Every Element is Directed and Every Two Elements are Included in Third Element implies Union is Directed

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Theorem

Let $P = \left({S, \preceq}\right)$ be an ordered set.

Let $A$ be a set of subsets of $S$.

Let

$\forall X \in A: X$ is directed.

Let

$\forall X, Y \in A: \exists Z \in A: X \cup Y \subseteq Z$


Then $\bigcup A$ is directed.


Proof

Let $x, y \in \bigcup A$.

By definition of union:

$\exists X \in A: x \in X$

and

$\exists Y \in A: y \in Y$

By assumption:

$\exists Z \in A: X \cup Y \subseteq Z$

By definition of union:

$x, y \in X \cup Y$

By definition of subset:

$x, y \in Z$

By assumption:

$Z$ is directed.

By definition of directed subset:

$\exists z \in Z: x \preceq z \land y \preceq z$

Thus by definition of union:

$z \in \bigcup A$

Thus

$x \preceq z \land y \preceq z$

Hence $\bigcup A$ is directed.

$\blacksquare$


Sources