Show that if A and B are posets and X \subseteq A \times B is directed (by the ordering in Construction [dt-000E]), then the subsets \pi _0(X) \subseteq A and \pi _1(X) \subseteq B (defined below) are also directed.
\begin {array}{lcl}
\pi _0(X) & = & \{ a \in A \mid \exists b \in B.\ (a,b) \in X \}\\
\pi _1(X) & = & \{ b \in B \mid \exists a \in A.\ (a,b) \in X \}\\
\end {array}
Given two elements x,y \in \pi _0(X), we know that there exists b_x, b_y \in B such that (x,b_x) \in X and (y,b_y) \in X. Because X is directed, we can conclude that there exists an upper bound(z, b_z) \in X where x \sqsubseteq z and y \sqsubseteq z. As z \in \pi _0(X), z is an upper bound in \pi _0(X) for both x and y and thus \pi _0(X) is directed. The proof for \pi _1 is analogous.
Give an example of a set X \subseteq \{\top ,\bot \}\times \{\top ,\bot \} such that \pi _0(X) and \pi _1(X) are directed, but X is not.
X = \{(\top ,\bot ), (\bot ,\top )\} is not directed as the two elements are not comparable, but \pi _0(X) = \pi _1(X) = \{\top ,\bot \} which is directed as \bot \sqsubseteq \top .
Show that if A and B are cpos and X \subseteq A \times B is directed, then \bigsqcup X = \left (\bigsqcup \pi _0(X), \bigsqcup \pi _1(X)\right )Note: Together with \bot _{A\times B} = (\bot _A, \bot _B) this shows that the Cartesian product of two cpos is itself a cpo.
Let (x,y) = \left (\bigsqcup \pi _0(X), \bigsqcup \pi _1(X)\right ). Because of the ordering in Construction [dt-000E], (x,y) is certainly aupper bound of the directed set X. To show it is the leastupper bound, assume for the sake of contradiction that (x,y) is not the least upper bound of X. Then, there must be a (v,w)\sqsubset (x,y) such that \forall x \in X.\ x \sqsubseteq (v,w). But then, v would be a lesser upper bound for \pi _0(X) than x, which is supposed to be the lub. Thus there is a contradiction, and (x,y) must be the lub of X.