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.

To facilitate further analysis of regular open sets, define the operation ${}^{\perp }$ as follows:

$$A^{\perp }:=X-\overline{A}.$$

Some of the properties of ${}^{\perp }$ and regular openness are listed and derived:

1. For any $A\subseteq X$, $A^{\perp }$ is open. This is obvious.

2. ${}^{\perp }$ reverses inclusion. This is also obvious.

3. ${\mathrm{\varnothing }}^{\perp }=X$ and $X^{\perp }=\mathrm{\varnothing }$. This too is clear.

4. $A\cap A^{\perp }=\mathrm{\varnothing }$, because $A\cap A^{\perp }\subseteq A\cap (X-A)=\mathrm{\varnothing }$.

5. $A\cup A^{\perp }$ is dense in $X$, because $X=\overline{A}\cup A^{\perp }\subseteq \overline{A}\cup \overline{A^{\perp }}=\overline{A\cup A^{\perp }}$.

6. $A^{\perp }\cup B^{\perp }\subseteq {(A\cap B)}^{\perp }$. To see this, first note that $A\cap B\subseteq A$, so that $A^{\perp }\subseteq {(A\cap B)}^{\perp }$. Similarly, $A^{\perp }\subseteq {(A\cap B)}^{\perp }$. Take the union of the two inclusions and the result follows.

7. $A^{\perp }\cap B^{\perp }={(A\cup B)}^{\perp }$. This can be verified by direct calculation:

   $$A^{\perp }\cap B^{\perp }=(X-\overline{A})\cap (X-\overline{B})=X-(\overline{A}\cup \overline{B})=X-\overline{A\cup B}={(A\cup B)}^{\perp }.$$

8. $A$ is regular open iff $A=A^{\perp \perp }$. See the remark at the end of this entry (http://planetmath.org/DerivationOfPropertiesOnInteriorOperation).

9. If $A$ is open, then $A^{\perp }$ is regular open.

   Proof.

   By the previous property, we want to show that $A^{\perp \perp \perp }=A^{\perp }$ if $A$ is open. For notational convenience, let us write $A^{-}$ for the closure of $A$ and $A^c$ for the complement of $A$. As ${}^{\perp }=^{-c}$, the equation now becomes $A^{-c-c-c}=A^{-c}$ for any open set $A$.

   Since $A\subseteq A^{-}$ for any set, $A^{-c}\subseteq A^c$. This means $A^{-c-}\subseteq A^{c-}$. Since $A$ is open, $A^c$ is closed, so that $A^{c-}=A^c$. The last inclusion becomes $A^{-c-}\subseteq A^c$. Taking complement again, we have

   $$A\subseteq A^{-c-c}.$$ (1)

   Since ${}^{\perp }=^{-c}$ reverses inclusion, we have $A^{-c-c-c}\subseteq A^{-c}$, 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, $A^{-c}\subseteq A^{-c-c-c}$, which is the other inclusion. ∎

10. If $A$ and $B$ are regular open, then so is $A\cap B$.

    Proof.

    Since $A,B$ are regular open, ${(A\cap B)}^{\perp \perp }={(A^{\perp \perp }\cap B^{\perp \perp })}^{\perp \perp }$, which is equal to ${(A^{\perp }\cup B^{\perp })}^{\perp \perp \perp }$ by property 7 above. Since $A^{\perp }\cup B^{\perp }$ is open, the last expression becomes ${(A^{\perp }\cup B^{\perp })}^{\perp }$ by property 9, or $A\cap B$ 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 $\overline{\mathrm{int}(X-A)}=\overline{X-\overline{A}}=X-\mathrm{int}(\overline{A})=X-A$. The converse is proved similarly. ∎

As a corollary, for example, we have: if $A$ is closed, then $\overline{X-A}$ is regular closed.

Proof.

If $A$ is closed, then $X-A$ is open, so that ${(X-A)}^{\perp }=X-\overline{X-A}$ is regular open by property 9 above, which implies that $X-{(X-A)}^{\perp }=\overline{X-A}$ 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