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) $

  • $\begingroup$ How do you know $Y^\circ=Y$? Do you know $Y$ is open? $\endgroup$ Feb 8, 2016 at 21:02
  • $\begingroup$ @GregoryGrant No it is not mentioned, thanks for pointing. $\endgroup$
    – user189013
    Feb 8, 2016 at 21:03
  • $\begingroup$ 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 $X\setminus Y^{\circ}$. $\endgroup$ Feb 8, 2016 at 21:05
  • $\begingroup$ @GregoryGrant How Can I show the other inclusion? $\endgroup$
    – user189013
    Feb 8, 2016 at 21:05
  • $\begingroup$ Good question, I'm thinking about it. $\endgroup$ Feb 8, 2016 at 21:05

1 Answer 1


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 $X \setminus Y$).


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