Compact Closure is Directed

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

Let $x \in S$.

Then $x^{\mathrm{compact} }$ is directed

where $x^{\mathrm{compact} }$ denotes the compact closure of $x$.

Proof
By Bottom in Compact Closure:
 * $\bot \in x^{\mathrm{compact} }$

where $\bot$ denotes the smallest element in $L$.

Thus by definition:
 * $x^{\mathrm{compact} }$ is non-empty.

Let $y, z \in x^{\mathrm{compact} }$

By definition of compact closure:
 * $y$ and $z$ are compact elements and $y \preceq x$ and $z \preceq x$

By definitions of supremum and upper bound:
 * $y \vee z \preceq x$

By definition of compact element:
 * $y \ll y$ and $z \ll z$

where $\ll$ denotes the way below relation.

By Join Succeeds Operands:
 * $y \preceq y \vee z$ and $z \preceq y \vee z$

By Preceding and Way Below implies Way Below and definition of reflexivity:
 * $y \ll y \vee z$ and $z \ll y \vee z$

By Join is Way Below if Operands are Way Below:
 * $y \vee z \ll y \vee z$

By definition:
 * $y \vee z$ is compact element.

By definition of compact closure:
 * $y \vee z \in x^{\mathrm{compact} }$

Thus
 * $\exists v \in x^{\mathrm{compact} }: y \preceq v \land z \preceq v$