On semiconvexity of open sets with smooth boundary in the plane

We study properties of classes of generalized convex sets in the plane known as 1-semiconvex and weakly 1-semiconvex. It is proved that an open, weakly 1-semiconvex but not 1-semiconvex set with smooth boundary in the plane consists of at least four connected components. Анотація. В даній роботі вивчаються властивості класів узагальнено опуклих множин на площині, які називаються 1-напівопуклими та слабко 1-напівопуклими. Відкрита множина багатовимірного дійсного евклідового простору R називається 1-напівопуклою, якщо для кожної точки із доповнення цієї множини до всього простору існує промінь, який починається в цій точці і не перетинає задану множину. Відкрита множина простору R називається слабко 1-напівопуклою, якщо для кожної точки межі множини існує промінь, який починається в цій точці та не перетинає задану множину. Ці поняття були введені Юрієм Борисовичем Зелінським. Усяка відкрита 1-напівопукла множина очевидно є слабко 1-напівопуклою. Ю. Б. Зелінський показав, що зворотнє твердження є невірним. Виявилось, що клас відкритих множин на площині, які є слабко 1-напівопуклими але не 1-напівопуклими, є досить широким і кожна множина з цього класу є незв’язною і складається не менше ніж з трьох компонент зв’язності. Дана робота присвячена переважно дослідженню нових властивостей слабко 1-напівопуклих але не 1-напівопуклих множин на площині. Тут вони, для зручності, називаються множинами, що мають Z-властивість. The work is dedicated to the memory of Yu. B. Zelinskii. The author expresses her gratitude to Professors Sergiy Maksymenko, Sergiy Plaksa, Oleksandr Bakhtin, Serhii Gryshchuk, Ruslan Salimov, Vitalii Shpakivskyi for valuable advice in planning the study and recommendations for the design of the paper. The author also is grateful to translators Julia Sheichenko and Anna Sheichenko who assisted in the preparation of the manuscript before being sent to the publisher. The author expresses her gratitude to all anonymous Referees of the paper for their comments. 2010 Mathematics Subject Classification: 32F17, 52A30 UDC 514.172


INTRODUCTION
A class of m-semiconvex sets is one of the classes of generalized convex sets. A semiconvexity notion was proposed by Yu. Zelinskii [9] and it was used in the formulation of a generalization of shadow problem. The shadow problem was proposed by G. Khudaiberganov [4] in 1982. It requires to find the minimal number of open (closed) balls in the real Euclidean space R n that are pairwise disjoint, centered on a sphere S n´1 (see [6]), do not contain the sphere center, and such that any straight line passing through the sphere center intersects at least one of the balls.
To formulate the generalized shadow problem, let us give at first the following definitions which we also use in our investigation.
Each m-dimensional affine subspace of the space R n , 0 ď m ă n, is called an m-dimensional plane. Definition 1.1. One of two parts of an m-dimensional plane, m ě 1, of the space R n , n ě 2, into which it is divided by its any (m´1)-dimensional plane (herewith, the points of the (m´1)-dimensional plane are included) is said to be an m-dimensional half-plane.
For instance, the 1-dimensional half-plane is a ray, the 2-dimensional half-plane is a half-plane, etc.
The generalized shadow problem requires to find the minimum number of pairwise disjoint closed (open) balls in R n (centered on a sphere S n´1 and whose radii are smaller than the radius of the sphere) such that any ray starting at the center of the sphere necessarily intersects at least one of these balls.
In the terms of m-semiconvexity this problem can be reformulated as follows: what is the minimum number of pairwise disjoint closed (open) balls in R n whose centers are located on a sphere S n´1 and the radii are smaller than the radius of this sphere such that the center of the sphere belongs to the 1-semiconvex hull of the family of these balls?
In [9] the generalized shadow problem is solved as n = 2 and only the sufficient number of balls is indicated for n = 3.
In the 60's L. Aizenberg and A. Martineau proposed their notions of a linearly convex set in the multi-dimensional complex space C n . The first author considered domains and their closures and used boundary points of the domains in his definition [1,2]. The second author used all points of the complement to a subset of the space C n , [5]. If one uses these definitions not only for domains and compact sets, then Aizenberg's definition isolates one connected component of a set which is linearly convex in the sense of Martineau.
Guided by similar ideas Yu. Zeliskii suggested to distinguish m-semiconvex and weakly m-semiconvex sets.
