Coverings: Variations on a result of Rogers and on the Epsilon-net theorem of Haussler and Welzl

Discrete Mathematics 341 (2018) 863–874

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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc

Coverings: Variations on a result of Rogers and on the Epsilon-net theorem of Haussler and Welzl Nóra Frankl a , János Nagy b , Márton Naszódi b a b

Department of Mathematics, London School of Economics and Political Science, United Kingdom Department of Geometry, Lorand Eötvös University, Pázmány Péter Sétány 1/C Budapest, 1117, Hungary

article

a b s t r a c t

info

Article history: Received 8 March 2017 Received in revised form 12 September 2017 Accepted 23 November 2017

Keywords: Covering Rogers’ bound Spherical strip Density Set-cover Epsilon-net theorem

We consider four problems. Rogers proved that for any convex body K , we can cover Rd by translates of K of density very roughly d ln d. First, we extend this result by showing that, if we are given a family of positive homothets of K of infinite total volume, then we can find appropriate translation vectors for each given homothet to cover Rd with the same (or, in certain cases, smaller) density. Second, we extend Rogers’ result to multiple coverings of space by translates of a convex body: we give a non-trivial upper bound on the density of the most economical covering where each point is covered by at least a certain number of translates. Third, we show that for any sufficiently large n, the sphere S2 can be covered by n strips of width 20n/ln n, where no point is covered too many times. Finally, we give another proof of the previous result based on a combinatorial observation: an extension of the Epsilon-net Theorem of Haussler and Welzl. We show that for a hypergraph of bounded Vapnik–Chervonenkis dimension, in which each edge is of a certain measure, there is a not-too large transversal set which does not intersect any edge too many times. © 2017 Elsevier B.V. All rights reserved.

1. Introduction Let F be a family of convex sets in Rd . The density of F is defined as

∑ lim

r →∞

F ∈F :F ∩B(0,r)̸ =∅

vol (F )

vol (B(0, r))

provided that this limit exists, where B(0, r) denotes the ball of radius r centered at the origin, and vol (·) is the volume (Lebesgue measure). Let K be a convex body, that is, a compact, convex set with non-empty interior. We denote its translative covering density, that is the supremum of the densities of the coverings of Rd by translates of K , by ϑ (K ), for properties of this quantity cf. [14]. Our starting point is Rogers’ estimate [15]:

ϑ (K ) ≤ d ln d + d ln ln d + 5d, which holds for any convex body K in Rd . E-mail addresses: [email protected] (N. Frankl), [email protected] (J. Nagy), [email protected] (M. Naszódi). https://doi.org/10.1016/j.disc.2017.11.017 0012-365X/© 2017 Elsevier B.V. All rights reserved.

(1)

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Our first result is an extension of (1). For a family F of sets in Rd , we say that F permits a translative covering of a subset ⋃ A of Rd with density ϑ , if we can select a translation vector xF ∈ Rd for each member F of F such that A ⊆ F ∈F xF + F , and the density of this covering is ϑ . Theorem 1.1. Let K be a convex body in Rd , and let F = {λ1 K , λ2 K , . . .}, with 0 < λi for all i, be a family of its homothets satisfying ∞ ∑

λdi = ∞.

i=1

Let Λ := {λ1 , λ2 , . . .}. (a) If Λ is bounded, and has a limit point other than zero, then F permits a covering of space of density ϑ (K ). (b) If Λ is bounded, and has no limit point other than zero, then F permits a covering of space of density one. (c) If Λ is unbounded, then F permits a covering of space with maximum multiplicity 4d + 1 (that is, where no point is covered by more than 4d + 1 sets). In case (c), we prove maximum multiplicity 2d + 1 in a special case which includes all smooth bodies, see Theorem 2.5. The proofs are in Section 2. In the proof of Theorem 1.1, we will use a result on covering K by homothets of K . Theorem 1.2. Let K ⊆ Rd be a convex body of volume one, and let F be a family of positive homothets of K with total volume at least

⎧ d ⎨(d3 · ln d · ϑ (K ) + ( e)2) , ⎩d · ln d · ϑ (K ) · 3

2d d

if K = −K ,

+ e · 4 , in general. d

Then F permits a translative covering of K . This result is a strengthening of a result of [12], which, in turn is a strengthening of a result of Januszewski [7]. We prove it in Section 2.1. We learned that a stronger bound was recently obtained by Livshyts and Tikhomirov [8] in the case when the dimension d is large. We note that when we apply Theorem 1.2, we will not make use of the actual bound, any bound that depends on the dimension only would suffice. As a corollary, we will show the following measure theoretic statement, which we will use in the proof of Theorem 1.1. Corollary 1.3. Let K be a convex body in Rd , let A ⊂ Rd be a set of zero Lebesgue measure. Then any family F of positive homothets of K with infinite total volume satisfying lim sup{vol (F ) : F ∈ F } = 0 permits a translative covering of A with density zero. Our second topic is multiple coverings of space. We denote the infimum of the densities of k-fold coverings of Rd by translates of K by ϑ (k) (K ). Apart from the estimate that follows from (1) using the obvious fact ϑ (k) (K ) ≤ kϑ (K ), no general estimate has been known. For the Euclidean ball Bd2 in Rd , G. Fejes Tóth [2,3] gave the non-trivial lower bound ϑ (k) (Bd2 ) > cd k for some cd > 1, see more in the survey [4]. We prove Theorem 1.4. Let K ⊆ Rd be a convex body and k ≤ d(2.5 ln d + ln ln d). Then

ϑ (k) (K ) ≤ 20d(2.5 ln d + ln ln d). This shows that G. Fejes Tóth’s bound (up to a constant factor) is sharp if k = d ln d. To prove Theorem 1.4, we present in Section 3 a more general statement, Theorem 3.3, which extends [1, Theorem 1.6] and [13, Theorem 1.2]. Our third topic is covering the sphere S2 := {x ∈ R3 : |x| = 1} by strips. For a given point x ∈ S2 , and 0 ≤ w ≤ 1, we call {v ∈ S2 : |⟨v, x⟩| ≤ w} the strip centered at x, of Euclidean half-width w . Theorem 1.5. For any sufficiently large integer N, there is a covering of S2 by N strips of Euclidean half-width point covered more than c ln N times, where c is a universal constant.

