Abstract.

The $\epsilon$-boundary of a set $A\subseteq {ℝ}^2$ is the set $\{p\in {ℝ}^2:\rho (p,A)=\epsilon \}$, where $\rho$ is the Euclidean distance. We prove that if $A,B\subseteq {ℝ}^2$ are nonempty, connected sets, $A$ is bounded, and , then the $\epsilon$-boundary of $A$ contains a simple closed curve (aka a Jordan curve) that separates $A$ and $B$. This statement follows from the theorem which says that if $\epsilon >0$ and $A\subseteq {ℝ}^2$ is a nonempty, bounded, connected set, then the boundary of each component of $\{p\in {ℝ}^2:\rho (p,A)>\epsilon \}$ is a simple closed curve. Another corollary of this theorem is that the $\epsilon$-boundary of a nonempty, bounded, connected set $A\subseteq {ℝ}^2$ contains a simple closed curve bounding the domain that contains the open $\epsilon$-neighbourhood of $A$. In all these statements the connectivity condition can be significantly weakened. We also show that, for all $\epsilon >0$, the $\epsilon$-boundary of a nonempty, bounded set $A\subseteq {ℝ}^2$ contains a simple closed curve.

1. Introduction

The $\epsilon$-boundary of a set $A\subseteq {ℝ}^2$ is the set $\{p\in {ℝ}^2:\rho (p,A)=\epsilon \}$, where $\rho$ is the Euclidean distance. The separation of planar sets by a simple closed curve lying in the $\epsilon$-boundary of one of these sets is used in engineering. But the algorithms for constructing such a curve are either based on heuristics or assume some smoothness of the boundary of the sets [7, 1]. Similarly, the line of the boundary of territorial waters is drawn heuristically [4]. No proof of the existence of a simple closed curve that separates arbitrary continua $A$ and $B$ and lies in the $\epsilon$-boundary of $A$ appears to be known.

It is known that two disjoint continua in the plane can be separated by some simple closed curve [9, Chap. 10, § 61.II, Theorem 5^′]. It is also apparently known that if $A,B\subseteq {ℝ}^2$ are nonempty, connected sets, $A$ is bounded, the closure of $A$ does not separate the plane, and , then the open $\epsilon$-neighbourhood of $A$, , contains a simple closed curve that separates $A$ and $B$. At least, if we understand R. L. Moore’s terminology correctly, this statement is a direct consequence of his theorem [11, Theorem 1], which states that if, in a plane $S$, $M$ is a closed point set and $K$ is a bounded maximal connected subset of $M$ which does not separate $S$, then, for every positive number $e$, there exists a simple closed curve which encloses $K$ and contains no point of $M$ and which is such that every point within it is at a distance less than $e$ from some point of $K$.

We prove that if $A,B\subseteq {ℝ}^2$ are nonempty, connected sets, $A$ is bounded, and , then the $\epsilon$-boundary of $A$ contains a simple closed curve that separates $A$ and $B$; see Corollary 5.3 and Remark 2.2(1). In the proof, we specify an example of such a curve explicitly: the boundary of that component of the set $\{p\in {ℝ}^2:\rho (p,A)>\epsilon \}$ which contains $B$.

It is known that, for all $\epsilon >0$, the open $\epsilon$-neighbourhood of a nonempty, bounded, connected set $A\subseteq {ℝ}^2$ contains a simple closed curve bounding the domain that contains $A$. This result directly follows from a theorem of L. Zoretti [17] which, according to [15, Corollary VI.3.11] (see also [11] and [10]), states that if $K$ is a component of a compact set $M$ and $\epsilon$ is any positive number, then there exists a simple closed curve $J$ which encloses $K$ and is such that $J\cap M=\mathrm{\varnothing }$, and every point of $J$ is at a distance less than $\epsilon$ from some point of K.

We prove that, for all $\epsilon >0$, the $\epsilon$-boundary of a nonempty, bounded, connected set $A\subseteq {ℝ}^2$ contains a simple closed curve bounding the region that contains the open $\epsilon$-neighbourhood of $A$; see Corollary 5.1 and Remark 2.2(1). This result admits the following mechanical interpretation: on such a set $A$ one can put a wheel of radius $\epsilon$ lying in the same plane and “roll it along (stretches of) the boundary of $A$” in such a way that at every moment the wheel touches $A$, its interior does not intersect $A$, and the center of the wheel eventually describes a simple closed curve bounding the domain that contains the open $\epsilon$-neighbourhood of $A$.

Sometimes there is a need to maximise the Euclidean distance from a simple closed curve to the sets it separates. We prove that if $A,B\subseteq {ℝ}^2$ are nonempty, connected sets, $A$ is bounded, and $\rho (A,B)>0$, then the $\rho (A,B)/2$-boundary of $A$ contains a simple closed curve $C$ that separates $A$ and $B$ and such that $\rho (C,A\cup B)=\rho (A,B)/2$; see Corollary 5.8 and Remark 2.2(1). Thus, among the simple closed curves separating $A$ and $B$, the curve $C$ is a curve maximally distant from $A\cup B$.

We also show that if among the simple closed curves separating nonempty sets $A$ and $B$ in ${ℝ}^2$, the curve $C$ is a curve maximally distant from $A\cup B$, then either $\rho (C,A\cup B)=0$ or $\rho (C,A\cup B)=\rho (A,B)/2$, see Remark 5.10. It is curious that there exist two sets $A$ and $B$ that can be separated by some simple closed curve, but the supremum of the distances from such curves to $A\cup B$ is not attained; see Subsection 6.5 for an example. However, we do not know whether there exist nonempty sets $A,B\subseteq {ℝ}^2$ that can be separated by a simple closed curve, with $\rho (A,B)=1$, and such that, for every simple closed curve $C$ that separates $A$ and $B$, $\rho (C,A\cup B)=0$ holds; see Question 5.11.

In all the above statements, the connectivity condition can be relaxed to the $\delta$-chainedness condition for some $\delta >0$ (a set $A$ is $\delta$-chained iff any two points of $A$ can be connected by a polygonal chain whose vertices belong to $A$ and whose segment lengths are all less than $\delta$). Specific values of the parameter $\delta$ are given in Corollaries 5.1, 5.3, and 5.8.

