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 $A\cap {A}^{\perp}\subseteq A\cap (XA)=\mathrm{\varnothing}$.

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 ${A}^{ccc}={A}^{c}$ 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
$$A\subseteq {A}^{cc}.$$ (1) Since ${}^{\perp}=^{c}$ reverses inclusion, we have ${A}^{ccc}\subseteq {A}^{c}$, 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, ${A}^{c}\subseteq {A}^{ccc}$, 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 $XA$ is regular closed.
Proof.
Suppose first that $A$ is regular open. Then $\overline{\mathrm{int}(XA)}=\overline{X\overline{A}}=X\mathrm{int}(\overline{A})=XA$. The converse^{} is proved similarly. ∎
As a corollary, for example, we have: if $A$ is closed, then $\overline{XA}$ is regular closed.
Proof.
If $A$ is closed, then $XA$ is open, so that ${(XA)}^{\perp}=X\overline{XA}$ is regular open by property 9 above, which implies that $X{(XA)}^{\perp}=\overline{XA}$ is regular closed by $(*)$. ∎
Title  derivation of properties of regular open set 

Canonical name  DerivationOfPropertiesOfRegularOpenSet 
Date of creation  20130322 17:59:24 
Last modified on  20130322 17:59:24 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  6 
Author  CWoo (3771) 
Entry type  Derivation 
Classification  msc 06E99 