10 ln N , N

with no

Our study of this question was motivated by a problem at the 2015 Miklós Schweitzer competition posed by András Bezdek, Ferenc Fodor, Viktor Vígh and Tamás Zarnócz on covering the two-dimensional sphere by strips of a given width such that no point is covered too many times. We note the following dual version of Theorem 1.5, and leave it to the reader to convince themselves that the two versions are equivalent: For any sufficiently large integer N, we can select N points of S2 such that each strip of Euclidean half-width 10 Nln N contains at least one and at most c ln N points, where c is a universal constant.

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In Section 4, we present a direct, probabilistic proof of Theorem 1.5. Our third topic, presented in Section 5, is studying variants of the Epsilon-net theorem of Haussler and Welzl [6]. A set X with a family H ⊆ 2X of some of its subsets is called a hypergraph, its Vapnik–Chervonenkis dimension (VC-dimension, for short) is defined in Section 5. Theorem 1.6. Let X be a set, H ⊂ 2X a hypergraph on X of VC-dimension at most d ≥ 2, and 0 < ε < 1. Let µ be a probability measure on X with µ(H) = ε for each⌊ H ∈ H. ⌋ C dε ln(1/ε ) elements of X (not necessarily distinct), such that each edge of H contains at (a) If ε ≤ 1d , then one can( choose ) 1 least one and at most C⌊ 1 d ln ε( chosen )⌋ points (with multiplicity), where C and C1 are universal constants. d (b) One can choose ( C ε ln ) 1ε + 1 elements of X (not necessarily distinct), such that each edge of H contains at least one and at most C1 d ln d ln 1ε + 1 chosen points (with multiplicity), where C and C1 are universal constants. The dual (and equivalent) version of Theorem 1.5 clearly follows from Theorem 1.6, since (using the uniform probability measure on S2 ) the measure of any strip of Euclidean half-width w is w , and the VC-dimension of strips on S2 is bounded, cf. [10, Chapter 10]. We also prove a similar result with essentially the same technique. Theorem 1.7. Let X be a set, H ⊂ 2X a hypergraph on X of VC-dimension at most d ≥ 2, and N ≥ 2 an integer. Let µ be a probability measure on X with µ(H) = 1/N for each H ∈ H. Then one can find a multi-subset of N elements of X (with multiplicity), such that each edge of H contains at most Cd lnlnlnNN chosen points, where C is a universal constant. It was pointed out to us by Nabil Mustafa that Theorems 1.6 and 1.7 can be obtained directly from results on epsilon approximations, see [11]. 2. Covering space with given homothets — Proof of Theorem 1.1 Remark 2.1. Let F denote a family of compact sets in Rd . We want to cover the space with translates of members of F . Among all these coverings, the minimum density that we can reach does not change whether we require that we use every member of F , or we may drop some members. Indeed, once we have a desired covering using a sub-family, we can take a zero-density arrangement of the rest of the members of F . In case (a) of Theorem 1.1, there is a subfamily of F which consists of essentially translates of K . The proof of case (a) now easily follows from Remark 2.1. We make some preparations for the proof of case (b), starting with the proof of Corollary 1.3. Proof of Corollary 1.3. For ε > 0 let G (ε ) be the(set )of those members λK ∈ F for which vol (λK ) ≤ ε . By the condition on F , there is a collection of disjoint subfamilies Fi,j i,j∈N of F such that for each i we have Fi,j ⊆ G (2−i−j ), and the total volume of Fi,j is infinite. ⋃ For each i ∈ N, let us define Ri = A ∩ (B(0, i) \ B(0, i − 1)). Then A = i∈N Ri . ( for ) each i we have λ(Ri ) = 0, and ⋃ Since Ri is of zero measure, one can find a collection of balls Bi,j j∈N of total volume at most 2−i such that Ri ⊆ j Bi,j . There is a constant c = c(K ) > 0 such that B(0, 1) ⊆ cK , that is for any r > 0 we have B(0, r) ⊆ rcK . Thus, by Theorem 1.2, there is a constant e = e(K ) such that if the total volume of a family G of homothets of K is at least e vol ((B(0), r)),∑ then G permits of B(0 ( ) covering ( ) a translative ⋃, r). For each i and j, choose a subfamily Hi,j⋃⊂ Fi,j such that −i−j . Now j Hi,j permits a translative covering of Ri and i,j Hi,j is of finite e vol Bi,j ≤ H ∈Hi,j vol Hi,j ≤ e vol Bi,j + 2 total volume, hence provides a translative covering of A of zero density. □ Definition 2.2. A collection V of Lebesgue-measurable subsets of Rd is a regular family if there is a constant C for which diam(V )d ≤ C vol (V ) holds for every V ∈ V . Definition 2.3. A collection V of subsets of Rd is a Vitali-covering of E ⊆ Rd , if for every x ∈ E and δ > 0, there is an element U of V such that x ∈ U and 0 < diam(U) < δ . We recall Vitali’s covering theorem [20]. Theorem 2.4. Let E ⊂ Rd be a measurable set with finite Lebesgue-measure, and let V be a regular family of closed subsets of Rd that is a Vitali covering for E. Then there is a finite or countably infinite subcollection {Uj } ⊆ V of disjoint sets such that

(

vol E \

⨆ j

Uj

)

= 0.

