How to prove a set $E$ is open if and only if $E = E^0$, where $E^0$ is the set of all the interior points of $E$? How to prove a set $E$ is open if and only if $E = E^0$, where $E^0$ is the set of all the interior points of $E$?

It's easy to prove that if $E = E^0$ then the set $E$ is open, because $E^0$ is open. But the reverse is much harder to me.

Suppose $p \in E^0$, there exists a neighborhood $N(p, r) \subset E$. And $E^0$ is open, so there also exists a neighborhood $N(p, r&#39;) \subset E^0$. Then, there exists a neighborhood $N(p, r&#39;&#39;) \subset E$, where $r&#39;&#39; = min(r, r&#39;)$. Then I do not know how to continue? 

Thanks.
Since the interior of a set is the largest open set contained in that set, if $E$ is open, then $E^0=E$. If $E=E^0$, then since $E^0$ is open, $E$ is open.
Suppose E is open it means it contains all its interior points,so that E is the subset of int(E). Also set of interior points of E is the subset of point E so that int(E) is the subset of E. thus E= int(E)