Compact Closure of Element is Principal Ideal on Compact Subset iff Element is Compact

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

Let $P = \left({K \left({L} \right), \precsim}\right)$ be an ordered subset of $L$

where $K \left({L} \right)$ denotes the compact subset of $L$.

Let $x \in S$.

Then $x^{\mathrm{compact} }$ is principal ideal in $P$ $x$ is a compact element.

Sufficient Condition
Assume that
 * $x^{\mathrm{compact} }$ is principal ideal in $P$.

By definitions of compact subset and compact closure:
 * $x^{\mathrm{compact} } \subseteq K \left({L} \right)$

By definition of principal ideal:
 * $\exists y \in x^{\mathrm{compact} }: y$ is upper bound for $x^{\mathrm{compact} }$ in $P$.

By definition of ordered subset:
 * $y$ is upper bound for $x^{\mathrm{compact} }$ in $L$.

By definition of supremum:
 * $\sup_L \left({x^{\mathrm{compact} } }\right) \preceq y$

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

By axiom of K-approximation:
 * $x \preceq y$

By definition of compact closure:
 * $y \preceq x$ and $y$ is a compact element.

Thus by definition of antisymmetry:
 * $x$ is a compact element.

Necessary Condition
Let $x$ be a compact element.

By definition of compact subset:
 * $x \in K\left({L}\right)$

We will prove that
 * $x^{\mathrm{compact} } = x^\precsim$

Let $y \in K\left({L}\right)$.


 * $y \in x^{\mathrm{compact} }$


 * $y \preceq x$ by definitions of compact closure and compact subset


 * $y \precsim x$ by definition of ordered subset


 * $y \in x^\precsim$

Hence $x^{\mathrm{compact} }$ is principal ideal in $P$.