We will use the following standard notations. For a subset G Ă R n let G be its closure, Int G be its interior, and BG = GzInt G be its boundary.
Say that a set A is approximated from the outside by a family of open sets A k , k = 1, 2, . . ., if A k+1 is contained in A k , and A = X k A k (see [2]). Thus, each weakly m-semiconvex set A is either open or closed. Among closed weakly m-semiconvex sets there also are sets with empty interior: The formulation of the following theorem is equivalent to Theorem 1.4 but will be also in use. The maximal connected subsets of a topological space A are called connected components (components) of A (see [2]).
In [8] it was constructed the example of an open, weakly 1-semiconvex, and not 1-semiconvex set (see Figure 2.3b). It was also conjectured that every open, weakly 1-semiconvex, and not 1-semiconvex set consists of at least three components. The latter statement was proved in [3]. Theorem 1.6. ( [3]) Let E Ă R 2 be an open, weakly 1-semiconvex, and not 1-semiconvex subset. Then E consists of at least three connected components.
We say that a component A of an open, bounded subset of the plane has smooth boundary if BA is the image of a C 1 -embedding of the unit circle. We say that an open, bounded subset of the plane has smooth boundary if each of its components has smooth boundary.
The present paper continues the research of Yu. Zelinskii by investigating properties of 1-semiconvex and weakly 1-semiconvex open sets with smooth boundary in the plane.
For the convenience, we give the following Definition 1.7. We say that a set in the plane has Z-property if the set is weakly 1-semiconvex and not 1-semiconvex.
The main result of thee present paper is the following

AUXILIARY RESULTS
In what follows, unless otherwise is specified, a point of the space R 2 will be denoted by a small or capital Latin letter and the ray starting at that point will be denoted by a small Greek letter with the index denoting this point. The straight segment between points x, y P R n will be denoted by xy and the distance between them will be denoted by |x´y|.
Let us provide several auxiliary statements.  Figure 2.1a) shows a set E consisting of four open rectangles with a common rays passing through their boundaries. By the construction, for any boundary point of E there exists a ray that does not intersect E, while for any point of the interior of the rhombus abcd such a ray can not be found.
Similarly, in Figure 2.1b), the set consisting of three open components is weakly 1-semiconvex. On the other hand it is not 1-semiconvex, since each ray η x starting at each point x of the interior of the triangle abc intersects that set.   Suppose that two disks of that system have the same radii r and their centers belong to the axis Ox and are symmetric with respect to the origin O, see Figure 2.2b). Assume that the other two disks have radii R and 74 T. Osipchuk their centers belong to the axis Oy. If we place the center of the fifth open disk with radius r at the point (R + r, 0), then the union of those five open disks will have Z-property. Adding more disks of radius r and centers on the axis Ox in the positive direction, we will get an open set with smooth boundary having Z-property and consisting of any finite or even countable number of components.
Further it will be shown that in the example of a set having Z-property and consisting of three components it is not possible to replace the components with those that have smooth boundary while maintaining Z-property.
The set of all points of rays starting at a point x P R n zA and passing through a subset A Ă R n is called the cone over the set A with respect to the point x and is denoted by C x A. The boundary of C x A consists of rays which are called boundary rays. The boundary rays of C x A are called supporting for the set A (with respect to the point x). We suppose that It can be proved that if A is bounded and open (closed), then C x A is open (closed) as well and C x A " C x A.

Lemma 2.5.
A ray γ x is supporting for an open bounded subset E Ă R n iff the following conditions are satisfied: We need to show that the ray γ x satisfying conditions 1) and 2) coincides with one of boundary rays of C For an open connected weakly 1-semiconvex subset E Ă R 2 there exists at least one but no more than two supporting rays starting at some point x P R 2 zE.
Proof. By Theorem 1.5 the set E is 1-semiconvex. Thus for any point x P R 2 zE there exists a ray γ x such that γ x X E = ∅. Since E is not empty, open and connected, C x E is a plane angle =α ą 0 not containing its sides.
If γ x is unique, i.e. γ x " R 2 zC x E, then =α = 2π and γ x is its unique side.
On the other hand, if γ x Ĺ R 2 zC x E, then 0 ă =α ă 2π with two sides. □ An example of open, connected, and weakly 1-semiconvex set E for which there exists a point x P R 2 zE admitting a unique supporting ray ξ x is shown in Figure 2.3a).
there is no m-dimensional half-plane that has x on its boundary and does not intersect A.

