Way Below iff Preceding Finite Supremum

Theorem
Let $\struct {S, \vee, \wedge, \preceq}$ be a complete lattice.

Let $x, y \in S$.

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

where
 * $\ll$ denotes the way below relation,
 * $\map {\it Fin} X$ 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:
 * $F := \set {\sup A: A \in \map {\it Fin} X}$

By definition of union:
 * $X = \bigcup \map {\it Fin} X$

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

We will prove that:
 * $F$ is directed

Let $a, b \in F$.

By definition of $F$:
 * $\exists A \in \map {\it Fin} X: a = \sup A$

and
 * $\exists B \in \map {\it Fin} X: b = \sup B$

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 ${\it Fin}$:
 * $A \cup B \in \map {\it Fin} X$

By definition of $F$:
 * $\map \sup {A \cup B} \in F$

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

By Supremum of Subset:
 * $a \preceq \map \sup {A \cup B}$ and $b \preceq \map \sup {A \cup B}$

Thus by definition:
 * $F$ is directed

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

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

Necessary Condition
Assume that:
 * $\forall X \subseteq S: y \preceq \sup X \implies \exists A \in \map {\it Fin} X: x \preceq \sup A$

Let $D$ be a directed subset of $S$ such that:
 * $y \preceq \sup D$

By assumption:
 * $\exists A \in \map {\it Fin} D: x \preceq \sup A$

By Directed iff Finite Subsets have Upper Bounds:
 * $\exists h \in D: \forall a \in A: a \preceq h$

By definition:
 * $h$ is upper bound for $A$

By definition of supremum:
 * $\sup A \preceq h$

Thus by definition of transitivity:
 * $x \preceq h$

Thus by definition of way below relation:
 * $x \ll y$