# 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 ${}^{\bot}$ as follows:

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

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

1. 1.

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

2. 2.

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

3. 3.

$\varnothing^{\bot}=X$ and $X^{\bot}=\varnothing$. This too is clear.

4. 4.

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

5. 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. 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. 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. 8.

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

9. 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

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

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. 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. ∎

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

Title derivation of properties of regular open set DerivationOfPropertiesOfRegularOpenSet 2013-03-22 17:59:24 2013-03-22 17:59:24 CWoo (3771) CWoo (3771) 6 CWoo (3771) Derivation msc 06E99