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

Let $Z$ such that
 * $Z \ne \varnothing \land Z \subseteq A \land \left({\forall X, Y \in Z: X \subseteq Y \lor Y \subseteq X}\right)$

We will prove that
 * $\bigcup Z$ is a chain of $L$

Let $a, b \in \bigcup Z$

By definition of union:
 * $\exists Y_1 \in Z: a \in Y_1$

and
 * $\exists Y_2 \in Z: b \in Y_2$

By assumption:
 * $Y_1 \subseteq Y_2$ or $Y_2 \subseteq Y_1$

By definition of subset:
 * $a, b \in Y_1$ or $a, b \in Y_2$

By definition of $A$:
 * $Y_1, Y_2 \in \mathit{Chains}\left({L}\right)$

Thus by definition of connected relation
 * $a \preceq b$ or $b \preceq a$

By definition of non-empty set:
 * $\exists Y: Y \in Z$

By definition of $A$:
 * $x \in Y$

By definition of union:
 * $x \in \bigcup Z$

By definition of $A$:
 * $\forall Y \in Z: Y \subseteq \complement_S\left({X}\right)$

By Union of Subsets is Subset/Set of Sets:
 * $\bigcup Z \subseteq \complement_S\left({X}\right)$

Thus by definition of $A$
 * $\bigcup Z \in A$

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

Aiming for a contradiction suppose that
 * $\exists y \in S: y \in \complement_S\left({X}\right) \land y \succ \sup Y$

By definition of antisymmetry:
 * $y \notin Y$

By definition of $\succ$
 * $\sup Y \preceq y$

We wiil prove that
 * $Y \cup \left\{ {y}\right\}$ is chain of $L$.

Let $a, b \in Y \cup \left\{ {y}\right\}$

Case $a, b \in Y$.

Thus by definition of connected relation:
 * $a \preceq b$ or $b \preceq a$

Case $a \in Y \land b \in \left\{ {y}\right\}$

By definition of singleton:
 * $b = y$

By definition of supremum:
 * $a \preceq \sup Y$

By definition of transitivity:
 * $a \preceq b$

Thus
 * $a \preceq b$ or $b \preceq a$

Case $a \in \left\{ {y}\right\} \land b \in Y$

Analogical case as previous.

Case $a, b \in \left\{ {y}\right\}$

By definition of singleton:
 * $ a = y$ and $b = y$

Be definition of reflexivity:
 * $a \preceq b$

Thus
 * $a \preceq b$ or $b \preceq a$

By definitions of singleton and subset:
 * $\left\{ {y}\right\} \subseteq \complement_S\left({X}\right)$

By Union of Subsets is Subset:
 * $Y \cup \left\{ {y}\right\} \subseteq \complement_S\left({X}\right)$

By definition of union:
 * $x \in Y \cup \left\{ {y}\right\}$

By definition of $A$:
 * $Y \cup \left\{ {y}\right\} \in A$

By Set is Subset of Union:
 * $Y \subseteq Y \cup \left\{ {y}\right\}$

Then
 * $Y = Y \cup \left\{ {y}\right\}$

By definitions of union and singleton:
 * $y \in Y$

This contradicts $y \notin Y$.

We will prove that
 * $\sup Y \in \complement_S\left({X}\right)$

Aiming for a contradiction suppose that
 * $\sup Y \in X$

By definition of way below open:
 * $\exists y \in X: y \ll \sup Y$

By Chain is Directed:
 * $Y$ is directed.

By definition of way below relation:
 * $\exists d \in Y: y \preceq d$

By definition of upper set:
 * $d \in X$

Thus it contradicts $d \in \complement_S\left({X}\right)$ by definition of subset.

By definition of maximal element
 * $\sup Y = \max \complement_S\left({X}\right)$

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