Up-Complete Product

Theorem
Let $\struct {S, \preceq_1}$, $\struct {T, \preceq_2}$ be non-empty ordered sets.

Let $\struct {S \times T, \preceq}$ be Cartesian product of $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$.

Then:
 * $\struct {S \times T, \preceq}$ is up-complete both $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ are also up-complete.

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.

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.