Intersection of Relation Segments of Approximating Relations equals Way Below Closure

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

Let $\mathit{App}\left({L}\right)$ be the set of all auxiliary approximating relations on $S$.

Let $x \in S$.

Then
 * $\displaystyle \bigcap \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} = x^\ll$

Proof
By Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element:
 * $\displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\} = x^\ll$

where $\mathit{Ids}$ denotes the set of all ideals in $L$.

For all $I \in \mathit{Ids}$ define a mapping $m_I: S \to \mathit{Ids}$:
 * $\forall x \in S: x \preceq \sup I \implies m_I\left({x}\right) = \left\{ {x \wedge i: i \in I}\right\}$

and
 * $\forall x \in S: x \npreceq \sup I \implies m_I\left({x}\right) = x^\preceq$

By Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element:
 * $\forall x \in S: \displaystyle \bigcap \left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\} = x^\ll$

We will prove that
 * $\left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\} \subseteq \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$

Let $a \in \left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\}$

Then
 * $\exists I \in \mathit{Ids}: a = m_I\left({x}\right)$

By Down Mapping is Generated by Approximating Relation:
 * $\exists \mathcal R \in \mathit{App}\left({L}\right): \forall s \in S: m_I(s) = s^{\mathcal R}$

Then
 * $a = x^{\mathcal R}$

Thus
 * $a \in \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$

By Intersection of Family is Subset of Intersection of Subset of Family:
 * $\displaystyle \bigcap \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} \subseteq x^\ll$

We will prove that
 * $\left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} \subseteq \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$

Let $a \in \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$

Then
 * $\exists \mathcal R \in \mathit{App}\left({L}\right): a = x^{\mathcal R}$

By definition of approximating relation:
 * $x = \sup a$

By Relation Segment of Auxiliary Relation is Ideal:
 * $a \in \mathit{Ids}$

Thus by definition of reflexivity:
 * $a \in \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$

By Intersection of Family is Subset of Intersection of Subset of Family:
 * $x^\ll \subseteq \displaystyle \bigcap \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$

Hence the result by definition of set equality.