Proof of (b) of Theorem 1.1. We may assume that vol (K ) = 1. For a subcollection G of F we denote by G (ε ) the subset of those elements of G in which the ratio of homothety does not exceed ε .

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Now, for every ε > 0, the total volume in F (ε ) is infinite. A bit more is true: for any subfamily G of F of infinite total volume, and for every ε > 0, the total volume in G (ε ) is infinite. We partition F =

(⨆ i

Ai

) ⨆( ) B

into countably many sub-families, so that the total volume in each Ai and in B is infinite. We will use these sub-families to cover different parts of space. Next partition Rd into countably many axis aligned unit cubes, and consider an enumeration Q1 , Q2 , . . . of them. We consider each Qi separately. Let εi be a sequence of positive reals that converges to zero. We will cover most of the cube Qi by a subfamily of Ai , in which the sum of the volumes is at most (1 + εi diam(K ))d . To cover most of Qi , we partition

⨆

Ai =

j

Cj

into countably many sub-families, so that the total volume in each Cj is infinite. Using Theorem 1.2 for every j ∈ N, we can cover the cube Qi by the translates of the elements (Cj )(εi ). Since we use homothets of a fixed convex body, K , the union (over j) of these coverings is clearly a regular Vitali covering. Therefore we can apply Theorem 2.4. There is a subcollection {Ul } of disjoint sets for which

(

vol Qi \

⨆ l

Ul

)

= vol (E )i = 0.

By removing those translates that are disjoint from Qi , we may assume that l Ul ⊆ Qi + εi diam K , providing the desired covering of Qi . ⋃ To finish the proof we only need to cover i Ei which is of measure zero. By Corollary 1.3, using B this is possible by a covering of zero density. □

⨆

Case (c) of Theorem 1.1 follows from the following statement. Theorem 2.5. Let K be a convex body, and let F = {λ1 K , λ2 K , . . .} be a family of its homothets so that the λi -s are not bounded. Then there is a subfamily of F that permits a translative covering of Rd so that every point is covered at most 4d + 1 times. Moreover, if K is smooth at the points of intersection of K with supporting hyperplanes that are parallel to one of the d coordinate hyperplanes, then there is a subfamily of F that permits a translative covering of Rd so that every point is covered at most 2d + 1 times. Clearly, if an affine image of K has the special property that ‘coordinate-hyperplane touching points’ are smooth, then the 2d + 1 bound on the covering multiplicity also follows. To prove case (c) of Theorem 1.1, after obtaining a covering of multiplicity at most 2d (resp. 4d) by a subfamily of F , we only have to arrange the leftover so that their contribution to the density is zero. However, with a more careful procedure achieving covering multiplicity at most 2d (resp. 4d) is also possible, as shown by the remark at the end of the section. Proof of Theorem 2.5 in the second case. Fix ε > 0. We may assume that F has an element µ0 K so that Q0 = [−ε, ε]d ⊆ µ0 K ⊆ [−1, 1]d . We present an algorithm to produce the desired covering. We will define inductively a sequence of cubes Qi (i ∈ N), which are centered at the origin and have side length at least i, a sequence of translation vectors 2d 2d 2d 1 1 1 2 1 1 x11 , x21 , . . . x2d 1 , x2 , . . . , x2 , x3 , . . ., and a sequence µ1 K , µ1 K , . . . µ1 K , µ2 K , . . . , µ2 K , µ3 K , . . . of elements of F so that the j j following hold with the convention x0 = 0 and µ0 = µ0 : (1) Qk ⊆

(⋃

j j=1 xi

⋃k ⋃2d i=0

k i=0

j j=1 xi

+ )µji K for k ∈ N

+ µji K + B(0, ε) ⊆ Qk+1 for k ∈ N ( ) ( ) j j j+d j+d (3) xi + µi K ∩ xi + µi K = ∅ for 1 ≤ j ≤ d ( ) ( ) j j m (4) xi + µi K ∩ xm l + µl K = ∅ if |i − l| ≥ 2. (2)

⋃2d

j

j

Indeed, assume that we found the xi -s, µi -s and Qi -s for i ≤ k. Choose Qk+1 so that

⎛

k 2d ⋃ ⋃

⎝

⎞ j xi

j iK

+ µ ⎠ + B(0, ε) ⊆ Qk+1 .

i=0 j=1

We enumerate the facets of Qk such that the ith and the (d + i)-th are opposite (see Fig. 1). Let Hi denote the support hyperplane of the ith facet of Qk , and Hi,+ the half-space bounded by Hi that does not contain Qk . Since the set of λi -s is

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Fig. 1. The squares with bold edges are Q1 , Q2 and Q3 (counting from inside out), and the squares without bold edges represent the homothets of K .

unbounded, by the smoothness of K at the touching points with the coordinate hyperplanes, we can choose a so far unused element µik+1 K of F , and a translation vector xik+1 such that Qk+1 ∩ Hi,+ ⊆ xik+1 + µik+1 K and j

j

(xik+1 + µik+1 K ) ∩ (xk−1 + µk−1 K ) = ∅ for all j. Also we have that if Hi (i ≤ d) and Hi+d support opposite sides of Qk+1 then (xik+1 + µik+1 K ) ∩ (xik++d1 + µik++d1 K ) = ∅, and Qk+1 \ Qk ⊆

2d ⋃

xik+1 + µik+1 K .

i=1

Hence we can find the desired Qi -s and translates. Since Qi has side length at least i,

Rd =

∞ ⋃ 2d ⋃

j

j

xi + µi K .

