Prove some identities involving interior, closure in topology I want to prove the following statements:
(i) $X \setminus Y ^ \circ $ = $cl ( X \setminus Y) $
I wrote down that $Y ^\circ$ is open so $Y ^\circ = Y $.  Therefore  $X \setminus Y ^ \circ $ is closed so is containing $cl ( X \setminus Y) $.  Is that the correct way? How can I show the other way that $X \setminus Y ^ \circ $ belongs to  $cl ( X \setminus Y) $
After some work I have found the other conclusion but I do not know if it is the correct way:
$cl(X \setminus Y)$ is closed so $(X \setminus cl(X \setminus Y)$ is open.
$X \setminus cl(X \setminus Y) \subset X \setminus X \setminus Y = Y$ Therefore $X \setminus cl(X \setminus Y) \subset Y$ ( which contains $Y ^ \circ$) therefore $ Y ^\circ \supset X \setminus cl(X \setminus Y) $ and hence $X \setminus Y ^ \circ  \subset cl ( X \setminus Y) $

If $X \setminus Y \subseteq C$ where $C$ is closed, then $X \setminus C \subseteq Y$. The set $X \setminus C$ is open and a subset of $Y$ so  $X \setminus C \subseteq Y^\circ$, as the interior is the largest open subset of $Y$. Taking complements we see that $X \setminus Y^\circ \subseteq C$. As $C$ is an arbitrary closed set containing $X \setminus Y$, by minimality of the closure we have that $\operatorname{cl}(X \setminus Y) = X \setminus Y^\circ$, as the latter set is closed and is minimally so (among the closed sets containing $X \setminus Y$).
