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

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $X$ be upper way below open subset of $S$.
Let $x \in S$ such that
:$x \in \relcomp S X$
Then
:$\exists m \in S: x \preceq m \land m = \max \relcomp S X$

Proof:
Define $A := \set {C \in \map {\mathit {Chains} } L: C \subseteq \relcomp S X \land x \in C}$
where $\map {\mathit {Chains} } L$ denotes the set of all chains of $L$.
We will prove that
:$\forall Z: Z \ne \O \land Z \subseteq A \land \paren {\forall X, Y \in Z: X \subseteq Y \lor Y \subseteq X} \implies \bigcup Z \in A$
Let $Z$ such that
:$Z \ne \O \land Z \subseteq A \land \paren {\forall X, Y \in Z: X \subseteq Y \lor Y \subseteq X}$
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 $A$:
:$Y_1, Y_2 \in \map {\mathit {Chains} } L$
Thus by definition of connected relation
:$a \preceq b$ or $b \preceq a$
{{qed|lemma}}
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 $A$:
:$\forall Y \in Z: Y \subseteq \relcomp S X$
By Union of Subsets is Subset/Set of Sets:
:$\bigcup Z \subseteq \relcomp S X$
Thus by definition of $A$
:$\bigcup Z \in A$
{{qed|lemma}}
By Singleton is Chain:
:$\set x$ is a chain of $L$.
By definition of $A$:
:$\set x \in A$
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 $A$:
:$Y \in \map {\mathit {Chains} } L \land Y \subseteq \relcomp S X \land x \in Y$
By definition of supremum:
:$\sup Y$ is upper bound for $Y$.
By definition of upper bound:
:$x \preceq \sup Y$
We will prove that
:$\lnot \exists y \in S: y \in \relcomp S X \land y \succ \sup Y$
{{AimForCont}}
:$\exists y \in S: y \in \relcomp S X \land y \succ \sup Y$
By definition of antisymmetry:
:$y \notin Y$
By definition of $\succ$
:$\sup Y \preceq y$
We wiil prove that
:$Y \cup \set y$ is a chain of $L$.
Let $a, b \in Y \cup \set y$
Case $a, b \in Y$.
Thus by definition of connected relation:
:$a \preceq b$ or $b \preceq a$
Case $a \in Y \land b \in \set y$
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 $a \in \set y \land b \in Y$
Analogical case as previous.
Case $a, b \in \set y$
By definition of singleton:
:$ a = y$ and $b = y$
Be definition of reflexivity:
:$a \preceq b$
Thus
:$a \preceq b$ or $b \preceq a$
{{qed|lemma}}
By definitions of singleton and subset:
:$\set y \subseteq \relcomp S X$
By Union of Subsets is Subset:
:$Y \cup \set y \subseteq \relcomp S X$
By definition of union:
:$x \in Y \cup \set y$
By definition of $A$:
:$Y \cup \set y \in A$
By Set is Subset of Union:
:$Y \subseteq Y \cup \set y$
Then
:$Y = Y \cup \set y$
By definitions of union and singleton:
:$y \in Y$
This contradicts $y \notin Y$.
{{qed|lemma}}
We will prove that
:$\sup Y \in \relcomp S X$
{{AimForCont}}
:$\sup Y \in X$
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 section:
:$d \in X$
Thus it contradicts $d \in \relcomp S X$ by definition of subset.
{{qed|lemma}}
By definition of maximal element
:$\sup Y = \max \relcomp S X$
Hence
:$\exists m \in S: x \preceq m \land m = \max \relcomp S X$
{{qed}}