i=1 j=1

⋃2d

Property (3) ensures that, the subfamily i=1 xik + µik K covers every point at most d times, and property (4) yields that ⋃2d every point of Rn is covered by at most two subfamilies i=1 xik + µik K , which finishes the proof. □ Remark 2.6. At first, one may believe that, by some approximation argument, the condition of smoothness can be dropped in Theorem 2.5. Unfortunately, this is not the case, the standard argument does not work. Let K be a convex body in Rd and F = {λ1 K , λ2 K , . . .} a family of its homothets, such that the coefficients λi -s are not bounded. Let L be a convex body with smooth boundary such that L ⊆ K ⊆ (1 + ε )L. Consider the family F ′ = {λ1 L, λ2 L, . . .}, and follow the steps of the proof of the smooth case in Theorem 2.5 for F ′ . j j j j We obtain a covering xi + λi L of Rd , where every point is covered at most 2d times. Then xi + (1 + ε )λi L is also a j j k covering. However, it may happen that xi + (1 + ε )λi L covers every point infinitely many times: If λi is sufficiently large, j j then xi + (1 + ε )λi K may contain B(0, i) for all i. To obtain the general case of Theorem 2.5, we will use the following observation. Lemma 2.7. Let K ⊆ ⋂ Rd be a convex body. Then there exist j ≤ 2d points {x1 , x2 , . . . xj } on the boundary of K , so that K is smooth in xi (1 ≤ i ≤ j) and i Hi+ = L is a bounded convex set with non-empty interior, where Hi+ is the half-space that contains K , bounded by the tangent hyperplane Hi at xi .

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Proof of Lemma 2.7. The statement is invariant under an affine transformation of K , thus, we may assume that the John ellipsoid, that is, the unique largest volume ellipsoid contained in K , is the ball B(0, 1) of radius one centered at the origin. Let A := bd K ∩ B(0, 1) denote the set of contact points of the ball with the boundary of K . Clearly, K is smooth at each point of A. Observe that the origin is in the interior of conv A, as otherwise, a small translate of B(0, 1) would also be contained in K contradicting the uniqueness of B(0, 1). It follows from a well known theorem of Steinitz, that the origin is in the interior of the convex hull of some j ≤ 2d members of A. Those are the desired points. □ Proof of Theorem 2.5 in the general case. Let L be the polytope obtained in Lemma 2.7. We may assume that L contains the origin. Let ε > 0 be fixed. We may also assume that F has an element µ0 K so that −L0 = −ε L ⊆ µ0 K ⊆ −L. Similarly to the proof of the smooth case, we can inductively define a sequence −α1 L, −α2 L, . . . of homothets of −L, a se2d 2d 2d 1 1 1 2 1 1 quence of translation vectors x11 , x21 , . . . x2d 1 , x2 , . . . , x2 , x3 , . . . and a sequence µ1 K , µ1 K , . . . µ1 K , µ2 K , . . . , µ2 K , µ3 K , . . . of members of F , so that αi ≥ i and the following hold:

⋃k ⋃2d j j x + µ K for k ∈ N (⋃k ⋃2di=0 j j=1 j i ) i µ K + B(0, ε) ⊆ −Lk+1 for k ∈ N (2) ( i=0 j=)1 xi + ( mi ) j j (3) xi + µi K ∩ xl + µm l K = ∅ if |i − l| ≥ 2.

(1) −Lk ⊆

Now, the general case of Theorem 2.5 easily follows. □ Remark 2.8. To obtain a covering using all members of F such that every point is covered by at most 2d (resp. 4d) sets, we modify the algorithm in the proof of Theorem 2.5 as follows. We choose Qk+1 so that the difference of the side length of Qk+1 and Qk is at least 2k. Then one can find a cube of side length at least k − ε in xik+1 µik+1 K ∩ Qk+1 which is covered only by xik+1 µik+1 K . Thus we can hide the members of the leftover one by one in sufficiently large cubes that are only covered once. We also note that we do not know how sharp case (c) of Theorem 1.1 is, however Theorem 2.5 is sharp up to a constant factor as shown by the family F = {B(0, 1), B(0, 2), B(0, 3), . . .} of Euclidean balls. 2.1. Covering K by its homothets Theorem 2.9. For any ε > 0, dimension d and any convex body K of volume one in Rd with o ∈ int K , if a family F of positive homothets of K has total volume at least d

2

⌈

⌉ ( vol (K ) vol (K − K ) ε )d d − log ε 2 ϑ (K ) + 1+ . log 1 + ε vol (K ) 2 vol (K ∩ (−K ))

Then F permits a translative covering of K . Theorem 1.2 clearly follows from this result. Indeed, we choose ε = 1d , and recall two facts. First, that there is a point (2d) x ∈ K such that vol (K ∩ (2x − K )) ≥ 21d vol (K ). And second, that, by [16], vol (K − K ) ≤ d vol (K ). Proof of Theorem 2.9. First, we restate [12, Theorem 1.3] in a slightly more general form than the original, which is easily obtained from the proof therein. The proof there easily yields this form. Theorem 2.10. Let K and L be convex bodies in Rd with o ∈ int K , and F = {λ1 K , λ2 K , . . .} be a family of its homothets with 0 < λi ≤ λ1 < 1. Assume that ∞ ∑

vol L + λ1

(

λdi ≥ 2d

i=1

K ∩(−K ) 2

vol (K ∩ (−K ))

) .

Then F permits a translative covering of L. We fix ε > 0. Now, we are given F = {λ1 K , λ2 K , . . .} with 0 < λi < 1 for all i. First, we consider the case when there is a subfamily F ′ = {µ1 K , µ2 K , . . .} of F in which (1 + ε )−1 ≤

µdi ≤ (1 + ε ) µdj

for all i and j, and ∞ ∑

µdi ≥ ϑ (K )(1 + ε )

vol (K − K ) vol (K )

i=1

In this case, F ′ has at least

1

µd1 (1+ε)

.

