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I am trying to show that the interior of a set of interior points is an interior point, that is, if $X$ is a subset of a metric space $M$, that $(X^\circ)^\circ = X^\circ$ for $X \subset M$. My approach is to show that each is a subset of the other. It is quite obvious that $(X^\circ)^\circ \subset X^\circ$. However, I am bogged down when trying to have an intuitive understanding how to show the other direction. Can anyone give me help? It would be greatly appreciated!

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    $\begingroup$ You can see $X^\circ=\bigcup_{U\subset X\text{ open}} U$. It should be clear from there. $\endgroup$ – Ian Coley Apr 22 '14 at 2:40
  • $\begingroup$ Hint Prove $A=A^\circ\iff A$ is open. Then prove $A^\circ $ is open. $\endgroup$ – Pedro Tamaroff Apr 22 '14 at 2:41
  • $\begingroup$ Thanks! Was able to solve it, just one thing, how did we know that the set of interior points equals the union of open sets on $X$? $\endgroup$ – user123276 Apr 22 '14 at 3:28
  • $\begingroup$ What does it mean to be in the interior of a set? Just check the definitions. $\endgroup$ – Andrés E. Caicedo Apr 22 '14 at 6:02
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One way to see this for a metric space, is to use that every point of an open ball $B(x,r) = \{y: d(x,y) < r \}$ is indeed an interior point of that open ball (justifying the name "open ball", one could say).

(Proof: if $y \in B(x,r)$, define $s = r - d(x,y) > 0$, and for any $z$, if $d(y,z) < s$, we know that $d(x,z) <= d(x,y) + d(y,z) < d(x,y) + s = d(x,y) + (r - d(x,y)) = r$, so indeed $B(y, s) \subset B(x,r)$. This $s$ shows that $y$ is an interior point of $B(x,r)$.)

Now, if $x \in X^\circ$, this means that there is some $r>0$ such that $B(x,r) \subset X$. For any $y \in B(x,r)$, there exists $s>0$ such that $B(y,s) \subset B(x,r) \subset X$, see above. But this just says that every point of $B(x,r)$ is itself an interior point of $X$, i.e. $B(x,r) \subset X^\circ$. And this by definition means $x \in (X^\circ)^\circ$, as required.

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