# The closure of the intersection of a closed set with a open set with compact closure

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.

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

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.
• 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. Sep 18, 2019 at 11:13