Continuous iff Way Below Closure is Ideal and Element Precedes Supremum

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

Then
 * $L$ is continuous


 * $\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \sup \left({x^\ll}\right)$ and
 * for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$

Sufficient Condition
Let $L$ be continuous.

Let $x \in S$.

By definition of continuous:
 * $x^\ll$ is directed.

By Way Below Closure is Lower Set:
 * $x^\ll$ is a lower set.

Thus by definition:
 * $x^\ll$ is an ideal in $L$.

By definition of continuous:
 * $L$ satisfies axiom of approximation.

By definition of axiom of approximation:
 * $x = \sup \left({x^\ll}\right)$

Thus by definition of reflexivity:
 * $x \preceq \sup \left({x^\ll}\right)$

Let $I$ be an ideal in $L$ such that
 * $x \preceq \sup I$

We will prove that
 * $x^\ll \subseteq I$

Let $y \in x^\ll$

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

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

Necessary Condition
Assume that
 * $\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \sup \left({x^\ll}\right)$ and
 * for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$

Let $x \in S$.

By assumption:
 * $x \preceq \sup \left({x^\ll}\right)$

By Way Below implies Preceding:
 * $x$ is upper bound for $x^\ll$

By definition of supremum:
 * $\sup \left({x^\ll}\right) \preceq x$

Thus by definition of antisymmetry:
 * $x = \sup \left({x^\ll}\right)$

Thus by definition:
 * $L$ satisfies axiom of approximation.

Thus by assumption and definition of ideal:
 * $\forall x \in S: x^\ll$ is directed.

Hence $L$ is continuous.