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Let $X$ be a topological space and $E$ be an open set.

If $A$ is open in $\overline{E}$ and $A\subset E$, then how do i prove that $A$ is open in $X$?

It seems trivial, but i'm stuck.. Thank you in advance.

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It is helpful to be aware of the fact that for subsets $E,D$ of $X$ such that $E\subset D$ the subspace topology of $E$ inherited from $X$ is the same as the subspace topology inherited from $D$. Thus $A$ being open in $D$ implies the openness of $A\cap E$ relative to $E$. – Stefan Hamcke Jan 18 '13 at 19:41

Since $A$ is open in $\overline{E}$, there exists an open set $W$ in $X$ such that $A=W\cap \overline{E}$. Since $A\subset E$, this implies that $W\cap E=W\cap \overline{E}$. Thus $A=W\cap E$, hence it is open in $X$.

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$A$ open in $\overline E$ means that there exists an open set $B\subseteq X$ such that $A=B\cap \overline E$. As $A\subseteq E$, you have that $A=B\cap E$ is the intersection of two open sets.

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Is enough to show that $A=G\cap E$ for an open set $G\subset X$.

Use the hypothesis and the fact that $A=G\cap \overline{E}\Rightarrow A=G\cap E$ (why?).

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