Union of Filtered Sets is Filtered

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Theorem

Let $\struct {S, \preceq}$ be a preordered set.

Let $A$ be a set of subsets of $S$ such that

$\forall X \in A: X$ is filtered

and

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


Then:

$\bigcup A$ is also filtered.


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 filtered.

By definition of filtered:

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

Thus by definition of union:

$z \in \bigcup A$

Hence

$\bigcup A$ is filtered.

$\blacksquare$


Sources