Meet-Irreducible Open Set Induces Completely Prime Filter
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Theorem
Let $\struct{S, \tau}$ be a topological space.
Let:
- $W \in \tau : W \ne S : W$ is meet-irreducible
Let:
- $\FF = \set{U \in \tau : U \nsubseteq W}$.
Then:
- $\FF$ is a completely prime filter in the complete lattice $\struct{\tau, \subseteq}$
Proof
$\FF$ satisfies Filter Axiom $\paren{1}$
We have by hypothesis:
- $W \ne S$
From Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology:
- $S \in \tau$
By definition of $\FF$:
- $S \in \FF$
By definition of empty set:
- $\FF \ne \O$
It follows that $\FF$ satisfies Filter Axiom $\paren{1}$.
$\Box$
$\FF$ satisfies Filter Axiom $\paren{2}$
Let $U, V \in \FF$.
By definition of $\FF$:
- $U \nsubseteq W$ and $V \nsubseteq W$
From the contrapositive statement of the definition of meet-irreducible open set:
- $U \cap V \nsubseteq W$
By definition of $\FF$:
- $U \cap V \in \FF$
From Intersection is Subset:
- $U \cap V \subseteq U$ and $U \cap V \subseteq V$
It follows that $\FF$ satisfies Filter Axiom $\paren{2}$.
$\Box$
$\FF$ satisfies Filter Axiom $\paren{3}$
Let $U \in \FF$.
Let $V \in \tau$ such that $U \subseteq V$.
From the contrapositive statement of Subset Relation is Transitive:
- $V \nsubseteq W$
By definition of $\FF$:
- $V \in \FF$
It follows that $\FF$ satisfies Filter Axiom $\paren{3}$.
$\Box$
$\FF$ is Proper
We have by hypothesis:
- $W \in \tau$
From Set is Subset of Itself:
- $W \notin \FF$
Hence:
- $\FF \ne \tau$
It follows that $\FF$ is a proper subset of $\tau$ by definition.
$\Box$
$\FF$ satisfies Complete Primeness
Let $\VV \subseteq \tau$:
- $\bigcup \VV \in \FF$
By definition of $\FF$:
- $\bigcup \VV \nsubseteq W$
By definition of subset:
- $\exists x \in \bigcup \VV : x \notin W$
By definition of set union:
- $\exists V \in \VV : x \in V$
By definition of subset:
- $\exists V \in \VV : V \nsubseteq W$
$\Box$
It has been shown that $\FF$ is a completely prime filter by definition.
$\blacksquare$