Up-Complete Product/Lemma 2

Theorem
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 $\pr_1$

Proof
Let $x, y \in \map {\pr_1^\to} X$.

By definitions of image of set and projections:
 * $\exists x' \in T: \tuple {x, x'} \in X$

and
 * $\exists y' \in T: \tuple {y, y'} \in X$

By definition of directed:
 * $\exists \tuple {a, b} \in X: \tuple {x, x'} \preceq \tuple {a, b} \land \tuple {y, y'} \preceq \tuple {a, b}$

By definition of simple order product:
 * $\exists a \in \map {\pr_1^\to} X: x \preceq_1 a \land y \preceq_1 a$

Thus by definition
 * $\map {\pr_1^\to} X$ is directed.

By mutatis mutandis:
 * $\map {\pr_2^\to} X$ is directed.