Element of Increasing Mappings Satisfying Inclusion in Lower Closure is Generated by Auxiliary Relation

Theorem
Let $R = \struct {S, \preceq}$ be a bounded below join semilattice.

Let $\map {\operatorname{Ids} } R$ be the set of all ideals in $R$.

Let $L = \struct {\map {\operatorname{Ids} } R, \precsim}$ be an ordered set where $\precsim \mathop = \subseteq\restriction_{\map {\operatorname{Ids} } R \times \map {\operatorname{Ids} } R}$.

Let $M = \struct {F, \preccurlyeq}$ be the ordered set of increasing mappings $g: S \to \map {\operatorname{Ids} } R$ satisfying $\forall x \in S: \map g x \subseteq x^\preceq$.

Let $f \in F$.

Then
 * there exists an auxiliary relation $\RR$ on $S$ such that:
 * $\forall x \in S: \map f x = x^\RR$

where $x^\RR$ denotes the $\RR$-segment of $x$.

Proof
Define relation $\RR$ on $S$:
 * $\forall x, y \in S: \tuple {x, y} \in \RR \iff x \in \map f y$

We will prove that:
 * $(i): \quad \forall x, y \in S: \tuple {x, y} \in \RR \implies x \preceq y$

Let $x, y \in S$ such that:
 * $\tuple {x, y} \in \RR$

By definitions of $\RR$ and $F$:
 * $x \in \map f y \subseteq y^\preceq$

By definition of subset;
 * $x \in y^\preceq$

Thus be definition of lower closure of element:
 * $x \preceq y$

We will prove that:
 * $(ii): \quad \forall x, y, z, u \in S: x \preceq y \land \tuple {y, z} \in \RR \land z \preceq u \implies \tuple {x, u} \in \RR$

Let $x, y, z, u \in S$ such that:
 * $x \preceq y \land \tuple {y, z} \in \RR \land z \preceq u$

By definition of $\RR$:
 * $y \in \map f z$

By definition of increasing mapping:
 * $\map f z \precsim \map f u$

By definition of $\precsim$:
 * $\map f z \subseteq \map f u$

By definition of subset:
 * $y \in \map f u$

By definition of $F$:
 * $\map f u$ is an ideal in $R$.

By definition of ideal:
 * $\map f u$ is a lower set.

By definition of lower set:
 * $x \in \map f u$

Thus by definition of $\RR$:
 * $\tuple {x, u} \in \RR$

We will prove that:
 * $(iii): \quad \forall x, y, z \in S: \tuple {x, z} \in \RR \land \tuple {y, z} \in \RR \implies \tuple {x \vee y, z} \in \RR$

Let $x, y, z \in S$ such that:
 * $\tuple {x, z} \in \RR \land \tuple {y, z} \in \RR$

By definition of $\RR$:
 * $x \in \map f z$ and $y \in \map f z$

By definition of $F$:
 * $\map f z$ is an ideal in $R$.

By definition of ideal:
 * $\map f z$ is a directed lower set.

By definition of directed subset:
 * $\exists d \in \map f z: x \preceq d \land y \preceq d$

By definition of supremum:
 * $x \vee y \preceq d$

By definition of lower set:
 * $x \vee y \in \map f z$

Thus by definition of $\RR$:
 * $\tuple {x \vee y, z} \in \RR$

We will prove that:
 * $(iv): \quad \forall x \in S: \tuple {\bot, x} \in \RR$

where $\bot$ denotes the smallest element in $R$.

Let $x \in S$.

By definition of $F$:
 * $\map f x$ is an ideal in $R$.

By definition of ideal:
 * $\map f z$ is a non-empty lower set.

By definition of non-empty set:
 * $\exists z: z \in \map f x$

By definition of smallest element:
 * $\bot \preceq z$

By definition of lower set:
 * $\bot \in \map f x$

Thus by definition of $\RR$:
 * $\tuple {\bot, x} \in \RR$

By definition:
 * $\RR$ is an auxiliary relation on $S$.

Thus by definitions of $\RR$ and $\RR$-segment:
 * $\forall x \in S: \map f x = x^\RR$