Continuous iff Meet-Continuous and There Exists Smallest Auxiliary Approximating Relation

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

Then:
 * $L$ is continuous


 * $L$ is meet-continuous and there exists the smallest auxiliary approximating relation on $S$
 * $L$ is meet-continuous and there exists the smallest auxiliary approximating relation on $S$

That is:
 * $L$ is continuous


 * $L$ is meet-continuous and there exists an auxiliary approximating relation $\RR$ on $S$
 * for every auxiliary approximating relation $\QQ$ on $S$: $\RR \subseteq \QQ$
 * for every auxiliary approximating relation $\QQ$ on $S$: $\RR \subseteq \QQ$

Sufficient Condition
Let $L$ be continuous.

Thus by Continuous Lattice is Meet-Continuous:
 * $L$ is meet-continuous.

Thus by Way Below is Approximating Relation and Way Below Relation is Auxiliary Relation:
 * $\ll$ is auxiliary approximating relation on $S$.

Thus by Continuous Lattice iff Auxiliary Approximating Relation is Superset of Way Below Relation:
 * for every auxiliary approximating relation $\QQ$ on $S$: $\ll \subseteq \QQ$

Necessary Condition
Let $L$ be meet-continuous.

Assume that:
 * there exists an auxiliary approximating relation $\RR$ on $S$
 * for every auxiliary approximating relation $\QQ$ on $S$: $\RR \subseteq \QQ$

Let $x \in S$.

By Intersection of Relation Segments of Approximating Relations equals Way Below Closure:
 * $\ds \bigcap \set {x^\QQ: \QQ \in \map {\operatorname{App} } L} = x^\ll$

where:
 * $x^\QQ$ denotes the $\QQ$-segment of $x$,
 * $\map {\operatorname{App} } L$ denotes the set of all auxiliary approximating relations on $S$.

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

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

We will prove that:
 * $\forall a \in \set {x^\QQ: \QQ \in \map {\operatorname{App} } L}: x^\RR \subseteq a$

Let $a \in \set {x^\QQ: \QQ \in \map {\operatorname{App} } L}$

Then:
 * $\exists \QQ \in \map {\operatorname{App} } L: a = x^\QQ$

By assumption:
 * $\RR \subseteq \QQ$

Thus by Relation Segment is Increasing:
 * $x^\RR \subseteq a$

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

By definition of set equality:
 * $x^\RR = x^\ll$

Thus
 * $x = \map \sup {x^\ll}$

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.