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Let $A,B$ be open subsets of $\mathbb{R}^n$.

Does the following equality hold?

$$\partial(A\cap B)= (\bar A \cap \partial B) \cup (\partial A \cap \bar B)$$

Edit: Thanks for showing me in the answers that above formula fails if $A$ and $B$ are disjoint but their boundaries still intersect. I was able to come up with a similar formula which avoids this case $$[\partial(A\cap B)]\setminus(\partial A \cap \partial B)= (A \cap \partial B) \cup (\partial A \cap B),$$ which I was able to prove and suffices for what I need to do.

However, when showing that $ (A \cap \partial B) \cup (\partial A \cap B)\subseteq \partial(A\cap B)$, I needed to assume that the topology is induced by a metric. I wonder if the formula still holds in an arbitrary topological space.

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  • $\begingroup$ Is the boundary defined as $\overline{A}\setminus A$? $\endgroup$ – Ben Nov 26 '16 at 16:31
  • $\begingroup$ As usual with equalities between sets, the most straight-forward way is to check that each set is a subset of the other. There might be other, cleverer methods out there, but this one, at least, will not fail. $\endgroup$ – Arthur Nov 26 '16 at 16:34
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It does not hold. Consider for example $$A=\{x\in \mathbb{R}^n:|x|<1\}$$ and $$B=\{x\in \mathbb{R}^n:|x-(2,0,\dots,0)|<1\}.$$ Since $A\cap B=\varnothing$, $\partial(A\cap B)=\varnothing$, but the RHS in your formula is the set $\{(1,0,\dots,0)\}$.

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  • $\begingroup$ Your example uses tow disjoint sets whose boundaries still intersect. It seems that this is only case when the formula does not hold. Please check the updated question. $\endgroup$ – J.Doe Nov 27 '16 at 11:56
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This is not true generally unless $\overline{A\cap B}=\overline{A}\cap \overline{B}$. \begin{align} \partial (A\cap B)&= \overline{A\cap B}-(A\cap B)^{o} \\ &=(\overline{A}\cap \overline{B})-(A^{o}\cap B^{o}) \\ &=(\overline{A}\cap \overline{B})\cap(A^{o}\cap B^{o})^c \\ &=(\overline{A}\cap \overline{B})\cap(A^{o^c}\cup B^{o^c}) \\ &=(\overline{A}\cap \overline{B}\cap A^{o^c})\cup(\overline{A}\cap \overline{B}\cap B^{o^c}) \\ &=(\overline{A}\cap \partial B)\cup (\overline{B}\cap \partial A) \end{align} $\overline{A\cap B}=\overline{A}\cap \overline{B}$ implies discrete space. Please read this post, When is the closure of an intersection equal to the intersection of closures?. So in this is impossible in $\Bbb{R}^n$. However, we can prove generally $$ \partial(A\cap B)\subset (\bar A \cap \partial B) \cup (\partial A \cap \bar B) $$ for it is always true that $\overline{A\cap B}\subset \overline{A}\cap \overline{B}$. This can be done easily by replacing "$=$" with "$\subset $" at 2nd line of above proof and rest follows.

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If $A$ is dense and co-dense in the non-empty space $X$ (that is, $X$ \ $A$ is also dense in $X$), suppose $B=X$ \ $A.$ Then $\emptyset=A\cap B=\partial (A\cap B)$ but $\bar A=\bar B=\partial A=\partial B=X\ne \emptyset.$

For example, with $X= \mathbb R^n$ let $A$ be the set of points with rational co-ordinates.

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