Lemma 2.8. Suppose that an open subset
where E j are its components, has Z-property. Let also x be a point of 1nonsemiconvexity of E. Then each component E j has exactly two supporting rays starting at x.
Proof. Since E has Z-property, it follows that each of its 1-nonsemiconvexity point x belongs to R 2 zE.
By Lemma 2.8 each component E j has either one or two supporting rays starting at x. Suppose some component E j 1 , j 1 P t1, 2, . . . , ku, has a unique supporting ray ξ x be the ray complementary to ξ x and y be the nearest point of BE j 1 to x along the ray ξ 1 x . Notice that any ray starting at y and not containing the ray ξ x intersects E j 1 . Indeed, suppose there exists a ray γ y not containing the ray ξ x and such that γ y X E j 1 = ∅. Let ξ y be the ray starting at y and containing ξ x . Then the polygonal chain γ y Y tyu Y ξ y cuts the plane into two open components (parts). Since E j 1 X (γ y Y tyu Y ξ y ) = ∅ and E j 1 is connected, it is completely contained in one of those parts. On the other hand, there exists a ray η x contained in the other part, which gives η x X E j 1 = ∅. Since η x differs from ξ x , it is clear that ξ x is not a unique supporting ray for E j 1 with respect to the point x. This contradicts to our assumption that E j 1 has a unique supporting ray.
Furthermore, since ξ x intersects EzE j 1 , the ray ξ y Ą ξ x also intersects EzE j 1 . Thus, y P BE is a point of 1-nonsemiconvexity of E, which contradicts to weak 1-semiconvexity of E and none of the components of E have a unique supporting ray. □ Lemma 2.11. Let E 1 , E 2 be two components of a subset E Ă R n and let x P R n zE. If a ray ξ x is inner supporting for E 1 Y E 2 and then ξ x is inner supporting for E.
Proof. Without loss of generality, suppose that ξ x is inner supporting for the set E 1 Y E 2 with respect to E 1 . Then ξ x is supporting for E 1 and has the following properties: a) ξ x X E 2 ‰ ∅, b) there exists a point a P ξ x X BE 1 such that |a´x| ă |b´x| for every b P ξ x X E 2 by Definition 2.9. Condition a) together with condition which together with condition b) gives |a´x| ă |b´x| for the point a P ξ x X BE 1 and for any point b P ξ x X (EzE 1 ). Thus, E 1 is a component of E such that ξ x is an inner supporting ray for E with respect to E 1 . □ be an open bounded set having Z-property, where E j are its components. Let x P R 2 zE be a point of 1-nonsemiconvexity of E, y P BE the nearest point of BE to x along some ray η x , and γ y any ray that does not intersect E.
Then there exists an inner supporting ray ξ x for E with respect to some component E j 0 , j 0 P t1, ku, such that Proof. Since x is a point of 1-nonsemiconvexity of E, γ y does not lie on the straight line that contains η x .
Choose the polar coordinate system (φ, ρ) in R 2 in which x is the pole, the ray η x = η x (0) is the polar axis, η x (φ) is a ray starting at x and constituting an angle φ with ray η x , and a positive angular coordinate φ is determined by a ray starting at x and intersecting ray γ y .
Let Φ be the set of all φ P (0, ϕ) such that xc(φ) X E ‰ ∅. Then Φ is non-empty. Indeed, since η x (ϕ) X E ‰ ∅ and E is open, there exists ε ą 0 small enough and such that xc(ϕ´ε) X E ‰ ∅.
Let J Ď t1, 2, . . . , ku be the set of all indexes j P t1, 2, . . . , ku such that This implies that E j X Int (△xyc(φ j )) = ∅, j P J. We claim that if η x (φ q ) X E q ‰ ∅ for some q P J, then E q does not have supporting rays starting at x, see Figure 2.4a). Indeed, suppose η x (φ 1 ), φ 1 ‰ φ q , is supporting for E q . Then η x (φ 1 ) X E q = ∅. Let γ c(φq) be the ray 78 T. Osipchuk starting at point c(φ q ) and lying on ray γ y . If η x (φ 1 ) X γ c(φq) = ∅, then the polygonal chain cuts the plane into two open components. Herewith, L X E q = ∅. One of the components of R 2 zL contains a part of the ray η x (φ q ) intersecting E q . Since E q is connected, it is entirely contained in this component. On the other hand, since the interval xc(φ), φ q ă φ ă ϕ, is contained in the second component of R 2 zL and xc(φ q + ε) X E q ‰ ∅, for ε ą 0 small enough, E q is contained in the second component. We have reached a contradiction.
