Mapping Assigning to Element Its Compact Closure Preserves Infima and Directed Suprema

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

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 $P = \left({\mathcal P\left({K\left({L}\right)}\right), \precsim}\right)$ be an inclusion ordered set of power set of $K\left({L}\right)$.

Then there exists $f:S \to \mathcal P\left({K\left({L}\right)}\right)$ such that $f$ preserves infima and directed suprema and is an injection and $\forall x \in S: f\left({x}\right) = x^{\mathrm{compact} }$

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

Proof
By definitions of compact subset, compact closure, and subset:
 * $\forall x \in S: x^{\mathrm{compact} } \subseteq K\left({L}\right)$

By definition of power set:
 * $\forall x \in S: x^{\mathrm{compact} } \in \mathcal P\left({K\left({L}\right)}\right)$

Define a mapping $f:S \to \mathcal P\left({K\left({L}\right)}\right)$ such that
 * $\forall x \in S: f\left({x}\right) = x^{\mathrm{compact} }$

By Compact Closure is Directed:
 * $\forall x \in S: x^{\mathrm{compact} }$ is directed.

By definition of ordered subset:
 * $\forall x \in S: x^{\mathrm{compact} }$ is directed in $C$.

We will prove that
 * $\forall x \in S: x^{\mathrm{compact} }$ is lower in $C$.

Let $x \in S$.

Let $y \in x^{\mathrm{compact} }, z \in K\left({L}\right)$ such that
 * $z \preceq y$

By definition of compact closure:
 * $y \preceq x$

By definition of transitivity:
 * $z \preceq x$

By definition of compact subset:
 * $z$ is compact.

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

By Bottom in Compact Closure:
 * $\forall x \in S: \bot_L \in x^{\mathrm{compact} }$

where $\bot_L$ denotes the bottom in $L$.

By definition:
 * $\forall x \in S: x^{\mathrm{compact} }$ is non-empty.

By definition of ideal in ordered set:
 * $\forall x \in S: x^{\mathrm{compact} } \in \mathit{Ids}\left({C}\right)$

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

Then
 * $f:S \to \mathit{Ids}\left({C}\right)$

Define $I = \left({\mathit{Ids}\left({C}\right), \precsim'}\right)$ as an inclusion ordered set.

By definitions of $\mathit{Ids}$ and power set:
 * $\mathit{Ids}\left({C}\right) \subseteq \mathcal P\left({K\left({L}\right)}\right)$