On the Beer index of convexity and its variants
OON THE BEER INDEX OF CONVEXITY AND ITS VARIANTS
MARTIN BALKO, VÍT JELÍNEK, PAVEL VALTR, AND BARTOSZ WALCZAK
Abstract.
Let S be a subset of R d with finite positive Lebesgue measure. The Beer index ofconvexity b( S ) of S is the probability that two points of S chosen uniformly independently atrandom see each other in S . The convexity ratio c( S ) of S is the Lebesgue measure of the largestconvex subset of S divided by the Lebesgue measure of S . We investigate the relationshipbetween these two natural measures of convexity.We show that every set S ⊆ R with simply connected components satisfies b( S ) (cid:54) α c( S )for an absolute constant α , provided b( S ) is defined. This implies an affirmative answer to theconjecture of Cabello et al. that this estimate holds for simple polygons.We also consider higher-order generalizations of b( S ). For 1 (cid:54) k (cid:54) d , the k -index of convexity b k ( S ) of a set S ⊆ R d is the probability that the convex hull of a ( k + 1)-tuple of points chosenuniformly independently at random from S is contained in S . We show that for every d (cid:62) β ( d ) > S ⊆ R d satisfies b d ( S ) (cid:54) β c( S ), providedb d ( S ) exists. We provide an almost matching lower bound by showing that there is a constant γ ( d ) > ε ∈ (0 ,
1) there is a set S ⊆ R d of Lebesgue measure 1 satisfyingc( S ) (cid:54) ε and b d ( S ) (cid:62) γ ε log /ε (cid:62) γ c( S )log / c( S ) . Introduction
For positive integers k and d and a Lebesgue measurable set S ⊆ R d , we use λ k ( S ) to denotethe k -dimensional Lebesgue measure of S . We omit the subscript k when it is clear from thecontext. We also write “measure” instead of “Lebesgue measure”, as we do not use any othermeasure in the paper.For a set S ⊆ R d , let smc( S ) denote the supremum of the measures of convex subsets of S .Since all convex subsets of R d are measurable [12], the value of smc( S ) is well defined. Moreover,Goodman’s result [9] implies that the supremum is achieved on compact sets S , hence it can bereplaced by maximum in this case. When S has finite positive measure, let c( S ) be defined assmc( S ) /λ d ( S ). We call the parameter c( S ) the convexity ratio of S .For two points A, B ∈ R d , let AB denote the closed line segment with endpoints A and B .Let S be a subset of R d . We say that points A, B ∈ S are visible one from the other or see eachother in S if the line segment AB is contained in S . For a point A ∈ S , we use Vis( A, S ) todenote the set of points that are visible from A in S . More generally, for a subset T of S , we useVis( T, S ) to denote the set of points that are visible in S from T . That is, Vis( T, S ) is the set ofpoints A ∈ S for which there is a point B ∈ T such that AB ⊆ S .Let Seg( S ) denote the set { ( A, B ) ∈ S × S : AB ⊆ S } ⊆ ( R d ) , which we call the segment set of S . For a set S ⊆ R d with finite positive measure and with measurable Seg( S ), we define the A journal version of this paper appeared in
Discrete Comput. Geom.
57 (1), 179–214, 2017.A preliminary version of this paper appeared in: Lars Arge and János Pach (eds.), , vol. 34 of
Leibniz International Proceedings in Informatics (LIPIcs) ,pp. 406–420, Leibniz-Zentrum für Informatik, Dagstuhl, 2015.The first three authors were supported by the grant GAČR 14-14179S. The first author acknowledges the supportof the Grant Agency of the Charles University, GAUK 690214 and the grant SVV-2015-260223. The last authorwas supported by the Ministry of Science and Higher Education of Poland
Mobility Plus grant 911/MOB/2012/0. a r X i v : . [ m a t h . M G ] D ec MARTIN BALKO, VÍT JELÍNEK, PAVEL VALTR, AND BARTOSZ WALCZAK parameter b( S ) ∈ [0 ,
1] by b( S ) := λ d (Seg( S )) λ d ( S ) . If S is not measurable, or if its measure is not positive and finite, or if Seg( S ) is not measurable,we leave b( S ) undefined. Note that if b( S ) is defined for a set S , then c( S ) is defined as well.We call b( S ) the Beer index of convexity (or just
Beer index ) of S . It can be interpreted asthe probability that two points A and B of S chosen uniformly independently at random seeeach other in S .1.1. Previous results.
The Beer index was introduced in the 1970s by Beer [1, 2, 3], whocalled it “the index of convexity”. Beer was motivated by studying the continuity propertiesof λ (Vis( A, S )) as a function of A . For polygonal regions, an equivalent parameter was laterindependently defined by Stern [20], who called it “the degree of convexity”. Stern was motivatedby the problem of finding a computationally tractable way to quantify how close a given setis to being convex. He showed that the Beer index of a polygon P can be approximated by aMonte Carlo estimation. Later, Rote [17] showed that for a polygonal region P with n edges theBeer index can be evaluated in polynomial time as a sum of O ( n ) closed-form expressions.Cabello et al. [6] have studied the relationship between the Beer index and the convexity ratio,and applied their results in the analysis of their near-linear-time approximation algorithm forfinding the largest convex subset of a polygon. We describe some of their results in more detailin Section 1.3.1.2. Terminology and notation.
We assume familiarity with basic topological notions suchas path-connectedness, simple connectedness, Jordan curve, etc. The reader can find thesedefinitions, for example, in Prasolov’s book [16].Let ∂S , S ◦ , and S denote the boundary, the interior, and the closure of a set S , respectively.For a point A ∈ R and ε >
0, let N ε ( A ) denote the open disc centered at A with radius ε . Fora set X ⊆ R and ε >
0, let N ε ( X ) := S A ∈ X N ε ( A ). A neighborhood of a point A ∈ R or a set X ⊆ R is a set of the form N ε ( A ) or N ε ( X ), respectively, for some ε > a and b is denoted by [ a, b ]. Intervals [ a, b ] with a > b areconsidered empty. For a point A ∈ R , we use x ( A ) and y ( A ) to denote the x -coordinate andthe y -coordinate of A , respectively.A polygonal curve Γ in R d is a curve specified by a sequence ( A , . . . , A n ) of points of R d suchthat Γ consists of the line segments connecting the points A i and A i +1 for i = 1 , . . . , n −
1. If A = A n , then the polygonal curve Γ is closed . A polygonal curve that is not closed is called a polygonal line .A set X ⊆ R is polygonally connected , or p-connected for short, if any two points of X can beconnected by a polygonal line in X , or equivalently, by a self-avoiding polygonal line in X . Fora set X , the relation “ A and B can be connected by a polygonal line in X ” is an equivalencerelation on X , and its equivalence classes are the p-components of X . A set S is p-componentwisesimply connected if every p-component of S is simply connected.A line segment in R d is a bounded convex subset of a line. A closed line segment includesboth endpoints, while an open line segment excludes both endpoints. For two points A and B in R d , we use AB to denote the open line segment with endpoints A and B . A closed line segmentwith endpoints A and B is denoted by AB .We say that a set S ⊆ R d is star-shaped if there is a point C ∈ S such that Vis( C, S ) = S .That is, a star-shaped set S contains a point which sees the entire S . Similarly, we say that aset S is weakly star-shaped if S contains a line segment ‘ such that Vis( ‘, S ) = S . N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 3 P (0 , ,
1) (1 ,
1) (2 ,
1) (3 ,
1) (2 n − ,
1) (2 n − , . . . Figure 1.