If η x (φ 1 ) X γ c(φq) ‰ ∅, then the polygonal chain L cuts the plane into three open components. The component of R 2 zL containing the part of the ray η x (φ q ) intersecting E q contains E q as well. Moreover, the component of R 2 zL bounded by the triangle generated by the intersection of rays η x (φ 1 ), γ c(φq) , and η x (φ q ) also contains E q , since xc(φ q + ε) X E q ‰ ∅, for ε ą 0 small enough. This contradicts to the fact that E q is connected. Thus, our assumption is incorrect and E q does not have supporting rays starting at x.
Similarly, it can be proved that if η x (φ q ) X E q = ∅ but there exists a point a P η x (φ q ) X BE q such that |a´x| ą |c(φ q )´x| for some q P J, then E q has a unique supporting ray starting at x coinciding with η x (φ q ) (see Figure 2.4 b).
The cases when component E q does not have supporting rays or has a unique supporting ray contradict to Lemma 2.8. Then by Lemma 2.5, for any j P J the ray η x (φ j ) is supporting for E j and has the following properties: a) η x (φ j ) X γ y = c(φ j ); b) |a´x| ď |c(φ j )´x| for any point a P η x (φ j ) X BE j ; c) E j X Int(△xyc(φ j )) = ∅. Thus, conditions 1) and 2) of our lemma are fulfilled for any ray η x (φ j ), j P J.
To finish the proof, we need only to show that among rays η x (φ j ), j P J, there is the one that is inner supporting for E and satisfying the lemma condition 3). Consider the angle φ j 0 := min j φ j , j 0 P J, and note that by the constructions E X △Int(xyc(φ j 0 )) = ∅. Thus, condition 3) holds for the ray η x (φ j 0 ). Since x is a point of 1-nonsemiconvexity of E, it follows that η x (φ j 0 ) X (EzE j 0 ) ‰ ∅. It then follows from properties 1) and 2), that |a´x| ă |b´x| for any point a P η x (φ j 0 ) X BE j 0 and any point b P η x (φ j 0 ) X (EzE j 0 ). Thus, ξ x := η x (φ j 0 ) is an inner supporting ray of E with respect to E j 0 satisfying the conditions 1)-3) of the lemma. □ Remark 2.13. Notice that if BE is not smooth at y, then the inner supporting ray ξ x can coincide with the ray η x . In this case the points c and y coincide and △xyc degenerates into the interval xy, see Figure 2.3b). Then all rays η x (φ) which are close enough to ξ x " η x and intersect ray γ y must also intersect the component E j 0 . If BE is smooth, then △xyc is non-degenerate, due to the fact that y is the nearest to x along η x . Definition 2.14. We say that a subset A Ă R n is projected from a point x P R n on a subset B Ă R n if any ray starting at point x and intersecting A intersects B as well.

Lemma 2.15. Suppose an open subset E Ă R 2 has Z-property and consists of three components. Then none of its components is projected on the union of the others from a point of 1-nonsemiconvexity of E.
Proof. Let E 1 , E 2 , E 3 be connected components of E and x P R 2 zE be a point of 1-nonsemiconvexity of E. Without loss of generality, suppose that E 1 is projected from x on E 2 Y E 3 , see Figure 2.5. Then the set, consisting only of components E 2 , E 3 , has Z-property, which contradicts to Theorem 1.6. □ Let E Ă R 2 be an open subset and γ a straight line passing through some boundary point x of E and does not intersect set E in some neighborhood U of x, i.e. γ X E X U = ∅. Notice that γ cuts the plane into two halfplanes. Let P be the half-plane such that P X U X E = ∅. We say that a ray η x starting at x lies above the straight line γ if η x Ă P . Proof. Let E j , j = 1, 2, 3, be components of E and let x P R 2 zE be a point of 1-nonsemiconvexity of the set. By Lemma 2.8 each E j , j = 1, 2, 3, has exactly two supporting rays which we denote by ξ j,1 x , ξ j,2 x , respectively. Notice that in general some of ξ j,k x , j = 1, 2, 3, k = 1, 2, can coincide, see Figure 2.7b).
