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

1. $\int(A)\subseteq A$ .
   Proof. If $a\in \int(A)$ , then $a\in U$ for some open set $U\subseteq A$ . So $a\in A$ .
2. $\int(A)$ is open.
   Proof. Since $\int(A)$ is a union of open sets, $\int(A)$ is open.
3. $\int(A)$ is the largest open set contained in $A$ .
   Proof. If $U$ is open set with $\int(A)\subseteq U\subseteq A$ , then $U\subseteq\bigcup \lbrace V\subseteq A\mid V\mbox{ open }\rbrace = \int(A)$ , so $U=\int(A)$ .
4. $A$ is open if and only if $A=\int(A)$ .
   Proof. If $A$ is open, then $A$ is the largest open set contained in $A$ , and so $\int(A)=A$ by property 3 above. On the other hand, if $\int(A)=A$ , then $A$ is open, since $\int(A)$ is, by property 2 above.
5. $\int(\int(A))=\int(A)$ .
   Proof. Since $\int(A)$ is open by property 2, $\int(A)=\int(\int(A))$ by property 4.
6. $\int(X)=X$ and $\int(\emptyset)=\emptyset$ .
   Proof. This is so because both $X$ and $\emptyset$ are open sets.
7. $\overline{A^\complement}=(\int(A))^\complement$ .
   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, $a\in (\int(A))^\complement$ , for $(\int(A))^\complement$ is the complement of an open set by property 2, and $A^\complement \subseteq (\int(A))^\complement$ by taking the complement of property 1.

   (RHS $\subseteq$ LHS). If $a\in (\int(A))^\complement$ , then $a\notin \int(A)$ . If $B$ is a closed set such that $A^\complement \subseteq B$ , then $B^\complement \subseteq A$ . Since $B^\complement$ is open, $B^\complement \subseteq \int(A)$ 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. $\overline{A}^\complement = \int(A^\complement)$ .
   Proof. Set $B=A^\complement$ , and apply property 7. So $\overline{A}^\complement = \overline{B^\complement}^\complement = (\int(B))^{\complement\complement}=\int(B)=\int(A^\complement)$ .
9. $A\subseteq B$ implies that $\int(A)\subseteq \int(B)$ .
   Proof. This is so because $\int(A)$ is open (property 2), contained in $A$ (and therefore contained in $B$ ), so contained in $\int(B)$ , as $\int(B)$ is the largest open set contained in $B$ (property 3).
10. $\int(A)=A\setminus \partial A$ , where $\partial A$ is the boundary of $A$ .
    Proof. Recall that $\partial A=\overline{A}\cap \overline{A^\complement}$ . So $\partial A = \overline{A}\cap (\int(A))^\complement$ by property 7. By direct computation, we have $A\setminus \partial A = A \setminus (\overline{A}\cap (\int(A))^\complement) = (A\setminus \overline{A})\cup (A\setminus (\int(A))^\complement)$ . Since $A\setminus \overline{A}=\varnothing$ and $A\setminus (\int(A))^\complement = A\cap (\int(A))^{\complement\complement}= A\cap \int(A)$ , which is $\int(A)$ by property 2.
11. $\overline{A} = \int(A)\cup \partial A$ .
    Proof. Again, by direct computation:

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12. $X=\int(A)\cup \partial A \cup \int(A^\complement)$ .
    Proof. By property 11, $\int(A)\cup \partial A \cup \int(A^\complement) = \overline{A} \cup \int(A^\complement)$ , which, by property 8, is $\overline{A} \cup \overline{A}^\complement$ , and the last expression is just $X$ .
13. $\int(A\cap B)=\int(A)\cap \int(B)$ .
    Proof. (LHS $\subseteq$ RHS). Let $C=\int(A\cap B)$ . Since $C$ is open and contained in both $A$ and $B$ , $C$ is contained in both $\int(A)$ and $\int(B)$ , since $\int(A)$ and $\int(B)$ are the largest open sets in $A$ and $B$ respectively. (RHS $\subseteq$ LHS). Let $D=\int(A)\cap \int(B)$ . 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 $\int(A\cap B)$ , 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: $$\int(A)=\overline{A^\complement}^\complement.$$