A star-shaped polygon P with b( P ) (cid:62) n − ε and c( P ) (cid:54) n . Thepolygon P is a union of n triangles (0 , i, i + 1 , i = 0 , . . . , n −
1, and ofa triangle (0 , , δ )((2 n − δ, δ ), where δ is very small.1.3. Results.
We start with a few simple observations. Let S be a subset of R such that Seg( S )is measurable. For every ε > S contains a convex subset K of measure at least (c( S ) − ε ) λ ( S ).Two points of S chosen uniformly independently at random both belong to K with probabilityat least (c( S ) − ε ) , hence b( S ) (cid:62) (c( S ) − ε ) . This yields b( S ) (cid:62) c( S ) . This simple lowerbound on b( S ) is tight, as shown by a set S which is a disjoint union of a single large convexcomponent and a large number of small components of negligible size.It is more challenging to find an upper bound on b( S ) in terms of c( S ), possibly underadditional assumptions on the set S . This is the general problem addressed in this paper.As a motivating example, observe that a set S consisting of n disjoint convex components ofthe same size satisfies b( S ) = c( S ) = n . It is easy to modify this example to obtain, for any ε > P with b( P ) (cid:62) n − ε and c( P ) (cid:54) n , see Figure 1. This shows thatb( S ) cannot be bounded from above by a sublinear function of c( S ), even for simple polygons S .For weakly star-shaped polygons, Cabello et al. [6] showed that the above example is essentiallyoptimal, providing the following linear upper bound on b( S ). Theorem 1.1 ([6, Theorem 5]) . For every weakly star-shaped simple polygon P , we have b( P ) (cid:54)
18 c( P ) . For polygons that are not weakly star-shaped, Cabello et al. [6] gave a superlinear bound.
Theorem 1.2 ([6, Theorem 6]) . Every simple polygon P satisfies b( P ) (cid:54)
12 c( P ) (cid:18) P ) (cid:19) . Moreover, Cabello et al. [6] conjectured that even for a general simple polygon P , b( P ) canbe bounded from above by a linear function of c( P ). (The question whether b( P ) = O (c( P )) forsimple polygons P was originally asked by Cabello and Saumell [7].) The next theorem, whichis the first main result of this paper, verifies this conjecture. Recall that b( S ) is defined for a set S if and only if S has finite positive measure and Seg( S ) is measurable. Recall also that a setis p-componentwise simply connected if its p-components are simply connected. In particular,every simply connected set is p-componentwise simply connected. Theorem 1.3.
Every p-componentwise simply connected set S ⊆ R whose b( S ) is definedsatisfies b( S ) (cid:54)
180 c( S ) . Clearly, every simple polygon satisfies the assumptions of Theorem 1.3. Hence we directlyobtain the following, which verifies the conjecture of Cabello et al. [6].
Corollary 1.4.
Every simple polygon P ⊆ R satisfies b( P ) (cid:54)
180 c( P ) . MARTIN BALKO, VÍT JELÍNEK, PAVEL VALTR, AND BARTOSZ WALCZAK
The main restriction in Theorem 1.3 is the assumption that S is p-componentwise simplyconnected. This assumption cannot be omitted, as shown by the set S := [0 , (cid:114) Q , where itis easy to verify that c( S ) = 0 and b( S ) = 1, see Proposition 3.7.A related construction shows that Theorem 1.3 fails in higher dimensions. To see this, consideragain the set S := [0 , (cid:114) Q , and define a set S ⊆ R by S := { ( tx, ty, t ) : t ∈ [0 ,
1] and ( x, y ) ∈ S } . Again, it is easy to verify that c( S ) = 0 and b( S ) = 1, although S is simply connected, evenstar-shaped.Despite these examples, we will show that meaningful analogues of Theorem 1.3 for higherdimensions and for sets that are not p-componentwise simply connected are possible. The key isto use higher-order generalizations of the Beer index, which we introduce now.For k ∈ { , . . . , d } and a set S ⊆ R d , we define the set Simp k ( S ) ⊆ ( R d ) k +1 bySimp k ( S ) := { ( A , . . . , A k ) ∈ S k +1 : Conv( { A , . . . , A k } ) ⊆ S } , where the operator Conv denotes the convex hull of a set of points. We call Simp k ( S ) the k -simplex set of S . Note that Simp ( S ) = Seg( S ).For k ∈ { , . . . , d } and a set S ⊆ R d with finite positive measure and with measurableSimp k ( S ), we define b k ( S ) by b k ( S ) := λ ( k +1) d (Simp k ( S )) λ d ( S ) k +1 . Note that b ( S ) = b( S ). We call b k ( S ) the k -index of convexity of S . We again leave b k ( S )undefined if S or Simp k ( S ) is non-measurable, or if the measure of S is not finite and positive.We can view b k ( S ) as the probability that the convex hull of k + 1 points chosen from S uniformly independently at random is contained in S . For any S ⊆ R d , we have b ( S ) (cid:62) b ( S ) (cid:62) · · · (cid:62) b d ( S ), provided all the b k ( S ) are defined.We remark that the set S := [0 , d (cid:114) Q d satisfies c( S ) = 0 and b ( S ) = b ( S ) = · · · =b d − ( S ) = 1, see Proposition 3.7. Thus, for a general set S ⊆ R d , only the d -index of convexitycan conceivably admit a nontrivial upper bound in terms of c( S ). Our next result shows thatsuch an upper bound on b d ( S ) exists and is linear in c( S ). Theorem 1.5.
For every d (cid:62) , there is a constant β = β ( d ) > such that every set S ⊆ R d with b d ( S ) defined satisfies b d ( S ) (cid:54) β c( S ) . We do not know if the linear upper bound in Theorem 1.5 is best possible. We can, however,construct examples showing that the bound is optimal up to a logarithmic factor. This is ourlast main result.
