1 Introduction and main results

It is well known that a line segment can be continuously moved in a planar set of arbitrarily small area such that it returns to its starting position, with its direction reversed. This fact was first proved by Besicovitch as a solution to the classical Kakeya problem [Reference Besicovitch1].

We may ask if there are other sets in the plane having a similar property. We shall say that a planar set $A$ has the Kakeya property (property (K) for short) if there exist two different positions of $A$ such that one can move $A$ continuously from the first position to the second such that the set of points touched by the moving set has an arbitrarily small area. Obviously, every line segment has property (K). More generally, if $A$ is a null set, and $A$ is the union of parallel lines, then $A$ has property (K). Indeed, sliding $A$ in the direction of the lines we only cover a set of measure zero. Similarly, if $A$ is a null set, and $A$ is the union of concentric circles, then $A$ has property (K), as rotating $A$ about the centre of the circles we only cover a set of measure zero. We shall say that a set $A\subset \mathbb{R}^{2}$ is a trivial (K)-set if $A$ can be covered by a null set which is either the union of parallel lines or the union of concentric circles. Our aim is to prove that if $A$ is closed and has the Kakeya property, then the union of the non-trivial connected components of $A$ is trivial.

Besicovitch’s construction used the ‘Pál joins’ in order to shift the line segment to an arbitrary parallel position using an arbitrarily small area. (There are solutions to the Kakeya problem without using Pál joins; see e.g. [Reference Cunningham3–Reference Davies5].) Using the Pál joins and Besicovitch’s theorem, it is clear that every line segment can be moved to arrive at any prescribed position within a set of arbitrarily small area.

We shall say that a planar set $A$ has the strong Kakeya property (property  $(\text{K}^{\text{s}})$ for short) if, for every prescribed position, $A$ can be continuously moved within a set of arbitrarily small area to the prescribed final position.

The question of whether or not circular arcs have property  $(\text{K}^{\text{s}})$ was asked by Cunningham (see [Reference Cunningham4, p. 591]). It is proved in [Reference Chang and Csörnyei2, Reference Héra and Laczkovich6] that the answer is affirmative, at least for circular arcs of angle short enough. In this note we shall prove that every connected closed set with property  $(\text{K}^{\text{s}})$ is a line segment or a circular arc.

Now we formulate these notions and results precisely. A rigid motion is an isometry of the plane preserving orientation; that is, a translation or a rotation. We denote the identity map by $j$ . We identify the plane with the complex plane $\mathbb{C}$ , and put $S^{1}=\{x\in \mathbb{C}:|x|=1\}$ . Then every rigid motion is a function of the form $x\mapsto ux+c$ , where $u,c\in \mathbb{C}$ and $u\in S^{1}$ . By a continuous movement $M$ we mean a map $t\mapsto M_{t}$   $(t\in [0,1])$ such that $M_{t}$ is a rigid motion for every $t\in [0,1]$ , the map $(t,x)\mapsto M_{t}(x)$ is continuous on $[0,1]\times \mathbb{R}^{2}$ and $M_{0}=j$ . If $M$ is a continuous movement, then the set of points touched by a moving set $A$ is

The two-dimensional Lebesgue measure of the set $A\subset \mathbb{R}^{2}$ is denoted by $\unicode[STIX]{x1D706}(A)$ .

The set $A$ has property (K) if there is a rigid motion $\unicode[STIX]{x1D6FC}\neq j$ such that for every $\unicode[STIX]{x1D700}>0$ there exists a continuous movement $M$ such that $M_{1}=\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D706}(W_{M}(A))<\unicode[STIX]{x1D700}$ . The set $A$ has property  $(\text{K}^{\text{s}})$ if for every rigid motion $\unicode[STIX]{x1D6FC}$ and for every $\unicode[STIX]{x1D700}>0$ there exists a continuous movement $M$ such that $M_{1}=\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D706}(W_{M}(A))<\unicode[STIX]{x1D700}$ . By a non-trivial connected component of a set $A$ we mean a connected component of $A$ having at least two points. Our main results are the following.

