Meet-Continuous iff Meet Preserves Directed Suprema

Theorem
Let $\mathscr S = \left({S, \vee, \wedge, \preceq}\right)$ be an up-complete lattice.

Let $\left({S \times S, \precsim}\right)$ be the Cartesian product of $\mathscr S$ and $\mathscr S$.

Let $f: S \times S \to S$ be a mapping such that
 * $\forall x, y \in S: f\left({x, y}\right) = x \wedge y$

Then
 * $\mathscr S$ is meet-continuous


 * $f$ preserves directed suprema

Sufficient Condition
Assume that
 * $\mathscr S$ is meet-continuous.

we will prove that
 * for every an element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x = \sup \left\{ {x \wedge d: d \in D}\right\}$

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

Thus

By definition of reflexivity:
 * for every an element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x \preceq \sup \left\{ {x \wedge d: d \in D}\right\}$

By Meet is Directed Suprema Preserving:
 * $f$ preserves directed suprema.

Necessary Condition
Assume that
 * $f$ preserves directed suprema.

By Meet is Directed Suprema Preserving implies Meet of Suprema equals Supremum of Meet of Directed Subsets:
 * $\left({\sup D_1}\right) \wedge \left({\sup D_2}\right) = \sup \left\{ {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}\right\}$

Thus by Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Directed Subsets:
 * $\mathscr S$ is meet-continuous.