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

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

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


 * for every directed subsets $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$
 * for every directed subsets $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$

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 $\mathscr S$: $\paren {\sup I} \wedge \paren {\sup J} = \sup \set {i \wedge j: i \in I, j \in J}$

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 $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$

By exemplification:
 * for every ideals $I_1, I_2$ of $S$: $\paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {d_1 \wedge d_2: d_1 \in I_1, d_2 \in I_2}$

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