By Lemma 2.15, none of the components of E is projected on the union of the others from x. Then for i = 1, 2, 3 there exists a ray τ i x starting at x and intersecting a unique component E i . In other words, Then the rays τ 1 x , τ 2 x and τ 3 x cut the plane into three open components G 1 , G 2 , G 3 such that Consider the closure of the domain G 3 between the rays τ 1 x , τ 2 x . We claim that G 3 does not contain points of E 3 . Indeed, BG 3 X E 3 = ∅ and BG 3 cuts R 2 ztxu into two components: G 3 and G 2 Y G 1 Y τ 3 x . Since E 3 is connected, it must be completely contained in one of those components.
By Lemma 2.5, ξ j,j x X BE j ‰ ∅, ξ j,j x X E j = ∅, j = 1, 2. Consider the points y j P ξ j,j x X BE j , j = 1, 2. By smoothness of BE and since ξ j,j x X E j = ∅, j = 1, 2, the rays ξ 1,1 x , ξ 2,2 x are tangent to BE 1 , BE 2 at the points y 1 , y 2 , respectively. Without loss of generality, suppose that there are points b 1 , b 2 P ξ 1 x X E 2 such that |b 1´x | ă |y 1´x | ă |b 2´x |, see Figure 2.6a). Since ray ξ 1,1 x is tangent to BE 1 at y 1 , all rays starting at y 1 and not intersecting E should lie above the straight containing ξ 1,1 x . On the other hand, all these rays intersect any curve in E 2 connecting points b 1 , b 2 , as E 2 is connected, which gives that these rays intersect E 2 and therefore y 1 is a point of 1-nonsemiconvexity of E. This contradicts weakly 1-semiconvexity of the set.
Thus, |b´x| ă |y 1´x | or |b´x| ą |y 1´x | for any point b P ξ 1,1 x XE 2 . These statements are also true for the ray ξ 2,2 x . Furthermore, |b 1´x | ą |y 2´x | for any point b 1 P ξ 2 x X E 1 if and only if |b 2´x | ă |y 1´x | for any point b 2 P ξ 1,1 x X E 2 . Swapping indices we will also get that |b 2´x | ą |y 1´x | for any point b 2 P ξ 1,1 x X E 2 if and only if |b 1´x | ă |y 2´x | for any point b 1 P ξ 2,2 x X E 1 , see Figure 2.6b). Otherwise, E 1 , E 2 would be overlapping. Hence, ξ j,j x is supporting for E j and ξ j,j x X ((E 1 Y E 2 )zE j ) ‰ ∅, j = 1, 2, but |b´x| ą |y j´x | for each y j P BE j and each b P ξ j,j x X ((E 1 Y E 2 )zE j ), if j = 1 or j = 2, which was necessary to prove. Thus, E has no more than three inner supporting rays which we denote by ξ k x , i = 1, 2, 3, such that ξ k x Ă G k . Herewith, if ξ k x Ă G k , k = 1, 2, 3, then the ray ξ k x is supporting for a unique of components E j , j = 1, 2, 3, j ‰ k, and therefore, it is inner supporting for E with respect to a unique component of E.
Since G i X G j = τ k x , (i, j, k) = (1, 2, 3), (2, 3, 1), (3, 1, 2), a ray ξ k x Ă G k , k = 1, 2, 3, can be inner supporting with respect to two components if and only if it coincides with a neighboring inner supporting ray on the common boundary of the respective sets G j , j = 1, 2, 3. Let us consider this case and prove that it is not possible under the lemma conditions. By this we will show that E has not less than three inner supporting rays starting at 82 T. Osipchuk x and ξ k x is inner supporting for E with respect to a single component of the set in BG k , k = 1, 2, 3, as well.
