Let $a \in \mathbb C$ and $r \gt 0$ be given; then there exists $\rho \gt 0$ such that $B_{\infty} (a,\rho) \subseteq B(a,r),$ where $$B_{\infty} (a,\rho) = \{z \in \mathbb C\ |\ d_{\infty} (z,a) \lt \rho \}$$
Here $$d_{\infty} (z,z') = \frac {2\ |z-z'|} {{\left [\left (1 + |z|^2 \right ) \left (1 + |a|^2 \right ) \right ]}^{\frac {1} {2}}}$$ if $z,z' \in \mathbb C$ and $$d_{\infty} (z,\infty) = \frac {2} {\left (1 + |z|^2 \right )^{\frac {1} {2}}}$$ if $z \in \mathbb C.$ By choosing $0 \lt \rho \lt \frac {1} {\left (1 + |a|^2 \right )^{\frac {1} {2}}}$ we can guarantee that $\infty \notin B_{\infty} (a,\rho).$ I am trying to show that $|z - z'| \leq C\ d_{\infty} (z,z'),$ for $C \gt 0$ but I didn't able to conclude that. Could anyone please help me in this regard?
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Let $a \in \Bbb C$ and $r > 0$. Choose $\rho > 0$ with $0 < \rho < d_\infty(a, \infty)$ and $$ \frac{2\rho}{(d_\infty(a, \infty) - \rho) d_\infty(a, \infty)} < r \, . $$
For all $z \in B_{\infty} (a,\rho)$ is $$ d_\infty(z, \infty) \ge d_\infty(a, \infty) - d_\infty(a, z) \ge d_\infty(a, \infty) - \rho > 0 $$ from the triangle inequality for the spherical metric. It follows that $$ |z-a| = \frac{2d_\infty(z,a)}{d_\infty(z, \infty) d(a, \infty)} \le \frac{2\rho}{(d_\infty(a, \infty) - \rho) d_\infty(a, \infty)} < r $$ and therefore $z \in B(a, r)$.