Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Ideals

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

Then
 * $\mathscr S$ is meet-continuous
 * for every ideals $I, J$ in $\mathscr S$: $\left({\sup I}\right) \wedge \left({\sup J}\right) = \sup \left\{ {i \wedge j: i \in I, j \in J}\right\}$

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

Define $\mathcal I$, the set of all ideals in $\mathscr S$

Define a mapping $f:\mathcal I \to S$ such that
 * $\forall I \in \mathcal I: f\left({I}\right) = \sup I$

By Meet-Continuous iff Ideal Supremum is Meet Preserving:
 * $f$ preserves meet.

Let $I, J \in \mathcal I$.

By definition of mapping preserves meet:
 * $f$ preserves the infimum of $\left\{ {I, J}\right\}$

Thus

Necessary Condition
Assume that
 * for every ideals $I, J$ in $\mathscr S$: $\left({\sup I}\right) \wedge \left({\sup J}\right) = \sup \left\{ {i \wedge j: i \in I, j \in J}\right\}$

By Meet of Suprema equals Supremum of Meet of Ideals implies Ideal Supremum is Meet Preserving:
 * $f$ preserves meet.

Thus by Meet-Continuous iff Ideal Supremum is Meet Preserving:
 * $\mathscr S$ is meet-continuous.