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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?


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Hint: does $p$ belong to $N(p,r)$? – Álvaro Lozano-Robledo Aug 20 '11 at 3:03
@Alvaro: According to my textbook, the neighborhood is set $\{q | d(p, q) < r, r > 0)\}$, becuase $d(p, p) = 0$, $p \subset N(q, r)$. I though $p$ was not in $N(p, r)$ before. Thanks. – Jichao Aug 20 '11 at 3:10
@Álvaro: Please post your hint as an answer so that it can be accepted and the question doesn't remain unanswered. – joriki Aug 20 '11 at 6:47
up vote 1 down vote accepted

Hint: does $p$ belong to $N(p,r)$?

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I posted my hint as an answer, as per @joriki's request. – Álvaro Lozano-Robledo Aug 20 '11 at 17:15

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.

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You're taking the wrong way. As you know, the set of all the interior point of $E$ is contained in $E$. You thus have not to show that points of $E^0$ belong to $E$ but that $E\subset E^0$.

And it's quite easy : if $E$ is open, then $E$ is a neighbourhood of each of its points. It follows that each point of $E$ is interior and therefore $E\subset E^0$.

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