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derivation of properties of regular open set∗ CWoo† 2013-03-22 0:52:36 Recall that a subset A of a topological space X is regular open if it is equal to the interior of the closure of itself. To facilitate further analysis of regular open sets, define the operation ⊥ as follows: A⊥ := X − A. Some of the properties of ⊥ and regular openness are listed and derived: 1. For any A ⊆ X, A⊥ is open. This is obvious. 2. ⊥ reverses inclusion. This is also obvious. 3. ∅⊥ = X and X ⊥ = ∅. This too is clear. 4. A ∩ A⊥ = ∅, because A ∩ A⊥ ⊆ A ∩ (X − A) = ∅. 5. A ∪ A⊥ is dense in X, because X = A ∪ A⊥ ⊆ A ∪ A⊥ = A ∪ A⊥ . 6. A⊥ ∪ B ⊥ ⊆ (A ∩ B)⊥ . To see this, first note that A ∩ B ⊆ A, so that A⊥ ⊆ (A ∩ B)⊥ . Similarly, A⊥ ⊆ (A ∩ B)⊥ . Take the union of the two inclusions and the result follows. 7. A⊥ ∩ B ⊥ = (A ∪ B)⊥ . This can be verified by direct calculation: A⊥ ∩ B ⊥ = (X − A) ∩ (X − B) = X − (A ∪ B) = X − A ∪ B = (A ∪ B)⊥ . 8. A is regular open iff A = A⊥⊥ . See the remark at the end of this entry. 9. If A is open, then A⊥ is regular open. ∗ hDerivationOfPropertiesOfRegularOpenSeti created: h2013-03-2i by: hCWooi version: h40501i Privacy setting: h1i hDerivationi h06E99i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 Proof. By the previous property, we want to show that A⊥⊥⊥ = A⊥ if A is open. For notational convenience, let us write A− for the closure of A and Ac for the complement of A. As ⊥ =−c , the equation now becomes A−c−c−c = A−c for any open set A. Since A ⊆ A− for any set, A−c ⊆ Ac . This means A−c− ⊆ Ac− . Since A is open, Ac is closed, so that Ac− = Ac . The last inclusion becomes A−c− ⊆ Ac . Taking complement again, we have A ⊆ A−c−c . (1) Since ⊥ =−c reverses inclusion, we have A−c−c−c ⊆ A−c , which is one of the inclusions. On the other hand, the inclusion (1) above applies to any open set, and because A−c is open, A−c ⊆ A−c−c−c , which is the other inclusion. 10. If A and B are regular open, then so is A ∩ B. Proof. Since A, B are regular open, (A ∩ B)⊥⊥ = (A⊥⊥ ∩ B ⊥⊥ )⊥⊥ , which is equal to (A⊥ ∪B ⊥ )⊥⊥⊥ by property 7 above. Since A⊥ ∪B ⊥ is open, the last expression becomes (A⊥ ∪ B ⊥ )⊥ by property 9, or A ∩ B by property 7 again. Remark. All of the properties above can be dualized for regular closed sets. If fact, proving a property about regular closedness can be easily accomplished once we have the following: (∗) A is regular open iff X − A is regular closed. Proof. Suppose first that A is regular open. Then int(X − A) = X − A = X − int(A) = X − A. The converse is proved similarly. As a corollary, for example, we have: if A is closed, then X − A is regular closed. Proof. If A is closed, then X − A is open, so that (X − A)⊥ = X − X − A is regular open by property 9 above, which implies that X − (X − A)⊥ = X − A is regular closed by (∗). 2