Completeness of Upper Half Plane 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.
 A: I think it likely that you want this: every unit speed geodesic, every one, is given by one of two formulas: with real constant $A,$ real constant $B > 0,$ and real parameter $t,$ either
$$ z(t) = A + i e^t,  $$ or
$$ z(t) = A + B \tanh t + i B\,\mathrm{sech}\,t. $$
Given any point and tangent direction, you can place one of these passing through that point in the desired direction, by taking appropriate values for $A,B.$ Furthermore the geodesic takes the variable $t$ from $-\infty$ to $\infty.$ The rest is called the Hopf-Rinow Theorem, you may need to read up on that.
A: $\def\eps{\varepsilon}$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.
A: Here is a variant approach, which might be of interest.  (It is a little different to Mariano's: instead of immediately trying to find a closed and bounded subset inside $\mathcal H$ which contains the Cauchy sequence, I will
instead use the fact that $\mathcal H$ sits inside a certain closed and 
bound subset of the Riemann sphere.  In short, instead of trying to work
only "from the inside" of $\mathcal H$, I will let myself work "from the
outside" instead.)
The upper half-plane $\mathcal H$ sits in the extended upper half-plane
$\overline{\mathcal H},$ which is the closure of $\mathcal H$ in the Riemann sphere.  So $\mathcal H$ is an open disk in the Riemann sphere (the open upper hemisphere, if you like), while $\overline{\mathcal H}$ is a closed disk (the closed upper hemisphere); it is the union of $\mathcal H$, the real line $\mathbb R$, and the point at infinty.
Let $x_n$ be a Cauchy sequence in the upper half-plane.  We want to show that it
has a limit $x$.  By general metric space arguments, it is enough to show
that some subsequence of $x_n$ has a limit.  
Now $\overline{\mathcal H}$ is compact (i.e. closed and bounded in the Riemann sphere, if you like), and so any sequence in $\mathcal H$
has a subsequence which converges in $\overline{\mathcal H}$.   By the remark of the preceding paragraph, it is suffices to show that this subsequence actually converges to a point of $\mathcal H$.
So, we have reduced to the following situation: If $x_n$ is a Cauchy sequence
in $\mathcal H$ converging to a point $x \in \overline{\mathcal H}$, then $x$ in fact lies in $\mathcal H$.
Suppose (with the goal of getting a contradiction) that $x \not\in \mathcal H$.
By applying an isometry (i.e. a point of $SL_2(\mathbb R)$) to the whole set-up, we may assume that $x$ is the point at infinity.  (This is not necessary, but simplifies the computation that follows.)
So now we have a $x_n$ in $\mathcal H$ converging to the point at infinity,
i.e. $x_n = a_n + b_n i$ with $b_n \to \infty$.
To get the desired contradiction, it suffices to show that this sequence is
not Cauchy.  This is an elementary computation using the definition of the
hyperbolic metric.
A: Let $p_n$ be a cauchy sequence in $(X, d)$, where $X$ is a closed and bounded subset of $H$ and $d$ is the hyperbolic metric in the upper half plane, defined as 
$d(x,y) = \log\frac{|x-\overline{y}|+|x-y|}{|x-\overline{y}|-|x-y|}$
Now as $p_n$ is cauchy, there exists a subsequence $p_{n_k}$ that converges to $p \in X$ in the standard euclidean metric. We note that the possibility that the point $p$ be such that $Im (p) = 0$ is excluded as this would contradict $p_n$ being a cauchy sequence in $(X,d)$. 
Now since $p_{n_k}$ converges to $p$ in the euclidean metric, it follows that $p_{n_k}$ converges to $p$ in the hyperbolic metric as well, using the formula above as $p_{n_k} \rightarrow p$ in the euclidean metric implies that $d(p, p_{n_k}) \rightarrow  \log \frac{|p-\overline{p_{n_k}}|}{|p-\overline{p_{n_k}}|} = 0$.
Another way to think about it would be that the function $\frac{ \log\frac{|x-\overline{y}|+|x-y|}{|x-\overline{y}|-|x-y|}}{|x-y|}$ is a continuous function and hence must achieve its maximum and minimum values on a compact set, so that the function 
$\log\frac{|x-\overline{y}|+|x-y|}{|x-\overline{y}|-|x-y|}$ 
is bounded above and below by $C|x-y|$ and $P|x-y|$, $C$ and $P$ some constants so by the squeeze theorem as $|x-y|$ goes to zero, $\log\frac{|x-\overline{y}|+|x-y|}{|x-\overline{y}|-|x-y|} \rightarrow 0$ as well.
So since a subsequence $p_{n_k}$ converges to $p$ in the hyperbolic metric, it follows that the original sequence $p_n$ converges to the $p$ as well in the hyperbolic metric, by a simple application of the triangle inequality.
My problem now is, I have proved this for a compact subset of $\mathbb{H}$. How do I prove this for the whole of the upper half of the complex plane?
Thanks.
