Let $G$ be a subgroup of $PSL(2,\mathbb{C})$, so that $G$ acts on $\mathbb{C}\cup\{\infty\}$ by linear fractional transformations. We say that $G$ acts properly discontinuously at a point $z\in \mathbb{C}\cup\{\infty\}$ if
- the stabilizer $G_z$ is finite, and
- there exists a neighborhood $U_z$ such that $g(U_z)=U_z$ for any $g$ in $G_z$ and $U_z\cap g(U_z)=\emptyset$ for any $g$ in $G\setminus G_z$.
Prove or disprove: $G$ acts properly discontinuously at least at one point on $\mathbb{C}\cup\{\infty\}$ $\iff$ $G$ is a discrete subgroup of $PSL(2,\mathbb{C})$.
My attempts:
$\implies$ Suppose $G$ is not discrete. Then there is a limit point $g_0\in PSL(2,\mathbb{C})$. So there is a sequence $\{g_n\}$ of elements of $G$ tending to $g_0$. Pick a point $z_0\in \mathbb{C}\cup\{\infty\}$ which lies in the set on which $G$ acts. Then $g_nz_0$ tends to $g_0z_0$ for all $n$, so that the sequence $g_nz_0$ is Cauchy. This means that given $\epsilon>0$, there is $N$ such that for all $m,n>N$ we have $|g_nz_0-g_mz_0|<\epsilon$. We claim that $G$ cannot act properly discontinuously at $z_0$. Indeed, assume the converse. Then there exists a neighbborhood $U_{z_0}$ whose image under all but finite number of elements of $G$ (viz. under all elements of $G\setminus G_{z_0}$) does not intersect itself. Since the sequence $\{g_n\}$ is infinite, there are $n_0, m_0>N$ such that $g_{n_0}$ and $g_{m_0}$ (which are assumed to be distinct) lie in $G\setminus G_{z_0}$. Then $g_{n_0}U_{z_0}\cap g_{m_0}U_{z_0}=\emptyset$. (Indeed, otherwise we would have $g_{n_0}u=g_{m_0}v$ for $u,v\in U_{z_0}$. This would imply $g_{m_0}^{-1}g_{n_0}u=v$ and consequently $g_{m_0}^{-1}g_{n_0}U_{z_0}=U_{z_0}$, which is impossible because $g_{m_0}\ne g_{n_0}$, neither $g_{n_0}$ nor $g_{m_0}$ lies in $G_{z_0}$, and $U_{z_0}$ is supposed to be displaced away from itself by any element of $G\setminus G_{z_0}$.) But on the other hand we must have $|g_{n_0}z_0-g_{m_0}z_0|<\epsilon$, a contradiction.
Is this a correct proof? The proof of Lemma 1.5.2.4.1 from here uses the countability of $G$, but I don't see how the proof above uses it. Also, the last sentence in the proof in the link looks mysterious for me, I didn't assume that $z_0\ne \infty$.
$\Longleftarrow$ Suppose $G$ is discrete, i.e. for any $g\in G$ there is an open subset $V_g$ of the topological space $PSL(2,\mathbb{C})$ that contains $g$ and such that $V_g\cap G=\{g\}$. We need to show that there is $z\in\mathbb{C}\cup\{\infty\}$ at which $G$ acts properly discontinuously. I have no idea what should I cook up this $z$ from...