Up-Complete Product

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Theorem

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

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


Then:

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


Proof

Lemma 1

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

Let $Y$ be a directed subset of $T$.

Then $X \times Y$ is also a directed subset of $S \times T$.

$\Box$


Lemma 2

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

Then

$\map {\pr_1^\to} X$ 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 $S \times T$
$\map {\pr_1^\to} X$ denotes the image of $X$ under $\pr_1$

$\Box$


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 simple order product:

$s \preceq_1 a$


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 simple order product:

$z \preceq_2 b$

By definition of upper bound:

$y \preceq_1 x$

Thus by definition of simple order product:

$\tuple {y, z} \preceq \tuple {x, b}$

This ends the proof of sublemma.

$\Box$


Then:

$\tuple {a, b} \preceq \tuple {x, b}$

Thus by definition of simple order product:

$x \preceq_1 a$


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 simple order product:

$\tuple {x, y} \preceq \tuple {a, b}$


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 simple order product:

$c \preceq_1 x$


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 simple order product:

$\tuple {a, b} \preceq \tuple {x, y}$

By definition

$X$ admits a supremum.

Thus by definition

$\struct {S \times T, \preceq}$ is up-complete.

$\blacksquare$


Sources