It is easy to see that full circles do not have property  $(\text{K}^{\text{s}})$ . Indeed, moving a circle $C$ to a circle disjoint from $C$ , the moving circle must touch every point inside $C$ . Since every circle has property (K), we can see that properties (K) and  $(\text{K}^{\text{s}})$ are not equivalent. As we mentioned earlier, all circular arcs of angle short enough have property  $(\text{K}^{\text{s}})$ . It is not clear, however, whether or not all circular arcs other than the full circles have property  $(\text{K}^{\text{s}})$ , while, obviously, they have property (K).

It is easy to see that every set of Hausdorff dimension less than 1 has property  $(\text{K}^{\text{s}})$ . We show that there are compact sets with property  $(\text{K}^{\text{s}})$ having Hausdorff dimension 2. Let $E\subset [1,2]$ and $F\subset [0,1/2]$ be compact sets of linear measure zero and of Hausdorff dimension 1. We put

It is easy to check that $A$ is a bi-Lipschitz image of $E\times F$ , and thus $A$ has Hausdorff dimension $2$ . We prove that $A$ has property  $(\text{K}^{\text{s}})$ . Let $\unicode[STIX]{x1D6FC}(x)=ux+c$ be a rigid motion, where $|u|=1$ . Let $c=r\cdot v$ , where $r\geqslant 0$ and $|v|=1$ . Then the following continuous movement brings $A$ onto $\unicode[STIX]{x1D6FC}(A)$ . First we rotate $A$ about the origin to obtain $v\cdot A$ , then we translate $v\cdot A$ by the vector $c$ to obtain $v\cdot A+c$ and then we rotate $v\cdot A+c$ about the point $c$ to obtain $u\cdot A+c=\unicode[STIX]{x1D6FC}(A)$ . If $B=\{(x,y):\sqrt{x^{2}+y^{2}}\in E\}$ , then the set of points touched by this continuous movement is a subset of

This is a set of plane measure zero, showing that $A$ has property  $(\text{K}^{\text{s}})$ .

We note that the union of finitely many parallel segments has property  $(\text{K}^{\text{s}})$ ; this was proved by Davies in [Reference Davies5, Theorem 3]. It is not clear if the union of finitely many concentric circular arcs can have property  $(\text{K}^{\text{s}})$ .

We sketch the construction of a totally disconnected compact set $A$ having property  $\text{(K)}$ such that $A$ has orthogonal projection of positive linear measure in every direction $e\in S^{1}$ , and the distance set

has positive linear measure for all $p\in \mathbb{R}^{2}$ . This easily implies that $A$ is not trivial. The construction is based on the so-called “Venetian blind” construction. Using this method, Talagrand proved in [Reference Talagrand8] that for every upper semi-continuous function $f$ on $S^{1}$ there exists a compact set $K\subset \mathbb{R}^{2}$ such that the measure of the projection of $K$ in the direction $e$ is $f(e)$ for every $e\in S^{1}$ . We say that a rectangle $R$ is in the direction $e$ if the longer side of $R$ is of direction  $e$ .

