Image of Directed Suprema Preserving Closure Operator is Algebraic Lattice

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

Let $c:S \to S$ be a closure operator that preserves directed suprema.

Let $C = \left({c\left[{S}\right], \precsim}\right)$ be an ordered subset of $L$.

Then $C$ is algebraic lattice.

Proof
By definition of algebraic ordered set:
 * $L$ is up-complete.

By Up-Complete Lower Bounded Join Semilattice is Complete:
 * $L$ is a complete lattice.

By definition of closure operator:
 * $c$ is idempotent.

By Image of Idempotent and Directed Suprema Preserving Mapping is Complete Lattice:
 * $C$ is a complete lattice.

We will prove that
 * $\forall x \in c\left[{S}\right]: x^{\mathrm{compact} }_C$ is directed.

Let $x \in c\left[{S}\right]$.

By Complete Lattice is Bounded:
 * $C$ is bounded.

By Bottom is Way Below Any Element:
 * $\bot_C \ll \bot_C$

where
 * $\bot_C$ denotes the bottom of $C$,
 * $\ll$ denotes the way below relation.

By definition:
 * $\bot_C$ is compact element in $C$.

By definition of the smalest element:
 * $\bot_C \precsim x$

By definition of compact closure:
 * $\bot_C \in x^{\mathrm{compact} }_C$

Thus by definition:
 * $x^{\mathrm{compact} }_C$ is non-empty.

Let $y, z \in x^{\mathrm{compact} }_C$.

By definition of compact closure:
 * $y$ is a compact element and $z$ is a compact element.

By definition of compact element:
 * $y \ll y$ and $z \ll z$

Define $v := y \vee_C z$

Thus by Join Succeeds Operands:
 * $y \precsim v$ and $z \precsim v$

By Preceding and Way Below implies Way Below:
 * $y \ll v$ and $z \ll v$

By Join is Way Below if Operands are Way Below:
 * $v \ll v$ and $v \precsim x$

Thus by definitions of compact element and compact closure:
 * $v \in x^{\mathrm{compact} }_C$

We will prove that
 * $\forall x \in c\left[{S}\right]: x = \sup_C x^{\mathrm{compact} }_C$

Let $x \in c\left[{S}\right]$.

By definition of image of set:
 * $\exists y \in S: x = c\left({y}\right)$

By definition of closure operator/idempotent:
 * $c\left({x}\right) = x$

By definition of algebraic ordered set:
 * $x^{\mathrm{compact} }_L$ is directed.

By definition of mapping preserves directed suprema:
 * $\sup_L \left({c\left[{x^{\mathrm{compact} }_L}\right]}\right) = c\left({\sup_L\left({x^{\mathrm{compact} }_L}\right)}\right)$

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

By axiom of K-approximation:
 * $\sup_L \left({c\left[{x^{\mathrm{compact} }_L}\right]}\right) = x$

By Supremum in Ordered Subset:
 * $\sup_C \left({c\left[{x^{\mathrm{compact} }_L}\right]}\right) = x$

We will prove that
 * $c\left[{x^{\mathrm{compact} }_L}\right] \subseteq x^{\mathrm{compact} }_C$

Let $a \in c\left[{x^{\mathrm{compact} }_L}\right]$.

By definition of image of set:
 * $\exists v \in x^{\mathrm{compact} }_L: a = c\left({v}\right)$

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

By definition of closure operator/increasing:
 * $c\left({v}\right) \preceq c\left({x}\right)$

By definition of ordered subset:
 * $a \precsim x$

By Supremum of Subset:
 * $x \precsim \sup_C\left({x^{\mathrm{compact} }_C}\right)$

By definition of compact closure:
 * $x$ is upper bound for $x^{\mathrm{compact} }_C$ in $C$.

By definition of supremum:
 * $\sup_C \left({x^{\mathrm{compact} }_C}\right) \precsim x$

Thus by definition of antisymmetry:
 * $x = \sup_C \left({x^{\mathrm{compact} }_C}\right)$

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

Hence $C$ is algebraic lattice.