Most of our results follow from Theorem 4.1, which says that if $\epsilon >0$ and $A\subseteq {ℝ}^2$ is a nonempty, bounded, $2\epsilon$-chained set, then the boundary of each component of the set $\{p\in {ℝ}^2:\rho (p,A)>\epsilon \}$ is a simple closed curve. Almost none of the conditions in Theorem 4.1 and Corollaries 5.1–5.8 can be relaxed. Furthermore, most of these statements cannot be transferred to the 3-dimensional case. We discuss these issues in Sections 6 and 7, respectively. But we do not know whether it is always possible to separate two disjoint continua $A,B\subseteq {ℝ}^3$ by a two-dimensional manifold lying at distance $\rho (A,B)/2$ from $A\cup B$; see Question 7.4.

Finally, we prove that, for all $\epsilon >0$, the $\epsilon$-boundary of a nonempty, bounded set $A\subseteq {ℝ}^2$ contains a simple closed curve, see Corollary 5.7. This statement complements previously known results about $\epsilon$-boundaries in the Euclidean spaces. In particular, M. Brown proved [2] that, for all $\epsilon >0$, the $\epsilon$-boundary of a compact subset of the plane is contained in the union of a finite number of simple closed curves. In the same paper he showed that if a nonempty, compact set $A\subseteq {ℝ}^n$ has a diameter smaller than $\epsilon$ and contains the origin $0$, then its $\epsilon$-boundary is the $(n-\mathrm{\hspace{0.17em}1})$-sphere. He also established that, for all but countable number of $\epsilon$, each component of the $\epsilon$-boundary of a compact subset of the plane is a point, a simple arc, or a simple closed curve. R. Gariepy and W. D. Pepe, answering M. Brown’s question, showed [8] that the $\epsilon$-boundary of a closed subset of the plane is a 1-manifold for almost every $\epsilon$. Four years later, S. Ferry proved [6] that, for $n$ equal to 2 or 3, the $\epsilon$-boundary of a set $A\subseteq {ℝ}^n$ is an $(n-\mathrm{\hspace{0.17em}1})$-manifold for almost all $\epsilon$. He also proved that if $A$ is a finite polyhedron in ${ℝ}^n$, then its $\epsilon$-boundary is an $(n-\mathrm{\hspace{0.17em}1})$-manifold for all sufficiently small values of $\epsilon$. In the same paper he constructed a set $B\subseteq {ℝ}^3$ such that the $\epsilon$-boundary of $B$ has components which are not 2-manifolds for uncountably many $\epsilon$. And he constructed a Cantor set in ${ℝ}^4$ whose $\epsilon$-boundary is not a 3-manifold for any $\epsilon$ between 0 and 1. P. Pikuta proved [13] that the $\epsilon$-boundary of a compact subset of the plane is a closed absolutely continuous curve for all sufficiently large values of $\epsilon$. Recently, J. Rataj and L. Zajíček extended [14] the results from [8] and [6] to sufficiently smooth normed linear spaces $X$ with $\mathrm{𝖽𝗂𝗆}X\in \{2,\mathrm{\hspace{0.17em}3}\}$.

2. Terminology and notation

We use terminology from the book [5]. A simple closed curve (also called a Jordan curve) is a set homeomorphic to a circle (i.e., a one-dimensional sphere). In metric spaces, simple closed curves are precisely the images of a circle under continuous injective mappings. According to the Jordan curve theorem, if $C$ is a simple closed curve in the plane ${ℝ}^2$, then its complement ${ℝ}^2\setminus C$ has exactly two components, the bounded and the unbounded, which we denote by $C^{-}$ and $C^{+}$, respectively. We say that a simple closed curve $C\subseteq {ℝ}^2$ separates sets $A$ and $B$ iff the sets $A$ and $B$ are contained in different components of the subspace ${ℝ}^2\setminus C$. We denote the range of a mapping $f$ by $\mathrm{𝗋𝖺𝗇}(f)$. A path is a continuous mapping whose domain equals the closed segment $[0,1]$. A path $f$ connects points $u$ and $v$ in a space $X$ iff $f(0)=u$, $f(1)=v$, and $\mathrm{𝗋𝖺𝗇}(f)\subseteq X$. A space $X$ is pathwise connected if any two of its points are connected by a path in $X$.

We define the distance between two sets in ${ℝ}^n$ as the infimum of pairwise Euclidean distances between points of these sets. The distance between a point $p$ and a set $A$ is the distance between sets $\{p\}$ and $A$. We denote all three distances by the symbol $\rho$. For $\epsilon >0$ and $A\subseteq {ℝ}^2$, the open $\epsilon$-neighbourhood of the set $A$, , is denoted by ${𝖮}_{\epsilon }(A)$; similarly the closed $\epsilon$-neighbourhood $\{p\in {ℝ}^2:\rho (p,A)⩽\epsilon \}$ and the $\epsilon$-boundary $\{p\in {ℝ}^2:\rho (p,A)=\epsilon \}$ of the set $A$ are denoted by ${𝖡}_{\epsilon }(A)$ and ${𝖲}_{\epsilon }(A)$, respectively. If $p$ is a point in ${ℝ}^2$, then ${𝖮}_{\epsilon }(p)≔{𝖮}_{\epsilon }(\{p\})$, ${𝖡}_{\epsilon }(p)≔{𝖡}_{\epsilon }(\{p\})$, and ${𝖲}_{\epsilon }(p)≔{𝖲}_{\epsilon }(\{p\})$ are the open and the closed disks and the circle of center $p$ and radius $\epsilon$, respectively. We denote the closure and boundary of a set $A$ in ${ℝ}^2$ by $\overline{A}$ and $\partial A$, respectively; we denote the boundary of a set $B$ in a space $X$ by ${\partial }_{X}B$.

Remark 2.1.

Suppose that $\epsilon >0$, $\mathrm{\varnothing }\ne A\subseteq {ℝ}^2$, $\mathrm{\varnothing }\ne B\subseteq {ℝ}^2$, and $p\in {ℝ}^2$. Then:

1. $\partial {𝖡}_{\epsilon }(A)\subseteq \partial {𝖮}_{\epsilon }(A)={𝖲}_{\epsilon }(A)$.

2. $\rho (p,A)=\rho (p,\overline{A})$ and $\rho (A,B)=\rho (\overline{A},\overline{B})$.

