=== Lemma 2 === {{:Meet Preserves Directed Suprema/Lemma 2}}{{qed|lemma}} Let $X$ be a directed subset of $S \times S$ such that :$X$ admits a supremum. By Up-Complete Product: :the Cartesian product of $\mathscr S$ and $\mathscr S$ is up-complete. By Up-Complete Product/Lemma 2: :$X_1 := \map {\pr_1^\to} X$ is directed and :$X_2 := \map {\pr_2^\to} X$ is directed where :$\pr_1$ denotes the first projection on $S \times S$ :$\pr_2$ denotes the second projection on $S \times S$ :$\map {\pr_1^\to} X$ denotes the image of $X$ We will prove that :$(1): \quad \set {x \wedge \sup X_2: x \in X_1} = \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$ '''First inclusion:''' Let $a \in \set {x \wedge \sup X_2: x \in X_1}$. Then :$\exists x \in X: a = x \wedge \sup X_2$ By Meet Precedes Operands: :$x \wedge \sup X_2 \preceq \sup X_2$ By assumption: :$x \wedge \sup X_2 \preceq \sup \set {x \wedge \sup X_2 \wedge y: y \in X_2}$ By definition of supremum: :$\forall y \in X_2: y \preceq \sup X_2$ By Preceding iff Meet equals Less Operand: :$\forall y \in X_2: \sup X_2 \wedge y = y$ By Meet is Associative: :$x \wedge \sup X_2 \preceq \sup \set {x \wedge y: y \in X_2}$ By Meet Semilattice is Ordered Structure: :$\forall y \in X_2: x \wedge y \preceq x \wedge \sup X_2$ By definition: :$x \wedge \sup X_2$ is upper bound for $\set {x \wedge y: y \in X_2}$ By definition of supremum: :$\sup \set {x \wedge y: y \in X_2} \preceq x \wedge \sup X_2$ By definition of antisymmetry: :$a = \sup \set {x \wedge y: y \in X_2}$ Thus :$a \in \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$ {{qed|lemma}} '''Second inclusion''' Let $a \in \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$. Then :$\exists x \in X_1: a = \sup \set {x \wedge y: y \in X_2}$ Analogically to first inclusion :$a = x \wedge \sup X_2$ Thus :$a \in \set {x \wedge \sup X_2: x \in X_1}$ {{qed|lemma}} We will prove as lemma that: :$\paren {\map {f^\to} X}$ is directed. Let $x, y \in \paren {\map {f^\to} X}$. By image of set: :$\exists \tuple {a, b} \in X: x = \map f {a, b}$ and :$\exists \tuple {c, d} \in X: y = \map f {c, d}$ By definition of $f$: :$x = a \wedge b$ and $y = c \wedge d$ By definition of directed subset: :$\exists \tuple {g, h} \in X: \tuple {a, b} \precsim \tuple {g, h} \land \tuple {c, d} \precsim \tuple {g, h}$ By Cartesian product of ordered sets: :$a \preceq g$, $b \preceq h$, $c \preceq g$, and $d \preceq h$ By Meet Semilattice is Ordered Structure: :$x \preceq g \wedge h$ and $y \preceq g \wedge h$ By definition of image of set: :$g \wedge h = \map f {g, h} \in \map {f^\to} X$ Thus: :$\exists z \in \map {f^\to} X: x \preceq z \land y \preceq z$ This ends the proof of lemma. Thus by definition of up-complete: :$\paren {\map {f^\to} X}$ admits a supremum. Thus {{begin-eqn}} {{eqn | l = \map \sup {\map {f^\to} X}       | r = \sup \set {x \wedge y: x \in X_1, y \in X_2}       | c = Supremum of Meet Image of Directed Set }} {{eqn | r = \sup \bigcup \set {\set {x \wedge y: y \in X_2}: x \in X_1} }} {{eqn | r = \sup \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}       | c = Supremum of Suprema }} {{eqn | r = \sup \set {x \wedge \sup X_2: x \in X_1}       | c = the equality $(1)$ }} {{eqn | r = \paren {\sup X_1} \wedge \paren {\sup X_2}       | c = Lemma 2 }} {{eqn | r = \map f {\sup X_1, \sup X_2}       | c = Definition of $f$ }} {{eqn | r = \map f {\sup X}       | c = Supremum by Suprema of Directed Set in Cartesian Product }} {{end-eqn}} {{qed}}