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}} that :$\exists y \in S: y \in \relcomp S X \land y \succ \sup Y$ 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 :$\sup Y \in \relcomp S X$ 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 $d \in \relcomp S X$ 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}}