Without loss of generality, assume that two inner supporting rays ξ 1 x , ξ 2 x coincide with the boundary ray τ 3 x , see Figure 2.7a). By the constructions, ξ 1 x can be supporting for E 2 or E 3 . But since τ 3 x intersects component E 3 and ξ 1 x coincides with τ 3 x , the ray ξ 1 x is supporting for E 2 . Similarly, the ray ξ 2 x is supporting for E 1 . Let y j be the nearest to x point of τ 3 x X BE j , j = 1, 2. Since τ 3 x X E j = ∅, j = 1, 2, and by smoothness of BE, the ray τ 3 x is tangent to BE j at the points y j , j = 1, 2. The points y 1 , y 2 do not coincide, since otherwise one can not draw a ray starting at the point y 1 = y 2 and does not intersect E, which contradicts to weak 1-semiconvexity of E. Assume for definiteness FIGURE 2.7. that x is closer to y 1 than to y 2 , i.e. |x´y 1 | ă |x´y 2 |. Then, let us draw any ray γ y 1 starting at y 1 and not intersecting E. It lies above the straight line containing the ray τ 3 x , since BE is smooth. The ray, complementary to τ 3 x , intersects BE at the point z 1 which is nearest to x along this ray. Let us draw a ray γ z 1 starting at z 1 and also not intersecting E. By Lemma 2.12, among all rays starting at point x and crossing rays γ y 1 and γ z 1 there exist two inner supporting rays different from τ 3 x . If they are distinct, this contradicts to the fact that E has only two inner supporting rays by our assumption. Such an inner supporting ray is unique only if rays γ y 1 and γ z 1 are intersecting and the inner supporting ray ξ 3 x passes through the point x 0 P γ y 1 X γ z 1 .
Since the polygonal chain γ y 1 Yty 1 uYy 1 z 1 Ytz 1 uYγ z 1 is self-intersecting, it cuts the plane into three components (parts). The first part contained inside the △y 1 z 1 x 0 does not have points of E, by condition 3) of Lemma 2.12. Since the ray τ 3 y 1 Ă τ 3 x is such that y 1 P τ 3 y 1 X BE 1 , y 2 P τ 3 y 1 X BE 2 , and τ 3 y 1 X E 3 ‰ ∅, the second part of the plane that holds τ 3 y 1 contains all three components of E. Thus, the third part also does not have points of E, which allows the ray ξ 3 x to not intersect E. This contradicts to non 1-semiconvexity of E. Lemma is proved. □ Remark 2.17. Lemma 2.16 fails for sets with non-smooth boundary. Figure 2.7b) contains an example of open bounded set having Z-property and consisting of three components which has only two inner supporting rays starting at a point of 1-nonsemiconvexity x P Int(△y 1 a 1 a 2 ).
In the following lemmas we will assume that inner supporting rays of the set start at some point of its 1-nonsemiconvexity.

Lemma 2.18.
There exists no open bounded subset E Ă R with smooth boundary having Z-property, consisting of three components E 1 , E 2 , E 3 , and such that two of inner supporting rays of E are inner supporting with respect to the same component.
Proof. Let us prove the lemma by the contradiction. Let x P R 2 zE be a point of 1-nonsemiconvexity of E. By Lemma 2.16, E has three inner supporting rays ξ i x , i = 1, 2, 3, and each of them is inner supporting with respect to a unique component of E.
Without loss of generality, suppose that the rays ξ 1 x and ξ 2 x are inner supporting for E with respect to the component E 1 and ξ 3 x is inner supporting with respect to E 3 .
Consider the intersections ξ 1 x X BE 1 , ξ 2 x X BE 1 , ξ 3 x X BE 3 . By Lemma 2.5, these intersections are not empty. Let y 1 P ξ 1 x X BE 1 , y 2 P ξ 2 x X BE 1 , y 3 P ξ 3 x X BE 3 be the nearest points to x. Since ξ i x , i = 1, 2, 3, does not intersect the component for which it is supporting and by smoothness of BE, it follows that ξ i x is tangent to the boundary of the correspondent component at the point y i , i = 1, 2, 3. This implies that all rays starting at y i and not intersecting E lie above the straight line containing the supporting ray ξ i x . Let S be the sector between rays ξ 1 x and ξ 2 x containing the component E 1 . We claim that then the ray ξ 3 x is not contained in S. Indeed, otherwise E 3 would be contained in the open component of R 2 zE bounded by the intervals xy 1 , xy 2 and the part of BE 1 between points y 1 , y 2 . Then E 3 would be projected on E 1 , which contradicts to Lemma 2.15. Therefore ξ 3 x is contained in the open sector that is complementary to S.
Let α be the angle of sector S. Consider two cases: a) α ě π and b) α ă π. a) Let z 3 P BE be the point contained in the ray complementary to ξ 3 x and closest to x. Then z 3 P BE 1 , see Figure 2.8. Let us draw a ray γ z 3 that does not intersect E. It passes through the interval xy 1 or xy 2 . It is sufficient to consider the case with interval xy 1 . For the interval xy 2 , the arguments will be the same.