Algebraic iff Continuous and For Every Way Below Exists Compact Between

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a lattice.

Then $L$ is algebraic
 * $L$ is continuous and
 * $\forall x, y \in S: x \ll y \implies \exists k \in K\left({L}\right): x \preceq k \preceq y$

where
 * $\ll$ denotes the way below relation,
 * $K\left({L}\right)$ denotes the compact subset of $L$.

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.

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:
 * $\sup \left({x^{\mathrm{compact} } }\right) \preceq \sup \left({ x^\ll}\right)$

By definition of algebraic:
 * $L$ satisfies axiom of K-approximation.

By definition of axiom of K-approximation:
 * $x \preceq \sup \left({ x^\ll}\right)$

By Operand is Upper Bound of Way Below Closure:
 * $x$ is upper bound for $x^\ll$.

By definition of supremum:
 * $\sup \left({ x^\ll}\right) \preceq x$

Thus by definition of antisymmetry:
 * $x = \sup \left({ x^\ll}\right)$

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 upper bound for $D$.

Thus by definition of upper bound:
 * $x \preceq d \preceq y$

Necessary Condition
Suppose that
 * $L$ is continuous and
 * $\forall x, y \in S: x \ll y \implies \exists k \in K\left({L}\right): x \preceq k \preceq y$

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 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 upper bound for $A$.

By definition of way below closure:
 * $b \ll x$

By assumption:
 * $\exists c \in K\left({L}\right): b \preceq c \preceq x$

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 $A$.

By Directed iff Finite Subsets have Upper Bounds:
 * $x^{\mathrm{compact} }$ is directed.

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 upper bound for $x^{\mathrm{compact} } \implies z$ is upper bound for $x^\ll$

Let $z \in S$ such that
 * $z$ is upper bound for $x^{\mathrm{compact} }$

Let $d \in x^\ll$.

By definition of way below closure:
 * $d \ll x$

By assumption:
 * $\exists k \in K\left({L}\right): d \preceq k \preceq x$

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:
 * $d \preceq z$

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:
 * $\sup\left({x^{\mathrm{compact} } }\right) = \sup\left({ x^\ll}\right)$

Thus by axiom of approximation:
 * $x = \sup\left({x^{\mathrm{compact} } }\right)$