(K −K ) ϑ (K )(1 + ε ) volvol = (K )

1

µd1

(K −K ) ϑ (K ) volvol members. (K )

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We may assume that µ1 is the smallest homothety ratio in F ′ . By the main result of [17], we can cover K by at most translates of µ1 K . The statement of the theorem in this case easily follows. Next, we assume that there is no such subfamily F ′ . Consider the intervals

vol(K −µ1 K )·ϑ (K ) vol(µ1 K )

(ε d , ε d (1 + ε )], (ε d (1 + ε ), ε d (1 + ε )2 ], . . . , (ε d (1 + ε )c(ε)−1 , ε d (1 + ε )c(ε) ], where c(ε ) = d

⌈

∑

− log ε log 1+ε

⌉

. Since ε d (1 + ε )c(ε) ≥ 1, we have

λdi < c(ε)d(1 + ε )ϑ (K )

λi K ∈F ,λdi >ε d

vol (K − K ) vol (K )

.

This implies that there exists a subfamily F ′ = {µ1 K , µ2 K , . . . }, in which

µdi ≤ ε d and ∞ ∑

( ε )d d µdi ≥ 1 + 2 2

i=1

vol (K ) vol (K ∩ (−K ))

.

Then, by Theorem 2.10, F ′ permits a translative covering of K .

□

3. Multiple covering — Proof of Theorem 1.4 Definition 3.1. Let F be a family of subsets of a base set X , and k ∈ Z+ . The k-fold covering number of F , denoted by τk (F ), is the minimum cardinality of a multi-subfamily of F such that each point of X is contained in at least k (with multiplicity) members of the subfamily. We recall that a fractional covering of X by F is a mapping w from F to R+ with x∈F ∈F w(F ) ≥ 1 for all x ∈ X . The total ∑ weight of a fractional covering is denoted by w (F ) := F ∈F w (F ), and its infimum is the fractional covering number of F :

∑

τ ∗ (X , F ) := inf{w (F ) : w : F → R+ is a fractional-covering of X }. For more on (fractional) coverings, cf. [5] in the abstract (combinatorial) setting and [14,10] in the geometric setting. We will use the following simple combinatorial statement. Lemma 3.2. Let F be a family of subsets of a base set X of fractional covering number τ ∗ := τ ∗ (F ), and k ∈ Z+ . Then

⌈ ( )⌉ 3 3√ ∗ τk ≤ τ k + ln|X | + (4k + ln|X |) ln|X | ≤ ⌈6τ ∗ max{ln|X |, k}⌉. 2

2

The proof is a standard probabilistic argument. Proof. Let w be a )⌉ fractional covering of X by F of total weight τ ∗ := τ ∗ (F ), and let m = τ ∗ k + 32 ln|X | + √ 3 (4k + ln | X | ) ln | X | . 2 We pick m members of F randomly, independently with the same distribution: at each draw, each member F of F is picked with probability w (F )/w (F ). For a fixed x ∈ X , the probability that x is not covered at least k times by the selected family is at most P(ξ < k), where ξ = ξ1 + · · · + ξm , with independent ( random Bernoulli (i.e., 0/1–valued) )

⌈ (

variables ξ1 , . . . , ξm , each of expectation 1/τ ∗ . By Chernoff’s inequality, P(ξ < k) ≤ exp −

(m−kτ ∗ )2 3mτ ∗

. Thus, P(there is an x ∈

( ) (m−kτ ∗ )2 X which is not covered) ≤ |X | exp − 3mτ ∗ . The lemma now clearly follows. □ For two sets K and L in Rd , we define Nk (L, K ), the k-fold covering number of L by K as the minimum number of translates of K that cover L k-fold. Note that Nk (L, K ) = τk (F ), where F = {(x + K ) ∩ L : x ∈ Rd }. We also define the fractional covering number of L by K as N ∗ (L, K ) = τ ∗ (F ). By [1, Theorem 1.7], we have

{ max

vol (L)

} vol (L − K ) , 1 ≤ N ∗ (L, K ) ≤ vol (K ) vol (K )

(2)

for any Borel measurable sets, K and L is Rd . The Minkowski difference of two sets K , T ⊆ Rd is defined as K ∼ T := {u ∈ Rd : u + T ⊆ K }. We will use the following statement.

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Theorem 3.3. Let K , L and T be bounded Borel measurable sets in Rd and let Λ ⊂ Rd be a finite set with L ⊆ Λ + T . Then Nk (L, K ) ≤ ⌈6N ∗ (L − T , K ∼ T ) max{ln|Λ|, k}⌉. If Λ ⊂ K , then we have Nk (L, K ) ≤ ⌈6N ∗ (L, K ∼ T ) max{ln|Λ|, k}⌉. Theorem 3.3 can be proved the same way as the corresponding result in the case of k = 1 in [13, Theorem 1.2] (or, [1, Theorem 1.6]). The only difference in the proof is that in [13] a result of Lovász [9] bounding τ in terms of τ ∗ is used, here we replace it by Lemma 3.2 which bounds τk in terms of τ ∗ . For completeness, we include this proof. Proof of Theorem 3.3. Consider the hypergraph with base set Λ∩(L−T ), and with set of edges F = {(K ∼ T +x)∩Λ∩(L−T ) : x ∈ Rd }. Clearly, Nk (L, K ) ≤ τk (F ). On the other hand, it is easy to see that τ ∗ (F ) ≤ N ∗ (L − T , K ∼ T ). Now, Lemma 3.2 yields the first formula of the theorem. For the second formula, observe that τ ∗ (F ) ≤ N ∗ (L, K ∼ T ) provided that Λ ⊆ L. □ Proof of Theorem 1.4. Using John’s theorem and the fact that the statement is invariant under affine transformations acting on K , we may assume that B(0, 1) ⊆ K ⊆ [−d, d]d . Let C = − 2a , Clearly,

[

a d 2

]

be a cube of edge length a, where we will set a later.