3. ${𝖮}_{\epsilon }(A)={𝖮}_{\epsilon }(\overline{A}),{𝖡}_{\epsilon }(A)={𝖡}_{\epsilon }(\overline{A})$,  and  ${𝖲}_{\epsilon }(A)={𝖲}_{\epsilon }(\overline{A})$. ∎

For $\epsilon >0$, we say that two points $p,q\in A$ are $\epsilon$-chained in $A$ iff there exists a finite sequence of points $r_0,r_1,\mathrm{\dots },r_n$ in $A$ such that $r_0=p$, $r_n=q$, and for all . A set $A$ is called $\epsilon$-chained iff any two of its points are $\epsilon$-chained in it [12, page 60, Definition 4.15]. For $p\in A$, the $\epsilon$-chained component of a point $p$ in a set $A$ is the set

$$\{q\in A:p\text{ and }q\text{ are }\epsilon \text{-chained in }A\}.$$

We say that a set $B$ is an $\epsilon$-chained component of a set $A$ iff $B$ equals the $\epsilon$-chained component of a point $p$ in $A$ for some $p\in A$.

Remark 2.2.

Suppose that $\mathrm{\varnothing }\ne A\subseteq {ℝ}^2$ and $\epsilon >0$. Then:

1. [12, Exercise 4.23(a)] If $A$ is connected, then it is $\epsilon$-chained.

2. If $A$ is $2\epsilon$-chained, then its open $\epsilon$-neighbourhood ${𝖮}_{\epsilon }(A)$ is pathwise connected.

3. If $A$ is $\epsilon$-chained and $\delta >\epsilon$, then $A$ is $\delta$-chained. ∎

Note that in the second clause of Remark 2.2 the reverse implication is also true. Thus, a nonempty subset of the plane is $2\epsilon$-chained if and only if its open $\epsilon$-neighbourhood is pathwise connected.

3. Auxiliary lemmas

To prove the main theorem, we need the following auxiliary statements.

Lemma 3.1.

Suppose that $E$ is a connected metric space, $U\subseteq E$ is open, $p\in U$, and $U\setminus \{p\}$ is connected. Then $E\setminus \{p\}$ is also connected.

Proof 3.2.

Suppose on the contrary that $E\setminus \{p\}$ equals the union of two disjoint nonempty sets $A$ and $B$ closed in $E\setminus \{p\}$. Then there exist two sets $C$ and $D$ closed in $E$ such that

$$A=C\cap (E\setminus \{p\})=C\setminus \{p\}\text{and}B=D\cap (E\setminus \{p\})=D\setminus \{p\}.$$

Clearly, $C=A$ or $C=A\cup \{p\}$, and, similarly, $D=B$ or $D=B\cup \{p\}$.

Case 1. $C=A$ or $D=B$. Let, without loss of generality, $C=A$. Note that $D\cup \{p\}$ is closed in $E$, so $E$ equals the union of two disjoint nonempty, closed sets $C$ and $D\cup \{p\}$. This contradicts the connectedness of $E$.

Case 2. $C=A\cup \{p\}$ and $D=B\cup \{p\}$. Note that in this case the sets $A$ and $B$ are open in $E$ as the complements of closed sets $D$ and $C$, respectively.

Consider the sets

$$A^{\prime }≔A\cap (U\setminus \{p\})\text{and}{B}^{\prime }≔B\cap (U\setminus \{p\}).$$

These sets are disjoint and open in $U\setminus \{p\}$ (because the sets $A$ and $B$ are disjoint and open in $E$) and $U\setminus \{p\}$ equals their union. And since, by assumption, $U\setminus \{p\}$ is connected, one of these sets is empty. Let, without loss of generality, $A^{\prime }$ be empty. Then $U\subseteq B\cup \{p\}$, and therefore $B\cup \{p\}=B\cup U$, since $p\in U$. Thus, the set $B\cup \{p\}$ is open in $E$ as the union of open sets. Hence, the space $E$ equals the union of two disjoint nonempty, open sets $A$ and $B\cup \{p\}$, which contradicts its connectedness.

Lemma 3.3.

Every open, connected set is a component of the complement of its boundary.

Proof 3.4.

Let $U$ be an open, connected set in a topological space $X$ and $C$ be the boundary of $U$. We need to show that $U$ is a $\subseteq$-maximal connected set in the subspace $X\setminus C$. Consider a nonempty set $V\subseteq X\setminus (U\cup C)$. It suffices to show that $U\cup V$ is not connected. Since $U$ is open in $X$, then $U$ is also open in $U\cup V$. The set $V$ is also open in $U\cup V$ because it equals the trace on $U\cup V$ of the open set $X\setminus (U\cup C)=X\setminus \overline{U}$. Thus, $U\cup V$ equals the union of two disjoint nonempty, open sets.

4. The main result

Recall that a neighbourhood of a point is a set whose interior contains the given point. A space is locally connected iff every neighbourhood of every point contains a connected neighbourhood of the same point. A continuum is a connected compact set, and a semi-continuum is a space such that any pair of points is contained in some continuum. A point is a cut point of a space iff the complement of this point is not a semi-continuum.

Theorem 4.1.

Suppose that $\epsilon >0$ and $A\subseteq {ℝ}^2$ is a nonempty, bounded, $2\epsilon$-chained set. Let $D$ be a component of ${ℝ}^2\setminus {𝖡}_{\epsilon }(A)$. Then the boundary of $D$ is a simple closed curve contained in the $\epsilon$-boundary of $A$. Moreover, we have:

• if $D$ is bounded, then $D={(\partial D)}^{-}$;

• if $D$ is unbounded, then $D={(\partial D)}^{+}$.

Proof 4.2.

Let us show that $\partial D\subseteq {𝖲}_{\epsilon }(A)$. Let $q\in \partial D$. If , then ${𝖮}_{\delta }(q)\subseteq {𝖡}_{\epsilon }(A)$ for any $\delta \in (0,\epsilon -\rho (q,A))$, which contradicts the fact that $q\in \partial D$. If $\rho (q,A)>\epsilon$, then ${𝖮}_{\delta }(q)\subseteq {ℝ}^2\setminus {𝖡}_{\epsilon }(A)$ for any $\delta \in (0,\rho (q,A)-\epsilon )$. Thus, the open neighbourhood ${𝖮}_{\delta }(q)$ is a connected subset of the subspace ${ℝ}^2\setminus {𝖡}_{\epsilon }(A)$ and intersects the component $D$ of this subspace, so it is contained in $D$ (a connected set is either disjoint with or contained in a component). Again we get a contradiction with the fact that $q\in \partial D$.

Let us add a new point $𝐩$ (the pole) to the plane ${ℝ}^2$ so that the new space $𝕊≔{ℝ}^2\cup \{𝐩\}$ is homeomorphic to the two-dimensional sphere. Being connected, the set $D$ is contained in some component $E$ of the subspace $𝕊\setminus {𝖡}_{\epsilon }(A)$.

