My question will rely on a clarification of a proof, which I simply don't understand.

Let us denote by $X$ a pseudo-riemannian symmetric space and define

$$
Z_{\mathrm{Iso}\left(X\right)}G(X) = \{\, f \in \mathrm{Iso}\!\left(X\right) \mid
gf = fg \text{ for all } g \in G(X) \,\}
$$
as the centralizer of the transvection group $G(X)$ of the space $X$ and let
$\mathrm{Iso}\left(X\right)$ be the isometry group of the regarding space.

In *Cahen,Parker,"Pseudo-riemannian symmetric spaces"* following assertion has been stated in Proposition 3.9:

**Assertion:** *Any discrete subgroup $A$ of $Z_{\mathrm{Iso}\left(X\right)}G(X)$* acts properly discontinuously on $X$.

**Proof:** Let $C \subseteq X$ be a compact subset of $X$ with the property that
$A_C =
\{\, g \in A \mid gC \cap C \neq \emptyset \,\}$ is infinite, $\lvert A_C \rvert = \infty$. There exists a sequence $(a_n) \subseteq A_C$ with distinct elements and a sequence $(c_n) \subseteq C$ such that $a_n c_n \in C$ for all $n \in \mathbb{N}$. By compactness of $C$ and therefore taking sub-sequences, we may assume that $(c_n)$ converges towards $c \in C$ and that $(a_nc_n)$ converges towards $c^\prime \in C$. Let us choose $c$ as a basepoint of $X$. Since the transvection group $G(X)$ acts transitively on $X$ we can find a $g \in G(X)$ such that $gc = c^\prime$ and let $V_1 \subseteq U_1 \subseteq W_1$ be three neighborhoods of the identity in $G(X)$ such that
$$
V_1=V_1^{-1}, g^{-1}V_1g\subseteq U_1 \text{ and } U_1^2 \subseteq W_1.
$$
Define;
$$
W_c = V_1 c, \\
\tilde{W}_c = W_1 c,\\ W_{c^\prime}=gW_c\\\tilde{W}_{c^\prime}=g\tilde{W}_c.
$$

**Question:** *Why does this define neighborhoods for the regarding points $c$ and $c^\prime$ and furthermore why does this define a neighborhoodbasis?*

There exists an integer $N$ such that for all $n \geq N$ we have $c_n \in W_c$ and $a_nc_n \in W_{c^\prime}$. Indeed, there exists $g_n \in V_1$ such that $c_n = g_nc$ for all $n \geq N$. We want to show that $\lim_{n \to \infty} (a_nc) = c^\prime$. Consider therefore $$ a_n c = a_ng_n^{-1}c_n = g_n^{-1} a_n c_n \in g_n^{-1} g W_c \\ = gg^{-1}g_n^{-1}gW_c \subseteq gU_1V_1c \subseteq gU_1^2c \subseteq gW_1c = \tilde{W}_{c^\prime}, $$ which shows the claimed result.

The sequence $(b_n = a_{n+1}^{-1} a_n)$ in $A$ is clearly such that $\lim_{n \to \infty}b_n c = c$.

**Question:** *Unfortunatley I have to redefine the term "clearly" into "highly unclearly". To me it is not obvious why this series converges towards $c$.*