Image of Directed Suprema Preserving Closure Operator is Algebraic Lattice

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

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

Let $C = \struct {c \sqbrk S, \precsim}$ 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
 * $\forall x \in c \sqbrk S: x^{\mathrm{compact} }_C$ is directed.

Let $x \in c \sqbrk S$.

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
 * $\forall x \in c \sqbrk S: x = \sup_C x^{\mathrm{compact} }_C$

Let $x \in c \sqbrk S$.

By definition of image of set:
 * $\exists y \in S: x = \map c y$

By definition of closure operator/idempotent:
 * $\map c x = x$

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

By definition of mapping preserves directed suprema:
 * $\map {\sup_L} {c \sqbrk {x^{\mathrm{compact} }_L} } = \map c {\map {\sup_L} {x^{\mathrm{compact} }_L} }$

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

By the axiom of $K$-approximation:
 * $\map {\sup_L} {c \sqbrk {x^{\mathrm{compact} }_L} } = x$

By Supremum in Ordered Subset:
 * $\map {\sup_C} {c \sqbrk {x^{\mathrm{compact} }_L} } = x$

We will prove that
 * $c \sqbrk {x^{\mathrm{compact} }_L} \subseteq x^{\mathrm{compact} }_C$

Let $a \in c \sqbrk {x^{\mathrm{compact} }_L}$.

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

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

By definition of closure operator/increasing:
 * $\map c v \preceq \map c x$

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

By Image of Compact Subset under Directed Suprema Preserving Closure Operator is Subset of Compact Subset:
 * $c \sqbrk {\map K L} \subseteq \map K C$

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

By definition of [Definition:Compact Subset of Lattice|compact subset]]:
 * $a$ is a compact element in $C$.

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

By Supremum of Subset:
 * $x \precsim \map {\sup_C} {x^{\mathrm{compact} }_C}$

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

By definition of supremum:
 * $\map {\sup_C} {x^{\mathrm{compact} }_C} \precsim x$

Thus by definition of antisymmetry:
 * $x = \map {\sup_C} {x^{\mathrm{compact} }_C}$

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

Hence $C$ is algebraic lattice.