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

Theorem
Let $L = \struct {S, \preceq, \tau}$ be a complete Scott topological lattice.

Let $D = \struct {\map \sigma L, \precsim}$ be an inclusion ordered set of the Scott sigma of $L$.

Let $x \in S$.

Then:
 * $\relcomp S {x^\preceq}$ is a prime element in $D$

and:
 * $\relcomp S {x^\preceq} \ne S$

Proof
By Scott Topology equals to Scott Sigma:
 * $\tau = \map \sigma L$

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

where $\set x^-$ denotes the topological closure of $\set x$.

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

Thus by Complement of Irreducible Topological Subset is Prime Element:
 * $\relcomp S {x^\preceq}$ 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:
 * $x \notin \relcomp S {x^\preceq}$

Hence by definition set equality:
 * $\relcomp S {x^\preceq} \ne S$