We fix a direction $e$ and choose a convergent sequence of distinct directions $e_{n}\rightarrow e$ . We shall construct a decreasing sequence of compact sets $K_{n}$ such that $K_{n}$ is the union of finitely many narrow rectangles in the direction $e_{n}$ for every $n$ . Let $K_{1}$ be a closed rectangle in the direction $e_{1}$ of width less than $1$ . Then the measure of the projection of $K_{1}$ in the direction $e_{1}$ is less than $1$ , and the measure of the distance set $D(K_{1};p)$ is positive for all $p\in \mathbb{R}^{2}$ . In the $(n+1)\text{th}$ step we replace each member of the finite system of closed rectangles whose union is $K_{n}$ by a finite union of closed rectangles such that the new rectangles have direction $e_{n+1}$ , and the union of these $(n+1)\text{th}$ -generation rectangles, $K_{n+1}$ , is a subset of $K_{n}$ . Moreover, by choosing these rectangles appropriately, the measure of the projection of $K_{n+1}$ in the direction $e_{n+1}$ will be less than $1/(n+1)$ , and the measure of the projection of $K_{n+1}$ in directions $e_{1},\ldots ,e_{n}$ will be only slightly smaller than that of the projection of $K_{n}$ in directions $e_{1},\ldots ,e_{n}$ , respectively. If we choose the new rectangles originating from a fixed previous rectangle to be close enough to each other, we can achieve that the set $D(K_{n+1};p)$ is only slightly smaller than $D(K_{n};p)$ for all $p\in \mathbb{R}^{2}$ , $|p|\leqslant n$ . We set $K=\bigcap _{n=1}^{\infty }K_{n}$ . We may also ensure that the projection of $K$ has positive linear measure in all directions (see [Reference Talagrand8]). The set $D(K;p)$ has positive measure for all $p\in \mathbb{R}^{2}$ . Indeed, for every $p$ there is an $n$ such that $|p|\leqslant n$ and, then, for all $m\geqslant n$ we have $\unicode[STIX]{x1D706}_{1}(D(K_{m};p))\geqslant \unicode[STIX]{x1D706}_{1}(D(K_{n};p))/2$ and thus $\unicode[STIX]{x1D706}_{1}(D(K;p))\geqslant \unicode[STIX]{x1D706}_{1}(D(K_{n};p))/2>0$ . It is easy to see that $K$ has property  $(\text{K})$ . Indeed, if $\unicode[STIX]{x1D6FC}$ is a translation by a vector of direction $e$ and length $R>0$ , then $K$ can be moved continuously to $\unicode[STIX]{x1D6FC}(K)$ within a set of arbitrarily small area. Then $K$ has the desired properties, and the proof is finished.

2 Some auxiliary results and the proofs of Theorems 1.1 and 1.2

In this section we formulate some auxiliary results needed for the proofs of Theorems 1.1 and 1.2.

The translation by the vector $c$ will be denoted by $T_{c}$ . Thus, $T_{c}(x)=x+c$ for every $x\in \mathbb{C}$ . The rotation about the point $c$ by angle $\unicode[STIX]{x1D719}$ is denoted by $R_{c,\unicode[STIX]{x1D719}}$ . Thus, $R_{c,\unicode[STIX]{x1D719}}(x)=\text{e}^{\text{i}\unicode[STIX]{x1D719}}(x-c)+c$ for every $x\in \mathbb{C}$ . If $\unicode[STIX]{x1D6FC}$ is a rigid motion and $\unicode[STIX]{x1D6FC}^{2}\neq j$ , then we define the elementary movement $E^{\unicode[STIX]{x1D6FC}}$ determined by $\unicode[STIX]{x1D6FC}$ as follows. If $\unicode[STIX]{x1D6FC}=T_{c}$ , then we put $E_{t}^{\unicode[STIX]{x1D6FC}}=T_{tc}$ for every $t\in [0,1]$ . If $\unicode[STIX]{x1D6FC}=R_{a,\unicode[STIX]{x1D719}}$ , then the condition $\unicode[STIX]{x1D6FC}^{2}\neq j$ implies that $\unicode[STIX]{x1D719}\not \equiv \unicode[STIX]{x1D70B}$ (mod $2\unicode[STIX]{x1D70B}$ ). Therefore, we may assume that $|\unicode[STIX]{x1D719}|<\unicode[STIX]{x1D70B}$ . Then we define $E_{t}^{\unicode[STIX]{x1D6FC}}=R_{a,t\unicode[STIX]{x1D719}}$ for every $t\in [0,1]$ .