ϑ (k) (K ) ≤ Nk (C , K )

vol (K ) vol (C )

,

(3)

thus, we are going to estimate Nk (C , K ). Let δ > 0 be fixed and let Λ ⊆ Rd be a finite set such that Λ + 2δ (K ∩ (−K )) is a saturated (i.e. maximal) packing of δ (K ∩ (−K )) in C − 2δ (K ∩ (−K )). Thus C ⊆ Λ + δ K ⊆ Λ + δ (K ∩ (−K )) ⊆ Λ + δ K . 2 By considering volume, and using that the cube of side length √2 is contained in B(0, 1), we have that d

vol C − 2δ (K ∩ (−K ))

(

|Λ| ≤

vol

(

)

(δ 2

(K ∩ (−K ))

)

≤

a+

) δd d 2

vol (B(0, 1))

2d

( δ )d ≤ 2

(

a+

) δ d d d/2

2 2d d

δ

d

(4)

Eq. (2) yields that N ∗ (C − δ (K ∩ (−K )), K ∼ δ (K ∩ (−K ))) ≤ N ∗ (C − δ K , (1 − δ )K ) ≤

vol (C − K ) vol ((1 − δ )K )

≤

(a + d)d . (1 − δ)d vol (K )

(5)

1 Next, we choose δ := d ln and a := d2 . Finally, combining Theorem 3.3 (using the bound on k) with Eqs. (4) and (5) yields d a bound on Nk (L, K ), which, when substituted into (3) yields the desired bound on the density ϑ (k) (K ). □

4. Covering the sphere by strips — a direct proof of Theorem 1.5 In this section, we present a direct, probabilistic proof of Theorem 1.5. We use the uniform probability measure on the sphere S2 , and recall that the measure of any strip of Euclidean half-width w is w . Let r = lnNN . By a standard saturated packing argument, we may fix a set of points v1 , . . . , vN 2 on S2 such that the caps around vi of radius r cover the sphere. Let Xi (i = 1, . . . , N) be independent random variables distributed uniformly on S2 . We prove that with positive probability, the points Xi will satisfy the conditions of the theorem. ⏐ ⏐ First consider the following probability: Q1⏐ = P(∃v ∈ S2 : ⏐⟨v, Xj ⟩⏐ ≥ 10 Nln N , for all j = 1, . . . , N). ⏐ Let Pi = P(∃v ∈ S2 : |v − vi | ≤ r , ⏐⟨v, Xj ⟩⏐ ≥ 10 Nln N , for all j = 1, . . . , N), where i = 1, . . . , N 2 . The union of the events corresponding to Pi covers the event corresponding to Q1 , because the caps around vi with radius r cover the sphere. On the other hand, clearly, Pi does not depend on i. We obtain that N 2 P1 ≥ Q1 . Assume that |v − vi | ≤ r.

⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐⟨v, Xj ⟩⏐ − ⏐⟨v − vi , Xj ⟩⏐ ≤ ⏐⟨vi , Xj ⟩⏐ , thus,

⏐ ⏐ ⏐ ⏐ ⏐⟨v, Xj ⟩⏐ − r ≤ ⏐⟨vi , Xj ⟩⏐ .

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Hence, we can estimate from above P1 as

( )N ⏐ ⏐ ln N N ⏐ ⏐ P1 ≤ P( ⟨v1 , Xj ⟩ ≥ 9r , j = 1, . . . , N) = (1 − 9r ) = 1 − 9 , N

thus, P1 ≤ e−9 ln n = N −9 , which yields Q1 ≤ N17 < 12 . ⏐ ⏐ ⏐ ⏐ Next, for any unit vector v , we denote the⏐number ⏐ of points in the strip ⟨v, Xj ⟩ ≤ 10r from the set X1 , . . . , XN by kv , and denote the number of points in the strip ⏐⟨v, Xj ⟩⏐ ≤ 11r from the set X1 , . . . , XN by hv . We will bound from above the probability Q2 = P(∃v ∈ S2 |kv ≥ c ln N), where we will fix c later. Let Ri = P(∃v ∈ S2 | |v − vi | ≤ r , kv ≥ c ln N) where i = 1, . . . , N 2 . Clearly, Ri does not depend on i, and N 2 R1 ≥ Q2 . So, we will estimate R1 from above. Assume that |v − v1 | ≤ r.

⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐⟨v, Xj ⟩⏐ + ⏐⟨v − v1 , Xj ⟩⏐ ≥ ⏐⟨v1 , Xj ⟩⏐ , thus,

⏐ ⏐ ⏐ ⏐ ⏐⟨v, Xj ⟩⏐ + r ≥ ⏐⟨v1 , Xj ⟩⏐ . We denote the integer part of c ln N by t, and let z = 11 ln N. R1 ≤ P(hvi ≥ c ln N) ≤

( z ) t (N ) N

t

.

By Stirling’s formula we easily get that there exists a universal constant D such that R1 ≤

Dz t · N N Nt

·

t t (N

−

t)(N −t)

≤

D(ez)t tt

(N ) t

≤

DN N . t t (N −t)(N −t)

Thus,

.

