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Homederivation of properties of regular open set

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# 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. 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

$A\subseteq A^{{-c-c}}.$ (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. 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 $(*)$. ∎

## Mathematics Subject Classification

06E99*no label found*

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