Meet-Irreducible Open Set iff Complement is Closed Irreducible Subspace/Necessary Condition

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Theorem

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

Let $U \in \tau$.

Let $F = S \setminus U$.

Let $F$ be a closed irreducible subspace.

Then:

$U$ is a meet-irreducible open set

Proof

By definition of closed set:

$F$ is a closed set

By definition of closed irreducible subspace there exists proper closed subsets $F_1, F_2$ of $F$:

$F = F_1 \cup F_2$

$F_1 \subsetneq F$

$F_2 \subsetneq F$

From Set Complement inverts Subsets and Equal Relative Complements iff Equal Subsets:

$S \setminus F \subsetneq S \setminus F_1$

$S \setminus F \subsetneq S \setminus F_2$

From De Morgan's Laws (Set Theory):

$S \setminus F = S \setminus F_1 \cap S \setminus F_2$

Consider:

$U_1 = S \setminus F_1$

$U_2 = S \setminus F_2$

From Complement of Closed Set is Open Set:

$U_1, U_2 \in \tau$

We have:

$U \subsetneq U_1$

$U \subsetneq U_2$

$U = U_1 \cap U_2$

Hence:

$U_1 \nsubseteq U$

$U_2 \nsubseteq U$

$U = U_1 \cap U_2$

It follows that $U$ is not a meet-irreducible open set by definition.

$\blacksquare$