If c ≥ 100 then t ≥ e2 z , so we have R1 ≤

D et

≤

1 , N3

if N is large enough. It follows that

1 . (6) 2 Overall, we obtained that Q1 + Q2 < 1, which means that if N is large enough, then with positive probability the points Xi will satisfy the conditions of the theorem, with the constant c = 100. This completes the proof of Theorem 1.5. Following the proof of Theorem 1.5, with very little modification in the calculation, one can easily get the following theorem: Q2 ≤ N 2 R1 ≤

Theorem 4.1. Let N be a positive integer. Then there exist N points xi (i = 1, . . . , N) on S2 , such that for any unit vector v , there are at most c lnlnlnNN chosen points in the strip |⟨v, x⟩| ≤ N1 , where c is a universal constant. We pose the following open questions: Conjecture 4.2. There is a function f on the positive integers tending to infinity such that, for any N points on S2 , there is a unit vector v , for which the strip |⟨v, x⟩| ≤ N1 contains at least f (N) of the given points. Conjecture 4.3. There is a function g on the positive integers tending to infinity such that for any N points in the unit disc on the plane, there is a strip of width N1 containing at least g(N) of the given points. Conjecture 4.4. There is a function h on the positive integers tending to infinity such that for any N points on S2 and any width w > 0, there is a unit vector v , for which there are at least h(N) given points in the strip |⟨v, x⟩| ≤ w, or there is no chosen point in the strip |⟨v, x⟩| ≤ w . Note that Conjecture 4.2 would imply Conjecture 4.4 with h = f . Indeed, if w ≥ N1 , then it follows from the definition of f . If w ≤ N1 , then computing the sum of the areas of the dual strips associated to the points xi , i = 1, . . . , N yields that they cannot cover the sphere. So, in that case, there is a unit vector v such that there is no chosen point in the strip |⟨v, x⟩| ≤ w . Remark 4.5. Notice that Conjecture 4.3 seems pretty hard, as it would easily imply that if c > 0 is an arbitrary constant, then there exists an integer n0 such that if there are n > n0 points in the unit disc, then there are 4 points among them, such that the area of their convex hull is less than nc . This problem is open. If one decreases the width of the strips to N12 in the conjecture, then a trivial counterexample is choosing N points uniformly distributed on the circle.

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5. Some analogues of the Epsilon-net Theorem In this section, we prove Theorems 1.6 and 1.7. Both proofs closely follow the double-sampling technique of Haussler and Welzl from [6]. We recall some basics notions from the theory of hypergraphs, for details, we refer to [10]. Definition 5.1. The shatter function of a hypergraph H on the set X is πH (m) = maxA⊂X ,|A|=m |H|A |. The Vapnik–Chervonenkis dimension (VC-dimension, in short) of H is the maximal m for which π (m) = 2m (if there is no maximum, the VC-dimension is infinite). Definition 5.2. Let us have a hypergraph H on the set X and let ε ∈ [0, 1] be a real number. A subset Y ⊂ X is called an

ε-net for (X , H) if Y ∩ H ̸= ∅ for all H ∈ H, whenever |H | ≥ ε|X |. We recall the Sauer–Shelah Lemma [18,19].

Lemma 5.3. Let H be a hypergraph of VC-dimension d. Then for any non-negative integer m, π (m) ≤

(m) (m) ( ) + 1 +· · ·+ md ≤ 2md . 0

Let us recall the original Epsilon-net theorem of Hansler and Welzl: Theorem 5.4. Suppose (X , H) is a set system of VC dimension d, and let ε , δ be real numbers, where ε ∈ [0, 1] and δ > 0. If we draw O( dε log dε + 1ε log 1δ ) points at random from X , then the resulting set Y is an ε -net with probability higher than δ . In the proof of Theorems 1.6 and 1.7 we assume that the measure of every singleton is 0, but this is not a restriction, because we can replace every singleton with measure greater than 0 with an interval having the same measure. Proof of Theorem 1.6. We will assume, that the measure of every singleton of X is 0, to ensure that in a random sample the probability of having some element more than once is zero. The general case will follow in the following way. If A := {p1 , p2 , . . .} is the set of elements of X that, as singletons, are of positive measure, then we replace each element, say pi , of A by a ‘labeled’ interval pi × [0, 1]. This way, we obtain the set Xˆ from X . We define the measure of µ ˆ on Xˆ in such a way that the measure µ(pi ) is uniformly distributed on the labeled interval pi × [0, 1]. The set family Fˆ on Xˆ is essentially F , where if F ∈ F contains pi , then the corresponding Fˆ ∈ Fˆ contains the entire labeled interval pi × [0, 1]. Next, with this assumption of having no positive-measure ⌊ ⌋ singletons, let X1 , . . . , X2N be independent random variables according to µ taking values in X , where N := C dε ln(1/ε ) . Set Q0 := {X1 , . . . , XN }, Q1 := {XN +1 , . . . , X2N } and Q := Q0 ∪ Q1 . The Epsilon-net theorem yields that the probability that Q0 is a transversal to H (that is, that each edge H ∈ H intersects Q0 ) is greater than 21 , if C is sufficiently large. ( ) For a given H ∈ H, let E0H be the event that |Q0 ∩ H | > C1 d ln 1ε , where C1 > 0 is to be chosen later. Let E1H be the event H that E0 holds and |Q1 ∩ H | ≤ 2ε N. Let (E0)be the union of the events E0H for all H ∈ H, that is, E0 is the event that, for some H ∈ H, we have |Q0 ∩ H | > C1 d ln 1ε . Let E1 be the union of the events E1H for all H ∈ H. We claim that P(E0 ) ≤ 2P(E1 ). Indeed,

P(E1 ) P(E0 )

= P(E1 |E0 ) ≥ min P(E1H |E0H ) > 1/2, H ∈H

by Markov’s inequality. Thus, it is sufficient to show that P(E1 ) < 14 to obtain the theorem. Next, we sample in a different way. We permute the indices of the variables X1 , . . . , X2N with a random permutation (taking each permutation with equal probability), and denote the resulting variables as Y1 , . . . , Y2N . They are again independent. We estimate the probability of the event E1 for these variables. We fix a 2N-element subset R of X , and let L := H ∩ R. We estimate the probability of the event E1H under the condition that Q = R. By Lemma 5.3, there are at most 2(2N)d possibilities for L, so we have

P(E1 |Q = R) ≤ 2(2N)d max P(E1H |Q = R).

