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.
Best Answer
If you use the topological definitions of closure and interior, it's very easy and natural: $$\overline A=\bigcap\,\{F\supseteq A\mid F\text{ is closed in }X\}$$ $$A^\circ=\bigcup\,\{G\subseteq A\mid G\text{ is open in }X\}$$ Now just use the rules for complements which turn the complement of a union into an intersection of the complements.