Arithmetic iff Compact Subset form Lattice in Algebraic Lattice

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

Then $L$ is arithmetic $\left({K\left({L}\right), \precsim}\right)$ is a lattice,

where $K\left({L}\right)$ denotes the compact subset of $L$,
 * $\mathord \precsim = \mathord \preceq \cap \left({K\left({L}\right) \times K\left({L}\right)}\right)$

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 $\precsim$:
 * $z \precsim v \vee_L y$

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 $\precsim$:
 * $x \vee_L y \precsim z$

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.

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

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:
 * $z \precsim \inf_K\left\{ {x, y}\right\}$

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:
 * $x \wedge_L y \in K\left({L}\right)$