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 $^{\bot}$ as follows: $$A^{\bot}:=X-\overline{A}.$$

Some of the properties of $^\bot$ and regular openness are listed and derived:

1. For any $A\subseteq X$ , $A^{\bot}$ is open. This is obvious.
2. $^\bot$ reverses inclusion. This is also obvious.
3. $\varnothing^{\bot}=X$ and $X^{\bot}=\varnothing$ . This too is clear.
4. $A\cap A^{\bot}=\varnothing$ , because $A\cap A^{\bot}\subseteq A\cap (X-A)=\varnothing$ .
5. $A\cup A^{\bot}$ is dense in $X$ , because $X=\overline{A}\cup A^{\bot} \subseteq \overline{A}\cup \overline{A^\bot} =\overline{A\cup A^\bot}$ .
6. $A^{\bot}\cup B^{\bot}\subseteq(A\cap B)^{\bot}$ . To see this, first note that $A\cap B\subseteq A$ , so that $A^\bot \subseteq (A\cap B)^\bot$ . Similarly, $A^\bot \subseteq (A\cap B)^\bot$ . Take the union of the two inclusions and the result follows.
7. $A^{\bot}\cap B^{\bot}=(A\cup B)^{\bot}$ . This can be verified by direct calculation: $$A^{\bot}\cap B^{\bot}= (X- \overline{A})\cap (X-\overline{B})=X-(\overline{A}\cup \overline{B})=X-\overline{A\cup B}=(A\cup B)^{\bot}.$$
8. $A$ is regular open iff $A=A^{\bot\bot}$ . See the remark at the end of this entry.
9. If $A$ is open, then $A^{\bot}$ is regular open.
   Proof. By the previous property, we want to show that $A^{\bot\bot\bot}=A^\bot$ 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 $^\bot=^{-c}$ , 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 \begin{equation} A\subseteq A^{-c-c}. \end{equation}Since $^\bot=^{-c}$ 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, $(A\cap B)^{\bot\bot}= (A^{\bot\bot}\cap B^{\bot\bot})^{\bot\bot}$ , which is equal to $(A^\bot \cup B^\bot)^{\bot\bot\bot}$ by property 7 above. Since $A^\bot \cup B^\bot$ is open, the last expression becomes $(A^\bot \cup B^\bot)^\bot$ 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 $\overline{\operatorname{int}(X-A)} = \overline{X-\overline{A}}=X-\operatorname{int}(\overline{A})=X-A$ . The converse is proved similarly.

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