Theorem 1.6.
For every d (cid:62) , there is a constant γ = γ ( d ) > such that for every ε ∈ (0 , ,there is a set S ⊆ R d satisfying c( S ) (cid:54) ε and b d ( S ) (cid:62) γ ε log /ε , and in particular, we have b d ( S ) (cid:62) γ c( S )log / c( S ) . The proof of Theorem 1.3 is given in Section 2. In Section 3, we will prove Theorems 1.5 and1.6. We conclude, in Section 4, with some further remarks and a collection of open problems.2.
Bounding the mutual visibility in the plane
The goal of this section is to prove Theorem 1.3. Since the proof is rather long and complicated,we first present a high-level overview of its main ideas.
N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 5
We first show that it is sufficient to prove the estimate from Theorem 1.3 for bounded opensimply connected sets. This is formalized by the next lemma, whose proof can be found inSection 2.2.
Lemma 2.1.
Let α > be a constant such that every bounded open simply connected set T ⊆ R satisfies b( T ) (cid:54) α c( T ) . It follows that every p-componentwise simply connected set S ⊆ R with b( S ) defined satisfies b( S ) (cid:54) α c( S ) . In the proof of Lemma 2.1, we first show that the set S can be reduced to a bounded openset S whose Beer index b( S ) can be arbitrarily close to b( S ) from below. This is done byconsidering a part S of S that is contained in a sufficiently large disc and by showing that allsegments in S are in fact contained in the interior of S , except for a set of measure zero. Theproof is then finished by choosing S as the interior of S and by applying the assumption of thelemma to every p-component of S .Suppose now that S is a bounded open simply connected set. We seek a bound of the formb( S ) = O (c( S )). This is equivalent to a bound of the form λ (Seg( S )) = O (smc( S ) λ ( S )). Wetherefore need a suitable upper bound on λ (Seg( S )).We first choose in S a diagonal ‘ (i.e., an inclusion-maximal line segment in S ), and showthat the set S (cid:114) ‘ is a union of two open simply connected sets S and S (Lemma 2.4). It isnot hard to show that the segments in S that cross the diagonal ‘ contribute to λ (Seg( S )) byat most O (smc( S ) λ ( S )) (Lemma 2.8). Our main task is to bound the measure of Seg( S i ∪ ‘ )for i = 1 ,
2. The two sets S i ∪ ‘ are what we call rooted sets . Informally, a rooted set is a unionof a simply connected open set S and an open segment r ⊆ ∂S , called the root.To bound λ (Seg( R )) for a rooted set R with root r , we partition R into levels L , L , . . . ,where L k contains the points of R that can be connected to r by a polygonal line with k segments,but not by a polygonal line with k − R is contained in a union L i ∪ L i +1 for some i (cid:62)
1. Thus, a bound of the form λ (Seg( L i ∪ L i +1 )) = O (smc( R ) λ ( L i ∪ L i +1 ))implies the required bound for λ (Seg( R )).We will show that each p-component of L i ∪ L i +1 is a rooted set, with the extra propertythat all its points are reachable from its root by a polygonal line with at most two segments(Lemma 2.5). To handle such sets, we will generalize the techniques that Cabello et al. [6] haveused to handle weakly star-shaped sets in their proof of Theorem 1.1. We will assign to everypoint A ∈ R a set T ( A ) of measure O (smc( R )), such that for every ( A, B ) ∈ Seg( R ), we haveeither B ∈ T ( A ) or A ∈ T ( B ) (Lemma 2.7). From this, Theorem 1.3 will follow easily.2.1. Proof of Theorem 1.3 for bounded open simply connected sets.
First, we need afew auxiliary lemmas.
Lemma 2.2.
For every positive integer d , if S is an open subset of R d , then the set Seg( S ) isopen and the set Vis(
A, S ) is open for every point A ∈ S .Proof. Choose a pair of points (
A, B ) ∈ Seg( S ). Since S is open and AB is compact, there is ε > N ε ( AB ) ⊆ S . Consequently, for any A ∈ N ε ( A ) and B ∈ N ε ( B ), we have A B ⊆ S , that is, ( A , B ) ∈ Seg( S ). This shows that the set Seg( S ) is open. If we fix A = A ,then it follows that the set Vis( A, S ) is open. (cid:3)
Lemma 2.3.
Let S be a simply connected subset of R and let ‘ and ‘ be line segments in S .It follows that the set Vis( ‘ , S ) ∩ ‘ is a (possibly empty) subsegment of ‘ .Proof. The statement is trivially true if ‘ and ‘ intersect or have the same supporting line, or ifVis( ‘ , S ) ∩ ‘ is empty. Suppose that these situations do not occur. Let A, B ∈ ‘ and A , B ∈ ‘ MARTIN BALKO, VÍT JELÍNEK, PAVEL VALTR, AND BARTOSZ WALCZAK be such that AA , BB ⊆ S . The points A, A , B , B form a (possibly self-intersecting) tetragon Q whose boundary is contained in S . Since S is simply connected, the interior of Q is containedin S . If Q is not self-intersecting, then clearly AB ⊆ Vis( ‘ , S ). Otherwise, AA and BB have apoint D in common, and every point C ∈ AB is visible in R from the point C ∈ A B such that D ∈ CC . This shows that Vis( ‘ , S ) ∩ ‘ is a convex subset and hence a subsegment of ‘ . (cid:3) Now, we define rooted sets and their tree-structured decomposition, and we explain how theyarise in the proof of Theorem 1.3.A set S ⊆ R is half-open if every point A ∈ S has a neighborhood N ε ( A ) that satisfies one ofthe following two conditions:(1) N ε ( A ) ⊆ S ,(2) N ε ( A ) ∩ ∂S is a diameter of N ε ( A ) splitting it into two subsets, one of which (including thediameter) is N ε ( A ) ∩ S and the other (excluding the diameter) is N ε ( A ) (cid:114) S .The condition (1) holds for points A ∈ S ◦ , while the condition (2) holds for points A ∈ ∂S .A set R ⊆ R is a rooted set if the following conditions are satisfied:(1) R is bounded,(2) R is p-connected and simply connected,(3) R is half-open,(4) R ∩ ∂R is an open line segment.The open line segment R ∩ ∂R is called the root of R . Every rooted set, as the union of anon-empty open set and an open line segment, is measurable and has positive measure.A diagonal of a set S ⊆ R is a line segment contained in S that is not a proper subset of anyother line segment contained in S . Clearly, if S is open, then every diagonal of S is an open linesegment. It is easy to see that the root of a rooted set R is a diagonal of R .The following lemma allows us to use a diagonal to split a bounded open simply connectedsubset of R into two rooted sets. It is intuitively clear, and its formal proof is postponed toSection 2.3. Lemma 2.4.
