Union of Filtered Sets is Filtered
Jump to navigation
Jump to search
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
- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices
- Mizar article WAYBEL_0:47