Lattice of Power Set is Algebraic

Theorem
Let $X$ be a set.

Let $L = \struct {\powerset X, \cup, \cap, \preceq}$ be the lattice of power set of $X$ where $\mathord \preceq = \mathord \subseteq \cap \paren {\powerset X \times \powerset X}$

Then $L$ is algebraic.

Proof
We will prove that
 * $\forall x \in \powerset X: x^{\mathrm{compact} }$ is directed.

Let $x \in \powerset X$.

By Empty Set is Bottom of Lattice of Power Set:
 * $\O = \bot$

where $\bot$ denotes the bottom of $L$.

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

where $\ll$ is the way below relation.

By definition:
 * $\bot$ is compact.

By definition of the smallest element:
 * $\bot \preceq x$

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

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

Thus by Non-Empty Compact Closure is Directed:
 * $x^{\mathrm{compact} }$ is directed.

By Power Set is Complete Lattice:
 * $L$ is complete lattice.

Thus by definition of complete lattice:
 * $L$ is up-complete.

It remains to prove that:
 * $L$ satisfies axiom of K-approximation.

Let $x \in \powerset X$.

We will prove that:
 * $\forall a \in \powerset X: a$ is upper bound for $x^{\mathrm{compact} } \implies x \preceq a$

Let $a \in \powerset X$ such that:
 * $a$ is upper bound for $x^{\mathrm{compact} }$

We will prove that:
 * $x \subseteq a$

Let $t \in x$.

By definition of power set:
 * $x \subseteq X$

By definitions of subset and singleton:
 * $\set t \subseteq X$ and $\set t \subseteq x$

By definition of power set:
 * $\set t \in \powerset X$

By Singleton is Finite:
 * $\set t$ is finite.

By definition of $\operatorname {Fin}$:
 * $\set t \in \map {\operatorname {Fin} } x$

where $\map {\operatorname {Fin} } x$ denotes the set of all finite subset of $x$.

By Compact Closure is Set of Finite Subsets in Lattice of Power Set:
 * $\set t \in x^{\mathrm{compact} }$

By definition of upper bound:
 * $\set t \preceq a$

By definition of $\preceq$:
 * $\set t \subseteq a$

Thus by definitions of subset and singleton:
 * $t \in a$

Thus by definition of $\preceq$:
 * $x \preceq a$

By definition of compact closure:
 * $\forall y \in x^{\mathrm{compact} }: y \preceq x$

By definition of upper bound:
 * $x$ is upper bound for $x^{\mathrm{compact} }$

Thus by definition of supremum:
 * $x = \map \sup {x^{\mathrm{compact} } }$