# 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 $X \setminus Y ^ \circ \subset cl ( X \setminus Y)$

• How do you know $Y^\circ=Y$? Do you know $Y$ is open? Feb 8, 2016 at 21:02
• @GregoryGrant No it is not mentioned, thanks for pointing. Feb 8, 2016 at 21:03
• Anyway you don't need $Y$ is open for your argument, you just need $Y^{\circ}$ is open. You have shown the one inclusion correctly, to answer your first question. You might point out though, for clarity, that $cl(X\setminus Y)$ is contained in $X\setminus Y^{\circ}$. Feb 8, 2016 at 21:05
• @GregoryGrant How Can I show the other inclusion? Feb 8, 2016 at 21:05
• Good question, I'm thinking about it. Feb 8, 2016 at 21:05

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 $X \setminus Y$).