Intersection of Upper Section with Directed Set is Directed Set
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Theorem
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $A, B$ be subsets of $S$ such that
- $A \cap B \ne \O$
and
- $A$ is an upper set,
- $B$ is a directed set.
Then $A \cap B$ is a directed set.
Proof
Let $x, y \in A \cap B$.
By definition of intersection:
- $x, y \in A$ and $x, y \in B$
By definition of directed subset:
- $\exists z \in B: x \preceq z \land y \preceq z$
By definition of upper section:
- $z \in A$
Thus by definition of intersection:
- $z \in A \cap B$
Thus:
- $\exists z \in A \cap B: x \preceq z$ and $y \preceq z$
$\blacksquare$
Sources
- Mizar article WAYBEL11:3