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If $X$ a topological space and $N$ is a subspace of $X$ and $\bar N$ its closure, is it true that $\partial \bar N= \partial (X-N)$?

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    @Rasmus: That’s false.2012-09-26

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No. For a counterexample, let $N$ and its complement be dense in $X$.

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    @palio: First, the complement is a red herring, since any subset and its complement have the same boundary. Second, we always have $\partial\bar N\subseteq\partial N$. When do we get inequality? If $x\in\partial N$ then $x\in\bar N$, so if it is not in the boundary of $\bar N$ then it must be in the interior. We conclude that the two boundaries are equal if and only if every interior point of $\bar N$ is already an interior point of $N$.2012-09-26
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Recall that $\partial A = \overline A \cap \overline {X\setminus A}$.

For $X=[0,1]$ and $N=(0,1)$ you have $\overline N=X$ and $\partial\overline N=\emptyset$.

But you have $X\setminus N=\{0,1\}$ and $\partial(X\setminus N)=X\setminus N=\{0,1\} \ne \partial \overline N$.

So the above claim is not true.

Directly from the definition of the boundary you can see that $\partial N=\partial(X\setminus N)$.

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    @palio I work in the topological space $X=[0,1]$. For $A=\overline N=[0,1]$ I get $X\setminus A=\emptyset$ and, consequently, $\partial A=\emptyset$.\\ If I tried to find the boundary of $N$ as a subset of the real line, I would indeed get $\{0,1\}$. (So we see that the boundary of a set depends on the topological space in which we study this set.)2012-09-26