Continuous Lattice is Meet-Continuous

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.

Then $L$ is meet-continuous.

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

By Way Below Closure is Directed in Bounded Below Join Semilattice:
 * $x^\ll$ is directed.

By definition of continuous:
 * $L$ is up-complete and satisfies the axiom of approximation.

By definition of up-complete:
 * $x^\ll$ admits a supremum.

By Lower Closure of Element is Ideal and definition of ideal:
 * $\paren {\sup \set {x \wedge d: d \in D} }^\preceq$ is directed.

By definition of up-complete:
 * $\paren {\sup \set {x \wedge d: d \in D} }^\preceq$ admits a supremum.

We will prove that
 * $x^\ll \subseteq \paren {\sup \set {x \wedge d: d \in D} }^\preceq$

Let $a \in x^\ll$

By definition of way below closure:
 * $a \ll x$

By Way Below implies Preceding:
 * $a \preceq x$

By definition of way below relation:
 * $\exists d \in D: a \preceq d$

By Meet is Idempotent and Meet Semilattice is Ordered Structure:
 * $a = a \wedge a \preceq x \wedge d \in \set {x \wedge y: y \in D}$

By definition of supremum:
 * $\sup \set {x \wedge y: y \in D}$ is upper bound for $\set {x \wedge y: y \in D}$

By definition of upper bound:
 * $x \wedge d \preceq \sup \set {x \wedge y: y \in D}$

By definition of transitivity:
 * $a \preceq \sup \set {x \wedge y: y \in D}$

Thus by definition of lower closure of element:
 * $a \in \paren {\sup \set {x \wedge d: d \in D} }^\preceq$

By Supremum of Subset:
 * $\map \sup {x^\ll} \preceq \map \sup {\paren {\sup \set {x \wedge d: d \in D} }^\preceq}$

By the axiom of approximation and Supremum of Lower Closure of Element:
 * $x \preceq \sup \set {x \wedge d: d \in D}$

By definition:
 * $x$ is upper bound for $\set {x \wedge d: d \in D}$

By definition of supremum:
 * $\sup \set {x \wedge d: d \in D} \preceq x$

Thus by definition of antisymmetry:
 * $x = \sup \set {x \wedge d: d \in D}$

Thus by Meet-Continuous iff if Element Precedes Supremum of Directed Subset then Element equals Supremum of Meet of Element by Directed Subset:
 * $L$ is meet-continuous.