Arithmetic iff Compact Subset form Lattice in Algebraic Lattice

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below algebraic lattice.

Then $L$ is arithmetic $\struct {\map K L, \precsim}$ is a lattice,

where $\map K L$ denotes the compact subset of $L$:
 * $\mathord \precsim = \mathord \preceq \cap \paren {\map K L \times \map K L}$

Sufficient Condition
Define $K = \struct {\map K L, \precsim}$.

Let $x, y \in \map K L$.

By Compact Subset is Join Subsemilattice:
 * $x \vee_L y \in \map K L$

We will prove that
 * $x \vee_L y$ is upper bound for $\set {x, y}$ in $K$

Let $z \in \set {x, y}$.

By Join Succeeds Operands:
 * $z \preceq x \vee_L y$

By definition of $\precsim$:
 * $z \precsim v \vee_L y$

We will prove that
 * $\forall z \in \map K L: z$ is upper bound for $\set {x, y}$ in $K \implies x \vee_L y \precsim z$

Let $z \in \map K L$ such that
 * $z$ is upper bound for $\set {x, y}$ in $K$.

Then by definition of $\precsim$:
 * $z$ is upper bound for $\set {x, y}$ in $L$.

By definition of supremum:
 * $x \vee_L y \preceq z$

By definition of $\precsim$:
 * $x \vee_L y \precsim z$

Thus by definition of supremum:
 * $\set {x, y}$ admits a supremum in $K$.

Thus analogically:
 * $\set {x, y}$ admits an infimum in $K$.

Hence $K$ is lattice.

Necessary Condition
Let $K$ be a lattice.

Thus $L$ is algebraic.

It remains to prove that
 * $\map K L$ is meet closed.

Let $x, y \in \map K L$.

Define $X := {\inf_L \set {x, y} }^{\mathrm{compact} }$

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

By axiom of K-approximation:
 * $\inf_L \set {x, y} = \sup_L X$

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:
 * $\set {x, y}$ admits an infimum in $K$.

By definition of infimum:
 * $\inf_K \set {x, y}$ is lower bound for $\set {x, y}$ in $K$.

We will prove that
 * $X \subseteq x^{\mathrm{compact} } \cap y^{\mathrm{compact} }$

Let $z \in X$.

By definition of compact closure:
 * $z \preceq \inf_L \set {x, y}$ and $z$ is compact.

By definition of infimum:
 * $\inf_L \set {x, y}$ is lower bound for $\set {x, y}$

By definition of lower bound:
 * $\inf_L \set {x, y} \preceq x$ and $\inf_L \set {x, y} \preceq y$

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:
 * $z \in x^{\mathrm{compact} } \cap y^{\mathrm{compact} }$

By Compact Closure is Intersection of Lower Closure and Compact Subset:
 * $X = \paren {\inf_L \set {x, y} }^\preceq \cap \map K L$

By Intersection is Subset:
 * $X \subseteq \map K L$

We will prove that:
 * $\inf_K \set {x, y}$ 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:
 * $z \precsim \inf_K \set {x, y}$

By definition of $\precsim$:
 * $\inf_K \set {x, y}$ is upper bound for $X$ in $L$.

By definition of supremum:
 * $\sup_L \set {x, y} \preceq \inf_K \set {x, y}$

By definition of supremum:
 * $\sup_K \set {x, y}$ is upper bound for $\set {x, y}$ in $K$.

By definition of upper bound:
 * $x \precsim \sup_K \set {x, y}$ and $y \precsim \sup_K \set {x, y}$

By definition of $\precsim$:
 * $x \preceq \sup_K \set {x, y}$ and $y \preceq \sup_K \set {x, y}$

By definition of infimum:
 * $\sup_K \set {x, y} \preceq \sup_L \set {x, y}$

By definition of antisymmetry:
 * $\sup_K \set {x, y} = \sup_L \set {x, y}$

Thus by definition of meet:
 * $x \wedge_L y \in \map K L$