=== Sufficient Condition === Let $L$ be algebraic. We will prove that :$\forall x \in S: x^\ll$ is directed where $x^\ll$ denotes way below closure of $x$. Let $x \in S$. By definition of algebraic: :$x^{\mathrm{compact} }$ is directed where $x^{\mathrm{compact} }$ denotes the compact closure of $x$. By Compact Closure is Subset of Way Below Closure: :$x^{\mathrm{compact} } \subseteq x^\ll$ By definitions of non-empty set and subset: :$x^\ll \ne \varnothing$ Thus by Non-Empty Way Below Closure is Directed in Join Semilattice: :$x^\ll$ is directed. {{qed|lemma}} Thus by definition of algebraic: :$L$ is up-complete. We will prove that: :$L$ satisfies axiom of approximation Let $x \in S$. By previous: :$x^\ll$ is directed. By definition of algebraic: :$x^{\mathrm{compact} }$ is directed. By definition of up-complete: :$x^\ll$ admits a supremum and :$x^{\mathrm{compact} }$ admits a supremum. By Compact Closure is Subset of Way Below Closure: :$x^{\mathrm{compact} } \subseteq x^\ll$ By Supremum of Subset: :$\map \sup {x^{\mathrm {compact} } } \preceq \map \sup {x^\ll}$ By definition of algebraic: :$L$ satisfies axiom of K-approximation. By definition of axiom of K-approximation: :$x \preceq \map \sup {x^\ll}$ By Operand is Upper Bound of Way Below Closure: :$x$ is an upper bound for $x^\ll$. By definition of supremum: :$\map \sup {x^\ll} \preceq x$ Thus by definition of antisymmetry: :$x = \map \sup {x^\ll}$ {{qed|lemma}} Hence $L$ is continuous. Let $x, y \in S$ such that: :$x \ll y$ By definition of algebraic: :$D := y^{\mathrm{compact} }$ is directed. By definition of axiom of K-approximation: :$y = \sup D$ By definition of way below relation: :$\exists d \in D: x \preceq d$ By definition of compact closure: :$d$ is compact. Thus by definition of compact subset: :$d \in K \left({L}\right)$ By definition of supremum: :$y$ is an upper bound for $D$. Thus by definition of upper bound: :$x \preceq d \preceq y$ {{qed|lemma}} === Necessary Condition === Suppose that: :$L$ is continuous and: :$\forall x, y \in S: x \ll y \implies \exists k \in \map K L: x \preceq k \preceq y$ We will prove that: :$\forall x \in S: x^{\mathrm{compact} }$ is directed. Let $x \in S$. We will prove that: :for every a finite subset $A$ of $x^{\mathrm{compact} }$: there exists $c \in x^{\mathrm{compact} }$: $c$ is an upper bound for $A$. Let $A$ be a finite subset of $x^{\mathrm{compact} }$. By Compact Closure is Subset of Way Below Closure: :$x^{\mathrm{compact} } \subseteq x^\ll$ By definition of subset: :$A$ is a finite subset of $x^\ll$ By definition of continuous: :$x^\ll$ is directed. By Directed iff Finite Subsets have Upper Bounds: :$\exists b \in x^\ll: b$ is an upper bound for $A$. By definition of way below closure: :$b \ll x$ By assumption: :$\exists c \in \map K L: b \preceq c \preceq x$ By definition of compact subset: :$c$ is compact. Thus by definition of compact closure: :$c \in x^{\mathrm{compact} }$ Thus by Preceding implies if Less Upper Bound then Greater Upper Bound: :$c$ is upper bound for $A$. {{qed|lemma}} By Directed iff Finite Subsets have Upper Bounds: :$x^{\mathrm{compact} }$ is directed. {{qed|lemma}} Thus by definition of continuous: :$L$ is up-complete. It remains to prove that: :$L$ satisfies axiom of K-approximation. Let $x \in S$. We will prove that: :$\forall z \in S: z$ is an upper bound for $x^{\mathrm{compact} } \implies z$ is an upper bound for $x^\ll$ Let $z \in S$ such that: :$z$ is an upper bound for $x^{\mathrm{compact} }$ Let $d \in x^\ll$. By definition of way below closure: :$d \ll x$ By assumption: :$\exists k \in \map K L: d \preceq k \preceq x$ By definition of compact subset: :$k$ is compact. By definition of compact closure: :$k \in x^{\mathrm{compact} }$ By definition of upper bound: :$k \preceq z$ Thus by definition of transitivity: :$d \preceq z$ {{qed|lemma}} By Compact Closure is Subset of Way Below Closure: :$x^{\mathrm{compact} } \subseteq x^\ll$ By Upper Bound is Upper Bound for Subset: :$\forall z \in S: z$ is upper bound for $x^\ll \implies z$ is upper bound for $x^{\mathrm{compact} }$ By definition of continuous: :$x^\ll$ is directed. By definition of up-complete: :$x^\ll$ admits a supremum. By Upper Bounds are Equivalent implies Suprema are equal: :$\map \sup {x^{\mathrm{compact} } } = \map \sup {x^\ll}$ Thus by axiom of approximation: :$x = \map \sup {x^{\mathrm{compact} } }$ {{qed}}