TITLE: Prove that the closure of complement, is the complement of the interior
QUESTION [19 upvotes]: Let $(X,d)$ a metric space and $A\subset X$. Prove that $$\overline{(A^c)}=\overset{\circ}{(A)}^c$$
i.e. the closure of complement, is  the complement of the interior.
Proof. If $x\in \overline{(A^c)}$ then $x\in A^c$ or $x\in (A^c)'$. Since $\overset{\circ}{A} \subset A \Rightarrow A^c \subset \overset{\circ}{(A)}^c$ so, if $x\in A^c$ then $x\in \overset{\circ}{(A)}^c$.
Now if $x \not\in A^c$ but $x\in (A^c)'$ then $\forall r>0, B(x;r)-\{x\}\cap A^c \not= \emptyset.$ And now I've to deduce that $x\not\in \overset{\circ}{A}$ so $x\in \overset{\circ}{(A)}^c$, but I can't figure out how.

REPLY [5 votes]: Here is a nice 'natural language proof', based on the facts that the interior of a set is the largest open set contained in it, and the closure of a set is the smallest closed set that contains it. By the way, this works for any topological space :
First, we note that $A^0$ is an open set contained in $A$, so $X\setminus A^0$ is a closed set that contains $X\setminus A$. So we know that $X\setminus A^0\supseteq \overline{X\setminus A}$. Now, assume that $X\setminus A^0$ is not the smallest closed set containing $X\setminus A$. Then, we would have some closed set $S$ such that $X\setminus A\subseteq S\subset X\setminus A^0$, which implies $A^0\subset X\setminus S\subseteq A$. Since $X\setminus S$ is open, this means that $A^0$ is not the smallest open set containing $A$, which is a contradiction.