Why is this sequence of isometries contained in a compact set? I'm reading through a proof of the Mostow rigidity theorem (pages 738-740) in https://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf
but I'm stuck at a certain part. 
In Step 2, we consider the vertical geodesic $L \subset \mathbb{H}^n$ emanating from $0$ (here we are viewing $\mathbb{H}^n$ in the upper half-space model), and we fix a base-point $y_0 \in L$. We have that $0$ is a conical limit point of $\Gamma$, which is a lattice in $\text{Isom}(\mathbb{H}^n)$. So there is a sequence of isometries $\gamma_i \in \Gamma$ such that 
$$\lim_{i\to\infty} \gamma_i(y_0) = 0$$
and
$$\text{dist}(\gamma_i(y_0),L) \le C$$
for some constant $C$. We let $y_i$ denote the nearest-point projection of $\gamma_i(y_0)$ to $L$, and we define $T_i$ to be the hyperbolic translation $y \mapsto t_iy$ along the axis $L$, such that $T_i(y_0) = y_i$. The author then proceeds to say that the sequence $k_i := \gamma_i^{-1} \circ T_i$ lies in a compact subset $C \subset \text{Isom}(\mathbb{H}^n)$. However it is not clear to me why this is the case.
I am hoping there is an argument which doesn't directly appeal to the explicit topology of $\text{Isom}(\mathbb{H}^n)$ (which, by the way, is what?), but rather I am hoping that this will follow from a general statement about group actions. 
Some observations I have (which may be irrelevant) are that 


*

*$\left\{k_i(y_0)\right\} \subset B_C(y_0)$ (i.e. the orbit of $y_0$ under $k_i$ is bounded)

*since $\text{Isom}(\mathbb{H}^n)$ acts transitively on $\mathbb{H}^n$, we have that $\mathbb{H}^n \cong \text{Isom}(\mathbb{H}^n) / \text{Stab}(y_0)$ (whether this isomorphism is more than a set bijection I'm not sure) 
 A: Let me write an answer under the assumption (which is implicit in your question) that you have some particular scheme for expressing $\text{Isom}(\mathbb H^n)$ as a matrix group, but I'll also say exactly what to do in the case $n=2$.
You have already observed that the orbit of $y_0$ under $k_i$ is bounded, in that $k_i$ moves the point $y_0$ a distance $\le C$. You can therefore write each $k_i$ as a product of matrices the form
$$k_i = P_i Q_i R_i
$$
where each of $P_i$ and $R_i$ is a matrix that fixes $y_0$ and where $Q_i$ is a matrix which translates $y_0$ upward on the axis $A$ by a distance at most $C$. So you just need to convince yourself of three things: 


*

*The set of matrices fixing $y_0$ has bounded entries; 

*The set of matrices which translates $y$ upward along $A$ by distance at most $C$ has bounded entries; 

*If one has a bound on the entries of three matrices $P,Q,R$ (of a fixed size) then one can derive a bound on the entries of their product $PQR$. 


For an example of the last point, if the matrices are $n \times n$ and the entries of $P,Q,R$ are $\le b$ in absolute value then a bound on the absolute values of the entries of the matrix $PQR$ is $n^2b^3$.
For the first two points, let me describe exactly what happens in the special case $n=2$, working in the upper half plane model with orientation preserving isometry group $\text{PSL}(2,\mathbb R)$ (which is probably not the matrices that you are working with, but it's the matrices I know best): 


*

*The set of matrices fixing $(0,1)$ is simply the compact set of rotation matrices 
$$SO(2,\mathbb R) = \begin{pmatrix} \cos \theta & \sin \theta \\ - \sin\theta & \cos\theta \end{pmatrix} \quad \text{where $0 \le \theta \le 2\pi$}
$$ 
which is obviously compact

*The set of matrices which translates $(0,1)$ upward along the line $x=0$ a distance at most $C$ is simply the compact set of translation matrices
$$\begin{pmatrix} e^{c/2} & 0 \\ 0 & e^{-c/2} \end{pmatrix} \quad\text{where $0 \le c \le C$}
$$