Let S be a bounded open simply connected subset of R , and let ‘ be a diagonalof S . It follows that the set S (cid:114) ‘ has two p-components S and S . Moreover, S ∪ ‘ and S ∪ ‘ are rooted sets, and ‘ is their common root. Let R be a rooted set. For a positive integer k , the k th level L k of R is the set of points of R that can be connected to the root of R by a polygonal line in R consisting of k segmentsbut cannot be connected to the root of R by a polygonal line in R consisting of fewer than k segments. We consider a degenerate one-vertex polygonal line as consisting of one degeneratesegment, so the root of R is part of L . Thus L = Vis( r, R ), where r denotes the root of R . A k -body of R is a p-component of L k . A body of R is a k -body of R for some k . See Figure 2 foran example of a rooted set and its partitioning into levels and bodies.We say that a rooted set P is attached to a set Q ⊆ R (cid:114) P if the root of P is subset of theinterior of P ∪ Q . The following lemma explains the structure of levels and bodies. Although itis intuitively clear, its formal proof requires quite a lot of work and can be found in Section 2.4. Lemma 2.5.
Let R be a rooted set and ( L k ) k (cid:62) be its partition into levels. It follows that (1) R = S k (cid:62) L k ; consequently, R is the union of all its bodies; (2) every body P of R is a rooted set such that P = Vis( r, P ) , where r denotes the root of P ; (3) L is the unique -body of R , and the root of L is the root of R ; (4) every j -body P of R with j (cid:62) is attached to a unique ( j − -body of R . N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 7 r RB = BA A d ( A, r ) d ( B ) = d ( B, r ) d ( A ) Figure 2.
Example of a rooted set R partitioned into six bodies. The threelevels of R are distinguished with three shades of gray. The segment A B ∪ { B } is the base segment of AB .Lemma 2.5 yields a tree structure on the bodies of R . The root of this tree is the unique1-body L of R , called the root body of R . For a k -body P of R with k (cid:62)
2, the parent of P inthe tree is the unique ( k − R that P is attached to, called the parent body of P . Lemma 2.6.
Let R be a rooted set, ( L k ) k (cid:62) be the partition of R into levels, ‘ be a closed linesegment in R , and k (cid:62) be minimum such that ‘ ∩ L k = ∅ . It follows that ‘ ⊆ L k ∪ L k +1 , ‘ ∩ L k is a subsegment of ‘ contained in a single k -body P of R , and ‘ ∩ L k +1 consists of atmost two subsegments of ‘ each contained in a single ( k + 1) -body whose parent body is P .Proof. The definition of the levels directly yields ‘ ⊆ L k ∪ L k +1 . The segment ‘ splits intosubsegments each contained in a single k -body or ( k + 1)-body of R . By Lemma 2.5, the bodiesof any two consecutive of these subsegments are in the parent-child relation of the body tree.This implies that ‘ ∩ L k lies within a single k -body P . By Lemma 2.3, ‘ ∩ L k is a subsegmentof ‘ . Consequently, ‘ ∩ L k +1 consists of at most two subsegments. (cid:3) In the setting of Lemma 2.6, we call the subsegment ‘ ∩ L k of ‘ the base segment of ‘ , and wecall the body P that contains ‘ ∩ L k the base body of ‘ . See Figure 2 for an example.The following lemma is the crucial part of the proof of Theorem 1.3. Lemma 2.7. If R is a rooted set, then every point A ∈ R can be assigned a measurable set T ( A ) ⊆ R so that the following is satisfied: (1) λ ( T ( A )) <
87 smc( R ) ; (2) for every line segment BC in R , we have either B ∈ T ( C ) or C ∈ T ( B ) ; (3) the set { ( A, B ) : A ∈ R and B ∈ T ( A ) } is measurable.Proof. Let P be a body of R with the root r . First, we show that P is entirely contained in oneclosed half-plane defined by the supporting line of r . Let h − and h + be the two open half-planesdefined by the supporting line of r . According to the definition of a rooted set, the sets { D ∈ r : ∃ ε > N ε ( D ) ∩ h − = N ε ( D ) ∩ ( P (cid:114) r ) } and { D ∈ r : ∃ ε > N ε ( D ) ∩ h + = N ε ( D ) ∩ ( P (cid:114) r ) } are open and partition the entire r , hence one of them must be empty. This implies that thesegments connecting r to P (cid:114) r lie all in h − or all in h + . Since P = Vis( r, P ), we conclude that P ⊆ h − or P ⊆ h + .According to the above, we can rotate and translate the set R so that r lies on the x -axisand P lies in the half-plane { B ∈ R : y ( B ) (cid:62) } . For a point A ∈ R , we use d ( A, r ) to denotethe y -coordinate of A after such a rotation and translation of R . We use d ( A ) to denote d ( A, r )where r is the root of the body of A . It follows that d ( A ) (cid:62) A ∈ R . MARTIN BALKO, VÍT JELÍNEK, PAVEL VALTR, AND BARTOSZ WALCZAK A = A HFD C B BrT ET Figure 3.
