How to prove $X \setminus A^\circ=\overline{X \setminus A}$ I am trying to proof $X \setminus A^\circ=\overline{X \setminus A}$
and $X\setminus \overline{A}=(X \setminus A)^\circ$
As hint I got the following equivalence which tried to use
(A open and A $\subseteq$ B) $\Leftrightarrow$ ($X \setminus A$ is closed and $X \setminus B \subseteq X \setminus A$)
as follows:
$A^\circ$ is open and $A^\circ \subseteq A$ , so $X \setminus A^\circ$ is closed and
$X \setminus A \subseteq X \setminus A^\circ$.
Taking the closure of $X \setminus A \subseteq X \setminus A^\circ$ gives $\overline{ X \setminus A} \subseteq \overline{ X \setminus A^\circ}=X \setminus A\circ$
This is how far I got. I can't figure out the other direction of the inclusion.
I tried to take a similar version of the equivalence that states
(A is closed and $B \subseteq A$) $\Leftrightarrow$ ($X \setminus A $ is open and $X \setminus A \subseteq X\setminus B$)
My second idea is to use de Morgans law to proof the equality:
$X \setminus A^\circ =\overline{X \setminus A}$ :
$$A^\circ:=\bigcup_{G \subseteq A, G \text{ open}} G$$
$$X \setminus A^\circ=\bigcap_{G \subseteq A, G\text{ open}}(X/G) = \bigcap_{G \subseteq A, (X \setminus G) \text{ closed}}(X/G)=\bigcap_{(X\setminus A )\subseteq (X \setminus G), (X \setminus G) \text{ closed}}(X/G)=\overline{X \setminus A}$$
I am not sure if how I changed the conditions under the cap/cup are legit. Would be nice if someone could tell me this.
 A: I'm assuming that your topological space is $X$.  Here's a pretty strong sketch, but requires the details to be proved.
Observe first that $A^\circ$ is open and $A^\circ\subseteq A$.  Then, $X\setminus A^\circ=(A^\circ)^c$ is closed as it is the complement of an open set.  In addition, $X\setminus A\subseteq X\setminus A^\circ$.  Therefore, $X\setminus A^\circ$ is a closed set containing $X\setminus A$, so it contains the closure of $X\setminus A$.  Hence $\overline{X\setminus A}\subseteq X\setminus A^\circ$.
On the other hand, suppose that $C$ is a closed set containing $X\setminus A$.  Then $X\setminus C\subseteq X\setminus (X\setminus A)=A$  Since $C$ is closed, $X\setminus C$ is open and $X\setminus C$ is an open set contained in $A$.  Therefore, $X\setminus C\subseteq A^\circ$.  Thus $C=X\setminus(X\setminus C)\supseteq X\setminus A^\circ$.  Since $\overline{X\setminus A}$ is the intersection of all closed sets containing $X\setminus A$ and each of these sets contains $X\setminus A^\circ$, it follows that $\overline{X\setminus A}\supseteq X\setminus A^\circ$.
