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 $x$.
Proof
$\subseteq$
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 $\operatorname{Fin}$:
- $y \in \map {\operatorname{Fin} } x$
$\Box$
$\supseteq$
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:
- $y \in x^{\mathrm{compact} }$
$\blacksquare$
Sources
- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices
- Mizar article WAYBEL_8:29