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Given a metric space $(X,d)$ with $A\subseteq X $, show that $\overline{X \setminus A} = X \setminus A^\circ$

I know that I have to show that $\overline{X \setminus A} \subseteq X \setminus A^\circ$ and that $X \setminus A^\circ \subseteq \overline{X \setminus A}$.

Already I have: since $A^\circ \subseteq A$, $X \setminus A \subseteq X \setminus A^\circ$ and that since $A^\circ$ is open, $X \setminus A^\circ$ is closed. Using the fact that $\overline{X\setminus A}$ is the smallest closed subset to contain $X \setminus A$, $$\overline{X \setminus A} \subseteq X \setminus A^\circ$$ How do I go about proving the other way?

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    $\begingroup$ $X \setminus A \subseteq A^\circ$ cannot be true.Since $\forall x \in X \setminus A, x \notin A$ and $\forall x \in A^\circ, x \in A$ $\endgroup$ – RJM Nov 12 '16 at 13:27
  • $\begingroup$ I had yped out my solution wrong. I now have $X \setminus A \subseteq X \setminus A^\circ$ $\endgroup$ – vanaghka Nov 12 '16 at 13:32
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    $\begingroup$ Since $X\setminus A^\circ$ is closed, this implies $\overline{X\setminus A}\subseteq X\setminus A^\circ$. $\endgroup$ – Alex Ravsky Nov 12 '16 at 13:36
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$X\setminus(\overline{X\setminus A})$ is an open set contained in $A$ (hence contained in $A^{\circ}$).

This leads to: $$X\setminus A^{\circ}\subseteq\overline{X\setminus A}$$

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Can't you start from the fact that $X \setminus A^\circ$ is closed, thus $X \setminus A^\circ = \overline{X \setminus A^\circ}$

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