Mapping Assigning to Element Its Compact Closure is Order Isomorphism

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

Let $C = \left({K\left({L}\right), \preceq'}\right)$ be an ordered subset of $L$

where $K\left({L}\right)$ denotes the compact subset of $L$.

Let $I = \left({\mathit{Ids}\left({C}\right), \precsim}\right)$ be an inclusion ordered set

where $\mathit{Ids}\left({C}\right)$ denotes the set of all ideals in $C$.

Let $f: S \to \mathit{Ids}\left({C}\right)$ br a mapping such that
 * $\forall x \in S: f\left({x}\right) = x^{\mathrm{compact} }$

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 $f$:
 * $f\left({x}\right) \subseteq f\left({y}\right)$

Thus by definition of inclusion ordered set:
 * $f\left({x}\right) \precsim f\left({y}\right)$

necessary condition

Assume that
 * $f\left({x}\right) \precsim f\left({y}\right)$

By definition of inclusion ordered set:
 * $f\left({x}\right) \subseteq f\left({y}\right)$

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

By Supremum of Subset:
 * $\sup \left({x^{\mathrm{compact} } }\right) \preceq \sup \left({y^{\mathrm{compact} } }\right)$

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 $y \in \mathit{Ids}\left({C}\right)$

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:
 * $d \in K\left({L}\right)$

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:
 * $d \in K\left({L}\right)$

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 $y = f\left({x}\right)$

Hence $f$ is order isomorphism.