Meet-Irreducible Open Set iff Complement is Closed Irreducible Subspace
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Theorem
Let $T = \struct{S, \tau}$ be a topological space.
Let $U \in \tau$.
Let $F = S \setminus U$.
Then:
- $U$ is a meet-irreducible open set if and only if $F$ is a closed irreducible subspace
Proof
We prove the contrapositive statement:
- $F$ is not a closed irreducible subspace if and only if $U$ is not a meet-irreducible open set
Necessary Condition
Let $F$ not be a closed irreducible subspace.
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.
$\Box$
Sufficient Condition
Let $U$ not be a meet-irreducible open set.
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.
$\blacksquare$