Mapping Assigning to Element Its Compact Closure is Order Isomorphism

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

Let $C = \struct {\map K L, \preceq'}$ be an ordered subset of $L$

where $\map K L$ denotes the compact subset of $L$.

Let $I = \struct {\map {\mathit {Ids} } C, \precsim}$ be an inclusion ordered set

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

Let $f: S \to \map {\mathit {Ids} } C$ be a mapping such that
 * $\forall x \in S: \map f x = x^{\mathrm {compact} }$

where $x^{\mathrm {compact} }$ denotes the compact closure of $x$.

Thus $f$ is order isomorphism between $L$ and $I$.

Proof
We will prove that
 * $f$ is an order embedding.

Let $x, y \in S$.

Sufficient condition

Assume that
 * $x \preceq y$

By Compact Closure is Increasing:
 * $x^{\mathrm {compact} } \subseteq y^{\mathrm {compact} }$

By definition of $f$:
 * $\map f x \subseteq \map f y$

Thus by definition of inclusion ordered set:
 * $\map f x \precsim \map f y$

Necessary condition

Assume that:
 * $\map f x \precsim \map f y$

By definition of inclusion ordered set:
 * $\map f x \subseteq \map f y$

By definition of $f$:
 * $x^{\mathrm {compact} } \subseteq y^{\mathrm{compact} }$

By Supremum of Subset:
 * $\map \sup {x^{\mathrm {compact} } } \preceq \map \sup {y^{\mathrm {compact} } }$

By definition of algebraic:
 * $L$ satisfies axiom of K-approximation.

Thus by axiom of K-approximation:
 * $x \preceq y$

We will prove that
 * $f$ is a surjection.

Let $y \in \map {\mathit {Ids} } C$

Define $x = \sup_L y$

Thus $x \in S$.

We will prove that:
 * $x^{\mathrm {compact} } \subseteq y$

Let $d \in x^{\mathrm {compact} }$

By definition of compact closure:
 * $d$ is a compact element and $d \preceq x$

By definition of compact subset:
 * $d \in \map K L$

By definition of compact element:
 * $d \ll d$

where $\ll$ denotes the way below relation.

By definition of way below relation:
 * $\exists z \in y: d \preceq z$

By definition of ordered subset:
 * $d \preceq' z$

Thus by definition of lower set:
 * $d \in y$

We will prove that
 * $y \subseteq x^{\mathrm {compact} }$

Let $d \in y$.

By definition of subset:
 * $d \in \map K L$

By definition of compact subset:
 * $d$ is a compact element.

By definition of supremum:
 * $x$ is upper bound for $y$.

By definition of upper bound:
 * $d \preceq x$

Thus by definition of compact closure:
 * $d \in x^{\mathrm {compact} }$

By definition of set equality:
 * $y = x^{\mathrm {compact} }$

Hence:
 * $y = \map f x$

Hence $f$ is order isomorphism.