Let us show that

$$𝐩\notin {\partial }_{𝕊}E\text{ and }E\setminus \{𝐩\}\text{ is connected}.$$

Since $𝐩\notin {𝖡}_{\epsilon }(A)$ and ${𝖡}_{\epsilon }(A)$ is closed in $𝕊$, there exists a connected open neighbourhood $U$ of the point $𝐩$ in $𝕊$ such that $U\cap {𝖡}_{\epsilon }(A)=\mathrm{\varnothing }$ and $U\setminus \{𝐩\}$ is connected. If the connected subset $U$ of the subspace $𝕊\setminus {𝖡}_{\epsilon }(A)$ intersects the component $E$ of that subspace, then $U\subseteq E$. In this case $𝐩\notin {\partial }_{𝕊}E$ and, by virtue of Lemma 3.1, $E\setminus \{𝐩\}$ is connected. If $U$ does not intersect $E$, then $𝐩\notin {\partial }_{𝕊}E$ and $E\setminus \{𝐩\}=E$, so $E\setminus \{𝐩\}$ is connected.

We have

$$D\subseteq E\setminus \{𝐩\}\subseteq {ℝ}^2\setminus {𝖡}_{\epsilon }(A),$$

that is, the component $D$ of the subspace ${ℝ}^2\setminus {𝖡}_{\epsilon }(A)$ is contained in the connected subset $E\setminus \{𝐩\}$ of this subspace. Therefore, $D=E\setminus \{𝐩\}$, and hence $D\subseteq E\subseteq D\cup \{𝐩\}$. Hence,

$$\text{either }E=D\text{ or }E=D\cup \{𝐩\}.$$

Let us prove that $\partial D={\partial }_{𝕊}E$. It suffices to show that an arbitrary point $q\in 𝕊$ either belongs or does not belong to both sets at the same time. If $q=𝐩$, then $q$ belongs neither to $\partial D$ nor to ${\partial }_{𝕊}E$. If $q\ne 𝐩$, then the point $q$ has a neighbourhood $U$ that does not contain $𝐩$. Since the sets $D$ and $E$ can differ only by the point $𝐩$, for every neighbourhood $V\subseteq U\subseteq 𝕊\setminus 𝐩$ of the point $q$, we have $V\cap D=V\cap E$ and $V\cap ({ℝ}^2\setminus D)=V\cap (𝕊\setminus E)$. Therefore, the point $q$ belongs to $\partial D$ if and only if it belongs to ${\partial }_{𝕊}E$.

On the two-dimensional sphere $𝕊$, according to Theorem 4 in [9, Chap. 10, § 61.II, p. 512], the following statement is true: if a locally connected continuum has no cut points, then the boundary of each component of its complement is a simple closed curve. Thus, if we show that ${𝖡}_{\epsilon }(A)$ is a locally connected continuum without cut points, then it follows that ${\partial }_{𝕊}E=\partial D$ is a simple closed curve. It is not difficult to show that the set $D$, being a component of an open subset of the plane, is open. Then, according to Lemma 3.3, $D$ is a component in ${ℝ}^2\setminus \partial D$. Hence, if $D$ is bounded, then $D={(\partial D)}^{-}$, and if $D$ is unbounded, then $D={(\partial D)}^{+}$. Thus, it remains to prove that ${𝖡}_{\epsilon }(A)$ is a locally connected continuum without cut points.

Let us show that ${𝖡}_{\epsilon }(A)$ is a locally connected, compact set. For every nonempty, compact set $K\subseteq {ℝ}^2$ of diameter smaller than $\epsilon$, its closed $\epsilon$-neighbourhood ${𝖡}_{\epsilon }(K)$ is homeomorphic [2, Lemma  1, (ii)-(iii)] to a closed disk in ${ℝ}^2$, so it is a locally connected continuum. Since the set $A$ is bounded, it can be represented as the union $A={\bigcup }_{i⩽n}{A}_i$ of a finite number of nonempty sets of diameter less than $\epsilon$. For all $i⩽n$, the closure ${\overline{A}}_i$ is a nonempty, compact set of diameter less than $\epsilon$. Then ${𝖡}_{\epsilon }({\overline{A}}_i)$ is a locally connected continuum. Hence, ${\bigcup }_{i⩽n}{𝖡}_{\epsilon }({\overline{A}}_i)$ is a locally connected compact [9, Chap. 6, § 49.II, p.230, Theorem 1]. Using Remark 2.1(3) and the definition of the closed $\epsilon$-neighbourhood we have

$$\bigcup _{i⩽n}{𝖡}_{\epsilon }({\overline{A}}_i)=\bigcup _{i⩽n}{𝖡}_{\epsilon }(A_i)={𝖡}_{\epsilon }(A).$$

Let us show that ${𝖡}_{\epsilon }(A)$ is connected and has no cut points. To do this, it suffices to show that for any point $r$ in ${𝖡}_{\epsilon }(A)$, the set ${𝖡}_{\epsilon }(A)\setminus \{r\}$ is pathwise connected. In this case, the set ${𝖡}_{\epsilon }(A)$ is also pathwise connected. Let $t$ and $s$ be two different points in ${𝖡}_{\epsilon }(A)\setminus \{r\}$; we will find a path connecting these points in ${𝖡}_{\epsilon }(A)\setminus \{r\}$.

Let $t^{\prime }$ and $s^{\prime }$ be points in $\overline{A}$ nearest, respectively, to $t$ and $s$. Since $r$ is different from $t$ and $s$, there are points $u$ and $v$ in the segments $[t,t^{\prime }]$ and $[s,s^{\prime }]$, respectively, such that and . According to Remark 2.2(2), the open $\epsilon$-neighbourhood ${𝖮}_{\epsilon }(A)$ is pathwise connected, so there exists a path $f$ connecting the points $u$ and $v$ in ${𝖮}_{\epsilon }(A)$. It is easy to show that then there exists a path $f^{\prime }$ connecting points $t$ and $s$ in ${𝖡}_{\epsilon }(A)$ such that $\mathrm{𝗋𝖺𝗇}(f^{\prime })\setminus \{t,s\}$ is contained in ${𝖮}_{\epsilon }(A)$.