The inverse of the rigid motion $\unicode[STIX]{x1D6FC}$ is denoted by $\unicode[STIX]{x1D6FC}^{-1}$ . It is clear that if $M$ is a continuous movement, then so is $M^{-1}$ , where we put $M_{t}^{-1}=(M_{t})^{-1}$ $(t\in [0,1])$ . The $\unicode[STIX]{x1D700}$ -neighbourhood of a set $E\subset \mathbb{R}^{2}$ is defined by

We denote by $L$ the space of functions $x\mapsto ux+v$   $(x\in \mathbb{C})$ , where $u,v\in \mathbb{C}$ . Clearly, $L$ is a linear space over $\mathbb{C}$ with pointwise addition and multiplication by constants. We endow $L$ with the norm $\Vert f\Vert =|u|+|v|$ . It is easy to check that $\Vert f\Vert =\sup \{|f(x)|:x\in \mathbb{C},|x|\leqslant 1\}$ for every $f\in L$ , and thus $\Vert \cdot \Vert$ is indeed a norm.

It is also easy to check that if $f_{1}(x)=u_{1}x+v_{1}$ and $f_{2}(x)=u_{2}x+v_{2}$ are rigid motions, then

By a continuum we mean a compact connected set. The set $A\subset \mathbb{R}^{2}$ is said to be irreducible between the distinct points $a$ and $b$ provided that $A$ is connected, $a,b\in A$ and these two points cannot be joined by any closed, connected, proper subset of $A$ . We shall need the following topological lemma on irreducible continuums.

The proofs of Lemmas 2.1 and 2.2 will be given in the next two sections.

In order to prove Theorem 1.2, we only have to show that halflines and full circles do not have property $(\text{K}^{\text{s}})$ . We already saw this for circles. The case of halflines is also clear, as a horizontal halfline cannot be moved continuously to a vertical halfline touching only a finite area.

3 Proof of Lemma 2.1

Let $\unicode[STIX]{x1D6FC}$ be a rigid motion and $\unicode[STIX]{x1D6FC}(x)=ux+a$ , where $|u|=1$ . Let $\unicode[STIX]{x1D6FC}^{n}$ denote the $n\text{th}$ iterate of $\unicode[STIX]{x1D6FC}$ . We would like to compare the distances $\Vert \unicode[STIX]{x1D6FC}^{n}-j\Vert$ and $\Vert \unicode[STIX]{x1D6FC}-j\Vert$ .

It is easy to check that if $\unicode[STIX]{x1D6FC}(x)=ux+a$ , where $|u|=1$ , then $\unicode[STIX]{x1D6FC}^{n}(x)=u^{n}x+(1+u+\cdots +u^{n-1})a$ , and thus

Since $|u|=1$ , it follows from (2) that $\Vert \unicode[STIX]{x1D6FC}^{n}-j\Vert \leqslant n\cdot \Vert \unicode[STIX]{x1D6FC}-j\Vert$ for every $\unicode[STIX]{x1D6FC}$ . Now we show that if $|u-1|\leqslant 1/n$ , then

Since $|u^{i}-u^{i-1}|=|u-1|$ for every $i=1,2,\ldots$ , we obtain $|u^{k}-1|\leqslant k\cdot |u-1|$ for every $k=1,2,\ldots \,$ . Then, assuming that $|u-1|\leqslant 1/n$ , we find that

Thus, $|1+u+\cdots +u^{n-1}|\geqslant n/2$ , and then (2) gives (3).

It is clear that if $\unicode[STIX]{x1D6FC}$ is a translation or a rotation of angle $\unicode[STIX]{x1D719}$ with $|\unicode[STIX]{x1D719}|<\unicode[STIX]{x1D70B}/n$ , then

We shall also need the following estimate.

In order to prove the lemma, one has to distinguish between the cases $u=1$ and $u\neq 1$ . In both cases the proof is an easy computation; we leave the details to the reader.

