Complement of Lower Closure is Prime Element in Inclusion Ordered Set of Scott Sigma

Theorem
Let $L = \left({S, \preceq, \tau}\right)$ be a complete Scott topological lattice.

Let $D = \left({\sigma\left({L}\right), \precsim}\right)$ be an inclusion ordered set of Scott sigma of $L$.

Let $x \in S$.

Then $\complement_S\left({x^\preceq}\right)$ is a prime element in $D$ and $\complement_S\left({x^\preceq}\right) \ne S$

Proof
By Scott Topology equals to Scott Sigma:
 * $\tau = \sigma\left({L}\right)$

By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice:
 * $x^\preceq = \left\{ {x}\right\}^-$

where $\left\{ {x}\right\}^-$ denotes the topological closure of $\left\{ {x}\right\}$.

By Topological Closure of Singleton is Irreducible:
 * $x^\preceq$ is topological irreducible

Thus by Complement of Irreducible Topological Subset is Prime Element:
 * $\complement_S\left({x^\preceq}\right)$ is a prime element in $D$.

By definitions of reflexivity and lower closure of element:
 * $x \in x^\preceq$

By definitions of relative complement and difference:
 * $x \notin \complement_S\left({x^\preceq}\right)$

Hence by definition set equality:
 * $\complement_S\left({x^\preceq}\right) \ne S$