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

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

Then:
 * $L$ is continuous


 * $L$ 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 $\mathcal R$ on $S$
 * for every auxiliary approximating relation $\mathcal Q$ on $S$: $\mathcal R \subseteq \mathcal Q$

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 $\mathcal Q$ on $S$: $\ll \subseteq \mathcal Q$

Necessary Condition
Let $L$ be meet-continuous.

Assume that
 * there exists an auxiliary approximating relation $\mathcal R$ on $S$
 * for every auxiliary approximating relation $\mathcal Q$ on $S$: $\mathcal R \subseteq \mathcal Q$

Let $x \in S$.

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

where
 * $x^{\mathcal Q}$ denotes the $\mathcal Q$-segment of $x$,
 * $\mathit{App}\left({L}\right)$ denotes the set of all auxiliary approximating relations on $S$.

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

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

We will prove that
 * $\forall a \in \left\{ {x^{\mathcal Q}: \mathcal Q \in \mathit{App}\left({L}\right)}\right\}: x^{\mathcal R} \subseteq a$

Let $a \in \left\{ {x^{\mathcal Q}: \mathcal Q \in \mathit{App}\left({L}\right)}\right\}$

Then
 * $\exists \mathcal Q \in \mathit{App}\left({L}\right): a = x^{\mathcal Q}$

By assumption:
 * $\mathcal R \subseteq \mathcal Q$

Thus by Relation Segment is Increasing:
 * $x^{\mathcal R} \subseteq a$

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

By definition of set equality:
 * $x^{\mathcal R} = x^\ll$

Thus
 * $x = \sup \left({x^\ll}\right)$

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.