# Poincaré disk with hyperbolic metric is a metric space

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 $$z_1=z_2$$?

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 $$d_D(z_1,z_2) \geq 2\delta > 0$$.
• 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 $z_1=z_2$. – vladr10 Mar 1 at 21:54
• 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 2 at 0:06
• It follows that the infimum of all these is $\ge 2\delta$, which is just $d_D(z_1,z_2)$. – Berci Mar 2 at 1:42