# derivation of properties on interior operation

Let $X$ be a topological space and $A$ a subset of $X$. Then

1. 1.

$\operatorname{int}(A)\subseteq A$.

###### Proof.

If $a\in\operatorname{int}(A)$, then $a\in U$ for some open set $U\subseteq A$. So $a\in A$. ∎

2. 2.

$\operatorname{int}(A)$ is open.

###### Proof.

Since $\operatorname{int}(A)$ is a union of open sets, $\operatorname{int}(A)$ is open. ∎

3. 3.

$\operatorname{int}(A)$ is the largest open set contained in $A$.

###### Proof.

If $U$ is open set with $\operatorname{int}(A)\subseteq U\subseteq A$, then $U\subseteq\bigcup\{V\subseteq A\mid V\mbox{ open }\}=\operatorname{int}(A)$, so $U=\operatorname{int}(A)$. ∎

4. 4.

$A$ is open if and only if $A=\operatorname{int}(A)$.

###### Proof.

If $A$ is open, then $A$ is the largest open set contained in $A$, and so $\operatorname{int}(A)=A$ by property 3 above. On the other hand, if $\operatorname{int}(A)=A$, then $A$ is open, since $\operatorname{int}(A)$ is, by property 2 above. ∎

5. 5.

$\operatorname{int}(\operatorname{int}(A))=\operatorname{int}(A)$.

###### Proof.

Since $\operatorname{int}(A)$ is open by property 2, $\operatorname{int}(A)=\operatorname{int}(\operatorname{int}(A))$ by property 4. ∎

6. 6.

$\operatorname{int}(X)=X$ and $\operatorname{int}(\varnothing)=\varnothing$.

###### Proof.

This is so because both $X$ and $\varnothing$ are open sets. ∎

7. 7.

$\overline{A^{\complement}}=(\operatorname{int}(A))^{\complement}$.

###### Proof.

(LHS $\subseteq$ RHS). If $a\in\overline{A^{\complement}}$, then $a\in B$ for every closed set $B$ such that $A^{\complement}\subseteq B$. In particular, $a\in(\operatorname{int}(A))^{\complement}$, for $(\operatorname{int}(A))^{\complement}$ is the complement of an open set by property 2, and $A^{\complement}\subseteq(\operatorname{int}(A))^{\complement}$ by taking the complement of property 1.

(RHS $\subseteq$ LHS). If $a\in(\operatorname{int}(A))^{\complement}$, then $a\notin\operatorname{int}(A)$. If $B$ is a closed set such that $A^{\complement}\subseteq B$, then $B^{\complement}\subseteq A$. Since $B^{\complement}$ is open, $B^{\complement}\subseteq\operatorname{int}(A)$ by property 3, so $a\notin B^{\complement}$, and thus $a\in B$. Since $B$ is arbitrary, $a\in\overline{A^{\complement}}$ as desired. ∎

8. 8.

$\overline{A}^{\complement}=\operatorname{int}(A^{\complement})$.

###### Proof.

Set $B=A^{\complement}$, and apply property 7. So $\overline{A}^{\complement}=\overline{B^{\complement}}^{\complement}=(% \operatorname{int}(B))^{\complement\complement}=\operatorname{int}(B)=% \operatorname{int}(A^{\complement})$. ∎

9. 9.

$A\subseteq B$ implies that $\operatorname{int}(A)\subseteq\operatorname{int}(B)$.

###### Proof.

This is so because $\operatorname{int}(A)$ is open (property 2), contained in $A$ (and therefore contained in $B$), so contained in $\operatorname{int}(B)$, as $\operatorname{int}(B)$ is the largest open set contained in $B$ (property 3). ∎

10. 10.

$\operatorname{int}(A)=A\setminus\partial A$, where $\partial A$ is the boundary of $A$.

###### Proof.

Recall that $\partial A=\overline{A}\cap\overline{A^{\complement}}$. So $\partial A=\overline{A}\cap(\operatorname{int}(A))^{\complement}$ by property 7. By direct computation, we have $A\setminus\partial A=A\setminus(\overline{A}\cap(\operatorname{int}(A))^{% \complement})=(A\setminus\overline{A})\cup(A\setminus(\operatorname{int}(A))^{% \complement})$. Since $A\setminus\overline{A}=\varnothing$ and $A\setminus(\operatorname{int}(A))^{\complement}=A\cap(\operatorname{int}(A))^{% \complement\complement}=A\cap\operatorname{int}(A)$, which is $\operatorname{int}(A)$ by property 2. ∎

11. 11.

$\overline{A}=\operatorname{int}(A)\cup\partial A$.

###### Proof.

Again, by direct computation:

 $\displaystyle\operatorname{int}(A)\cup\partial A$ $\displaystyle=\operatorname{int}(A)\cup(\overline{A}\cap(\operatorname{int}(A)% )^{\complement})$ $\displaystyle\qquad\mbox{because }\partial A=\overline{A}\cap(\operatorname{% int}(A))^{\complement}$ $\displaystyle=(\operatorname{int}(A)\cup\overline{A})\cap(\operatorname{int}(A% )\cup(\operatorname{int}(A))^{\complement})$ $\displaystyle\qquad\cap\mbox{ distributes over }\cup$ $\displaystyle=\overline{A}\cap X=\overline{A}.$ $\displaystyle\qquad\operatorname{int}(A)\subseteq A\subseteq\overline{A}$

12. 12.

$X=\operatorname{int}(A)\cup\partial A\cup\operatorname{int}(A^{\complement})$.

###### Proof.

By property 11, $\operatorname{int}(A)\cup\partial A\cup\operatorname{int}(A^{\complement})=% \overline{A}\cup\operatorname{int}(A^{\complement})$, which, by property 8, is $\overline{A}\cup\overline{A}^{\complement}$, and the last expression is just $X$. ∎

13. 13.

$\operatorname{int}(A\cap B)=\operatorname{int}(A)\cap\operatorname{int}(B)$.

###### Proof.

(LHS $\subseteq$ RHS). Let $C=\operatorname{int}(A\cap B)$. Since $C$ is open and contained in both $A$ and $B$, $C$ is contained in both $\operatorname{int}(A)$ and $\operatorname{int}(B)$, since $\operatorname{int}(A)$ and $\operatorname{int}(B)$ are the largest open sets in $A$ and $B$ respectively. (RHS $\subseteq$ LHS). Let $D=\operatorname{int}(A)\cap\operatorname{int}(B)$. So $D$ is open and is a subset of both $A$ and $B$, hence a subset of $A\cap B$, and therefore a subset of $\operatorname{int}(A\cap B)$, since it is the largest open set contained in $A\cap B$. ∎

Remark. Using property 7, we see that an alternative definition of interior can be given:

 $\operatorname{int}(A)=\overline{A^{\complement}}^{\complement}.$
Title derivation of properties on interior operation DerivationOfPropertiesOnInteriorOperation 2013-03-22 17:55:28 2013-03-22 17:55:28 CWoo (3771) CWoo (3771) 9 CWoo (3771) Derivation msc 54-00