Why is this discrete subgroup of $PSL(2,\mathbb{C})$ not Kleinian? Definition:  For a subgroup $G$ of the group $PSL(2,\mathbb{C})$ acting on  $\mathbb{P}^1$, its domain of discontinuity is the set of all points, $z$, with the following properties:
$1.$ The stabilizer $G_z$ of $z$ is finite.
$2.$ $\exists U$, a neighbourhood of $z$ such that, 
$\space \space\space\space(a)\space $$\forall g\in G_z$, $gU=U$.
$\space \space\space\space(b)\space $$\forall g\in G - G_z$, $gU \cap U=\emptyset$.
Definition: A Kleinian group is defined to be a subgroup $G$ of $PSL(2,\mathbb{C})$ such that its domain of discontinuity is not empty.
Question: Consider the subgroup $\{M \in PSL(2,\mathbb{C}) \mid  m_{ij}\in \mathbb{Z}[i], det {M}=1\}$. Why is this not Kleinian?
Is this clear from the above definition or would I have to prove some special property of Kleinian groups that this so called 'Picard Modular Group' does not satisfy?
 A: The domain of discontinuity of this group is indeed empty. Hence by the definition of Kleinian group that you give this group is not Kleinian. That's an out-of-date definition, however; in most usages a Kleinian group is simply a discrete subgroup of $PSL_2(\mathbb{C})$, and then one uses the phrase "Type 2 Kleinian group" to refer to those which have nonempty domain of discontinuity.
Terminological evolutionary issues aside, for any discrete subgroup $G$ of $PSL(2,\mathbb{C})$, its action on $\mathbb{CP}^1$ subdivides that sphere into two sets: its domain of discontinuity which you have defined; and the complement of the domain of discontinuity called the limit set. One can characterize the limit set as the set of points of $\mathbb{CP}^1$ which are contained in the closure of the $G$-orbit of every point of $\mathbb{CP}^1$. 
Your subgroup, also denoted $PSL_2(\mathbb{Z}[i])$, has limit set equal to the whole of $\mathbb{CP}^1$, and hence its domain of discontinuity is empty. 
One way to prove this is consider the extended action to the upper half space $\{(x,y,z) \in \mathbb{R}^3 \mid z > 0\}$, thought of as hyperbolic 3-space $\mathbb{H}^3$, whose 2-sphere at infinity is identified with $\mathbb{CP}^1$. What one does is to construct a fundamental domain $D \subset \mathbb{H}^3$ for the action of $PSL_2(\mathbb{Z}[i])$ on $\mathbb{H}^3$. Then one computes the volume of $D$. It turns out that the volume of $D$ is finite. Then one invokes a theorem which says that if the volume of a fundamental domain of a discrete subgroup of $PSL(2,\mathbb{C})$ is finite then that subgroup has limit set equal to the whole of $\mathbb{CP}^1$.
I think there is most likely another way to prove this, using a little number theory. It turns out that the limit set of a discrete subgroup $G < PSL(2,\mathbb{C})$ is equal to the whole of $\mathbb{CP}^1$ if and only if the action of $G$ on $\mathbb{CP}^1$ has a dense orbit. I'm pretty sure that, when one identifies $\mathbb{CP}^1 = \mathbb{C} \cup \infty$, the orbit of $0=0+0i$ under the action of $PSL(2,\mathbb{Z}[i])$ is equal to $\mathbb{Q}[i] \cup \{\infty\}$, which is clearly dense in $\mathbb{C} \cup \infty$.
