Show $A^o=A\setminus\partial A$.
Attempt:
I need to show inclusion on both sides, so:
(a) $A^o\subseteq A\setminus\partial A$
(b) $A\setminus\partial A \subseteq A^o$
Attempt at (a):
If $x\in A^o$ then $x\in A$ by definition. So we need to show $x\not\in\partial A$. Well, $A^o$ is open so $\exists\epsilon>0$ : $B_{\epsilon}(x)\subseteq A^o \subseteq A$. For this $\epsilon$, $B_{\epsilon}(x)\cap A^c=\emptyset$ so $x\not\in\partial A$.
Attempt at (b): This is trickier. Let $x\in A$ : $x\not\in\partial A$. That means $\exists\epsilon>0$ : $B_{\epsilon}(x)\cap A^c=\emptyset$. Does this imply $x\in\ A^o$?