Up-Complete Product

Theorem
Let $\left({S, \preceq_1}\right)$, $\left({T, \preceq_2}\right)$ be ordered sets.

Let $\left({S \times T, \preceq}\right)$ be Cartesian product of $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$.

Then
 * $\left({S \times T, \preceq}\right)$ is up-complete both $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ are also up-complete.

Sufficient Condition
Assume that
 * $\left({S \times T, \preceq}\right)$ is up-complete.

Let $X$ be a directed subset of $S$.

By assumption:
 * $T \ne \varnothing$

By definition by non-empty set:
 * $\exists t: t \in T$

By Singleton is Directed and Filtered Subset:
 * $\left\{ {t}\right\}$ is directed

By Lemma 1:
 * $X \times \left\{ {t}\right\}$ is a directed subset of $S \times T$

By definition of up-complete:
 * $X \times \left\{ {t}\right\}$ admits a supremum

By definition of supremum:
 * $\exists \left({a, b}\right) \in S \times T: \left({a, b}\right)$ is upper bound for $X \times \left\{ {t}\right\}$ and
 * $\forall \left({c, d}\right) \in S \times T: \left({c, d}\right)$ is upper bound for $X \times \left\{ {t}\right\} \implies \left({a, b}\right) \preceq \left({c, d}\right)$

We will prove that
 * $a$ is upper bound for $X$

Let $s \in X$.

By definition of Cartesian product:
 * $\left({s, t}\right) \in X \times \left\{ {t}\right\}$

By definition of upper bound:
 * $\left({s, t}\right) \preceq \left({a, b}\right)$

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
 * $\left({x, b}\right)$ is upper bound for $X \times \left\{ {t}\right\}$

Let $\left({y, z}\right) \in X \times \left\{ {t}\right\}$.

By definition of Cartesian product:
 * $y \in X$

By definition of upper bound:
 * $\left({y, z}\right) \preceq \left({a ,b}\right)$

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:
 * $\left({y, z}\right) \preceq \left({x, b}\right)$

This ends the proof of sublemma.

Then:
 * $\left({a, b}\right) \preceq \left({x, b}\right)$

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
 * $\left({S, \preceq_1}\right)$ is up-complete

By mutatis mutandis
 * $\left({T, \preceq_2}\right)$ is up-complete

Necessary Condition
Assume that $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ are up-complete.

Let $X$ be directed subset of $S \times T$.

By Lemma 2
 * $\operatorname{pr}_1^\to\left({X}\right)$ and $\operatorname{pr}_2^\to\left({X}\right)$ are directed

where
 * $\operatorname{pr}_1$ denotes the first projection on $S \times T$
 * $\operatorname{pr}_2$ denotes the second projection on $S \times T$

By definition of up-complete:
 * $\operatorname{pr}_1^\to\left({X}\right)$ and $\operatorname{pr}_2^\to\left({X}\right)$ admit suprema

By definition of supremum:
 * $\exists a \in S: a$ is upper bound for $\operatorname{pr}_1^\to\left({X}\right)$ and
 * $\forall c \in S: c$ is upper bound for $\operatorname{pr}_1^\to\left({X}\right) \implies a \preceq_1 c$

and
 * $\exists b \in T: b$ is upper bound for $\operatorname{pr}_2^\to\left({X}\right)$ and
 * $\forall c \in T: c$ is upper bound for $\operatorname{pr}_2^\to\left({X}\right) \implies b \preceq_2 c$