Set interiors and closures in topological space Let $(X,\mathcal{T})$ be a topological space and let $S\subset X$. Show that $X\setminus \overset{\circ}{S}=\overline{X\setminus S}$.
Also show that $X\setminus\bar{S}=(X\setminus S)^{\circ}$
I want to show these two facts in topological terms rather than metric like I am most used to. Can someone help me at least get started here?
 A: In general, $A^°$ is the largest open set contained in $A$ and $\overline{A}$ is the smallest closed set containing $A$.
Now, $X\setminus S^°$ is a closed set (complement of open set) and clearly $X \setminus S \subseteq X \setminus S^°$. Thus by the remark above we get $\overline{X\setminus S} \subseteq X \setminus S^°$.
On the other hand, $X\setminus(\overline{X\setminus S})$ is an open set contained in $S$, so we get $X \setminus (\overline{X\setminus S}) \subseteq S^°$. Taking complements, 
$$ X \setminus S^°\subseteq\overline{X \setminus S}$$
and we have proven the desired equality.
A: $X \setminus \overset{\circ}{S}$ is closed (as $\overset{\circ}{S}$ is open by definition) and as $\overset{\circ}{S} \subseteq S$ we also have that 
$$ X \setminus S \subseteq X \setminus \overset{\circ}{S}$$ and as $\overline{X \setminus S}$ is by definition the smallest closed set containing $X \setminus S$ we have 
$$\overline{X \setminus S} \subseteq X \setminus \overset{\circ}{S}$$
On the other hand, $X \setminus \overline{X \setminus S}$ is open, as $\overline{X \setminus S}$ is closed and 
$$X \setminus \overline{X \setminus S} \subseteq X \setminus (X \setminus S)=S$$
and as $\overset{\circ}{S}$ is the largest open set contained in $S$, we have 
$$X \setminus \overline{X \setminus S} \subseteq  \overset{\circ}{S}$$ which implies
$$X\setminus \overset{\circ}{S} \subseteq \overline{X \setminus S}$$ and the two inclusions finish the job.
