Way Above Closures Form Basis

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

Then $\set {x^\gg: x \in S}$ is an (analytic) basis of $L$.

Proof
Define $B = \set {x^\gg: x \in S}$.

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

We will prove that:
 * for all $x \in S$: there exists a local basis $Q$ of $x$: $Q \subseteq B$

Let $x \in S$.

By Way Above Closures that Way Below Form Local Basis:
 * $Q := \set {g^\gg: g \in S \land g \ll x}$ is a local basis at $x$.

Thus by definition of subset:
 * $Q \subseteq B$

Thus by Characterization of Analytic Basis by Local Bases:
 * $B$ is an (analytic) basis of $L$.