I am trying to prove that the upper half plane, defined as $\mathbb{H} = \{z \in \mathbb{C} : \Im(z)>0 \}$, is complete with respect to the hyperbolic metric.
First I note that if I have some closed and bounded subset $X$ of $\mathbb{H}$, it is complete. However, when dealing with $\mathbb{H}$, I would like to use the nice property that if I am in $\mathbb{R}^n$, and have some Cauchy sequence $x_n$ in a closed and bounded subset of $\mathbb{R}^n$, then it has a subsequence say $x_{n_k}$ which converges in my closed and bounded in the euclidean metric, viz. if I am in $\mathbb{R}^2$ it is just $|\mathbf{x} – \mathbf{y}|$, where $\mathbf{x}, \mathbf{y}$ some points in my set.
How do I deal with the fact that at the boundary, my euclidean metric remains bounded but the hyperbolic metric defined as
$$d(z_1,z_2) = \ln \left[ \frac{|z_1 – \bar{z_2}| + |z_1 – z_2 | }{|z_1 – \bar{z_2}| – |z_1 – z_2 | }\right]$$
goes to infinity?
In addition, the upper half of the complex plane is not closed, so how can I use nice properties like convergence of subsequences and stuff to prove that it is complete?
Thanks.
Best Answer
$\def\eps{\varepsilon}$Suppose $(p_i)_{i\geq1}$ is a Cauchy sequence in $\mathbb H$. Show first there exist $\eps>0$ and $R>0$ such that $\operatorname{Im}p_i\geq\eps$ and $|p_i|\leq R$ for all $i\geq1$. It follows that the sequence does not leave the compact subset $$K=\{z\in\mathbb H:\operatorname{Im}z\geq\eps, |z|\leq R\}\subset\mathbb H.$$ Therefore you can use your initial observation to conclude in the general situation.