Now we turn to the proof of Lemma 2.1. Since $A\subset \mathbb{R}^{2}$ has property (K), there is a rigid motion $\unicode[STIX]{x1D6FD}\neq j$ such that for every $\unicode[STIX]{x1D700}>0$ there exists a continuous movement $M$ with $M_{1}=\unicode[STIX]{x1D6FD}$ and $\unicode[STIX]{x1D706}(W_{M}(A))<\unicode[STIX]{x1D700}$ . Let $\unicode[STIX]{x1D6FD}(x)=v_{0}x+b_{0}$ , where $|v_{0}|=1$ . We put $\unicode[STIX]{x1D702}=\min (|v_{0}-1|+|b_{0}|,1)$ . Since $\unicode[STIX]{x1D6FD}\neq j$ , we have $\unicode[STIX]{x1D702}>0$ .

We choose, for every $n$ , a continuous movement $M^{n}$ such that $M_{1}^{n}=\unicode[STIX]{x1D6FD}$ and $\unicode[STIX]{x1D706}(W_{M^{n}}(A))<1/n^{2}$ . Let $M_{t}^{n}(x)=v_{n}(t)x+b_{n}(t)$ for every $x\in \mathbb{C}$ and $t\in [0,1]$ , where $|v_{n}|=1$ . Since the maps $t\mapsto v_{n}(t)$ , $t\mapsto b_{n}(t)$ are continuous and $v_{n}(1)=v_{0},~b_{n}(1)=b_{0}$ , there is a $0<t_{n}\leqslant 1$ such that $|v_{n}(t_{n})-1|+|b_{n}(t_{n})|=\unicode[STIX]{x1D702}/n$ , and $|v_{n}(t)-1|+|b_{n}(t)|<\unicode[STIX]{x1D702}/n$ for every $t\in [0,t_{n} )$ . Replacing $M^{n}$ by $M^{n}\circ \unicode[STIX]{x1D713}$ , where $\unicode[STIX]{x1D713}$ is a suitable homeomorphism of $[0,1]$ onto itself, we may assume that $t_{n}=1/n$ .

We put $\unicode[STIX]{x1D6FD}_{n}=M_{1/n}^{n}$ and $\unicode[STIX]{x1D6FC}_{n}=\unicode[STIX]{x1D6FD}_{n}^{n}$ . Then $\Vert \unicode[STIX]{x1D6FD}_{n}-j\Vert =\unicode[STIX]{x1D702}/n$ and $\unicode[STIX]{x1D702}/2\leqslant \Vert \unicode[STIX]{x1D6FC}_{n}-j\Vert \leqslant \unicode[STIX]{x1D702}$ by (3). Let $\unicode[STIX]{x1D6FC}_{n}(x)=u_{n}x+a_{n}$ , where $|u_{n}|=1$ . Since $\unicode[STIX]{x1D702}/2\leqslant |u_{n}-1|+|a_{n}|\leqslant \unicode[STIX]{x1D702}$ for every $n$ , the sequences $(u_{n})$ and $(a_{n})$ have convergent subsequences. Turning to a suitable subsequence, we may assume that the sequences $(u_{n})$ and $(a_{n})$ converge. Let $u_{n}\rightarrow u$ and $a_{n}\rightarrow a$ . Then $|u|=1$ , and thus $\unicode[STIX]{x1D6FC}(x)=ux+a$ defines a rigid motion. Our aim is to show that $\unicode[STIX]{x1D6FC}$ satisfies the requirements. Since $\Vert \unicode[STIX]{x1D6FC}-j\Vert =|u-1|+|a|\geqslant \unicode[STIX]{x1D702}/2>0$ , we can see that $\unicode[STIX]{x1D6FC}\neq j$ . Also, by $\Vert \unicode[STIX]{x1D6FC}-j\Vert \leqslant \unicode[STIX]{x1D702}\leqslant 1$ , we have $\unicode[STIX]{x1D6FC}^{2}\neq j$ . Indeed, $\unicode[STIX]{x1D6FC}^{2}=j$ would imply that $u=-1$ and $\Vert \unicode[STIX]{x1D6FC}-j\Vert \geqslant 2$ .

