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Let $ (X,d)$ be a metric space and $A \subset X$. How can I show that $(A^\circ)^\circ = A^\circ $? $A^\circ$ denotes the set of all interior points of $A$.

I know that $x \in A$ is interior if there exists an $ r>0$ such that $B_r(x) \subset A$ . So i think to try to prove that $(A ^\circ)^\circ \subset A^\circ$ and $A^\circ \subset (A^\circ)^\circ$ , but I cannot think how to start.

user189013
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  • What are your thoughts? What have you tried? Do you know the definition of the interior of a set? – Greg Martin Oct 09 '15 at 19:29
  • I know that $x \in A$ is interior if there exists an $ r>0$ such that $B_r(x) \in A$ . So i think to try to prove that $(A ^\circ)^\circ \subset A^\circ$ and $A^\circ \subset (A^\circ)^\circ$ , but I cannot think how to start. – user189013 Oct 09 '15 at 19:34
  • Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you [edit] your question to include some motivation, and an explanation of your own attempts. – Lord_Farin Oct 09 '15 at 23:02

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Which sets $B$ have the property that $B^\circ = B$, meaning every point is interior to them? They have a name: open sets.

So, you are asking how to prove that the interior of any set is an open set. Luckily, this was done before.

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