Open implies There Exists Way Below Element

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

Let $p \in S, A \subseteq S$ such that
 * $A$ is open and $p \in A$.

Then
 * $\exists q \in A: q \ll p$

Proof
By definition of continuous ordered set:
 * $p^\ll$ is directed

and
 * $L$ satisfies axiom of approximation.

By axiom of approximation:
 * $p = \sup \left({p^\ll}\right)$

By definition of Scott topology:
 * $A$ is inaccessible by directed suprema.

By definition of inaccessible by directed suprema:
 * $A \cap p^\ll \ne \varnothing$

By definition of non-empty set:
 * $\exists q: q \in A \cap p^\ll$

Thus by definition of intersection:
 * $q \in A$

By definition of intersection:
 * $q \in p^\ll$

By definition of way below closure:
 * $q \ll p$