If $r\notin \mathrm{𝗋𝖺𝗇}(f^{\prime })$, then $f^{\prime }$ is the path connecting $t$ and $s$ in ${𝖡}_{\epsilon }(A)\setminus \{r\}$, so we are done. If $r\in \mathrm{𝗋𝖺𝗇}(f^{\prime })$, then, since $r$ is distinct from $t$ and $s$, we have $r\in \mathrm{𝗋𝖺𝗇}(f^{\prime })\setminus \{t,s\}\subseteq {𝖮}_{\epsilon }(A)$. Then there exists $\delta >0$ such that

$${𝖡}_{\delta }(r)\subseteq {𝖮}_{\epsilon }(A)\setminus \{t,s\}.$$

Let $\tilde{t}$ and $\tilde{s}$ be the <<first>> and <<last>> points in the compact set ${𝖡}_{\delta }(r)\cap \mathrm{𝗋𝖺𝗇}(f^{\prime })$ <<on the path $f^{\prime }$ from $t$ to $s$>>. Then if we replace the segment of path $f^{\prime }$ between points $\tilde{t}$ and $\tilde{s}$ with one of the arcs of the circle ${𝖲}_{\delta }(r)$ connecting $\tilde{t}$ and $\tilde{s}$, then we get a new path connecting $t$ and $s$, but now in ${𝖡}_{\epsilon }(A)\setminus \{r\}$.

5. Corollaries of the theorem

Corollary 5.1.

Suppose that $\epsilon >0$ and $A\subseteq {ℝ}^2$ is a nonempty, bounded, $2\epsilon$-chained set. Then the $\epsilon$-boundary of $A$ contains a simple closed curve $C$ such that

$${𝖮}_{\epsilon }(A)\subseteq C^{-}\text{ and }{𝖡}_{\epsilon }(A)\subseteq C^{-}\cup C.$$

Moreover, if the closed $\epsilon$-neighbourhood ${𝖡}_{\epsilon }(A)$ is simply connected, then its boundary $E$ is a simple closed curve and ${𝖡}_{\epsilon }(A)=E^{-}\cup E$.

Note that the formula ${𝖮}_{\epsilon }(A)\subseteq C^{-}$ does not turn into the equality even if the set $B_{\epsilon }(A)$ is simply connected. For example, in the case $A={𝖲}_{\epsilon }(p)$, $p\in {ℝ}^2$.

Proof 5.2.

The number $\epsilon$ and the set $A$ satisfy the conditions of Theorem 4.1. Let $D$ be an unbounded component of the subspace ${ℝ}^2\setminus {𝖡}_{\epsilon }(A)$. By Theorem 4.1, $C≔\partial D$ is a simple closed curve, $C\subseteq {𝖲}_{\epsilon }(A)$, and $D=C^{+}$.

Let us show that ${𝖮}_{\epsilon }(A)\subseteq C^{-}$ and ${𝖡}_{\epsilon }(A)\subseteq C^{-}\cup C$. It is true that $C^{+}=D\subseteq {ℝ}^2\setminus {𝖡}_{\epsilon }(A)$, so ${𝖡}_{\epsilon }(A)\cap C^{+}=\mathrm{\varnothing }$, hence

$${𝖮}_{\epsilon }(A)\subseteq {𝖡}_{\epsilon }(A)\subseteq C^{-}\cup C\subseteq C^{-}\cup {𝖲}_{\epsilon }(A).$$

The only thing left to recall is that ${𝖮}_{\epsilon }(A)\cap {𝖲}_{\epsilon }(A)=\mathrm{\varnothing }$.

Let ${𝖡}_{\epsilon }(A)$ be simply connected. We show that $D={ℝ}^2\setminus {𝖡}_{\epsilon }(A)$. If not, then there exists a component $F$ in ${ℝ}^2\setminus {𝖡}_{\epsilon }(A)$ different from $D$; in particular, $F\cap D=\mathrm{\varnothing }$. Then

$$F\subseteq {ℝ}^2\setminus D={ℝ}^2\setminus C^{+}=C^{-}\cup C.$$

Hence, the component $F$ is bounded. According to Theorem 4.1, the boundary $\partial F\subseteq {𝖲}_{\epsilon }(A)\subseteq {𝖡}_{\epsilon }(A)$ is a simple closed curve and ${(\partial F)}^{-}\cap {𝖡}_{\epsilon }(A)=F\cap {𝖡}_{\epsilon }(A)=\mathrm{\varnothing }$. Thus, ${𝖡}_{\epsilon }(A)$ contains a simple closed curve $\partial F$ such that ${(\partial F)}^{-}$ is disjoint with ${𝖡}_{\epsilon }(A)$ — a contradiction with simple connectedness of ${𝖡}_{\epsilon }(A)$.

Thus, $D$ and ${𝖡}_{\epsilon }(A)$ are disjoint and their union equals ${ℝ}^2$. Consequently, $\partial D=\partial ({𝖡}_{\epsilon }(A))$, i.e., $C=E$, and hence $E$ is a simple closed curve. Finally, note that:

$${𝖡}_{\epsilon }(A)={ℝ}^2\setminus D={ℝ}^2\setminus C^{+}=C^{-}\cup C=E^{-}\cup E.$$

Corollary 5.3.

Suppose that $A\subseteq {ℝ}^2$ is a nonempty, bounded, $2\epsilon$-chained set, $B\subseteq {ℝ}^2$ is a nonempty, $2\left(\rho (A,B)-\epsilon \right)$-chained set, and . Then the $\epsilon$-boundary of $A$ contains a simple closed curve separating $A$ and $B$.

Note that for some $B$ (for example, a straight line) no $\delta$-boundary of $B$ contains a simple closed curve.

Proof 5.4.

Put $\delta ≔\rho (A,B)-\epsilon >0$. Note that, for every $p\in {ℝ}^2$, we have

$$\rho (p,A)+\rho (p,B)\ge \rho (A,B)=\epsilon +\delta ,$$

therefore ${𝖮}_{\delta }(B)\subseteq {ℝ}^2\setminus {𝖡}_{\epsilon }(A)$. The set $B$ is $2\delta$-chained, hence, according to Remark 2.2(2), ${𝖮}_{\delta }(B)$ is pathwise connected. Consider the component $D$ of the subspace ${ℝ}^2\setminus {𝖡}_{\epsilon }(A)$ that contains ${𝖮}_{\delta }(B)$. The number $\epsilon$ and the sets $A$ and $D$ satisfy the conditions of Theorem 4.1, so $C≔\partial D$ is a simple closed curve, $C\subseteq {𝖲}_{\epsilon }(A)$, and $D\in \{C^{+},C^{-}\}$. By construction,

$$A\subseteq {𝖮}_{\epsilon }(A)={𝖡}_{\epsilon }(A)\setminus {𝖲}_{\epsilon }(A)\subseteq {𝖡}_{\epsilon }(A)\setminus C\subseteq ({ℝ}^2\setminus D)\setminus C=({ℝ}^2\setminus C)\setminus D=(C^{+}\cup C^{-})\setminus D.$$

Thus, either $A\subseteq C^{-}$ or $A\subseteq C^{+}$, so $C$ separates $A$ and $D$. Then $C$ separates $A$ and $B$ because $B\subseteq D$.