Illustration for the proof of Claim 1 in the proof of Lemma 2.7.Let γ ∈ (0 ,
1) be a fixed constant whose value will be specified at the end of the proof. For apoint A ∈ R , we define sets V ( A ) := { B ∈ Vis(
A, R ) : | A B | (cid:62) γ | AB | , A ∈ Vis( r , R ) , d ( A , r ) (cid:62) d ( B , r ) } , V ( A ) := { B ∈ Vis(
A, R ) : | A B | (cid:62) γ | AB | , A / ∈ Vis( r , R ) , d ( A , r ) (cid:62) d ( B , r ) } , V ( A ) := { B ∈ Vis(
A, R ) : | A B | < γ | AB | , | AA | (cid:62) | BB |} , where r denotes the root of the base body of AB and A and B denote the endpoints ofthe base segment of AB such that | AA | < | AB | . For every A ∈ R , the sets V ( A ), V ( A ),and V ( A ) are pairwise disjoint. Moreover, we have A ∈ S i =1 V i ( B ) or B ∈ S i =1 V i ( A ) forevery line segment AB in R . If for some B ∈ S i =1 V i ( A ) the point A lies on r , then we have B ∈ V ( A ) and V ( A ) ⊆ r .For the rest of the proof, we fix a point A ∈ R . We show that the union S i =1 V i ( A ) iscontained in a measurable set T ( A ) ⊆ R with λ ( T ( A )) <
87 smc( R ) that is a union of threetrapezoids. We let P be the body of A and r be the root of P . If P is a k -body with k (cid:62) r to denote the root of the parent body of P . Claim . V ( A ) is contained in a trapezoid T ( A ) with area γ − smc( R ) . Let H be a point of r such that AH ⊆ R . Let T be the r -parallel trapezoid of height d ( A )with bases of length R ) d ( A ) and R ) d ( A ) such that A is the center of the larger base and H is the center of the smaller base. The homothety with center A and ratio γ − transforms T into the trapezoid T := A + γ − ( T − A ). Since the area of T is 6 smc( R ), the area of T is6 γ − smc( R ). We show that V ( A ) ⊆ T . See Figure 3 for an illustration.Let B be a point in V ( A ). Using a similar approach to the one used by Cabello et al.[6] in the proof of Theorem 1.1, we show that B ∈ T . Let A B be the base segment of AB such that | AA | < | AB | . Since B ∈ V ( A ), we have | A B | (cid:62) γ | AB | , A ∈ Vis( r , R ), and d ( B, r ) (cid:54) d ( A, r ), where r denotes the root of the base level of AB . Since A is visible from r in R , the base body of AB is the body of A and thus A = A and r = r . As we have observed,every point C ∈ { A } ∪ AB satisfies d ( C, r ) = d ( C ) (cid:62) ε >
0. There is a point E ∈ AB such that | B E | < ε . Since E lies on the base segmentof AB , there is F ∈ r such that EF ⊆ R . It is possible to choose F so that AH and EF have apoint C in common where C = F, H . Let D be a point of AH with d ( D ) = d ( E ). The point D exists, as d ( H ) = 0 (cid:54) d ( E ) (cid:54) d ( A ). The points A, E, F, H form a self-intersecting tetragon Q whose boundary is contained in R . Since R is simply connected, the interior of Q is contained in R and the triangles ACE and
CF H have area at most smc( R ).The triangle ACE is partitioned into triangles
ADE and
CDE with areas ( d ( A ) − d ( D )) | DE | and ( d ( D ) − d ( C )) | DE | , respectively. Therefore, we have ( d ( A ) − d ( C )) | DE | = λ ( ACE ) (cid:54) N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 9
F H T Tp + r A A DCG B E Brp q H q + D O Figure 4.
Illustration for the proof of Claim 2 in the proof of Lemma 2.7.smc( R ). This implies | DE | (cid:54) R ) d ( A ) − d ( C ) . For the triangle
CF H , we have d ( C ) | F H | = λ ( CF H ) (cid:54) smc( R ). By the similarity of thetriangles CF H and
CDE , we have | F H | = | DE | d ( C ) / ( d ( E ) − d ( C )) and therefore | DE | (cid:54) R ) d ( C ) ( d ( E ) − d ( C )) . Since the first upper bound on | DE | is increasing in d ( C ) and the second is decreasing in d ( C ), theminimum of the two is maximized when they are equal, that is, when d ( C ) = d ( A ) d ( E ) / ( d ( A ) + d ( E )). Then we obtain | DE | (cid:54) R ) d ( A ) ( d ( A ) + d ( E )). This and 0 (cid:54) d ( E ) (cid:54) d ( A ) imply E ∈ T .Since ε can be made arbitrarily small and T is compact, we have B ∈ T . Since | AB | (cid:62) γ | AB | ,we conclude that B ∈ T . This completes the proof of Claim 1. Claim . V ( A ) is contained in a trapezoid T ( A ) with area − γ ) − γ − smc( R ) . We assume the point A is not contained in the first level of R , as otherwise V ( A ) is empty.Let p be the r -parallel line that contains the point A and let q be the supporting line of r . Let p + and q + denote the closed half-planes defined by p and q , respectively, such that r ⊆ p + and A / ∈ q + . Let O be the intersection point of p and q .Let T ⊆ p + ∩ q + be the trapezoid of height d ( A, r ) with one base of length R )(1 − γ ) d ( A,r ) on p ,the other base of length R )(1 − γ ) d ( A,r ) on the supporting line of r , and one lateral side on q . Thehomothety with center O and ratio γ − transforms T into the trapezoid T := O + γ − ( T − O ).Since the area of T is 3(1 − γ ) − smc( R ), the area of T is 3(1 − γ ) − γ − smc( R ). We show that V ( A ) ⊆ T . See Figure 4 for an illustration.Let B be a point of V ( A ). We use A B to denote the base segment of AB such that | AA | < | AB | . By the definition of V ( A ), we have | A B | (cid:62) γ | AB | , A / ∈ Vis( r , R ), and d ( B, r ) (cid:54) d ( A, r ), where r denotes the root of the base body of AB . By Lemma 2.6 and thefact that A / ∈ Vis( r , R ), we have r = r . The bound d ( A, r ) (cid:62) d ( B, r ) thus implies A ∈ r ∩ p + and B ∈ q + . We have d ( C, r ) = d ( C ) (cid:62) C ∈ A B . Observe that (1 − γ ) d ( A, r ) (cid:54) d ( A , r ) (cid:54) d ( A, r ). The upper bound is trivial, as d ( B, r ) (cid:54) d ( A, r ) and A lies on AB . For the lower bound, we use the expression A = tA + (1 − t ) B for some t ∈ [0 , d ( A , r ) = td ( A, r ) + (1 − t ) d ( B , r ). By the estimate | A B | (cid:62) γ | AB | , we have | AA | + | BB | (cid:54) (1 − γ ) | AB | = (1 − γ )( | AB | + | BB | ) . This can be rewritten as | AA | (cid:54) (1 − γ ) | AB | − γ | BB | . Consequently, | BB | (cid:62) γ > | AA | (cid:54) (1 − γ ) | AB | . This implies t (cid:62) − γ . Applying the bound d ( B , r ) (cid:62)
