aa r X i v : . [ m a t h . M G ] J un A CAP COVERING THEOREM
A. POLYANSKII
Abstract. A cap of spherical radius α on a unit -sphere S is the set of points withinspherical distance α from a given point on the sphere. Let S be a finite set of caps lyingon S . We prove that if there is no great circle non-intersecting caps of S and dividing S into two non-empty subsets, then there is a cap of radius equal to the total radius ofcaps of S covering all caps of S provided that the total radius is less π/ .This is the spherical analog of the so-called Circle Covering Theorem by Goodmanand Goodman and the strengthening of Fejes Tóth’s zone conjecture proved by Jiangand the author. Introduction
A finite collection K of convex bodies in R d is called non-separable if any hyperplaneintersecting conv S K meets a convex body of K . The following theorem was conjecturedby Erdős and proved by Goodman and Goodman [8]. Theorem 1 (A. W. Goodman, R. E. Goodman, 1945) . If a finite collection of disks ofradii r , . . . , r n is non-separable, then it is possible to cover them by a disk of radius r + · · · + r n , Recently, Akopyan, Balitskiy, and Grigorev [1] proved a generalization of Theorem 1about covering of a non-separable collection of positive homothets of a convex body (nonnecessary o -symmetric); see also the work [6] of Bezdek and Lángi for a weaker result.The main goal of the current note is to prove the spherical analog of Theorem 1.In order to state our result, we need several definitions. Denote by S the unit d -sphereembedded in R d +1 centered at the origin. A cap of spherical radius α on S is defined asthe set of points within spherical distance α from a given point on the sphere. A greatsphere is the intersection of S and a hyperplane passing through the origin. We call agreat sphere avoiding for a collection of caps if it does not intersect the caps. A finitecollection of caps is called non-separable if no avoiding great sphere divides the collectioninto two non-empty sets. Theorem 2.
Let S be a non-separable collection of caps of spherical radii α , . . . , α n . If α + · · · + α n < π/ , then S can be covered by one cap of radius α + · · · + α n . Recall that a pair of antipodal caps can be viewed as the dual of a zone, where a zone of width α on S is the set of points within spherical distance α from a given great sphere.(The projective duality in R d +1 interchanges a line through the origin with its orthogonalhyperplane through the origin, that is, a pair of antipodal points is the dual of a greatsphere.) It is worth mentioning that Theorem 2 for α + · · · + α n = π/ under sometechnical assumptions is a corollary of so-called Fejes Tóth’s zone conjecture [13] that isproved in [9, Theorem 1, Corollary 3]; see also the recent work [11] of Ortega-Moreno,where the conjecture is confirmed for zones of the same width. The author acknowledges the financial support from the Ministry of Education and Science of theRussian Federation in the framework of MegaGrant no 075-15-2019-1926. heorem 3 (Jiang, Polyanskii, 2017) . The total width of any collection of zones covering S is at least π . The proof of Theorem 2 heavily relies on ideas developed in the context of studyingplanks [4, 5, 12] covering a convex body, where a plank (or slab , or strip ) of width w is a setof all points lying between two parallel hyperplanes in R d at distance w . The connectionbetween planks and zones is obvious: A zone of width α is the intersection of S andthe o -symmetric plank of width α . Another key idea of our proof is considering thefarthest point Voronoi diagram of the so-called Bang set (see (1)). This idea appeared inthe very recent work [3] of Balitskiy, where trying to understand the proof of the theoremof Kadets [10] on total inradius of convex bodies covering a unit ball, he introduced anew concept of multiplanks . Since we do not use this involved concept in its greatestgenerality, we decided to give a short remark in the discussion section about relation ofour proof to this concept. Nevertheless, we highly recommend an interested reader tounderstand it: We believe that it will be very helpful in proving new results on coverings. Acknowledgements.
The author thanks Alexey Balitskiy for the fruitful discussions ofhis work [3].
Covering zones instead of caps
Let us introduce useful notation. For a plank P ⊂ R d of width w , denote by w ( P ) anyof two vectors orthogonal to the boundary hyperplanes of P of length w/ . For a zone Z , set w ( Z ) = w ( P ) , where P is the open plank such that the closure of S ∩ P is Z .We apply the following lemma, which is the dual reformulation of [9, Lemma 4]. Lemma 4.
Let Z , . . . , Z n ⊂ S be zones of width α , . . . , α n , respectively, such that α = α + · · · + α n ≤ π/ . Set w i := w ( Z i ) . If w = P ni =1 w i satisfies the followinginequalities | w | ≥ sin α and | w − w i | ≤ sin( α − α i ) for all i ∈ [ n ] , then there is a zone of width α covering Z , . . . , Z n . Also, we use the following well-known observation.
Lemma 5.
Let P , . . . , P n ⊂ R d be open o -symmetric planks. Set w i := w ( P i ) . Then,for t ∈ R d , the vector with the maximum norm among elements of R = { t + P ni =1 ± w i } lies in R d \ ∪ ni =1 P i .Moreover, if t − P ni =1 ε i w i has the maximum norm, then it lies in ∩ ni =1 { x : h x , ε i w i i ≥ h w i , w i i} . Proof.
Without loss of generality assume that w = t + P ni =1 w i is of maximum normamong vectors of R . Suppose that w ∈ { x : h x , w i i < h w i , w i i} for some i ∈ [ n ] . Thenthe vector w − w i ∈ R is longer than w (see Figure 1), a contradiction. (cid:3) w − w i w | w i | Figure 1.
