derivation of properties of regular open set

Recall that a subset A of a topological spaceMathworldPlanetmath X is regular open if it is equal to the interior of the closureMathworldPlanetmathPlanetmath of itself.

To facilitate further analysis of regular open sets, define the operationMathworldPlanetmath as follows:


Some of the properties of and regularPlanetmathPlanetmathPlanetmathPlanetmath openness are listed and derived:

  1. 1.

    For any AX, A is open. This is obvious.

  2. 2.

    reverses inclusion. This is also obvious.

  3. 3.

    =X and X=. This too is clear.

  4. 4.

    AA=, because AAA(X-A)=.

  5. 5.

    AA is dense in X, because X=A¯AA¯A¯=AA¯.

  6. 6.

    AB(AB). To see this, first note that ABA, so that A(AB). Similarly, A(AB). Take the union of the two inclusions and the result follows.

  7. 7.

    AB=(AB). This can be verified by direct calculation:

  8. 8.

    A is regular open iff A=A. See the remark at the end of this entry (

  9. 9.

    If A is open, then A is regular open.


    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 complementPlanetmathPlanetmath of A. As =-c, the equation now becomes A-c-c-c=A-c for any open set A.

    Since AA- for any set, A-cAc. 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

    AA-c-c. (1)

    Since =-c reverses inclusion, we have A-c-c-cA-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-cA-c-c-c, which is the other inclusion. ∎

  10. 10.

    If A and B are regular open, then so is AB.


    Since A,B are regular open, (AB)=(AB), which is equal to (AB) by property 7 above. Since AB is open, the last expression becomes (AB) by property 9, or AB 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.


Suppose first that A is regular open. Then int(X-A)¯=X-A¯¯=X-int(A¯)=X-A. The converseMathworldPlanetmath is proved similarly. ∎

As a corollary, for example, we have: if A is closed, then X-A¯ is regular closed.


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