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

Theorem
Let $X, E$ be sets.

Let $P = \left({\mathcal P\left({X}\right), \precsim}\right)$ be an inclusion ordered set

where
 * $\mathcal P\left({X}\right)$ denotes the power set of $X$,
 * $\mathord\precsim = \mathord\subseteq \cap \left({\mathcal{P}\left({X}\right) \times \mathcal{P}\left({X}\right)}\right)$

Let $L = \left({S, \preceq}\right)$ be a continuous lattice subframe of $P$.

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

where $\mathit{Fin}\left({X}\right)$ 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:
 * $\operatorname{operator}\left({L}\right)\left[{\mathcal P\left({X}\right)}\right] = S$

where $\operatorname{operator}\left({L}\right)$ denotes the operator generated by $L$.

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

By Image of Compact Subset under Directed Suprema Preserving Closure Operator:
 * $\operatorname{operator}\left({L}\right)\left[{K\left({P}\right)}\right] = K\left({\left({\operatorname{operator}\left({L}\right)\left[{\mathcal P\left({X}\right)}\right], \precsim'}\right)}\right)$

where $K\left({P}\right)$ denotes the compact subset of $P$.

By Image of Operator Generated by Closure System is Set of Closure System:
 * $ = K\left({L}\right)$

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

By definition of compact subset:
 * $E \in K\left({L}\right)$

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

By definition of closure operator/inflationary:
 * $x \precsim \operatorname{operator}\left({L}\right)\left({x}\right)$

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
 * $x^\succsim \cap S = \left\{ {I \in S: x \subseteq I}\right\}$

Let $y \in \mathcal P\left({X}\right)$
 * $y \in x^\succsim \cap S$


 * $y \in x^\succsim$ 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 $\precsim$


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

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

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

Hence $\exists F \in \mathit{Fin}\left({X}\right): E = \bigcap \left\{ {I \in S: F \subseteq I}\right\} \land F \subseteq E$

Necessary Condition
Assume that
 * $\exists F \in \mathit{Fin}\left({X}\right): E = \bigcap \left\{ {I \in S: F \subseteq I}\right\} \land F \subseteq E$

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

By definition of compact subset:
 * $F \in K\left({P}\right)$

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

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

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

By definition of image of set:
 * $E \in K\left({L}\right)$

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