Supremum by Suprema of Directed Set in Simple Order Product

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

Let $\struct {S \times S, \precsim}$ be the Cartesian 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 Cartesian product of ordered sets:
 * $x \preceq d_2$

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 Cartesian product of ordered sets:
 * $\tuple {\sup D_1, \sup D_2} \precsim \sup D$

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 Cartesian Product:
 * $\sup D \precsim \tuple {\sup D_1, \sup D_2}$

Thus by definition of antisymmetry:
 * $\sup D = \tuple {\sup D_1, \sup D_2}$