Upper Closure of Element is Way Below Open Filter iff Element is Compact

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.

Let $x \in S$.

Then
 * $x^\succeq$ is a way below open filter on $L$


 * $x$ is compact

Sufficient Condition
Suppose
 * $x^\succeq$ is a way below open filter on $L$

By definitions of upper closure of element and reflexivity:
 * $x \in x^\succeq$

By definition of way below open:
 * $\exists y \in x^\succeq: y \ll x$

By definition of upper closure of element:
 * $x \preceq y$

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

By definition of compact:
 * $x$ is compact.

Necessary Condition
Let $x$ be compact.

By Upper Closure of Element is Filter:
 * $x^\succeq$ is filter on $L$.

It remains to show that
 * $x^\succeq$ is way below open.

Let $u \in x^\succeq$.

By definition of upper closure of element:
 * $x \preceq u$

and by definition of reflexivity:
 * $x \in x^\succeq$

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

By Preceding and Way Below implies Way Below:
 * $x \ll u$

Thus
 * $\exists z \in x^\succeq: z \ll u$