# 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

- Mizar article WAYBEL_0:46