Completely Prime Filter Induces Meet-Irreducible Open Set

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Theorem

Let $\struct{S, \tau}$ be a topological space.

Let $\FF$ be a completely prime filter in the complete lattice $\struct{\tau, \subseteq}$.

Let $W = \bigcup \set{U \in \tau : U \notin \FF}$.

Then:

$W$ is a meet-irreducible open set

Proof

By definition of completely prime filter:

$\FF$ is a proper subset of $\tau$

Hence:

$\set{U \in \tau : U \notin \FF} \ne \O$

By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:

$W \in \tau$

From Filter Contains Greatest Element:

$S \in \FF$

By definition of completely prime filter:

$W \notin \FF$

and so:

$W \ne S$

Let $U_1, U_2 \in \tau$:

$U_1 \cap U_2 \subseteq W$

By the contrapositive statement of Meet Semilattice Filter Axiom $\paren{\text {MSF} 1 }$: Upper Section of Meet Semilattice:

$U_1 \cap U_2 \notin \FF$

as $W \notin \FF$

By the contrapositive statement of Meet Semilattice Filter Axiom $\paren{\text {MSF} 2 }$: Subsemilattice of Meet Semilattice:

either $U_1 \notin \FF$ or $U_2 \notin \FF$

Hence:

either $U_1 \subseteq W$ or $U_2 \subseteq W$

It follows that $W$ is a meet-irreducible open set by definition.

$\blacksquare$

Also see

• Topology forms Complete Lattice