Up-Complete Product/Lemma 1

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Theorem

Let $\left({S, \preceq_1}\right)$, $\left({T, \preceq_2}\right)$ be non-empty ordered sets.

Let $\left({S \times T, \preceq}\right)$ be Cartesian product of $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$.


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 $S \times T$.


Proof

Let $\left({s_1, t_1}\right)$, $\left({s_2, t_2}\right) \in X \times Y$.

By definition of Cartesian product:

$s_1, s_2 \in X$ and $t_1, t_2 \in Y$

By definition of directed subset:

$\exists h_1 \in X: s_1 \preceq_1 h_1 \land s_2 \preceq_1 h_1$

and

$\exists h_2 \in X: t_1 \preceq_2 h_2 \land t_2 \preceq_2 h_2$

By definition of Cartesian product of ordered sets:

$\exists \left({h_1, h_2}\right) \in X \times Y: \left({s_1, t_1}\right) \preceq \left({h_1, h_2}\right) \land \left({s_2, t_2}\right) \preceq \left({h_1, h_2}\right)$

Thus by definition:

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

$\blacksquare$

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