We define the continuous movement $F^{n}$ as follows: we replace the movement $E_{t}^{\unicode[STIX]{x1D6FC}_{n}}$ between the moments $t=(i-1)/n$ and $t=i/n$ by suitable copies of the movement $M_{t}$   $(t\in [0,1/n])$ . More precisely, we put

for $t\in [(i-1)/n,i/n]$ and $i=1,\ldots ,n$ . Since $M_{0}^{n}=j$ and $M_{1/n}^{n}=\unicode[STIX]{x1D6FD}_{n}$ , it follows that $\unicode[STIX]{x1D6FD}_{n}^{i-1}\circ M_{1/n}^{n}=\unicode[STIX]{x1D6FD}_{n}^{i}\circ M_{0}^{n}$ , and thus the definition makes sense. This also proves that $F^{n}$ is a continuous movement. Let $W=\{M_{t}^{n}(x):x\in A,t\in [0,\,1/n]\}$ . Since $W\subset W_{M}(A)$ , we have $\unicode[STIX]{x1D706}(W)<1/n^{2}$ . Also, $W_{F^{n}}(A)$ is the union of $n$ congruent copies of $W$ ; therefore, we have $\unicode[STIX]{x1D706}(W_{F^{n}}(A))<1/n$ .

Next we show that for every $x\in \mathbb{C}$ , $|x|\leqslant 1$ and $t\in [0,1]$ , we have

If $t\in [0,1/n]$ , then $|v_{n}(t)-1|+|b_{n}(t)|\leqslant \unicode[STIX]{x1D702}/n\leqslant 1/n$ , and thus

for every $x\in \mathbb{C}$ . Let $1\leqslant i\leqslant n$ and $t\in [(i-1)/n,i/n]$ be given. Then we have

where we used the fact that $\unicode[STIX]{x1D6FD}_{n}^{i-1}$ is an isometry. On the other hand, by Lemma 3.1 and (4), we have

Note that $\Vert \unicode[STIX]{x1D6FD}_{n}-j\Vert =\unicode[STIX]{x1D702}/n<1/n$ implies that either $\unicode[STIX]{x1D6FD}_{n}$ is a translation or it is a rotation by an angle $\unicode[STIX]{x1D719}$ with $\unicode[STIX]{x1D719}|<\unicode[STIX]{x1D70B}/n$ , and thus (4) can be applied. Then, comparing with (8), we have (7). Since $u_{n}\rightarrow u$ and $a_{n}\rightarrow a$ , it follows from Lemma 3.2 that $E_{t}^{\unicode[STIX]{x1D6FC}_{n}}(x)\rightarrow E_{t}^{\unicode[STIX]{x1D6FC}}(x)$ uniformly on the set $\{(t,x):t\in [0,1],|x|\leqslant 1\}$ . Let $\unicode[STIX]{x1D700}>0$ be given. Then we can choose an $n$ such that $8/n<\unicode[STIX]{x1D700}/2$ and

for every $t\in [0,1]$ and $|x|\leqslant 1$ . Then, by (7), we find that $|F_{t}^{n}(x)-E_{t}^{\unicode[STIX]{x1D6FC}}(x)|<\unicode[STIX]{x1D700}$ for every $t\in [0,1]$ and $|x|\leqslant 1$ . Thus, $\Vert F_{t}^{n}-E_{t}^{\unicode[STIX]{x1D6FC}}\Vert <\unicode[STIX]{x1D700}$ for every $t\in [0,1]$ . Since $\unicode[STIX]{x1D706}(W_{F^{n}}(A))<1/n<\unicode[STIX]{x1D700}$ , this completes the proof.◻

