Compact Element iff Existence of Finite Subset that Element equals Intersection and Includes Subset

Theorem
Let $X, E$ be sets.

Let $P = \struct {\powerset X, \precsim}$ be an inclusion ordered set

where
 * $\powerset X$ denotes the power set of $X$
 * $\mathord \precsim = \mathord \subseteq \cap \paren {\powerset X \times \powerset X}$

Let $L = \struct {S, \preceq}$ be a continuous lattice subframe of $P$.

Then $E$ is compact element in $L$ :
 * $\exists F \in \map {\mathit {Fin} } X: E = \bigcap \set {I \in S: F \subseteq I} \land F \subseteq E$

where $\map {\mathit {Fin} } X$ denotes the set of all finite subsets of $X$.

Proof
By Power Set is Complete Lattice:
 * $P$ is a complete lattice.

By Infima Inheriting Ordered Subset of Complete Lattice is Complete Lattice:
 * $L$ is a complete lattice.

By Image of Operator Generated by Closure System is Set of Closure System:
 * $\map {\operatorname {operator} } L \sqbrk {\powerset X} = S$

where $\map {\operatorname {operator} } L$ denotes the operator generated by $L$.

By Closure Operator Preserves Directed Suprema iff Image of Closure Operator Inherits Directed Suprema:
 * $\map {\operatorname {operator} } L$ preserves directed suprema.

By Image of Compact Subset under Directed Suprema Preserving Closure Operator:
 * $\map {\operatorname {operator} } L \sqbrk {\map K P} = \map K {\paren {\map {\operatorname {operator} } L \sqbrk {\powerset X}, \precsim'} }$

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

By Image of Operator Generated by Closure System is Set of Closure System:
 * $ = \map K L$

Sufficient Condition
Assume that
 * $E$ is compact element in $L$.

By definition of compact subset:
 * $E \in \map K L$

By definition of image of set:
 * $\exists x \in \map K P: E = \map {\map {\operatorname {operator} } L} x$

By definition of closure operator/inflationary:
 * $x \precsim \map {\map {\operatorname {operator} } L} x$

By definition of $\precsim$:
 * $x \subseteq E$

By definition of compact subset:
 * $x$ is compact in $P$.

By Element is Finite iff Element is Compact in Lattice of Power Set:
 * $x$ is finite.

we will prove that
 * $x^\succsim \cap S = \set {I \in S: x \subseteq I}$

Let $y \in \powerset X$
 * $y \in x^\succsim \cap S$


 * $y \in x^\succsim$ and $y \in S$ by definition of intersection


 * $y \succsim x$ and $y \in S$ by definition of upper closure of element


 * $x \subseteq y$ and $y \in S$ by definition of $\precsim$


 * $y \in \set {I \in S: x \subseteq I}$

By definition of operator genereted by ordered subset:
 * $\map {\map {\operatorname {operator} } L} x = \map {\inf_P} {x^\succsim \cap S}$

By the proof of Power Set is Complete Lattice:
 * $E = \bigcap \set {I \in S: x \subseteq I}$

Hence $\exists F \in \map {\mathit {Fin} } X: E = \bigcap \set {I \in S: F \subseteq I} \land F \subseteq E$

Necessary Condition
Assume that:
 * $\exists F \in \map {\mathit {Fin} } X: E = \bigcap \set {I \in S: F \subseteq I} \land F \subseteq E$

By Element is Finite iff Element is Compact in Lattice of Power Set:
 * $F$ is compact element in $P$,

By definition of compact subset:
 * $F \in \map K P$

Then
 * $F^\succsim \cap S = \set {I \in S: F \subseteq I}$

By definition of operator generated by ordered subset:
 * $\map {\map {\operatorname {operator} } L} F = \map {\inf_P} {F^\succsim \cap S}$

By the proof of Power Set is Complete Lattice:
 * $\map {\map {\operatorname {operator} } L} F = E$

By definition of image of set:
 * $E \in \map K L$

Thus by definition of compact subset:
 * $E$ is a compact element in $L$.