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

Theorem
Let $X, E$ be sets.

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

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

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.

Sufficient Condition
By :
 * $\operatorname{operator}\left({L}\right)$ preserves directed suprema.

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