Continuous iff For Every Element There Exists Ideal Element Precedes Supremum

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

Then
 * $L$ is continuous


 * for every element $x$ of $S$ there exists ideal $I$ in $L$:
 * $x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$

Sufficient Condition
Let $L$ be continuous.

By Continuous iff Way Below Closure is Ideal and Element Precedes Supremum:
 * $\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \map \sup {x^\ll}$ and
 * for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$

where $x^\ll$ denotes the way below closure of $x$.

Thus
 * for every element $x$ of $S$ there exists ideal $I$ in $L$:
 * $x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$

Necessary Condition
Assume that
 * for every element $x$ of $S$ there exists ideal $I$ in $L$:
 * $x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$

Let $x \in S$.

There exists ideal $I$ in $L$:
 * $x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$

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

where $x^\ll$ denotes the way below closure of $x$.

Let $y \in I$.

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

Then $I \subseteq J$

Thus by definition of subset:
 * $y \in J$

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

Thus by definition of way below closure:
 * $y \in x^\ll$

By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal
 * $\forall y \in x^\ll: y \in I$

By definition of subset:
 * $x^\ll \subseteq I$

Thus by definition of set equality:
 * $\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \map \sup {x^\ll}$ and
 * for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$

Hence by Continuous iff Way Below Closure is Ideal and Element Precedes Supremum:
 * $L$ is continuous