Way Below in Ordered Set of Topology

Theorem
Let $\left({S, \tau}\right)$ be a topological space.

Let $\left({\tau, \preceq}\right)$ be an ordered set where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$

Let $x, y \in \tau$.

Then
 * $x \ll y$ in $\left({\tau, \preceq}\right)$


 * for every set $F$ of open subsets of $S$: if $y \subseteq \bigcup F$, then
 * there exists a finite subset $G$ of $F$: $x \subseteq \bigcup G$

where $\ll$ denotes the way below relation.

Sufficient Condition
Let
 * $x \ll y$

Let $F$ be a set of open subsets of $S$ such that
 * $y \subseteq \bigcup F$

By definition of subset:
 * $F \subseteq \tau$

By proof of Topology forms Complete Lattice:
 * $\sup F = \bigcup F$

By assumption:
 * $y \preceq \sup F$

By Topology forms Complete Lattice:
 * $\left({\tau, \preceq}\right)$ is complete lattice.

By Way Below in Complete Lattice:
 * $\exists G \in {\it Fin}\left({F}\right): x \preceq \sup G$

where ${\it Fin}\left({F}\right)$ denotes the set of all finite subsets of $F$.

Thus by proof of Topology forms Complete Lattice:
 * $x \subseteq \bigcup G$

Necessary Condition
Assume that
 * for every set $F$ of open subsets of $S$: if $y \subseteq \bigcup F$, then
 * there exists a finite subset $G$ of $F$: $x \subseteq \bigcup G$

Let $I$ be an ideal in $\left({\tau, \preceq}\right)$ such that
 * $y \preceq \sup I$

By proof of Topology forms Complete Lattice:
 * $y \subseteq \bigcup I$

By assumption:
 * there exists a finite subset $G$ of $I$: $x \subseteq \bigcup G$

By proof of Topology forms Complete Lattice:
 * $x \preceq \sup I$

By definition of ideal:
 * $I$ is directed and lower.

By Directed iff Finite Subsets have Upper Bounds:
 * $\exists u \in I: u$ is upper bound for $G$

By definition of supremum:
 * $\sup G \preceq u$

By definition of transitivity:
 * $x \preceq u$

Thus by definition of lower set:
 * $x \in I$

Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
 * $x \ll y$