=== Lemma 1 === {{:Up-Complete Product/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Up-Complete Product/Lemma 2}}{{qed|lemma}} === Sufficient Condition === Assume that: :$\struct {S \times T, \preceq}$ is up-complete. Let $X$ be a directed subset of $S$. By assumption: :$T \ne \O$ By definition by non-empty set: :$\exists t: t \in T$ By Singleton is Directed and Filtered Subset: :$\set t$ is directed By Lemma 1: :$X \times \set t$ is a directed subset of $S \times T$ By definition of up-complete: :$X \times \set t$ admits a supremum By definition of supremum: :$\exists \tuple {a, b} \in S \times T: \tuple {a, b}$ is upper bound for $X \times \set t$ and: :$\forall \tuple {c, d} \in S \times T: \tuple {c, d}$ is upper bound for $X \times \set t \implies \tuple {a, b} \preceq \tuple {c, d}$ We will prove that :$a$ is upper bound for $X$ Let $s \in X$. By definition of Cartesian product: :$\tuple {s, t} \in X \times \set t$ By definition of upper bound: :$\tuple {s, t} \preceq \tuple {a, b}$ Thus by definition of Cartesian product of ordered sets: :$s \preceq_1 a$ We will prove that :$\forall x \in S: x$ is upper bound for $X \implies a \preceq_1 x$ Let $x \in S$ such that :$x$ is upper bound for $X$ We will prove as sublemma that: :$\tuple {x, b}$ is upper bound for $X \times \set t$ Let $\tuple {y, z} \in X \times \set t$. By definition of Cartesian product: :$y \in X$ By definition of upper bound: :$\tuple {y, z} \preceq \tuple {a, b}$ By definition of Cartesian product of ordered sets: :$z \preceq_2 b$ By definition of upper bound: :$y \preceq_1 x$ Thus by definition of Cartesian product of ordered sets: :$\tuple {y, z} \preceq \tuple {x, b}$ This ends the proof of sublemma. {{qed|lemma}} Then: :$\tuple {a, b} \preceq \tuple {x, b}$ Thus by definition of Cartesian product of ordered sets: :$x \preceq_1 a$ Thus by definition of supremum: :$X$ admits a supremum: Thus by definition: :$\struct {S, \preceq_1}$ is up-complete. By mutatis mutandis :$\struct {T, \preceq_2}$ is up-complete === Necessary Condition === Assume that :$\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ are up-complete. Let $X$ be directed subset of $S \times T$. By Lemma 2: :$\map {\pr_1^\to} X$ and $\map {\pr_2^\to} X$ are directed where  :$\pr_1$ denotes the first projection on $S \times T$ :$\pr_2$ denotes the second projection on $S \times T$ By definition of up-complete: :$\map {\pr_1^\to} X$ and $\map {\pr_2^\to} X$ admit suprema By definition of supremum: :$\exists a \in S: a$ is upper bound for $\map {\pr_1^\to} X$ and: :$\forall c \in S: c$ is upper bound for $\map {\pr_1^\to} X \implies a \preceq_1 c$ and :$\exists b \in T: b$ is upper bound for $\map {\pr_2^\to} X$ and: :$\forall c \in T: c$ is upper bound for $\map {\pr_2^\to} X \implies b \preceq_2 c$ We will prove that :$\tuple {a, b}$ is upper bound for $X$ Let $\tuple {x, y} \in X$. By definitions of image of set and projections: :$x \in \map {\pr_1^\to} X$ and $y \in \map {\pr_2^\to} X$ By definition of upper bound: :$x \preceq_1 a$ and $y \preceq_2 b$ Thus by definition of Cartesian product of ordered sets: :$\tuple {x, y} \preceq \tuple {a, b}$ We will prove that :$\forall \tuple {x, y} \in S \times T: \tuple {x, y}$ is upper bound for $X \implies \tuple {a, b} \preceq \tuple {x, y}$ Let $\tuple {x, y} \in S \times T$ such that: :$\tuple {x, y}$ is upper bound for $X$ We will prove as sublemma that :$x$ is upper bound for $\map {\pr_1^\to} X$ Let $c \in \map {\pr_1^\to} X$. By definitions of image of set and projections: :$\exists d: \tuple {c, d} \in X$ By definition of upper bound: :$\tuple {c, d} \preceq \tuple {x, y}$ Thus by definition of Cartesian product of ordered sets: :$c \preceq_1 x$ By mutatis mutandis :$y$ is upper bound for $\map {\pr_2^\to} X$ Then: :$a \preceq_1 x$ and $b \preceq_2 y$ Thus by definition of Cartesian product of ordered sets: :$\tuple {a, b} \preceq \tuple {x, y}$ By definition :$X$ admits a supremum. Thus by definition :$\struct {S \times T, \preceq}$ is up-complete. {{qed}}