Meet-Continuous iff if Element Precedes Supremum of Directed Subset then Element equals Supremum of Meet of Element by Directed Subset

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

Then
 * $L$ is meet-continuous


 * $\forall x \in S$, the directed subset $D$ of $S: x \preceq \sup D \implies x = \sup \set {x \wedge d: d \in D}$
 * $\forall x \in S$, the directed subset $D$ of $S: x \preceq \sup D \implies x = \sup \set {x \wedge d: d \in D}$

Sufficient Condition
Let $L$ be meet-continuous.

Let $x$ be an element of $S$, $D$ be a directed subset of $S$ such that
 * $x \preceq \sup D$

Thus

Necessary Condition
Let:
 * $\forall x \in S$, directed subset $D$ of $S: x \preceq \sup D \implies x = \sup \set {x \wedge d: d \in D}$

By definition of reflexivity:
 * $\forall x \in S$, directed subset $D$ of $S: x \preceq \sup D \implies x \preceq \sup \set {x \wedge d: d \in D}$

By Meet is Directed Suprema Preserving:
 * $\wedge$ preserves directed suprema as a mapping from the simple order product $\struct {S \times S, \precsim}$ of $L$ and $L$ into $L$.

Thus by Meet-Continuous iff Meet Preserves Directed Suprema
 * $L$ is meet-continuous.