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 $y \in \mathcal P\left({X}\right)$ :$y \in x^\succsim \cap S$ {{iff}} :$y \in x^\succsim$ and $y \in S$ by definition of intersection {{iff}} :$y \succsim x$ and $y \in S$ by definition of upper closure of element {{iff}} :$x \subseteq y$ and $y \in S$ by definition of $\precsim$ {{iff}} :$y \in \left\{ {I \in S: x \subseteq I}\right\}$ {{qed|lemma}} 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 $\exists F \in \mathit{Fin}\left({X}\right): E = \bigcap \left\{ {I \in S: F \subseteq I}\right\} \land F \subseteq E$ {{qed|lemma}} === 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 $L$. {{qed}}