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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 $z_1=z_2$?

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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 $d_D(z_1,z_2) \geq 2\delta > 0$.

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

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