Continuous Lattice iff Auxiliary Approximating Relation is Superset of Way Below Relation

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

Let $\mathit{App}\left({L}\right)$ be the set of all auxiliary approximating relation on $S$.

Then
 * $L$ is continuous


 * $\forall \mathcal R \in \mathit{App}\left({L}\right): \ll \subseteq \mathcal R$ and $\ll$ is an approximating relation

Sufficient Condition
Let $L$ be continuous.

Let $\mathcal R \in \mathit{App}\left({L}\right)$

Let $\left({a, b}\right) \in S \times S$ such that
 * $a \ll b$

By definition of way below closure:
 * $a \in b^\ll$

where $b^\ll$ denotes the way below closure of $b$.

By Complete Lattice is Bounded and Continuous Lattice is Meet-Continuous: $L$ is a bounded below meet-continuous lattice.

By Intersection of Relation Segments of Approximating Relations equals Way Below Closure:
 * $\displaystyle \bigcap \left\{ {b^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} = b^\ll$

where $b^{\mathcal R}$ denotes the $\mathcal R$-segment of $b$.

By Intersection is Subset/General Result:
 * $b^\ll \subseteq b^{\mathcal R}$

By definition of subset:
 * $a \in b^{\mathcal R}$

By definition of $\mathcal R$-segment:
 * $\left({a, b}\right) \in \mathcal R$

Thus by definition of subset:
 * $\ll \subseteq \mathcal R$

Thus by Way Below is Approximating Relation:
 * $\ll$ is an approximating relation.

Necessary Condition
Let
 * $\forall \mathcal R \in \mathit{App}\left({L}\right): \ll \subseteq \mathcal R$ and $\ll$ is an approximating relation

Define $\mathcal R := \mathord \ll$.

By definition of approximating relation:
 * $\forall x \in S: x = \sup \left({x^{\mathcal R} }\right)$

By definitions of way below closure and $\mathcal R$-segment:
 * $\forall x \in S: x = \sup \left({x^\ll}\right)$

By definition:
 * $L$ satisfies axiom of approximation.

By Way Below Closure is Directed in Bounded Below Join Semilattice:
 * $\forall x \in S: x^\ll$ is directed

By definition of complete lattice:
 * $L$ is up-complete.

Hence $L$ is continuous.