=== Sufficient Condition === Define $K = \left({K\left({L}\right), \precsim}\right)$. Let $x, y \in K\left({L}\right)$. By Compact Subset is Join Subsemilattice: :$x \vee_L y \in K\left({L}\right)$ We will prove that :$x \vee_L y$ is upper bound for $\left\{ {x, y}\right\}$ in $K$ Let $z \in \left\{ {x, y}\right\}$. By Join Succeeds Operands: :$z \preceq x \vee_L y$ By definition of $\precsim$: :$z \precsim v \vee_L y$ {{qed|lemma}} We will prove that :$\forall z \in K\left({L}\right): z$ is upper bound for $\left\{ {x, y}\right\}$ in $K \implies x \vee_L y \precsim z$ Let $z \in K\left({L}\right)$ such that :$z$ is upper bound for $\left\{ {x, y}\right\}$ in $K$. Then by definition of $\precsim$: :$z$ is upper bound for $\left\{ {x, y}\right\}$ in $L$. By definition of supremum: :$x \vee_L y \preceq z$ By definition of $\precsim$: :$x \vee_L y \precsim z$ {{qed|lemma}} Thus by definition of supremum: :$\left\{ {x, y}\right\}$ admits a supremum in $K$. Thus analogically: :$\left\{ {x, y}\right\}$ admits an infimum in $K$. Hence $K$ is lattice. {{qed|lemma}} === Necessary Condition === Let $K$ be a lattice. Thus $L$ is algebraic. It remains to prove that :$K\left({L}\right)$ is meet closed. Let $x, y \in K\left({L}\right)$. Define $X := {\inf_L\left\{ {x, y}\right\} }^{\mathrm{compact} }$ By definition of algebraic: :$L$ satisfies axiom of K-approximation. By axiom of K-approximation: :$\inf_L\left\{ {x, y}\right\} = \sup_L X$ By definition of algebraic: :$X$ is directed and $L$ is up-complete. By definition of up-complete: :$X$ admits a supremum in $L$. By definition of lattice: :$\left\{ {x, y}\right\}$ admits an infimum in $K$. By definition of infimum: :$\inf_K\left\{ {x, y}\right\}$ is lower bound for $\left\{ {x, y}\right\}$ in $K$. We will prove that :$X \subseteq x^{\mathrm{compact} } \cap y^{\mathrm{compact} }$ Let $z \in X$. By definition of compact closure: :$z \preceq \inf_L\left\{ {x, y}\right\}$ and $z$ is compact. By definition of infimum: :$\inf_L\left\{ {x, y}\right\}$ is lower bound for $\left\{ {x, y}\right\}$ By definition of lower bound: :$\inf_L\left\{ {x, y}\right\} \preceq x$ and $\inf_L\left\{ {x, y}\right\} \preceq y$ By definition of transitivity: :$z \preceq x$ and $z \preceq y$ By definition of compact closure: :$z \in x^{\mathrm{compact} }$ and $z \in y^{\mathrm{compact} }$ Thus by definition of intersection: :$z \in x^{\mathrm{compact} } \cap y^{\mathrm{compact} }$ {{qed|lemma}} By Compact Closure is Intersection of Lower Closure and Compact Subset: :$X = \left({\inf_L\left\{ {x, y}\right\} }\right)^\preceq \cap K\left({L}\right)$ By Intersection is Subset: :$X \subseteq K\left({L}\right)$ We will prove that :$\inf_K \left\{ {x, y}\right\}$ is upper bound for $X$ in $K$. Let $z \in X$. By definition of intersection: :$z \in x^{\mathrm{compact} }$ and $z \in y^{\mathrm{compact} }$ By definition of compact closure: :$z \preceq x$ and $z \preceq y$ By definition of $\precsim$: :$z \precsim x$ and $z \precsim y$ Thus by definition of infimum: :$z \precsim \inf_K\left\{ {x, y}\right\}$ {{qed|lemma}} By definition of $\precsim$: :$\inf_K \left\{ {x, y}\right\}$ is upper bound for $X$ in $L$. By definition of supremum: :$\sup_L\left\{ {x, y}\right\} \preceq \inf_K \left\{ {x, y}\right\}$ By definition of supremum: :$\sup_K\left\{ {x, y}\right\}$ is upper bound for $\left\{ {x, y}\right\}$ in $K$. By definition of upper bound: :$x \precsim \sup_K\left\{ {x, y}\right\}$ and $y \precsim \sup_K\left\{ {x, y}\right\}$ By definition of $\precsim$: :$x \preceq \sup_K\left\{ {x, y}\right\}$ and $y \preceq \sup_K\left\{ {x, y}\right\}$ By definition of infimum: :$\sup_K\left\{ {x, y}\right\} \preceq \sup_L\left\{ {x, y}\right\}$ By definition of antisymmetry: :$\sup_K\left\{ {x, y}\right\} = \sup_L\left\{ {x, y}\right\}$ Thus by definition of meet: :$x \wedge_L y \in K\left({L}\right)$ {{qed}}