38,108
  • 15
  • 85
  • 133
wwinters57
  • 735

1 Answers1

4

Suppose $z_1 \neq z_2$; let $\delta$ be the Euclidean distance from $z_1$ to $z_2$. The integrand $\frac{2}{1-|z|^2}$ is bounded below by $2$ on $D$, so $L_D(\gamma) > 2\delta$ for any $\gamma$ joining $z_1$ and $z_2$. It follows that $d_D(z_1,z_2) \geq 2\delta > 0$.

Micah
  • 38,108
  • 15
  • 85
  • 133
  • Thank you, but I have already proved that. What I don't understand is how to show that if $d_D(z_1,z_2)=0$ then $z_1=z_2$. – wwinters57 Mar 01 '19 at 21:54
  • 2
    It's the same thing. My answer shows that $(z_1 \neq z_2) \rightarrow (d(z_1, z_2) \neq 0)$. What you're asking for is $(d(z_1, z_2)=0) \rightarrow (z_1 = z_2)$: the contrapositive. – Micah Mar 02 '19 at 00:06
  • It follows that the infimum of all these is $\ge 2\delta$, which is just $d_D(z_1,z_2)$. – Berci Mar 02 '19 at 01:42