0, weconclude that d ( A , r ) (cid:62) (1 − γ ) d ( A, r ).Let ( G n ) n ∈ N be a sequence of points from A B that converges to A . For every n ∈ N , thereis a point H n ∈ r such that G n H n ⊆ R . Since r is compact, there is a subsequence of ( H n ) n ∈ N that converges to a point H ∈ r . We claim that H ∈ q . Suppose otherwise, and let q = q bethe supporting line of A H . Let ε > N ε ( A ) ⊆ R . For n large enough, G n H n is contained in an arbitrarily small neighborhood of q . Consequently, for n large enough,the supporting line of G n H n intersects q at a point K n such that G n K n ⊆ N ε ( A ), which implies K n ∈ r ∩ Vis( r , R ), a contradiction.Again, let ε >
0. There is a point E ∈ A B such that | B E | < ε . Let D be a point of q with d ( D , r ) = d ( E ), and let δ >
0. There are points G ∈ A B and H ∈ r such that G ∈ N δ ( A )and GH ⊆ R ∩ N δ ( q ). If δ is small enough, then d ( E ) (cid:54) d ( A , r ) − δ (cid:54) d ( G ) (cid:54) d ( A , r ). Let D be the point of GH with d ( D ) = d ( E ). The point E lies on A B and thus it is visible from apoint F ∈ r . Again, we can choose F so that the line segments EF and GH have a point C in common where C = F, H . The points
E, F, H, G form a self-intersecting tetragon Q whoseboundary is in R . The interior of Q is contained in R , as R is simply connected. Therefore, thearea of the triangles CEG and
CF H is at most smc( R ).The argument used in the proof of Claim 1 yields | DE | (cid:54) R ) d ( G ) ( d ( G ) + d ( E )) (cid:54) R )( d ( A , r ) − δ ) ( d ( A , r ) + d ( E )) . This and the fact that δ (and consequently | D D | ) can be made arbitrarily small yield | D E | (cid:54) R ) d ( A ,r ) ( d ( A , r )+ d ( E )). This and d ( A , r ) (cid:62) (1 − γ ) d ( A, r ) yield | D E | (cid:54) R )(1 − γ ) d ( A,r ) ( d ( A, r )+ d ( E )). This and 0 (cid:54) d ( E ) (cid:54) d ( A ) imply E ∈ T .Since ε can be made arbitrarily small and T is compact, we have B ∈ T . Since | A B | (cid:62) γ | AB | (cid:62) γ | A B | , we conclude that B ∈ T . This completes the proof of Claim 2. Claim . V ( A ) is contained in a trapezoid T ( A ) with area (4(1 − γ ) − −
1) smc( R ) . By Lemma 2.3, the points of r that are visible from A in R form a subsegment CD of r .The homothety with center A and ratio 2(1 − γ ) − transforms the triangle T := ACD into thetriangle T := A + 2(1 − γ ) − ( T − A ). See Figure 5 for an illustration. We claim that V ( A ) isa subset of the trapezoid T := T (cid:114) T .Let B be an arbitrary point of V ( A ). Consider the segment AB with the base segment A B such that | AA | < | AB | . Since B ∈ V ( A ), we have | A B | < γ | AB | and | AA | (cid:62) | BB | . Thisimplies | AA | (cid:62) − γ | AB | > A = A and B / ∈ P . From the definition of C and D ,we have A ∈ CD . Since | AA | (cid:62) − γ | AB | and B / ∈ P , we have B ∈ T .The area of T is (4(1 − γ ) − − λ ( T ). The interior of T is contained in R , as all points ofthe open segment CD are visible from A in R . The area of T is at most smc( R ), as its interioris a convex subset of R . Consequently, the area of T is at most (4(1 − γ ) − −
1) smc( R ). Thiscompletes the proof of Claim 3. N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 11 AA B B T DCT r Figure 5.
Illustration for the proof of Claim 3 in the proof of Lemma 2.7.To put everything together, we set T ( A ) := S i =1 T i ( A ). Then, it follows that S i =1 V i ( A ) ⊆ T ( A ) for every A ∈ R . Clearly, the set T ( A ) is measurable. Summing the three estimates onareas of the trapezoids, we obtain λ ( T ( A )) (cid:54) (cid:0) γ − + 3(1 − γ ) − γ − + 4(1 − γ ) − − (cid:1) smc( R )for every point A ∈ R . We choose γ ∈ (0 ,
1) so that the value of the coefficient is minimized.For x ∈ (0 , x x − + 3(1 − x ) − x − + 4(1 − x ) − − . <
87 at x ≈ . λ ( T ( A )) <
87 smc( R ) for every A ∈ R .It remains to show that the set { ( A, B ) : A ∈ R and B ∈ T ( A ) } is measurable. For every body P of R and for i ∈ { , , } , the definition of the trapezoid T i ( A ) in Claim i implies that theset { ( A, B ) : A ∈ P and B ∈ T i ( A ) } is the intersection of P × R with a semialgebraic (hencemeasurable) subset of ( R ) and hence is measurable. There are countably many bodies of R , aseach of them has positive measure. Therefore, { ( A, B ) : A ∈ R and B ∈ T ( A ) } is a countableunion of measurable sets and hence is measurable. (cid:3) Let S be a bounded open subset of the plane, and let ‘ be a diagonal of S that lies on the x -axis. For a point A ∈ S , we define the set S ( A, ‘ ) := { B ∈ Vis(
A, S ) : AB ∩ ‘ = ∅ and | y ( A ) | (cid:62) | y ( B ) |} . The following lemma is a slightly more general version of a result of Cabello et al. [6].
Lemma 2.8.
Let S be a bounded open simply connected subset of R , and let ‘ be its diagonal.It follows that λ ( S ( A, ‘ )) (cid:54) S ) for every A ∈ S .Proof. We can assume without loss of generality that ‘ lies on the x -axis. Using an argumentsimilar to the proof of Lemma 2.2, we can show that the set { B ∈ Vis(
A, S ) : AB ∩ ‘ = ∅} isopen. Therefore, S ( A, ‘ ) is the intersection of an open set and the closed half-plane { ( x, y ) ∈ R : y (cid:54) − y ( A ) } or { ( x, y ) ∈ R : y (cid:62) − y ( A ) } , whichever contains A . Consequently, the set S ( A, ‘ ) is measurable for every A ∈ S .We clearly have λ ( S ( A, ‘ )) = 0 for points A ∈ S (cid:114) Vis( ‘, S ). By Lemma 2.3, the setVis(
A, S ) ∩ ‘ is an open subsegment CD of ‘ . The interior T ◦ of the triangle T := ACD iscontained in S . Since T ◦ is a convex subset of S , we have λ ( T ◦ ) = | CD | · | y ( A ) | (cid:54) smc( S ).Therefore, every point B ∈ S ( A, ‘ ) is contained in a trapezoid of height | y ( A ) | with bases oflength | CD | and 2 | CD | . The area of this trapezoid is | CD | · | y ( A ) | (cid:54) S ). Hence we have λ ( S ( A, ‘ )) (cid:54) S ) for every point A ∈ S . (cid:3) Proof of Theorem 1.3.