4 Proof of Lemma 2.2

If $\unicode[STIX]{x1D6FE}:[a,b]\rightarrow \mathbb{C}$ is a continuous curve and $p\in \mathbb{C}\setminus \unicode[STIX]{x1D6FE}([a,b])$ , then we denote by $w(\unicode[STIX]{x1D6FE};p)$ the increment of the argument of $\unicode[STIX]{x1D6FE}(t)-p$ , as $t$ goes from $a$ to $b$ . More precisely, if $\unicode[STIX]{x1D719}:[a,b]\rightarrow \mathbb{R}$ is a continuous function such that $\unicode[STIX]{x1D719}(t)$ is one of the arguments of $\unicode[STIX]{x1D6FE}(t)-p$ for every $t\in [a,b]$ , then we put $w(\unicode[STIX]{x1D6FE};p)=\unicode[STIX]{x1D719}(b)-\unicode[STIX]{x1D719}(a)$ . It is easy to see that $w(\unicode[STIX]{x1D6FE};p)$ does not depend on the choice of the continuous function $\unicode[STIX]{x1D719}$ . If $\unicode[STIX]{x1D6FE}$ is a closed curve, then $w(\unicode[STIX]{x1D6FE};p)/(2\unicode[STIX]{x1D70B})$ is the winding number of $\unicode[STIX]{x1D6FE}$ with respect to the point  $p$ .

Let $p$ be an arbitrary element of $A\cap D$ , and let $U\subset D$ be an open disc of centre $p$ . We prove that $\text{cl}\,U$ intersects at least two connected components of $D\setminus A$ .

Since $A$ is irreducible between the points $a,b\in A\setminus U$ , it follows that $A\setminus U$ is not connected; moreover, $a$ and $b$ belong to different connected components of $A\setminus U$ . Let $C_{b}$ denote the connected component of $A\setminus U$ containing the point  $b$ .

By [Reference Kuratowski7, Theorem 2, §47, II, p. 169], $C_{b}$ is a quasi-component of $A\setminus U$ ; that is, $C_{b}$ is the intersection of all (relative) clopen subsets of $A\setminus U$ containing $b$ . Since $a\notin C_{b}$ , there is a relative clopen set $G\subset A\setminus U$ such that $C_{b}\subset G$ and $a\notin G$ . Then $(A\setminus U)\setminus G$ and $G$ are disjoint compact sets.

Since $A$ is irreducible between $a$ and $b$ , it follows that $A$ is nowhere dense. Indeed, it is clear that otherwise it would contain proper subcontinuums containing $a$ and  $b$ .

By assumption, the set $\mathbb{R}^{2}\setminus A$ is connected. This implies that $\mathbb{R}^{2}\setminus C$ is connected for every $C\subset A$ . Indeed, $\mathbb{R}^{2}\setminus C$ contains the connected set $\mathbb{R}^{2}\setminus A$ , and is contained in its closure, so must be connected itself. Thus, $(A\setminus U)\setminus G$ and $G$ are disjoint compact sets such that they do not cut the plane. Indeed, $\mathbb{R}^{2}\setminus (A\setminus U)$ and $\mathbb{R}^{2}\setminus G$ are semicontinuums, being connected open sets.

Therefore, by [Reference Kuratowski7, Theorem 9, §61, II, p. 514], $(A\setminus U)\setminus G$ and $G$ can be separated by a simple closed curve. Let $J$ be a simple closed curve separating them. Then $J\cap (A\setminus U)=\emptyset$ and $J$ separates $a$ and $b$ . By symmetry, we may assume that $a\in \text{Int}\,J$ and $b\in \text{Ext}\,J$ .

If $J\subset D$ , then $\text{Int}\,J\subset D$ and $a\notin \text{Int}\,J$ , which is impossible. Therefore, $J$ is not covered by $D$ , and we can pick a point $a_{0}\in J\setminus D$ .

Since $A$ is connected and contains the points $a,b$ , $A$ must intersect $J$ . Now we have $J\cap (A\setminus U)=\emptyset$ , and thus $J\cap U\neq \emptyset$ . Since $a_{0}\notin U$ , it follows that $J\cap \unicode[STIX]{x2202}U\neq \emptyset$ .

