Meet-Continuous iff Ideal Supremum is Meet Preserving

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

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

where
 * $\map {\it Ids} {\mathscr S}$ denotes the set of all ideals in $\mathscr S$

Then
 * $\mathscr S$ is meet-continuous


 * $f$ preserves meet as a mapping from $\struct {\map {\it Ids} {\mathscr S}, \subseteq}$ 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 $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$

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 $x = \sup \set {x \wedge d: d \in D}$

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

Thus

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

Define a mapping $g: S \times S \to S$:
 * $\forall \tuple {s, t} \in S \times S: \map g {s, t} = s \wedge t$

By Meet is Directed Suprema Preserving:
 * $g$ preserves directed suprema as a mapping from Cartesian product $\struct {S \times S, \precsim}$ 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:
 * $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$

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 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 $\mathscr S$: $\paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}$

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

Thus

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

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:
 * $\set x$ is directed.

Thus