derivation of properties of regular open set
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.
Some of the properties of ⊥ and regular openness are listed and derived:
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1.
For any A⊆X, A⊥ is open. This is obvious.
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2.
⊥ reverses inclusion. This is also obvious.
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3.
∅⊥=X and X⊥=∅. This too is clear.
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4.
A∩A⊥=∅, because A∩A⊥⊆A∩(X-A)=∅.
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5.
A∪A⊥ is dense in X, because X=ˉA∪A⊥⊆ˉA∪¯A⊥=¯A∪A⊥.
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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.
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7.
A⊥∩B⊥=(A∪B)⊥. This can be verified by direct calculation:
A⊥∩B⊥=(X-ˉA)∩(X-ˉB)=X-(ˉA∪ˉB)=X-¯A∪B=(A∪B)⊥. -
8.
A is regular open iff A=A⊥⊥. See the remark at the end of this entry (http://planetmath.org/DerivationOfPropertiesOnInteriorOperation).
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9.
If A is open, then A⊥ is regular open.
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. ∎
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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-ˉA=X-int(ˉA)=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 (*). ∎
Title | derivation of properties of regular open set |
---|---|
Canonical name | DerivationOfPropertiesOfRegularOpenSet |
Date of creation | 2013-03-22 17:59:24 |
Last modified on | 2013-03-22 17:59:24 |
Owner | CWoo (3771) |
Last modified by | CWoo (3771) |
Numerical id | 6 |
Author | CWoo (3771) |
Entry type | Derivation |
Classification | msc 06E99 |