Way Above Closures that Way Below Form Local Basis

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

Let $p \in S$.

Then $\left\{ {q^\gg: q \in S \land q \ll p}\right\}$ is local basis at $p$.

Proof
Define $B := \left\{ {q^\gg: q \in S \land q \ll p}\right\}$

By Way Above Closure is Open:
 * $B \subseteq \tau$

By definition of way above closure:
 * $\forall X \in B: p \in X$

Thus by definition:
 * $B$ is set of open neighborhoods.

Let $U$ be an open subset of $S$ such that
 * $p \in U$

By Open implies There Exists Way Below Element:
 * $\exists u \in U: u \ll p$

Thus by definition of $B$:
 * $u^\gg \in B$

By definition of Scott topology:
 * $U$ is upper.

We will prove that
 * $u^\gg \subseteq U$

Let $z \in u^\gg$

By definition of way above closure:
 * $u \ll z$

By Way Below implies Preceding:
 * $u \preceq z$

Thus by definition of upper set:
 * $z \in U$