In view of Lemma 2.1, we can assume without loss of generality that S is a bounded open simply connected set. Let ‘ be a diagonal of S . We can assume without lossof generality that ‘ lies on the x -axis.According to Lemma 2.4, the set S (cid:114) ‘ has exactly two p-components S and S , the sets S ∪ ‘ and S ∪ ‘ are rooted sets, and ‘ is their common root. By Lemma 2.7, for i ∈ { , } , every point A ∈ S i ∪ ‘ can be assigned a measurable set T i ( A ) so that λ ( T i ( A )) <
87 smc( S i ∪ ‘ ) (cid:54)
87 smc( S ),every line segment BC in S i ∪ ‘ satisfies B ∈ T i ( C ) or C ∈ T i ( B ), and the set { ( A, B ) : A ∈ S i ∪ ‘ and B ∈ T i ( A ) } is measurable.We set S ( A ) := T i ( A ) ∪ S ( A, ‘ ) for every point A ∈ S i with i ∈ { , } . We set S ( A ) := T ( A ) ∪ T ( A ) for every point A ∈ ‘ = S (cid:114) ( S ∪ S ). Let S := { ( A, B ) : A ∈ S and B ∈ S ( A ) } ∪ { ( B, A ) : A ∈ S and B ∈ S ( A ) } ⊆ ( R ) . It follows that the set S is measurable.Let AB be a line segment in S , and suppose | y ( A ) | (cid:62) | y ( B ) | . Then either A and B are indistinct p-components of S (cid:114) ‘ or they both lie in the same component S i with i ∈ { , } . In thefirst case, we have B ∈ S ( A ), since AB intersects ‘ and S ( A, ‘ ) ⊆ S ( A ). In the second case, wehave B ∈ T i ( A ) ⊆ S ( A ) or A ∈ T i ( B ) ⊆ S ( B ). Therefore, we have Seg( S ) ⊆ S . Since bothSeg( S ) and S are measurable, we have λ (Seg( S )) (cid:54) λ ( S ) (cid:54) Z A ∈ S λ ( S ( A )) , where the second inequality is implied by Fubini’s Theorem. The bound λ ( S ( A )) (cid:54)
90 smc( S )implies λ (Seg( S )) (cid:54) Z S
90 smc( S ) = 180 smc( S ) λ ( S ) . Finally, this bound can be rewritten as b( S ) = λ (Seg( S )) λ ( S ) − (cid:54)
180 c( S ). (cid:3) Proof of Lemma 2.1.
In this section, we prove Lemma 2.1, which reduces the generalsetting of Theorem 1.3 to the case that S is a bounded open simply connected subset of R . Lemma 2.9.
Let S ⊆ R be a set whose b( S ) is defined. For every ε > , there is a bounded set S ⊆ S such that λ ( S ) (cid:62) (1 − ε ) λ ( S ) and b( S ) (cid:62) b( S ) − ε . Moreover, if S is p-componentwisesimply connected, then so is S .Proof. Let B be an open ball in R centered at the origin. Consider the sets S = S ∩ B and S = S (cid:114) B partitioning the set S . Fix the radius of B large enough, so that S has measure atmost ελ ( S ) /
2. We claim that S has the properties stated in the lemma.Clearly λ ( S ) (cid:62) (1 − ε/ λ ( S ) > (1 − ε ) λ ( S ). Moreover, Seg( S ) = Seg( S ) (cid:114) (( S × S ) ∪ ( S × S )),and hence Seg( S ) is measurable and we have λ (Seg( S )) (cid:62) λ (Seg( S )) − ελ ( S ) . Therefore,b( S ) = λ (Seg( S )) λ ( S × S ) (cid:62) λ (Seg( S )) λ ( S × S ) (cid:62) λ (Seg( S )) − ελ ( S ) λ ( S × S ) = b( S ) − ε, as claimed. It is clear from the construction that if S is p-componentwise simply connected,then so is S . (cid:3) Lemma 2.10.
Let S ⊆ R be a bounded p-componentwise simply connected measurable set withmeasurable segment set. Then λ (Seg( S ) (cid:114) Seg( S ◦ )) = 0 . In other words, all the segments in S are in fact contained in S ◦ , except for a set of measure zero. N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 13
Proof.
Let B denote the set Seg( S ) (cid:114) Seg( S ◦ ), that is, B is the set of segments in S containingat least one point of ∂S . Note that B is measurable, since Seg( S ) is measurable by assumptionand Seg( S ◦ ) is an open set by Lemma 2.2, hence it is measurable as well.Let AB be a segment contained in S , and let C be a point of AB . We say that C is an isolated boundary point of the segment AB , if C ∈ ∂S , but there is an ε > AB ∩ N ε ( C ) belongs to ∂S .We partition the set B into four parts as follows: B | := { ( A, B ) ∈ B : A = B or AB is a vertical segment } , B (cid:67) := { ( A, B ) ∈ B (cid:114) B | : A is an isolated boundary point of AB } , B (cid:66) := { ( A, B ) ∈ B (cid:114) ( B | ∪ B (cid:67) ) : B is an isolated boundary point of AB } , B • := B (cid:114) ( B | ∪ B (cid:67) ∪ B (cid:66) ) . We claim that each of these sets has measure zero. For B | , this is clear, since B | is a subset of { ( A, B ) ∈ R × R : A = B or AB is a vertical segment } , which clearly has λ -measure zero.Consider now the set B (cid:67) . We first argue that it is measurable. For a set α ⊆ [0 , A, B ), define AB [ α ] := { tB + (1 − t ) A : t ∈ α } , and let S ( α ) be theset { ( A, B ) ∈ R × R : AB [ α ] ⊆ S ◦ } . In particular, if α = [0 ,
1] then AB [ α ] = AB and S ( α ) = Seg( S ◦ ). If α is a closed interval, then AB [ α ] is a segment, and it is not hard to seethat S ( α ) is an open set, and, in particular, it is measurable. If α is an open interval, say α = ( s, t ) ⊆ [0 , S ( α ) = T n ∈ N S ([ s + n − , t − n − ]), and hence S ( α ) is measurable aswell. We then see that B (cid:67) = B ∩ ( ∂S × S ) ∩ (cid:16) [ n ∈ N S ((0 , n − )) (cid:17) , showing that B (cid:67) is measurable. An analogous argument shows that B (cid:66) is measurable, andhence B • is measurable as well.In the rest of the proof, we will use two basic facts of integral calculus, which we now stateexplicitly. Fact 1 (see [18, Lemma 7.25 and Theorem 7.26]) . Let
