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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?

Thanks.

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    @Álvaro: Please post your hint as an answer so that it can be accepted and the question doesn't remain unanswered.2011-08-20

4 Answers 4

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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)

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    This "answer" is not much more than repeating the 4 year old question. The OP has already said that the direction $E=E^o$ $\implies$ $E$ s open is "easy to prove... because $E^o$ is open". So if you have new light to shed on the other direction, you would help your Readers by highlighting it.2016-07-29
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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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Hint: does $p$ belong to $N(p,r)$?

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    I posted my hint as an answer, as per @joriki's request.2011-08-20
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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 $E\subset E^0$.