Let $J$ be the range of the continuous map $\unicode[STIX]{x1D6FE}:[0,1]\rightarrow \mathbb{R}^{2}$ , where $\unicode[STIX]{x1D6FE}(0)=\unicode[STIX]{x1D6FE}(1)\in \unicode[STIX]{x2202}U$ , and $\unicode[STIX]{x1D6FE}$ is injective on $[\!0,1\!)$ . The set $F=\unicode[STIX]{x1D6FE}^{-1}(\unicode[STIX]{x2202}U)$ is a closed subset of $[0,1]$ such that $0,1\in F$ . If $(x,y)$ is a connected component of $[0,1]\setminus F$ , then $\unicode[STIX]{x1D6FE}((x,y))\cap \unicode[STIX]{x2202}U=\emptyset$ , and thus $\unicode[STIX]{x1D6FE}((x,y))$ is covered by one of $U$ and $\text{ext}\,U$ . By the uniform continuity of $\unicode[STIX]{x1D6FE}$ , there exists a $\unicode[STIX]{x1D6FF}>0$ such that whenever $x,y\in F$ and $|x-y|<\unicode[STIX]{x1D6FF}$ , then $\unicode[STIX]{x1D6FE}((x,y))\subset D$ . This easily implies that there is a partition $0=t_{0}<t_{1}<\cdots <t_{n}=1$ such that $t_{0},\ldots ,t_{n}\in F$ and, for every $i=1,\ldots ,n$ , either $\unicode[STIX]{x1D6FE}((t_{i-1},t_{i}))\subset D$ or $\unicode[STIX]{x1D6FE}((t_{i-1},t_{i}))\cap \text{cl}\,U=\emptyset$ .

Let $J_{i}=\unicode[STIX]{x1D6FE}([t_{i-1},t_{i}])$ for every $i=1,\ldots ,n$ . If $L_{i}$ denotes the subarc of the circle $\unicode[STIX]{x2202}U$ with end points $\unicode[STIX]{x1D6FE}(t_{i})$ and $\unicode[STIX]{x1D6FE}(t_{i-1})$ in this order, then $J_{i}\cup L_{i}$ is a closed curve for every $i=1,\ldots ,n$ . Since the points $a$ and $b$ are outside the closed disc $\text{cl}\,U$ , we have

Thus,

and

Therefore, we can choose an index $i$ such that $w(J_{i}\cup L_{i};a)\neq w(J_{i}\cup L_{i};b)$ .

We prove that $\unicode[STIX]{x1D6FE}(t_{i-1})$ and $\unicode[STIX]{x1D6FE}(t_{i})$ belong to different connected components of $D\setminus A$ . Since $\unicode[STIX]{x1D6FE}(t_{i-1})$ , $\unicode[STIX]{x1D6FE}(t_{i})\in \text{cl}\,U$ , this will prove the statement of the lemma.

Suppose that $\unicode[STIX]{x1D6FE}(t_{i-1})$ and $\unicode[STIX]{x1D6FE}(t_{i})$ belong to the same connected component of $D\setminus A$ . Then there is an arc $L\subset D\setminus A$ with end points $\unicode[STIX]{x1D6FE}(t_{i-1})$ and $\unicode[STIX]{x1D6FE}(t_{i})$ .

We have $w(L_{i}\cup L;a)=w(L_{i}\cup L;b)=0$ , since the closed curve $L_{i}\cup L$ is in $D$ , while $a$ and $b$ are not in $D$ . Therefore, we have $w(L_{i};a)=-w(L;a)=w(\overline{L};a)$ and $w(L_{i};b)=-w(L;b)=w(\overline{L};b)$ , where $\overline{L}$ is obtained from $L$ by changing its direction. Then we have

Consequently, $a$ and $b$ belong to different connected components of the open set $\mathbb{R}^{2}\setminus (J_{i}\cup \overline{L})$ . This, however, is impossible, as $a,b\in A$ , $(J_{i}\cup \overline{L})\cap A=\emptyset$ and $A$ is connected.◻