Way Below iff Preceding Finite Supremum

Theorem
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.

Let $x, y \in S$.

Then $x \ll y$
 * $\forall X \subseteq S: y \preceq \sup X \implies \exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$

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

Sufficient Condition
Let $x \ll y$

Let $X \subseteq S$ such that
 * $y \preceq \sup X$

Define $F := \left\{ {\sup A: A \in {\it Fin}\left({X}\right)}\right\}$

By definition of union:
 * $X = \bigcup {\it Fin}\left({X}\right)$

By Supremum of Suprema:
 * $\sup X = \sup F$

We will prove that
 * $F$ is directed

Let $a, b \in F$

By definition of $F$:
 * $\exists A \in {\it Fin}\left({X}\right): a = \sup A$

and
 * $\exists B \in {\it Fin}\left({X}\right): b = \sup B$

By Union of Subsets is Subset:
 * $A \cup B \subseteq X$

By Finite Union of Finite Sets is Finite:
 * $A \cup B$ is finite

By definition of {\it Fin}:
 * $A \cup B \in {\it Fin}\left({X}\right)$

By definition of $F$:
 * $\sup\left({A \cup B}\right) \in F$

By Set is Subset of Union:
 * $A \subseteq A \cup B$ and $B \subseteq A \cup B$

By Supremum of Subset:
 * $a \preceq \sup\left({A \cup B}\right)$ and $b \preceq \sup\left({A \cup B}\right)$

Thus by definition
 * $F$ is directed

By definition of way below relation:
 * $\exists d \in F: x \preceq d$

Thus by definition of $F$:
 * $\exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$