Up-Complete Product/Lemma 2

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

Then
 * $\operatorname{pr}_1^\to\left({X}\right)$ and $\operatorname{pr}_2^\to\left({X}\right)$ are directed

where
 * $\operatorname{pr}_1$ denotes the first projection on $S \times T$
 * $\operatorname{pr}_2$ denotes the second projection on $S \times T$
 * $\operatorname{pr}_1^\to\left({X}\right)$ denotes the image of $X$ under $\operatorname{pr}_1$

Proof
Let $x, y \in \operatorname{pr}_1^\to\left({X}\right)$.

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

and
 * $\exists y' \in T: \left({y, y'}\right) \in X$

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

By definition of Cartesian product of ordered sets:
 * $\exists a \in \operatorname{pr}_1^\to\left({X}\right): x \preceq_1 a \land y \preceq_1 a$

Thus by definition
 * $\operatorname{pr}_1^\to\left({X}\right)$ is directed.

By mutatis mutandis:
 * $\operatorname{pr}_2^\to\left({X}\right)$ is directed.