By definition of algebraic ordered set: :$L$ is up-complete. By Up-Complete Lower Bounded Join Semilattice is Complete: :$L$ is a complete lattice. By definition of closure operator: :$c$ is idempotent. By Image of Idempotent and Directed Suprema Preserving Mapping is Complete Lattice: :$C$ is a complete lattice. We will prove that :$\forall x \in c\left[{S}\right]: x^{\mathrm{compact} }_C$ is directed. Let $x \in c\left[{S}\right]$. By Complete Lattice is Bounded: :$C$ is bounded. By Bottom is Way Below Any Element: :$\bot_C \ll \bot_C$ where :$\bot_C$ denotes the bottom of $C$, :$\ll$ denotes the way below relation. By definition: :$\bot_C$ is compact element in $C$. By definition of the smalest element: :$\bot_C \precsim x$ By definition of compact closure: :$\bot_C \in x^{\mathrm{compact} }_C$ Thus by definition: :$x^{\mathrm{compact} }_C$ is non-empty. Let $y, z \in x^{\mathrm{compact} }_C$. By definition of compact closure: :$y$ is a compact element and $z$ is a compact element. By definition of compact element: :$y \ll y$ and $z \ll z$ Define $v := y \vee_C z$ Thus by Join Succeeds Operands: :$y \precsim v$ and $z \precsim v$ By Preceding and Way Below implies Way Below: :$y \ll v$ and $z \ll v$ By Join is Way Below if Operands are Way Below: :$v \ll v$ and $v \precsim x$ Thus by definitions of compact element and compact closure: :$v \in x^{\mathrm{compact} }_C$ {{qed|lemma}} We will prove that :$\forall x \in c\left[{S}\right]: x = \sup_C x^{\mathrm{compact} }_C$ Let $x \in c\left[{S}\right]$. By definition of image of set: :$\exists y \in S: x = c\left({y}\right)$ By definition of closure operator/idempotent: :$c\left({x}\right) = x$ By definition of algebraic ordered set: :$x^{\mathrm{compact} }_L$ is directed. By definition of mapping preserves directed suprema: :$\sup_L \left({c\left[{x^{\mathrm{compact} }_L}\right]}\right) = c\left({\sup_L\left({x^{\mathrm{compact} }_L}\right)}\right)$ By definition of algebraic ordered set: :$L$ satisfies axiom of K-approximation. By axiom of K-approximation: :$\sup_L \left({c\left[{x^{\mathrm{compact} }_L}\right]}\right) = x$ By Supremum in Ordered Subset: :$\sup_C \left({c\left[{x^{\mathrm{compact} }_L}\right]}\right) = x$ We will prove that :$c\left[{x^{\mathrm{compact} }_L}\right] \subseteq x^{\mathrm{compact} }_C$ Let $a \in c\left[{x^{\mathrm{compact} }_L}\right]$. By definition of image of set: :$\exists v \in x^{\mathrm{compact} }_L: a = c\left({v}\right)$ By definition of compact closure: :$v \preceq x$ By definition of closure operator/increasing: :$c\left({v}\right) \preceq c\left({x}\right)$ By definition of ordered subset: :$a \precsim x$ By Image of Compact Subset under Directed Suprema Preserving Closure Operator is Subset of Compact Subset: :$c\left[{K\left({L}\right)}\right] \subseteq K\left({C}\right)$ where $K\left({L}\right)$ denotes the compact subset of $L$. By definition of [Definition:Compact Subset of Lattice|compact subset]]: :$a$ is a compact element in $C$. Thus by definition of compact closure: :$a \in x^{\mathrm{compact} }_C$ {{qed|lemma}} By Supremum of Subset: :$x \precsim \sup_C\left({x^{\mathrm{compact} }_C}\right)$ By definition of compact closure: :$x$ is upper bound for $x^{\mathrm{compact} }_C$ in $C$. By definition of supremum: :$\sup_C \left({x^{\mathrm{compact} }_C}\right) \precsim x$ Thus by definition of antisymmetry: :$x = \sup_C \left({x^{\mathrm{compact} }_C}\right)$ {{qed|lemma}} By definition: :$C$ satisfies axiom of K-approximation. Hence $C$ is algebraic lattice. {{qed}}