Compact Closure is Intersection of Lower Closure and Compact Subset

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

Let $x \in S$.

Then $x^{\mathrm{compact} } = x^\preceq \cap K\left({L}\right)$

where
 * $x^{\mathrm{compact} }$ denotes the compact closure of $x$,
 * $x^\preceq$ denotes the lower closure of $x$,
 * $K\left({L}\right)$ 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 $y \in K\left({L}\right)$ by definition of compact subset


 * $y \in x^\preceq \cap K\left({L}\right)$ by definition of intersection

Thus by definition of set equality:
 * the result holds.