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

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Theorem

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

Let $U \in \tau$ not be a meet-irreducible open set.

Let $F = S \setminus U$.

Then:

$F$ is not a closed irreducible subspace

Proof

By definition of meet-irreducible open set there exists $V_1, V_2\in \tau$:

$V_1 \nsubseteq U$

$V_2 \nsubseteq U$

$V_1 \cap V_2 \subseteq U$

Let:

$U_1 = V_1 \cup U$

$U_2 = V_2 \cup U$

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

$U_1, U_2 \in \tau$

From Set is Subset of Union:

$U \subsetneq U_1$

$U \subsetneq U_2$

We have:

\(\ds U_1 \cap U_2\)

\(=\)

\(\ds \paren{V_1 \cup U} \cap \paren{V_2 \cup U}\)

\(\ds \)

\(=\)

\(\ds \paren{U \cap V_1} \cup \paren{U \cap U} \cup \paren{V_2 \cap U} \cup \paren{V_2 \cap V_1}\)

Intersection Distributes over Union

\(\ds \)

\(=\)

\(\ds \paren{U \cap V_1} \cup U \cup \paren{V_2 \cap U} \cup \paren{V_2 \cap V_1}\)

Set Intersection is Idempotent

\(\ds \)

\(=\)

\(\ds U\)

Union with Superset is Superset

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

$S \setminus U_1 \subsetneq S \setminus U$

$S \setminus U_2 \subsetneq S \setminus U$

From De Morgan's Laws (Set Theory):

$S \setminus U = S \setminus U_1 \cup S \setminus U_2$

Consider:

$F_1 = S \setminus U_1$

$F_2 = S \setminus U_2$

By definition of closed set:

$F_1, F_2$ and $F$ are closed sets

We have:

$F_1 \subsetneq F$

$F_2 \subsetneq F$

$F_1 \cup F_2 = F$

It follows that $F$ is not a closed irreducible subspace by definition.

$\blacksquare$