(7)

H ∈H

We fix H ∈ H. If t := |L| < C1 d ln ε , then E1H does not hold. We consider the case when t ≥ C1 d ln ε . In order to bound P(E1H |Q = R), we first note the following simple combinatorial fact. Let V be a subset of L with 0 ≤ m := |V | ≤ t. Then,

(1)

(1)

) ( ) ( ) ( ) 3 2N 2N − t 2N DN P(Q1 ∩ H = V |Q = R) = / ≤ / , N −m N N − ⌊t /2⌋ N 2t (t ) where D is a universal constant. Since, for any 0 ≤ m ≤ t, we have m ways to choose an m-element subset of L. Thus, (( ) ( ) ( )) ( ) t t t t P(E1H |Q = R) ≤ D · + + ··· + 2−t N 3 ≤ 3Dε N 2−t N 3 , 0 1 ⌊2ε N ⌋ ⌊2ε N ⌋ (

2N − t

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using t > 4ε N if C1 is large enough compared to C . By Stirling’s approximation, with the notations l := DN 3 , k := ⌊2ε N ⌋ and r := t /k, if C1 is large enough compared to C , then r > 10, and the right hand side is less than 3lε N

t

·

2t

t

≤

kk (t − k)t −k

2lkr

k

( ·

2rk

)(r −1)k

r r −1

<

( )(r −1)k 3

4

·l<

( ) C1 d ln2 3

4

( ) 1 ε

· l,

if C1 (and hence, r) is sufficiently large. So, by (7), it is sufficient to show, that if C1 is large enough, then 2(2N)d

( ) C1 d ln2

( ) 1 ε

3

· DN 3 < 1/4.

4

Clearly, it follows from d

D(4C

ε

( ) C1 d ln2

( ) 1 ε

3

ln(1/ε ))d+3

< 1/4.

4

The latter holds by the restriction ε ≤ 1/d, if C1 is large enough, because:

√

D(4Cd)d+3

( ) C1 d ln4

( ) 1 ε

3

4

<

1 4

,

and

√

( D

ln(1/ε )

)d+3 ( ) C1 d ln4

( ) 1 ε

3

ε

< 1.

4

Thus, by (7), we have that P(E1 |Q = R) < 1/4. Since R was arbitrary, we obtain P(E1 ) < 1/4 completing the proof of the first part of Theorem 1.6. The second part follows by the same calculation and the inequality (

( D 4C

d

ε

( ln

1

ε

1

))d+3 ( ) d ln d ln2 ε +1

)

3

+1

4

< 1/4.

□

Proof of Theorem 1.7. Similarly to the proof of Theorem 1.6, we will assume, that the measure of every singleton of X is 0. Let X1 , . . . , XN(ln N +1) be independent random variables according to µ taking values in X . Set Q0 := {X1 , . . . , XN }, Q1 := {XN +1 , . . . , XN(ln N +1) } and Q := Q0 ∪ Q1 . Denote by r = Cd lnlnlnNN , where C is to be chosen later. For a given H ∈ H, let E0H be the event that |Q0 ∩ H | > r. Let E1H be the event that E0H holds and |Q1 ∩ H | = ∅. Let E0 be the union of the events E0H for all H ∈ H, that is, E0 is the event that, for some H ∈ H, we have |Q0 ∩ H | > r. Let E1 be the union of the events E1H for all H ∈ H. We claim that P(E0 ) ≤ N 2 P(E1 ). Indeed,

P(E1 ) P(E0 )

= P(E1 |E0 ) ≥

min P(E1H H ∈H

|

E0H )

>

(

N −1

)N ln N ≥

N

1 N2

,

where, in the last inequality we used the fact that (1 − x/2) ≥ e−x for 0 < x < 1.59. Thus, it is sufficient to show that P(E1 ) < N12 to obtain the theorem. Next, we sample in a different way. We permute the indices of the variables X1 , . . . , XN(ln N +1) with a random permutation (taking each permutation with equal probability), and let the resulting variables be Y1 , . . . , YN(ln N +1) . They are again independent random variables. We will estimate the probability of the event E1 for these variables. We fix a subset R of X of N(ln N + 1) elements, and let L := H ∩ R. We estimate the probability of the event E1H under the condition that Q = R. By Lemma 5.3, there are at most 2(N ln N)d possibilities for L = H ∩ R, so we have

P(E1 |Q = R) ≤ 2(N ln N)d max P(E1H |Q = R).

(8)

H ∈H

We fix H ∈ H. If t := |L| ≤ r, then E1H does not hold. If t > r, then

(N ) P(E1H

|Q = R) ≤ (N(ln tN +1)) ≤ t

(

N N ln N

)t

( ≤

1 ln N

)r

= N −Cd .

If C is large enough, then by (8), P(E1 |Q = R) < 1/N 2 . Since this bound does not depend on the choice of R, we have P(E1 ) < 1/N 2 finishing the proof of Theorem 1.7. □

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Acknowledgments We are grateful for the illuminating conversations that we had with Nabil Mustafa on epsilon nets, and with Gábor Fejes Tóth on geometric covering problems. We thank the reviewers whose suggestions made the presentation much cleaner. M.N. thanks the support of János Pach’s Chair of Discrete and Computational Geometry at EPFL, partly supported by the Swiss National Science Foundation grants Nos. 200020-162884 and 200021-175977. M.N. was also supported by the János Bolyai Research Scholarship of the Hungarian Academy of Sciences, the National Research, Development, and Innovation Office, NKFIH Grant PD-104744 and by the ÚNKP-17-4 New National Excellence Program of the Ministry of Human Capacities. The three authors were supported by the National Research, Development, and Innovation Office, NKFIH Grant K119670. Part of this work is part of the MSc thesis of N. Frankl. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]

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