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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.
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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 (∗).
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