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Show $A$ is open $\iff$ $A\cap\partial A=\emptyset$.


($\rightarrow)$ $A$ open $\implies A\cap\partial A= \emptyset$.

$x\in A$ open $\implies\exists\epsilon>0:B_{\epsilon}(x)\subseteq A$ $\implies$ $B_{\epsilon}(x)\cap A^c=\emptyset\implies$ $x\not\in\partial A$. Then $A\cap\partial A= \emptyset$.

($\leftarrow$) $A\cap\partial A= \emptyset$ $\implies A$ open.

$A\cap\partial A= \emptyset\implies$ for $x\in A$, $\exists\epsilon>0:B_{\epsilon}(x)\cap A^c=\emptyset\implies B_{\epsilon}(x)\subseteq A\implies A$ open.

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Assuming that you’re working in a metric space, it looks fine. – Brian M. Scott Feb 19 '12 at 7:35
Why don’t you go ahead and write up your attempt as an answer; after waiting a few hours, you’ll be able to accept your own answer, and we’ll have one problem fewer on the Unanswered list. – Brian M. Scott Feb 19 '12 at 8:04
up vote 2 down vote accepted

If you define the interior, boundary and exterior of a set in a topological space $X$ by

  • $int(A)=\{x \in A : \text{ there exists an open set }U\text{ such that }x \in U \subset A\}$

  • $ext(A)=\{x \notin A : \text{ there exists an open set }U\text{ such that }x \in U \subset X\setminus A\}$

  • $\partial A= \{x \in X : \text{for any an open set }U\text{ such that }x \in U \text{ we have } U \cap A\neq \emptyset,\ U \cap (X \setminus A)\neq\emptyset \}$

It is easy to see that $X=int(A) \cup \partial A \cup ext(A)$ and the sets are disjoint. $A$ is open if and only if $A=int(A)$. Moreover for every $A$ we have $int(A) \subset A$. From the definitions above it is easy to see that $int(A) \cap \partial A=\emptyset$ and $A \subset int(A)\cup \partial A$.

Then if $A$ is open, $A=int(A)$ and $A \cap \partial A=\emptyset$.

If $A \cap \partial A=\emptyset$, then $A \subset int(A)$ and furthermore $A=int(A)$, which means that $A$ is open.

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In your previous question, you had established the equality $$A^{\circ} = A\setminus\partial A.$$ Since $A$ is open if and only if $A=A^{\circ}$, this gives: $$\begin{align*} A\text{ is open}&\Longleftrightarrow A=A^{\circ}\\ &\Longleftrightarrow A=A\setminus \partial A\\ &\Longleftrightarrow A\cap\partial A = \varnothing. \end{align*}$$

Moral: Use previous results when possible!

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+1 for the moral. It's for me as well. :) – user21436 Feb 19 '12 at 22:20

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