Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Directed Subsets

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

Then:
 * $\mathscr S$ is meet-continuous


 * for every directed subsets $D_1, D_2$ of $S$: $\left({\sup D_1}\right) \wedge \left({\sup D_2}\right) = \sup \left\{ {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}\right\}$
 * for every directed subsets $D_1, D_2$ of $S$: $\left({\sup D_1}\right) \wedge \left({\sup D_2}\right) = \sup \left\{ {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}\right\}$

Sufficient Condition
Let $\mathscr S$ be meet-continuous.

By Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Ideals:
 * for every ideals $I, J$ in $\mathscr S$: $\left({\sup I}\right) \wedge \left({\sup J}\right) = \sup \left\{ {i \wedge j: i \in I, j \in J}\right\}$

Let $D_1, D_2$ directed subsets of $S$.

By definition of up-complete:
 * $D_1$ and $D_2$ admit suprema

By Supremum of Lower Closure of Set:
 * $D_1^\preceq$ and $D_2^\preceq$ admit suprema

where
 * $D_1^\preceq$ denotes the lower closure of $D_1$.

Thus

Necessary Condition
Assume that
 * for every directed subsets $D_1, D_2$ of $S$: $\left({\sup D_1}\right) \wedge \left({\sup D_2}\right) = \sup \left\{ {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}\right\}$

By exemplification:
 * for every ideals $I_1, I_2$ of $S$: $\left({\sup I_1}\right) \wedge \left({\sup I_2}\right) = \sup \left\{ {d_1 \wedge d_2: d_1 \in I_1, d_2 \in I_2}\right\}$

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