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Show $A^o=A\setminus\partial A$.

I need to show inclusion on both sides, so:
(a) $A^o\subseteq A\setminus\partial A$
(b) $A\setminus\partial A \subseteq A^o$

Attempt at (a):

If $x\in A^o$ then $x\in A$ by definition. So we need to show $x\not\in\partial A$. Well, $A^o$ is open so $\exists\epsilon>0$ : $B_{\epsilon}(x)\subseteq A^o \subseteq A$. For this $\epsilon$, $B_{\epsilon}(x)\cap A^c=\emptyset$ so $x\not\in\partial A$.

Attempt at (b): This is trickier. Let $x\in A$ : $x\not\in\partial A$. That means $\exists\epsilon>0$ : $B_{\epsilon}(x)\cap A^c=\emptyset$. Does this imply $x\in\ A^o$?

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Note: $/$ denotes quotient. You want $\setminus$ (\setminus). – Arturo Magidin Feb 19 '12 at 5:34
You labeled the question "general topology"; but your argument uses epsilons, which works only in a metric space. Which is it? General topology, or metric spaces? – Arturo Magidin Feb 19 '12 at 5:36
whoops -- it's definitely metric spaces. – Emir Feb 19 '12 at 5:38
@Emir You can get around the balls by saying "There is an open neighbourhood $U$ about $x$ such that...." – user38268 Feb 19 '12 at 5:38
up vote 3 down vote accepted

For a general topology setting:

Let $x\in A^{\circ}$. Then there exists an open set $U$ such that $x\in U\subseteq A$; in particular, $U\cap A^c=\varnothing$, so $x\notin\partial A$ (as $x\in\partial A$ if and only if for every open set $V$, if $x\in V$ then $V\cap A\neq\varnothing$ and $V\cap A^c\neq\varnothing$). Therefore, $x\in A$, $x\notin \partial A$, so $x\in A\setminus\partial A$.

Now let $x\in A\setminus\partial A$. Since $x\notin\partial A$, there exists an open set $U$ such that $x\in U$ and either $U\cap A=\varnothing$, or $U\cap A^c=\varnothing$. Since $x\in A$, then $x\in U\cap A$, hence $U$ must satisfy $A\cap A^c=\varnothing$. Hence $x\in U\subseteq A$, so $x\in A^{\circ}$.

(Obviously, the same argument works for metric spaces; you can replace the open sets $U$ and $V$ mentioned above with open balls centered at $x$, as needed).

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For (b): Suppose that $x \in A$ but is not in the boundary. Then by definition of the boundary, viz.

$$\partial A = \overline{A} \setminus \operatorname{int}(A)$$

it is plain that $x \notin {\partial A}^{c}$ implies that $x \in {\overline{A}}^{c}$ or $x \in \operatorname{int}(A)$.

Now $x$ cannot be in the complement of the closure for otherwise this would contradict $x \in A$. Hence $x \in \operatorname{int}(A)$ and you are done.

$\textbf{Edit:}$ I see you have edited your question to it being in the context of metric spaces alone. I believe the proof here is equally adaptable to both metric spaces and general topological spaces.

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