Let $E$ be a nonempty subset of $\mathbb{R}^{n}$.The distance from a point $\mathbb{x}\in\mathbb{R}^{n}$ to set $E$ is defined by $$d(\mathbb{x},E)=\inf\{||\mathbb{x}-\mathbb{y}||:\mathbb{y}\in E\}.$$ $E^{\circ}$ (respectively,$\overline{E}, \partial{E}$) is the interior (respectively,closure,boundary) of a set $E$;

Vol$(E)$ is the jordan content of $E.$

We have a set $S\subset \mathbb{R}^{n},$ such that $S=\overline{D}$ for some nonempty,open, connected jordan region $D$ in $\mathbb{R}^{n}.$ For each sufficiently small positive value of $\varepsilon $,let $$D_{\varepsilon}=\{\mathbb{x}\in D:d(\mathbb{x},\partial{D})\geq \varepsilon\}.$$ Whether $\overline{D}_{\varepsilon}$ is a connected Jordan region in $\mathbb{R}^{n}?$

The following is my effort: $$F_{1}=\{\mathbb{x}\in\mathbb{R}^{n}:d(\mathbb{x},\partial{D})\geq \varepsilon\}\text{ is closed},\quad F_{2}=\{\mathbb{x}\in\mathbb{R}^{n}:d(\mathbb{x},\partial{D})>\varepsilon\}\text{ is open}.$$

$$F^{\circ}_{1}=F_{2},D=D^{\circ},D^{\circ}_{\varepsilon}=F_{1}\cap D \Longrightarrow D^{\circ}_{\varepsilon}=(F_{1}\cap D)^{\circ}=F^{\circ}_{1}\cap D^{\circ}=F_{2}\cap D.$$ $$\partial F_{1}=\{\mathbb{x}\in\mathbb{R}^{n}:d(\mathbb{x},\partial{D})= \varepsilon\}.$$ Since $D$ is connected open set,it follows that $D^{\circ}_{\varepsilon}$ is connected open set. $$\overline{D^{\circ}_{\varepsilon}}=\overline{D}_{\varepsilon}=D^{\circ}_{\varepsilon}\cup \partial {D_{\varepsilon}},\quad\partial {D_{\varepsilon}}\subseteq \partial F_{1}\cup \partial{D},\quad\text{Vol}{(\partial{D})}=0.$$ If Vol$(\partial F_{1})=0 $ holds,then Vol$(\partial{D_{\varepsilon}})=0 $ holds.

In my opinion ,If we can prove Vol$(\partial F_{1})=0 $,then $\overline{D}_{\varepsilon}$ is a connected Jordan region in $\mathbb{R}^{n}.$ But how can I prove Vol$(\partial F_{1})=0$?

$$\overline{B}_{\varepsilon }(\mathbf{p})=\{\mathbf{x}\in\mathbb{R}^{n}:||\mathbf{x}-\mathbf{p}||\leq \varepsilon\} ,\\\overline{B}_{\varepsilon }(\mathbf{p})\cap \partial D=[\partial B_{\varepsilon }(\mathbf{p})\cup B^{\circ}_{\varepsilon }(\mathbf{p})]\cap\partial D=[\partial B_{\varepsilon }(\mathbf{p})\cap\partial D]\cup[ B^{\circ}_{\varepsilon }(\mathbf{p})\cap\partial D],\\ [\partial B_{\varepsilon }(\mathbf{p})\cap\partial D]\cap[ B^{\circ}_{\varepsilon }(\mathbf{p})\cap\partial D]=\emptyset.$$

$\mathbf{A}.$ $$S_{1}=\{\mathbf{p}\in D^{\circ}:\overline{B}_{\varepsilon }(\mathbf{p})\cap \partial D=\emptyset\}=\{\mathbf{p}\in D^{\circ}:d(\mathbf{p},\partial D)> \varepsilon\}=D_{\varepsilon}^{\circ}$$


$$S_{2}=\{\mathbf{p}\in D^{\circ}:\overline{B}_{\varepsilon }(\mathbf{p})\cap \partial D\ne\emptyset\}$$

$\mathbf(1).\partial B_{\varepsilon }(\mathbf{p})\cap\partial D\ne\emptyset\Longrightarrow \mathbf{p}\in \{\mathbf{p}\in D^{\circ}:d(\mathbf{p},\partial D)=\varepsilon\};$

$\mathbf(2).B^{\circ}_{\varepsilon }(\mathbf{p})\cap\partial D\ne\emptyset\Longrightarrow \mathbf{p}\in \{\mathbf{p}\in D^{\circ}:d(\mathbf{p},\partial D)<\varepsilon\}.$

$$\mathbf{A}+\mathbf{B}\Longrightarrow \partial D_{\varepsilon}=\{\mathbf{p}\in D^{\circ}:\partial B_{\varepsilon }(\mathbf{p})\cap\partial D\ne\emptyset \quad \wedge \quad B^{\circ}_{\varepsilon }(\mathbf{p})\cap\partial D=\emptyset \}.$$

Conversely,$\overline{D}_{\varepsilon}$ may not be a connected Jordan region in $\mathbb{R}^{n}$.But how can I find a counterexample of that?

If you have some good ideas about how to solve this question ,please give me some hints.Any help is going to be appreciated!


Results of this type, under further regularity assumptions on $E$, are established by D.Carballo "Areas of level sets of distance functions...", for almost all $\epsilon$ (with estimates on the Hausdorff $n-1$-measure of the boundary of $D_{\epsilon}$). Take a look also at the earlier paper by Almgren et al, Carballo refers to. Maybe one of the references proves the exact theorem you want. As for connectivity, it will be false even in the plane. Take a bounded sequence of disjoint circles $C_i$ in the plane, whose radii are $i^{-2}$ and which are all tangent to a line segment $L$. Next, remove an arc of length $i^{-10}$ from each circle $C_i$, such that the missing arcs are disjoint from $L$; call the remaining arcs $C_i'$. Now, take $D$ to be the union of $L$ with the union of arcs $C_i'$, $i>0$.

  • $\begingroup$ ${D}^{\circ}_{\varepsilon}$ is something like Tubular neighborhoods. $\endgroup$ – user250236 Aug 29 '15 at 4:02
  • $\begingroup$ @user250236: It depends on what you mean by "something like" and "tubular neighborhood". Tubular neighborhoods are usually understood as disk bundles. This is completely false in the case of open $\epsilon$-neighborhoods. $\endgroup$ – Moishe Kohan Aug 29 '15 at 7:35

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