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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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Did you mean $\partial N$ instead of $\partial\bar N$? –  Rasmus Sep 26 '12 at 11:04
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no i want the boundary of the closure –  palio Sep 26 '12 at 11:06
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@Rasmus: That’s false. –  Brian M. Scott Sep 26 '12 at 11:07

2 Answers 2

up vote 5 down vote accepted

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

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@palio: For a concrete example, take $N=\Bbb Q$ and $X=\Bbb R$. –  Brian M. Scott Sep 26 '12 at 11:08
    
then how does $\partial \bar N$ relate to $\partial N$ and in what conditions the formula i gave is true? –  palio Sep 26 '12 at 11:20
    
@Avatar: Far from it. Take $N=[0,1]\cap\mathbb{Q}$ and $X=\mathbb{R}$. Then $\partial\bar N=\partial[0,1]=\{0,1\}$, while $\partial(X\setminus N)=[0,1]$. –  Harald Hanche-Olsen Sep 26 '12 at 12:10
    
@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 $N$. –  Harald Hanche-Olsen Sep 26 '12 at 12:22

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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In the second line you say that $\partial \bar N=\emptyset$ why? don't we have $\partial \[0,1\]=\{0,1\}$? –  palio Sep 26 '12 at 11:16
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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.) –  Martin Sleziak Sep 26 '12 at 11:19

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