A property of $\partial(A)$

Question

$\textbf{Problem:}$ Let $(X,\tau)$ a topological space and $A\subseteq X$. Prove that : $$ \overline{int(\partial(A))} = \overline{int(\partial(A)) \cap A} $$

$\textbf{My proof:}$

$(\subseteq)$ Let $x \in int(\partial(A))$ so existe a open set $U$ such that $x\in U \subseteq \partial(A)$. Let´s consider a open set $V$ such that $x \in V$ and put $W=U\cap V$. So $W\subseteq \partial(A)$ how $x \in W$ we have to :

$$ \exists \hspace{0.7mm} y \in W\cap A \implies y \in W \subseteq \partial(A) $$

Then $y \in int(\partial(A)) \cap A \implies W\cap (int(\partial(A))\cap A) \neq \emptyset $ and :

$$ \implies V \cap (int(\partial(A))\cap A) \neq \emptyset $$

So $ x\in \overline{int(\partial(A))\cap A}$ and $int(\partial(A))\subseteq \overline{int(\partial(A))\cap A} \implies \overline{int(\partial(A))} \subseteq \overline{int(\partial(A))\cap A}$

$(\supseteq)$ How $int(\partial(A))\cap A \subseteq int(\partial(A)) \implies \overline{int(\partial(A))\cap A} \subseteq \overline{int(\partial(A))} $

Answer 1

The reverse is trivial: $A \subseteq B \implies \overline{A} \subseteq \overline{B}$ always holds. (If $x \in \overline{A}$ and $U$ is an open set containing $x$, $U \cap A \neq \emptyset$ and so also $U \cap B \neq \emptyset$, so indeed $x \in \overline{B}$)

Answer 2

Assume x in int $\partial$A.
Let U be any open nhood of x. x in open U $\cap$ int $\partial$A.
As x in $\bar A,$ exists y in U $\cap$ int $\partial$A $\cap$ A.
Thus x in closure of int $\partial$A $\cap$ A.
Whence int $\partial$A subset $\overline {int. \partial A \cap A.}$

The rest of this proof for the diligent student, is simple.