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In Example 6.17 in Section 6.3 of Linear Integral Equations by Rainer Kress, Kress uses that $$\partial ((B_r(x))^C\cap D)=(\partial B_r(x)\cap D)\cup(\partial D\cup B_r(x))\qquad \text{(EQ 1)}$$ for an arbitrary bounded domain $D\in \mathbb{R}^n$ and $x\in \partial D$ and without proving this statement.

Drawing a picture when $D$ is 2-dimensional the result seems obvious but I want to prove the result rigorously.

I started the proof by rewriting the left-hand side of (EQ 1) and got

$$\partial(B_r(x)^C\cap D)=\overline{B_r(x)^C\cap D}\setminus(B_r(x)^C\cap D)^0$$ $$=(\overline{B_r(x)^C\cap D})\cap((B_r(x)^C\cap D)^0)^C$$ $$=(\overline{B_r(x)^C\cap D})\cap((B_r(x)^C)^0\cap D)^C$$ $$=(\overline{B_r(x)^C\cap D})\cap(((B_r(x)^C)^0)^C\cup D^C)$$ $$=(\overline{B_r(x)^C\cap D})\cap(\overline{(B_r(x)^C)^C}\cup D^C)$$ $$=(\overline{B_r(x)^C\cap D})\cap(\overline{B_r(x)}\cup D^C)$$ $$=(\overline{B_r(x)^C\cap D})\cap(\overline{B}_r(x)\cup D^C)\qquad \text{(EQ 2)}$$

Then I rewrote the right-hand side to get $$(\partial B_r(x)\cap D)\cup (\partial D\cap B_r(x))=((\overline{B}_r(x)\setminus B_r(x))\cap D)\cup((\overline{D}\setminus D)\cap B_r(x)^C)$$ $$=(\overline{B}_r(x)\cap B_r(x)^C\cap D)\cup (\overline{D}\cap D^C\cap B_r(x)^C)\qquad \text{(EQ 3)}$$

Show that the left-hand side of (EQ 1) is included in the right-hand side of (EQ 1), I use the property that contains the intersection of the closures of two sets contains the closure of an intersection of the two sets and EQ1-EQ 2:

$$\partial(B_r(x)^C\cap D)=(\overline{B_r(x)^C\cap D})\cap(\overline{B}_r(x)\cup D^C)$$ $$\subset (\overline{B_r(x)^C}\cap \overline{D})\cap (\overline{B}_r(x)\cup D^C)$$ $$=(B_r(x)^C\cap \overline{D})\cap(\overline{B}_r(x)\cup D^C)$$ $$=B_r(x)^C\cap \overline{D}\cap \overline{B}_r(x)\cup B_r(x)^C\cap \overline{D}\cap D^C$$ $$=(\partial B_r(x)\cap D)\cup (\partial D\cap B_r(x))$$

For the reverse inclusion, I would need $$(\overline{B_r(x)^C\cap D})\cap(\overline{B}_r(x)\cup D^C)\supset (\overline{B_r(x)^C}\cap \overline{D})\cap (\overline{B}_r(x)\cup D^C).$$ This would imply that $$(\overline{B_r(x)^C\cap D})\supset (\overline{B_r(x)^C}\cap \overline{D}) \qquad \text{(EQ 4)}$$

It not generally true that the closure of the intersection of the closures of two sets is contained in the closure of their intersection (for example, consider the sets $(0,1)$ and $(1,2)$).

What separates this from the most general case are that $B_r(x)^C$ is closed, $D$ is open, and $\bar{D}$ is compact (since any closed and bounded subset of $R^n$ is compact).

I'm really hoping that these two assumptions make EQ 4 true, but I don't see a way to prove this.

I didn't know what to title this question but the point of the question essentially boils down to analyzing the closure of the intersection of a closed set and an open set which has compact closure.

EDIT: What I am trying to prove is trivially false if $D=B_r(x)$. This means that I must use the assumption that $x\in \partial D$ somehow.

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  • $\begingroup$ Shouldn't (EQ1) be $∂((B_r(x))^C∩D) = (∂B_r(x) ∩ D) ∪ (∂D ∩ B_r(x)^C)$? $\endgroup$
    – user87690
    Sep 17, 2019 at 8:13
  • $\begingroup$ How exactly is domain defined here? $\endgroup$
    – user87690
    Sep 18, 2019 at 11:02

1 Answer 1

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Take D to be a punctured disc, missing a point p, and B to be a ball centered on the boundary of D and with p on its boundary. p is in the LHS but not the RHS, whether you rewrite the equation as user87690 suggests, or just by changing the last $\cup$ to $\cap$, or leave it as it is.

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  • $\begingroup$ Maybe the definition of domain includes contractibility here. $\endgroup$
    – user87690
    Sep 17, 2019 at 14:36
  • $\begingroup$ Any D which is not equal to the interior of its own closure will fail in the same way. Such can be homeomorphic to an open disk - e.g. an open disk with a radius removed. Any sufficent condition will need to address how the domain is embedded in Euclidean space, not just its intrinsic topology. Perhaps "equal to the interior of its closure" works. $\endgroup$ Sep 18, 2019 at 1:55
  • $\begingroup$ Yes, I would consider a domain to be regular open as well. I'll ask the OP about the used definition of domain. $\endgroup$
    – user87690
    Sep 18, 2019 at 11:02
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    $\begingroup$ I've just tracked down the reference. The domains in that section are all assumed to be what the author calls of class $C^2$, which appears to be very restrictive (points on the boundary have nbhds diffeomorphic to a half ball, and more). See slideshare.net/EduardoEspinosaPerez/… pages 30 and 88. Not a topic I know anything about. $\endgroup$ Sep 18, 2019 at 11:13

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