derivation of properties on interior operation

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

1. $\mathrm{int}(A)\subseteq A$.

   Proof.

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

2. $\mathrm{int}(A)$ is open.

   Proof.

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

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

   Proof.

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

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

   Proof.

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

5. $\mathrm{int}(\mathrm{int}(A))=\mathrm{int}(A)$.

   Proof.

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

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

   Proof.

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

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

   Proof.

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

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

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

   Proof.

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

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

   Proof.

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

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

    Proof.

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

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

    Proof.

    Again, by direct computation:

    $\mathrm{int}(A)\cup \partial A$$=\mathrm{int}(A)\cup (\overline{A}\cap {(\mathrm{int}(A))}^{\mathrm{\complement }})$$\mathrm{\hspace{1em}\hspace{1em}}\text{because }\partial A=\overline{A}\cap {(\mathrm{int}(A))}^{\mathrm{\complement }}$

    $=(\mathrm{int}(A)\cup \overline{A})\cap (\mathrm{int}(A)\cup {(\mathrm{int}(A))}^{\mathrm{\complement }})$$\mathrm{\hspace{1em}\hspace{1em}}\cap \text{ distributes over }\cup$

    $=\overline{A}\cap X=\overline{A}.$$\mathrm{\hspace{1em}\hspace{1em}}\mathrm{int}(A)\subseteq A\subseteq \overline{A}$

    ∎

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

    Proof.

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

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

    Proof.

    (LHS $\subseteq$ RHS). Let $C=\mathrm{int}(A\cap B)$. Since $C$ is open and contained in both $A$ and $B$, $C$ is contained in both $\mathrm{int}(A)$ and $\mathrm{int}(B)$, since $\mathrm{int}(A)$ and $\mathrm{int}(B)$ are the largest open sets in $A$ and $B$ respectively. (RHS $\subseteq$ LHS). Let $D=\mathrm{int}(A)\cap \mathrm{int}(B)$. 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 $\mathrm{int}(A\cap B)$, 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:

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