Compact Closure is Intersection of Lower Closure and Compact Subset

Theorem
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.

Let $x \in S$.

Then $x^{\mathrm {compact} } = x^\preceq \cap \map K L$

where
 * $x^{\mathrm {compact} }$ denotes the compact closure of $x$,
 * $x^\preceq$ denotes the lower closure of $x$,
 * $\map K L$ denotes the compact subset of $L$.

Proof

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


 * $y \preceq x$ and $y$ is compact by definition of compact closure


 * $y \in x^\preceq$ and $y$ is compact by definition of lower closure of element


 * $y \in x^\preceq$ and $y \in \map K L$ by definition of compact subset


 * $y \in x^\preceq \cap \map K L$ by definition of intersection

Hence the result, by definition of set equality.