Way Below in Meet-Continuous Lattice

Theorem
Let $\mathscr S = \left({S, \vee, \wedge, \preceq}\right)$ be a meet-continuous bounded below lattice.

Let $x, y \in S$.

Then $x \ll y$
 * $\forall I \in {\it Ids}\left({\mathscr S}\right): y = \sup I \implies x \in I$

where
 * $\ll$ denotes the way below relation,
 * ${\it Ids}\left({\mathscr S}\right)$ denotes the set of all ideals in $\mathscr S$.

Sufficient Condition
Let $x \ll y$

Let $I \in {\it Ids}\left({\mathscr S}\right)$ such that
 * $y = \sup I$

By definition of reflexivity:
 * $y \preceq \sup I$

By definition of meet-continuous:
 * $\mathscr S$ is up-complete.

Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
 * $x \in I$

Necessary Condition
Assume that
 * $\forall I \in {\it Ids}\left({\mathscr S}\right): y = \sup I \implies x \in I$

We will prove that
 * $\forall I \in {\it Ids}\left({\mathscr S}\right): y \preceq \sup I \implies x \in I$

Let $I \in {\it Ids}\left({\mathscr S}\right)$ such that
 * $y \preceq \sup I$

By definition of ideal:
 * $I$ is directed and lower.

Define $Y := \left\{ {y \wedge d: d \in I}\right\}$

By Set of Finite Suprema is Directed:
 * $\left\{ {\sup A: A \in {\it Fin}\left({Y}\right)}\right\}$ is directed.

where ${\it Fin}\left({Y}\right)$ denotes the set of all finite subsets of $Y$.

By definition of up-complete:
 * $\left\{ {\sup A: A \in {\it Fin}\left({Y}\right)}\right\}$ admits a supremum.

By definition of union:
 * $Y = \bigcup {\it Fin}\left({Y}\right)$

where $^\preceq$ denotes the lower closure of set.

By Lower Closure of Directed Subset is Ideal:
 * $\left\{ {\sup A: A \in {\it Fin}\left({Y}\right)}\right\}^\preceq$ is an ideal in $\mathscr S$

By assumption:
 * $x \in \left\{ {\sup A: A \in {\it Fin}\left({Y}\right)}\right\}^\preceq$

By definition of lower closure of set:
 * $\exists z \in \left\{ {\sup A: A \in {\it Fin}\left({Y}\right)}\right\}: x \preceq z$

Then
 * $\exists A \in {\it Fin}\left({Y}\right): z = \sup A$

We will prove as sublemma that
 * $A \subseteq I$

Let $a \in A$.

By definition of subset:
 * $a \in Y$

By definition of $Y$:
 * $\exists i \in I: a = y \wedge i$

By Meet Precedes Operands:
 * $a \preceq i$

Thus by definition of lower set:
 * $a \in I$

This ends the proof of sublemma.

By Directed iff Finite Subsets have Upper Bounds:
 * $\exists h \in I: \forall a \in A: a \preceq h$

By definition:
 * $h$ is upper bound for $A$

By definition of supremum:
 * $z = \sup A \preceq h$

By definition of transitivity:
 * $x \preceq h$

Thus by definition of lower set:
 * $x \in I$

Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
 * $x \ll y$