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

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

Let $\map {\operatorname{App} } L$ be the set of all auxiliary approximating relation on $S$.

Then
 * $L$ is continuous


 * $\forall \RR \in \map {\operatorname{App} } L: \ll \subseteq \RR$ and $\ll$ is an approximating relation

Sufficient Condition
Let $L$ be continuous.

Let $\RR \in \map {\operatorname{App} } L$

Let $\tuple {a, b} \in S \times S$ 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:
 * $\ds \bigcap \set {b^\RR: \RR \in \map {\operatorname{App} } L} = b^\ll$

where $b^\RR$ denotes the $\RR$-segment of $b$.

By Intersection is Subset/General Result:
 * $b^\ll \subseteq b^\RR$

By definition of subset:
 * $a \in b^\RR$

By definition of $\RR$-segment:
 * $\tuple {a, b} \in \RR$

Thus by definition of subset:
 * $\ll \subseteq \RR$

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

Necessary Condition
Let
 * $\forall \RR \in \map {\operatorname{App} } L: \ll \subseteq \RR$ and $\ll$ is an approximating relation

Define $\RR := \mathord \ll$.

By definition of approximating relation:
 * $\forall x \in S: x = \map \sup {x^\RR}$

By definitions of way below closure and $\RR$-segment:
 * $\forall x \in S: x = \map \sup {x^\ll}$

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.