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') \subset E^0. Then, there exists a neighborhood N(p, r'') \subset E, where r'' = min(r, r'). Then I do not know how to continue?
Thanks.