Corollary 5.5.

Suppose that $\epsilon >0$ and $M\subseteq {ℝ}^2$ is a nonempty, bounded set. Then each pair of different $2\epsilon$-chained components of $M$ are separated by some simple closed curve contained in the $\epsilon$-boundary of $M$.

Proof 5.6.

Let $A$ and $B$ be two different $2\epsilon$-chained components of $M$. Note that $\rho (A,B)\ge 2\epsilon$. Then $2(\rho (A,B)-\epsilon )\ge 2\epsilon$, and therefore, by Remark 2.2(3), $B$ is $2(\rho (A,B)-\epsilon )$-chained. Thus, the number $\epsilon$ and the sets $A$ and $B$ satisfy the conditions of Corollary 5.3, so ${𝖲}_{\epsilon }(A)$ contains a simple closed curve separating $A$ and $B$. Finally note that ${𝖲}_{\epsilon }(A)\subseteq {𝖲}_{\epsilon }(M)$.

From Corollaries 5.1 and 5.5 the following curious result follows, which complements Morton Brown’s claim [2] that the $\epsilon$-boundary of a compact subset of the plane is contained in the union of a finite number of simple closed curves:

Corollary 5.7.

The $\epsilon$-boundary of a nonempty, bounded subset of the plane contains a simple closed curve for all $\epsilon >0$. ∎

Corollary 5.8.

Suppose that $A,B\subseteq {ℝ}^2$ are nonempty, $\rho (A,B)$-chained sets, $A$ is bounded, and $\rho (A,B)>0$. Then the $\rho (A,B)/2$-boundary of $A$ contains a simple closed curve $C$ that separates $A$ and $B$ and such that

$$\rho (C,A\cup B)=\rho (C,A)=\rho (C,B)=\rho (A,B)/2.$$

In particular, among the simple closed curves separating $A$ and $B$, the curve $C$ is a curve maximally distant from $A\cup B$.

Proof 5.9.

The sets $A$ and $B$ and the number $\epsilon ≔\rho (A,B)/2$ satisfy the conditions of Corollary 5.3. Hence ${𝖲}_{\epsilon }(A)$ contains a simple closed curve $C$ separating $A$ and $B$. In particular, $\rho (C,A)=\epsilon$.

Let us show that $\rho (C,B)\ge \epsilon$. If this is not true, then there are points $p\in C$, $q\in B$ and a number $\delta >0$ such that . Since $p\in C\subseteq {𝖲}_{\epsilon }(A)$, there exists a point $r\in A$ such that . But then

a contradiction.

Now we show that $\rho (C,B)\le \epsilon$. Since $\rho (C,A)>0$ and $\rho (C,B)>0$, the simple closed curve $C$ separates $\overline{A}$ and $\overline{B}$. The set $\overline{A}$ is compact, so there are points $s\in \overline{A}$ and $t\in \overline{B}$ such that $\rho (s,t)=\rho (A,B)=2\epsilon$. Thus the segment $[s,t]$ intersects both components of the complement of $C$, and so it intersects $C$ as well. Let $u\in C\cap [s,t]$. Since $u\in C\subseteq {𝖲}_{\epsilon }(A)$, then $\rho (u,s)\ge \epsilon$. Then $\rho (u,t)=\rho (s,t)-\rho (s,u)\le \rho (s,t)-\epsilon =\epsilon$. Therefore, $\rho (C,B)=\rho (C,\overline{B})\le \rho (u,t)\le \epsilon$.

Thus, $\rho (C,A\cup B)=\rho (C,A)=\rho (C,B)=\rho (A,B)/2$. It remains to show that

$$\rho (C,A\cup B)=\max \{\rho (C^{\prime },A\cup B):C^{\prime }\text{ is a simple closed curve separating }A\text{ and }B\}.$$

Suppose that a simple closed curve $C^{\prime }$ separates $A$ and $B$. Let $u^{\prime }\in C^{\prime }\cap [s,t]$. Then

$$\rho (u^{\prime },\{s,t\})\le \rho (s,t)/2=\rho (A,B)/2=\rho (C,A\cup B).$$

Thus,

$$\rho (C,A\cup B)\ge \rho (u^{\prime },\{s,t\})\ge \rho (C^{\prime },\overline{A\cup B})=\rho (C^{\prime },A\cup B).$$

Remark 5.10.

Suppose that a simple closed curve $C$ separates nonempty sets $A$ and $B$ in ${ℝ}^2$ and . Then there exists a simple closed curve $C^{\prime }$ that separates $A$ and $B$ and such that $\rho (C^{\prime },A\cup B)>\rho (C,A\cup B)$.

In light of Corollary 5.8 and Remark 5.10, it is interesting to note that there exist two sets, $A$ and $B$, which can be separated by some simple closed curve, but the supremum of the distances from such curves to $A\cup B$ is not attained. See Subsection 6.5 for an example.

Question 5.11.

Suppose that $A,B\subseteq {ℝ}^2$ are nonempty sets, $\rho (A,B)=1$, and there exists a simple closed curve that separates $A$ and $B$. Is there a simple closed curve $C$ that separates $A$ and $B$ and such that $\rho (C,A\cup B)>0$?

Proof 5.12 (Proof of Remark 5.10.).

We may assume that $A\subseteq C^{-}$. Put $\delta ≔\rho (C,A\cup B)>0$. We have ${𝖮}_{\delta }(A)=\bigcup \{{𝖮}_{\delta }(p):p\in A\}\subseteq C^{-}$, so, for every $q\in {𝖮}_{\delta }(A)$, there is $p\in A$ such that $q\in {𝖮}_{\delta }(p)\subseteq {𝖮}_{\delta }(A)$. It follows that the Lebesgue measure of each component of ${𝖮}_{\delta }(A)$ is greater than $\pi {\delta }^2$. Consequently, the number of components of ${𝖮}_{\delta }(A)$ is finite; denote them by $A_1,\mathrm{\dots },A_n$. For every $m\le n$, there is a path that connects in $C^{-}$ some point from $A_1$ and some point from $A_m$ because $C^{-}$ is pathwise connected. Then we can find a compact set $D\subseteq C^{-}$ such that the set

$$A^{\prime }≔{𝖮}_{\delta }(A)\cup D\subseteq C^{-}$$

is connected.