Proof of Lemma 5The following theorem implies Theorem 2. heorem 6. Let Z , . . . , Z n ⊂ S be zones of width α , . . . , α n , respectively, such that α + · · · + α n < π/ . If S \ S ni =1 Z i is at most one pair of two antipodal connected sphericalcomponents, then the zones Z , . . . , Z n can be covered by a zone of width α + · · · + 2 α n .Theorem 6 implies Theorem 2. Suppose that spherical caps D , . . . , D n satisfy the con-ditions of Theorem 2. Let D ′ i be an open cap cocentric with D i of spherical radius π/ − α i . By projective duality, the center of an open hemisphere lies in D ′ i if and only ifthis hemisphere covers D i . Since no avoiding great sphere divides { D , . . . , D n } into twonon-empty sets, by projective duality, we get n \ i =1 ε i D ′ i = ∅ unless all ε i ∈ {± } are the same . Hence the zones S \ ( D ′ ∪ ( − D ′ )) ,. . . , S \ ( D ′ n ∪ ( − D ′ n )) satisfy the conditions of The-orem 6. Therefore, they can be covered by a zone Z of width α + · · · + 2 α n . Since S \ Z is the union of two antipodal open caps D ′ and − D ′ of radii π/ − ( α + · · · + α n ) ,without loss of generality we obtain D ′ ⊆ n \ i =1 D ′ i ! . By projective duality, the closed cap cocentric with D ′ of radius α + · · · + α n covers caps D , . . . , D n . (cid:3) Remark.
Theorem 2 also easily implies Theorem 6.
Proof of Theorem 6.
Denote by P i the open plank such that the intersection of its closurewith S is Z i . Set w i = w ( P i ) for all i ∈ [ n ] . Without loss of generality let us assume that w = w + · · · + w n has the maximum norm among vectors of the Bang’s set L = ( n X i =1 ± w i ) . (1)First, let us show that we can assume | w | ≤ sin( α + · · · + α n ) . Indeed, suppose that | w | > sin( α + · · · + α n ) . Thus the family of all subsets J ⊂ [ n ] such that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X i ∈ J w i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > sin X i ∈ J α i ! is non-empty. Choose among them a minimal subset I . Since | w i | = sin α i , we have | I | > , and so we can apply Lemma 4 to I and cover zones Z i , i ∈ I , by the zone Z . Replacing the zones Z i , i ∈ I , by Z , we obtain a new collection of zones with thesame total width covering the original collection of zones. Put S \ Z i = D i ∪ ( − D i ) and S \ Z = D ∪ ( − D ) , where D i and D are open caps such that D ⊂ ∩ i ∈ I D i . By theconditions of the theorem, we can assume that \ i ∈ I ε i D i = ∅ unless all ε i are the same.Since \ i ∈ [ n ] \ I ε i D i ∩ εD ⊆ \ i ∈ [ n ] \ I ε i D i ∩ \ i ∈ I εD i ! , we get that the new collection of zones satisfies the conditions of the theorem. Therefore,we can reduce the number of zones and assume that | w | ≤ sin( α + · · · + α n ) . enote by B the unit open ball bounded by S . Consider the following set of points T = \ x ∈ L ( B + x ) that can be also defined in the following way T = (cid:8) t ∈ R d : t − x ⊆ B for all x ∈ L (cid:9) . (2)Our next goal is to show that T is the intersection of only two balls B + w and B − w .For that we consider the farthest point Voronoi diagram of L and its intersection with T .For a point x ∈ L , set A x := (cid:8) y ∈ R d : | y − x | ≥ | y − x ′ | for all x ′ ∈ L (cid:9) and T x := T ∩ A x . If t ∈ T x ⊂ T , then, by (2), the point t − x lies in the open ball B . Moreover, if t ∈ T x ⊂ A x , the point t − x has the maximum norm among { t − x ′ : x ′ ∈ L } , and thus,by Lemma 5, it does not belong to any of the planks P i . Moreover, if x = P ni =1 ε i w i , then t − x must belong to P x := ∩ ni =1 { y ∈ R d : h y , − ε i w i i ≥ h w i , w i i} . Hence T x − x ⊆ B ∩ P x .Therefore, the central projection of T x − x onto S also lies in P x . T − w T w T − w + w T w − ww − w Figure 2.
Proof of Theorem 6Since any two sets P x for x ∈ L are strictly separated by one of the hyperplanes { y ∈ R d : h y , w i i = 0 } and there are at most two connected regions in S \ ∩ ni =1 Z i , weobtain that there are at most two non-empty sets among T x for x ∈ L . Thus, by themaximality of the norm of w , we have that T w and T − w are the only two non-empty setsamong T x for x ∈ L ; see Figure 2. Hence T is the intersection of only two balls B + w and B − w because every point of T is closer to one of the points of L \ {± w } than to w or − w .Next, we easily finish the proof of the theorem. Since | w | ≤ sin( α + · · · + α n ) < ,the projection of the set T w − w is an open cap X radius at least π/ − ( α + · · · + α n ) ,which does not intersect ∪ ni =1 Z i ; see Figure 2. Therefore, the zone Z := S \ ( X ∪ ( − X )) of width at most α + · · · + 2 α n covers Z , . . . , Z n . (cid:3) Discussion
First, we discuss the connection of our proof with the concept of multiplank. emark. Using the terminology [3, Definition 2.2] of Balitskiy, it is easy to see that theset P L := R d \ [ x ∈ L ( A x − x ) is the open multiplank of the set L covering ∪ ni =1 P i ; see [3, Example 2.7]. In some senseour proof of Theorem 2 relies on the fact P L ∩ B = P ∩ B , where P is the plank with w ( P ) = w ; see the stratification of a multiplank in the general case in [3, Theorem 2.9].We recall the following problem resembling Theorem 2. It was proved for α ≥ π/ in [7, Theorem 6.1] but still open for α < π/ ; see also [2, Problem 6.2]. Conjecture 7.
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