Meet-Continuous iff Ideal Supremum is Meet Preserving

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

Let $f: {\it Ids}\left({\mathscr S}\right) \to S$ be a mapping such that
 * $\forall I \in {\it Ids}\left({\mathscr S}\right): f\left({I}\right) = \sup_{\mathscr S} I$

where
 * ${\it Ids}\left({\mathscr S}\right)$ denotes the set of all ideals in $\mathscr S$

Then
 * $\mathscr S$ is meet-continuous


 * $f$ preserves meet as a mapping from $\left({ {\it Ids}\left({\mathscr S}\right), \subseteq}\right)$ into $\mathscr S$

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

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

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

we will prove as sublemma 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 $x \preceq \sup \left\{ {x \wedge d: d \in D}\right\}$

Define a mapping $g: S \times S \to S$:
 * $\forall \left({s, t}\right) \in S \times S: g\left({s, t}\right) = s \wedge t$

By Meet is Directed Suprema Preserving:
 * $g$ preserves directed suprema as a mapping from Cartesian product $\left({S \times S, \precsim}\right)$ of $\mathscr S$ and $\mathscr S$ into $\mathscr S$.

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

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

Necessary Condition
Assume that
 * $f$ preserves meet

Thus
 * $\mathscr S$ is up-complete

We will prove that
 * for every ideals $I_1, I_2$ in $\mathscr S$: $\left({\sup I_1}\right) \wedge \left({\sup I_2}\right) = \sup\left\{ {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}\right\}$

Let $I_1, I_2$ be ideals in $\mathscr S$.

Thus

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

Let $D_1, D_2$ be 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

Thus

It remains to prove (MC) of definition of meet-continuous.

Let $x$ be an element of $S$.

Let $D$ be a directed subset of $S$.

By Singleton is Directed and Filtered Subset:
 * $\left\{ {x}\right\}$ is directed.

Thus