Supremum by Suprema of Directed Set in Simple Order Product

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Theorem

Let $\struct {S, \preceq}$ be an up-complete meet semilattice.

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

Let $D$ be a directed subset of $S \times S$.

Then:

$\sup D = \tuple {\map \sup {\map {\pr_1^\to} D}, \map \sup {\map {\pr_2^\to} D} }$

where

$\pr_1$ denotes the first projection on $S \times S$
$\pr_2$ denotes the second projection on $S \times S$
$\map {\pr_1^\to} D$ denotes the image of $D$ under $\pr_1$


Proof

By Up-Complete Product:

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

By definition of up-complete:

$D$ admits a supremum.

By definition of Cartesian product:

$\exists d_1, d_2 \in S: \sup D = \tuple {d_1, d_2}$

By Up-Complete Product/Lemma 2:

$D_1 := \map {\pr_1^\to} D$ is directed

and

$D_2 := \map {\pr_2^\to} D$ is directed.

By definition of up-complete:

$D_1$ admits a supremum

and

$D_2$ admits a supremum

We will prove that

$d_2$ is upper bound for $D_2$

Let $x \in D_2$.

By definition of image of set:

$\exists \tuple {a, b} \in D: \map {\pr_2} {a, b} = x$

By definition of second projection: $b = x$

By definition of supremum:

$\tuple {d_1, d_2}$ is upper bound for $D$.

By definition of upper bound:

$\tuple {a, x} \precsim \tuple {d_1, d_2}$

Thus by definition of simple order product:

$x \preceq d_2$

$\Box$


Analogically we have that

$d_1$ is upper bound for $D_1$

By definition of supremum:

$\sup D_1 \preceq d_1$ and $\sup D_2 \preceq d_2$

By definition of simple order product:

$\tuple {\sup D_1, \sup D_2} \precsim \sup D$

By Up-Complete Product/Lemma 1:

$D_1 \times D_2$ is directed.

By definition of up-complete:

$D_1 \times D_2$ admits a supremum.

By definition of subset:

$D \subseteq D_1 \times D_2$

By Supremum of Subset:

$\sup D \precsim \map \sup {D_1 \times D_2}$

By Supremum of Simple Order Product:

$\sup D \precsim \tuple {\sup D_1, \sup D_2}$

Thus by definition of antisymmetry:

$\sup D = \tuple {\sup D_1, \sup D_2}$

$\blacksquare$


Sources