Compact Closure is Set of Finite Subsets in Lattice of Power Set

Theorem
Let $X$ be a set.

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

Let $x \in \powerset X$.

Then $x^{\mathrm{compact} } = \map {\operatorname{Fin} } x$

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

$\subseteq$
Let $y \in x^{\mathrm{compact} }$.

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

By definition of $\preceq$:
 * $y \subseteq x$

By Element is Finite iff Element is Compact in Lattice of Power Set: "$y$ is a finite set.

Thus by definition of $\operatorname{Fin}$:
 * $y \in \map {\operatorname{Fin} } x$

$\supseteq$
Let $y \in \map {\operatorname{Fin} } x$.

By definition of $\operatorname{Fin}$:
 * $y \subseteq x$ and $y$ is finite.

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

By Element is Finite iff Element is Compact in Lattice of Power Set:
 * $y$ is compact.

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