There exists an open ball $O\subseteq {ℝ}^2$ that contains $C\cup C^{-}$. Put $E≔{ℝ}^2\setminus O$. We have $E\subseteq C^{+}$ and ${𝖮}_{\delta }(B)\subseteq C^{+}$. The Lebesgue measure of each component of ${𝖮}_{\delta }(B)$ is greater than $\pi {\delta }^2$. So, the number of components of ${𝖮}_{\delta }(B)$ that are disjoint from $E$ is finite. Then we can find a compact set $F\subseteq C^{+}$ such that the set

$$B^{\prime }≔E\cup {𝖮}_{\delta }(B)\cup F\subseteq C^{+}$$

is connected. There exists $\epsilon >0$ such that

Note that $\rho (A,B)>2(\epsilon +\delta )$.

We show that $\rho (A^{\prime },B^{\prime })⩾\epsilon$. Let $p\in A^{\prime }$ and $q\in B^{\prime }$. The segment $[p,q]$ intersects $C$ because $p\in C^{-}$ and $q\in C^{+}$. Let $r\in [p,q]\cap C$. If $p\in D$, then $\rho (p,r)>\epsilon$. If $q\in E\cup F$, then $\rho (q,r)>\epsilon$. In both cases, $\rho (p,q)>\epsilon$. It remains to consider the case when $p\in {𝖮}_{\delta }(A)$ and $q\in {𝖮}_{\delta }(B)$. There are $\dot{p}\in A$ and $\dot{q}\in B$ such that and . By the triangle inequality, we have

$$\rho (\dot{p},p)+\rho (p,q)+\rho (q,\dot{q})\ge \rho (\dot{p},\dot{q})\ge \rho (A,B)>\mathrm{\hspace{0.25em}2}(\epsilon +\delta ),$$

therefore

$$\rho (p,q)>\mathrm{\hspace{0.25em}2}\epsilon +2\delta -\rho (\dot{p},p)-\rho (q,\dot{q})>\mathrm{\hspace{0.25em}2}\epsilon >\epsilon .$$

The sets $A^{\prime },B^{\prime }\subseteq {ℝ}^2$ are nonempty and connected, $A^{\prime }$ is bounded, and $\rho (A^{\prime },B^{\prime })>0$. Then, by Corollary 5.8, there is a simple closed curve $C^{\prime }$ that separates $A^{\prime }$ and $B^{\prime }$ and such that

$$\rho (C^{\prime },A^{\prime }\cup B^{\prime })=\rho (A^{\prime },B^{\prime })/2\ge \epsilon /2.$$

Then $C^{\prime }$ that separates $A$ and $B$ because $A\subseteq A^{\prime }$ and $B\subseteq B^{\prime }$.

It remains to show that $\rho (C^{\prime },A\cup B)>\rho (C,A\cup B).$ We shall prove

$$\rho (C^{\prime },A\cup B)\ge \rho (C,A\cup B)+\epsilon /2.$$

Let $p\in C^{\prime }$ and $q\in A\cup B$. First consider the case when $q\in A$. We have $\rho (p,q)\ge \delta$ — otherwise, $p\in {𝖮}_{\delta }(A)$, and then

$$0=\rho (C^{\prime },{𝖮}_{\delta }(A))\ge \rho (C^{\prime },A^{\prime }\cup B^{\prime })\ge \epsilon /2$$

because ${𝖮}_{\delta }(A)\subseteq A^{\prime }$. Then there is a point $r\in [p,q]$ such that $\rho (r,q)=\delta$. We have $r\in \overline{{𝖮}_{\delta }(A)}$, so again, now using Remark 2.1(2), we have

$$\rho (p,r)\ge \rho (C^{\prime },\overline{{𝖮}_{\delta }(A)})=\rho (C^{\prime },{𝖮}_{\delta }(A))\ge \rho (C^{\prime },A^{\prime }\cup B^{\prime })\ge \epsilon /2.$$

Therefore,

$$\rho (p,q)=\rho (p,r)+\rho (r,q)\ge \epsilon /2+\delta =\rho (C,A\cup B)+\epsilon /2.$$

The case when $q\in B$ is identical.

6. Necessity of conditions in the theorem and its corollaries

What if $\epsilon =0$? In this case the $\epsilon$-boundary ${𝖲}_{\epsilon }(A)$ of a set $A$ coincides with its closure. Therefore, questions about simple closed curves lying in ${𝖲}_0(A)$ are far from the topic of this paper. Also note that the example of Lakes of Wada [3, 16] shows that even if $A$ and $B$ are disjoint, simply connected domains and $\rho (A,B)=0$, there may not exist a simple closed curve separating them.

The following examples show the necessity of the assumptions of Theorem 4.1 and its corollaries.

6.1. Boundedness of the set $A$

The boundedness condition in Theorem 4.1 and Corollaries 5.1, 5.3, 5.5, and 5.7 is essential. Indeed, if the set $A$ is a straight line, then its $\epsilon$-boundary does not contain a simple closed curve. In Corollary 5.8, the boundedness condition is also essential, since two unbounded subsets of the plane cannot be separated by a simple closed curve.

6.2. $2\epsilon$-chainedness of the set $A$

The $2\epsilon$-chainedness condition cannot be weakened to the $(2\epsilon +\delta )$-chainedness condition in neither Theorem 4.1 nor in Corollaries 5.1 and 5.3, for no $\delta >0$. Indeed, if the set $A$ consists of two points at distance $2\epsilon$, then its $\epsilon$-boundary does not contain a simple closed curve $C$ such that $A\subseteq C^{-}$ or $A\subseteq C^{+}$.

6.3. Simply connectedness of the closed $\epsilon$-neighbourhood of the set $A$

The simply connectedness condition in Corollary 5.1 is essential. Indeed, if the set $A$ is a circle of radius greater than $\epsilon$, then its closed $\epsilon$-neighbourhood is not simply connected. And the boundary of that $\epsilon$-neighbourhood is not a simple closed curve.

