$int(E)=\bar{E}\setminus\partial{E}$ [homework] Can you help me to figure how can I demonstrate the following : 
$$int(E)=\bar{E}\setminus\partial{E}$$
I don't know how to start.
 A: A point $x$ in $\operatorname{int}(E)$ has a neighbourhood $U_x$ such that $x \in U_x \subseteq E$. In particular, $x \in E \subseteq \overline{E}$. But $U_x$ also witnesses that $x \notin \partial E$ (why?). 
And reversely, if $x \in \overline{E}\setminus \partial E$, every neighbourhood of $x$ intersects $E$, but at least some neighbourhood $U_x$ cannot intersect $E^c$ as well (or it would be in the boundary of $E$). So $U_x$ must be a subset of $E$ and so $x \in \operatorname{int}(E)$.
A: $x \in{ \bar{E} \setminus \partial{E} } $ if
$$ \exists \delta(x) > 0 \text{ such as } \left] x-\delta, x+\delta \right[ \subset{E}  
\\ \land{} \lnot \big(\forall \delta(x) > 0 \text{ such as } \left] x-\delta, x+\delta \right[ \cap{E} \land \forall \delta(x) > 0 \text{ such as } \left] x-\delta, x+\delta \right[ \cap{E^C} \big) 
\\ \equiv 
\\ \exists \delta(x) > 0 \text{ such as } \left] x-\delta, x+\delta \right[ \subset{E}  
\\ \land{} \big(\exists\delta(x) > 0 \text{ such as } \left] x-\delta, x+\delta \right[ \subset{E} \lor \exists\delta(x) > 0 \text{ such as } \left] x-\delta, x+\delta \right[ \subset{E^C} \big) 
\\ \equiv
\\ x \in \bar{E} \cap{\big( int(E) \cup int(E^C) \big)}
\\ \equiv 
\\ x \in \bar{E} \cap{int(E)}
\\ \equiv
\\ x \in int(E)$$  
