Way Below in Complete Lattice

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
 * $\ll$ denotes the way below relation,
 * ${\it Fin}\left({X}\right)$ denotes the set of all finite subsets of $X$.

Sufficient Condition
Let
 * $x \ll y$

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

By Set of Finite Suprema is Directed:
 * $\left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ is directed.

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

By Supremum of Suprema:
 * $\sup X = \sup \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$

By definition of way below relation:
 * $\exists z \in \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}: x \preceq z$

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

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

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

By assumption:
 * $\exists A \in {\it Fin}\left({D}\right): x \preceq \sup A$

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

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

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

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