Compact Closure is Subset of Way Below Closure

Theorem
Let $L = \left({S, \preceq}\right)$ be an ordered set.

Let $x \in S$.

Then $x^{\mathrm{compact} } \subseteq x^\ll$

where
 * $x^{\mathrm{compact} }$ denotes the compact closure of $x$,
 * $x^\ll$ denotes the way below closure of $x$.

Proof
Let $y \in x^{\mathrm{compact} }$.

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

By definition of compact:
 * $y \ll y$

where $\ll$ denotes the way below relation.

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

Thus by definition of way below closure:
 * $y \in x^\ll$