Upper Way Below Open Subset Complement is Non Empty implies There Exists Maximal Element of Complement

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

Let $X$ be upper way below open subset of $S$.

Let $x \in S$ such that
 * $x \in \complement_S\left({X}\right)$

Then
 * $\exists m \in S: x \preceq m \land m = \max \complement_S\left({X}\right)$

Proof
Define $A := \left\{ {C \in \mathit{Chains}\left({L}\right): C \subseteq \complement_S\left({X}\right) \land x \in C}\right\}$

where $\mathit{Chains}\left({L}\right)$ denotes the set of all chains of $L$.

We will prove that
 * $\forall Z: Z \ne \varnothing \land Z \subseteq A \land \left({\forall X, Y \in Z: X \subseteq Y \lor Y \subseteq X}\right) \implies \bigcup Z \in A$

By Singleton is Chain:
 * $\left\{ {x}\right\}$ is a chain of $L$.

By definition of $A$:
 * $\left\{ {x}\right\} \in A$

By Zorn's Lemma:
 * $\exists Y \in A: Y$ is a maximal element of $A$.

By definition of maximal element:
 * $\forall Z \in A: Y \subseteq Z \implies Y = Z$

By definition of $A$:
 * $Y \in \mathit{Chains}\left({L}\right) \land Y \subseteq \complement_S\left({X}\right) \land x \in Y$

By definition of supremum:
 * $\sup Y$ is upper bound for $Y$.

By definition of upper bound:
 * $x \preceq \sup Y$

We will prove that
 * $\lnot \exists y \in S: y \in \complement_S\left({X}\right) \land y \succ \sup Y$