Way Above Closure is Open

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

Let $x \in S$.

Then $x^\gg$ is open

where $x^\gg$ denotes the way above closure of $x$.

Proof
By Way Above Closure is Upper:
 * $x^\gg$ is upper.

We will prove that
 * $x^\gg$ is inaccessible by directed suprema.

Let $D$ be a directed subset of $S$ such that
 * $\sup D \in x^\gg$

By definition of way above closure:
 * $x \ll \sup D$

By Way Below iff Second Operand Preceding Supremum of Directed Set There Exists Element of Directed Set First Operand Way Below Element:
 * $\exists d \in D: x \ll d$

By definition of way above closure:
 * $d \in x^\gg$

By definitions of intersection and non-empty set:
 * $x^\gg \cap D \ne \varnothing$

Thus by definition of Scott topology:
 * $x^\gg$ is open.