Show that if $\operatorname{int}(M\backslash A) = \emptyset$, then $\overline A = M$. Let $(M, d)$ be any metric space.
Question: How do I show that if int($M\backslash A) = \emptyset$, then $\overline A = M$?
If tried to use proof by contradiction, assuming that int$(M\backslash A) \neq \emptyset$, but it didn't get me very far.
Thanks in advance!
 A: Suppose that $m\in\operatorname{int}(M\setminus A)$. That means that there is a $r>0$ such that $B_r(m)\subset M\setminus A$. Therefore, $m\notin\overline A$, since there is a neighborhood of $m$ wich does not intersect $A$.
A: Take $x \in M$. To see that $x \in \overline{A}$ we need to show that for every $r>0$, $B(x,r) \cap A \neq \emptyset$. Suppose this would fail, then there exists $r_0 > 0$ such that $B(x,r_0) \cap A = \emptyset$. But then 
$$B(x,r_0) \subseteq M \setminus A$$
and this $r_0$ would show that $x \in \operatorname{int}(M\setminus A)$ which cannot be. This contradiction shows that indeed $x \in \overline{A}$.
A: Suppose $ (M \setminus A)^{\circ} = \emptyset $. Clearly $ \overline{A} \subset M $ so we just need to show that $ M \subset \overline{A} $. 
Well, take $ x \in M $. If $ x \not\in \overline{A} $ then there is some open set $ U \subset M $ with $ x \in U, \ U \cap A = \emptyset $. But then $ U \subset M \setminus A $ and so $ x \in (M \setminus A)^{\circ} $, a contradiction. Hence $ x \in \overline{A} $ and we are done.
A: Prove the contrapositive.
Suppose $\bar{A}\ne M$ and let $x\in M\setminus \bar{A}$. Then there exists a neighborhood $U$ of $x$ (an open ball, if you wish) such that $U\cap A=\emptyset$, which is equivalent to $U\subseteq M\setminus A$. Therefore $x\in\operatorname{int}(M\setminus A)$.