6.4. $2(\rho (A,B)-\epsilon )$-chainedness of the set $B$

The $2(\rho (A,B)-\epsilon )$-chainedness condition of the set $B$ in Corollary 5.3 cannot be relaxed to the $(2(\rho (A,B)-\epsilon )+\delta )$–chainedness for no $\delta >0$. Indeed, consider a circle of radius $2\epsilon$. Choose points $p$ and $q$ on it such that $\rho (p,q)=2\epsilon$. Let $A$ be the closed arc of the circle between $p$ and $q$ whose length is greater than half the length of the circle. Let $B$ consists of two different points of the perpendicular bisector of the line segment $[p,q]$, which are at distance $\mathrm{𝗆𝗂𝗇}\{\delta /2,\epsilon \}$ from the segment $[p,q]$. The set $B$ is $(2(\rho (A,B)-\epsilon )+\delta )$–chained, but ${𝖲}_{\epsilon }(A)$ does not contain a simple closed curve separating $A$ and $B$.

6.5. $\rho (A,B)$-chainedness of the sets $A$ and $B$

The $\rho (A,B)$-chainedness condition of $A$ in Corollary 5.8 is essential. Indeed, let $B$ be the closed longer arc of a circle with ends at points $p$ and $q$ such that $\rho (p,q)$ equals the radius of the circle. Let the point $s$ be the center of the circle and the point $t$ be the point symmetric to $s$ with respect to the segment $[p,q]$. Put $A≔\{s,t\}$. We have

$$\rho (p,q)=\rho (s,p)=\rho (s,q)=\rho (p,t)=\rho (q,t)=\rho (A,B).$$

In particular, the set $A$ is not $\rho (A,B)$-chained.

Suppose that a simple closed curve $C$ separates $A$ and $B$. Then $C$ intersects the open segments $(s,p)$ and $(p,t)$. Let $u\in C\cap (s,p)$ and $v\in C\cap (p,t)$. Then the simple closed curve $B\cup (p,q)$ separates the points $u$ and $v$. It follows that $C$ intersects $(p,q)$ at least twice. Therefore,

In other words, there is no simple closed curve $C$ separating $A$ and $B$ such that $\rho (C,A\cup B)=\rho (A,B)/2$.

For every $\epsilon >0$, there is a simple closed curve $C_{\epsilon }$ separating $A$ and $B$ such that

$$\rho (C_{\epsilon },A\cup B)>(\rho (A,B)/2)-\epsilon .$$

It follows that among the simple closed curves separating $A$ and $B$, there is no curve maximally distant from $A\cup B$.

If we swap the sets $A$ and $B$ in this example, we get an example showing that the $\rho (A,B)$-chainedness condition of the set $B$ in Corollary 5.8 is also essential.

7. Similar questions in ${ℝ}^3$

In 1976 S. Ferry showed [6] that the $\epsilon$-boundary of a set $A\subseteq {ℝ}^3$ is a 2-manifold for almost all $\epsilon$. He also proved that if $A$ is a finite polyhedron in ${ℝ}^n$, then its $\epsilon$-boundary is an $(n-\mathrm{\hspace{0.17em}1})$-manifold for all sufficiently small values of $\epsilon$. In the same paper he constructed a set $B\subseteq {ℝ}^3$ such that the $\epsilon$-boundary of $B$ has components which are not 2-manifolds for uncountably many $\epsilon$. And he constructed a Cantor set in ${ℝ}^4$ whose $\epsilon$-boundary is not a 3-manifold for any $\epsilon$ between 0 and 1.

A subset of the plane is homeomorphic to a circle if and only if it is a compact, connected, one-dimensional manifold. Thus, in ${ℝ}^3$, there are two different analogues of the concept of a simple closed curve, the “spherical” and the “topological”:

• a set that is homeomorphic to the two-dimensional sphere and

• a compact, connected, two-dimensional manifold.

The following example shows that in both cases the ${ℝ}^3$ variants of Theorem 4.1 and Corollaries 5.1–5.7 do not hold.

Example 7.1.

Consider two linked circles, each with a small open arc removed:

$$A≔\{(x,y,0)\in {ℝ}^3:x^2+y^2=1\text{ and }x\ge -1+\delta \}\text{and}$$

$$B≔\{(x,0,z)\in {ℝ}^3:{(x-1)}^2+z^2=1\text{ and }x\le 1-\delta \},$$

where . Let $\epsilon ≔\rho (p,q)/2$, where $p$ and $q$ are the ends of the closed arc $A$.

The $\epsilon$-boundary of $A$ resembles the surface of a “sausage” bent so that its ends touch each other. It can be shown that the $\epsilon$-boundary of $A$ contains no nonempty, compact, connected, two-dimensional manifold. In particular, it does not contain any set homeomorphic to the two-dimensional sphere. The same is true about $B$.

If we modify the above example by taking $B$ to be a line such that $\rho (A,B)/2=\epsilon$, then we can see that neither the “spherical” nor the “topological” ${ℝ}^3$ version of Corollary 5.8 holds. It is therefore interesting to consider a slightly weaker version of Corollary 5.8:

Corollary 7.2.

Let $A,B\subseteq {ℝ}^2$ be disjoint nonempty continua. Then there is a simple closed curve $C$ that separates $A$ and $B$ and such that $\rho (C,A\cup B)=\rho (A,B)/2$.

The following example shows that the “spherical” ${ℝ}^3$ analogue of Corollary 7.2 also fails.

Example 7.3.

Consider two linked circles

$$A^{\prime }≔\{(x,y,0)\in {ℝ}^3:x^2+y^2=1\}\text{and}{B}^{\prime }≔\{(x,0,z)\in {ℝ}^3:{(x-1)}^2+z^2=1\}.$$

The sets $A^{\prime }$ and $B^{\prime }$ cannot be separated by a set $C$ homeomorphic to the 2-dimensional sphere and such that $\rho (C,A^{\prime }\cup B^{\prime })=\rho (A^{\prime },B^{\prime })/2$.

This example does not refute the “topological” ${ℝ}^3$ analogue of Corollary 7.2: the sets $A^{\prime }$ and $B^{\prime }$ can be separated by a compact, connected, two-dimensional manifold $C$ such that $\rho (C,A^{\prime }\cup B^{\prime })=\rho (A^{\prime },B^{\prime })/2$. So the question of the validity of the “topological” ${ℝ}^3$ version of Corollary 7.2 remains open:

Question 7.4.

Let $A,B\subseteq {ℝ}^3$ be disjoint (simply connected) nonempty continua. Is there a (compact, connected) two-dimensional manifold $C$ such that $A$ and $B$ lie in different components of ${ℝ}^3\setminus C$ and $\rho (C,A\cup B)=\rho (A,B)/2$ ?