Closure of Singleton is Lower Closure of Element in Scott Topological Lattice

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

Let $x \in S$.

Then:
 * $\set x^- = x^\preceq$

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

Proof
By Lower Closure of Element is Closed under Directed Suprema:
 * $x^\preceq$ is closed under directed suprema.

By Lower Closure of Singleton:
 * $\set x^\preceq = x^\preceq$

By Lower Closure is Lower Set:
 * $x^\preceq$ is a lower set.

By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
 * $x^\preceq$ is closed.

By definitions of lower closure of element and reflexivity:
 * $x \in x^\preceq$

By Singleton of Element is Subset:
 * $\set x \subseteq x^\preceq$

We will prove that:
 * for every a closed subset $C$ of $S$: $\set x \subseteq C \implies x^\preceq \subseteq C$

Let $C$ be a closed subset of $S$ such that:
 * $\set x \subseteq C$

By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
 * $C$ is a lower set.

Let $y \in x^\preceq$.

By definition of lower closure of element:
 * $y \preceq x$

By definitions of subset and singleton:
 * $x \in C$

Thus by definition of lower set:
 * $y \in C$

Thus by definition of topological closure:
 * $\set x^- = x^\preceq$