X, Y ⊆ R d be two open sets, and let σ : X → Y be a bijection such that both σ and σ − are continuous and differentiable on X and Y , respectively. Then, for any X ⊆ X , the set X is measurable if and only if σ ( X ) ismeasurable. Moreover, λ ( X ) = 0 if and only if λ ( σ ( X )) = 0 . Fact 2 (Fubini’s Theorem, see [18, Theorem 8.12]) . Let M ⊆ R k × R ‘ be a measurable set. For x ∈ R k , define M x := { y ∈ R ‘ : ( x, y ) ∈ M } . Then, for almost every x ∈ R k , the set M x is λ ‘ -measurable, and λ k + ‘ ( M ) = Z x ∈ R k λ ‘ ( M x ) . Let us prove that λ ( B (cid:67) ) = 0. The basic idea is as follows: suppose that we have fixed a non-vertical line L and a point B ∈ L . It can be easily seen that there are at most countably manypoints A ∈ L such that ( A, B ) ∈ B (cid:67) . Since a line L with a point B ∈ L can be determined bythree parameters, we will see that B (cid:67) has λ -measure zero.Let us describe this reasoning more rigorously. Let L a,b denote the line { ( x, y ) ∈ R : y = ax + b } . Define a mapping σ : R → R × R as follows: σ ( a, b, x, x ) = ( A, B ), where A = ( x, ax + b )and B = ( x , ax + b ). In other words, σ ( a, b, x, x ) is the pair of points on the line L a,b whosehorizontal coordinates are x and x , respectively. For every non-vertical segment AB , there is aunique quadruple ( a, b, x, x ) with x = x , such that σ ( a, b, x, x ) = ( A, B ). In particular, σ is a bijection from the set { ( a, b, x, x ) ∈ R : x = x } to the set { ( A, B ) ∈ R × R : A, B not on thesame vertical line } .Define d B (cid:67) = σ − ( B (cid:67) ). Note that σ satisfies the assumptions of Fact 1, and therefore d B (cid:67) ismeasurable. Moreover, λ ( d B (cid:67) ) = 0 if and only if λ ( B (cid:67) ) = 0.For a fixed triple ( a, b, x ) ∈ R , let X a,b,x denote the set { x ∈ R : ( a, b, x, x ) ∈ d B (cid:67) } . We claimthat X a,b,x is countable. To see this, choose a point x ∈ X a,b,x and define ( A, B ) := σ ( a, b, x, x ).Since ( A, B ) ∈ B (cid:67) , we know that A is an isolated boundary point of AB , which implies thatthere is a closed interval β ⊆ R of positive length such that β ∩ X a,b,x = { x } . This implies that X a,b,x is countable and thus of measure zero.Since d B (cid:67) is measurable, we can apply Fubini’s Theorem to get λ ( d B (cid:67) ) = Z ( a,b,x ) ∈ R λ ( X a,b,x ) . Therefore λ ( d B (cid:67) ) = 0 as claimed. A similar argument shows that λ ( d B (cid:66) ) = 0.It remains to deal with the set B • . We will use the following strategy: we will fix two parallelnon-horizontal lines L , L , and study the segments orthogonal to these two lines, with oneendpoint on L and the other on L . Roughly speaking, our goal is to show that for “almostevery” choice of L and L , there are “almost no” segments of this form belonging to B • .Let L a,b denote the (non-horizontal) line { ( ay + b, y ) : y ∈ R } . Let us say that a pair of distinctpoints ( A, B ) has type ( a, b, c ), if A ∈ L a,b , B ∈ L a,c , and the segment AB is orthogonal to L a,b (and therefore also to L a,c ). The value a is then called the slope of the type t = ( a, b, c ).Note that every pair of distinct points ( A, B ) defining a non-vertical segment has a uniquetype ( a, b, c ), with b = c . Define a mapping τ : R → R × R , where τ ( a, b, c, y ) is the pair ofpoints ( A, B ) of type ( a, b, c ) such that A = ( ay + b, y ). Note that τ is a bijection from the set { ( a, b, c, y ) ∈ R : b = c } to the set { ( A, B ) ∈ R × R : A, B not on the same vertical line } . Wecan easily verify that τ satisfies the assumptions of Fact 1.Define f B • = τ − ( B • ). From Fact 1, it follows that f B • is measurable, and λ ( B • ) = 0 ifand only if λ ( f B • ) = 0. For a type t = ( a, b, c ) ∈ R , define Y t = { y ∈ R : ( a, b, c, y ) ∈ f B • } .Furthermore, for a set α ⊆ [0 , B • ( α ) = B • ∩ S ( α ), f B • ( α ) = τ − ( B • ( α )), and Y t ( α ) = { y ∈ R : ( a, b, c, y ) ∈ f B • ( α ) } . In our applications, α will always be an interval (in fact,an open interval with rational endpoints), and in such case we already know that B • ( α ) ismeasurable, hence f B • ( α ) is measurable.By Fubini’s Theorem, we have( ∗ ) λ ( f B • ) = Z a ∈ R Z ( b,c ) ∈ R λ ( Y ( a,b,c ) ) , and Y t is measurable for all t ∈ R up to a set of λ -measure zero. An analogous formula holdsfor f B • ( α ) and Y t ( α ) for any open interval α ⊆ [0 ,
1] with rational endpoints. Since there areonly countably many such intervals, and a countable union of sets of measure zero has measurezero, we know that there is a set T ⊆ R of measure zero, such that for all t ∈ R (cid:114) T the set Y t is measurable, and moreover for any rational interval α the set Y t ( α ) is measurable as well.Our goal is to show that there are at most countably many slopes a ∈ R for which there is a( b, c ) ∈ R such that λ ( Y ( a,b,c ) ) >
0. From ( ∗ ) it will then follow that e λ ( B • ) = 0. To achievethis goal, we will show that to any type t for which λ ( Y t ) >
0, we can assign a set R t ⊆ ∂S ofpositive λ -measure (the region of t ), so that if t and t have different slopes and if Y t and Y t both have positive measure, then R t and R t are disjoint. Since there cannot be uncountablymany disjoint sets of positive measure, this will imply the result. N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 15 L a,b L a,c A A B B S y ∈ Y ∗ t y ∈ Y t Figure 6.
An illustration for the proof of Lemma 2.10. The element y of Y t ishalf-isolated while y is not.Let us fix a type t = ( a, b, c ) ∈ R (cid:114) T such that λ ( Y t ) >
0. Let us say that an element y ∈ Y t is half-isolated if there is an ε > y, y + ε ] ∩ Y t = { y } or [ y − ε, y ] ∩ Y t = { y } .Clearly, Y t has at most countably many half-isolated elements. Define Y ∗ t := { y ∈ Y t : y is nothalf-isolated } . Of course, λ ( Y ∗ t ) = λ ( Y t ). See Figure 6 for an illustration.Choose y ∈ Y ∗ t , and define ( A y , B y ) := τ ( a, b, c, y ). We claim that A y B y ∩ S ◦ is either empty ora single interval. Let us choose any two points C, D ∈ A y B y ∩ S ◦ . We will show that the segment CD is inside S ◦ . For ε > N ε ( C ) and N ε ( D ) are subsets of S .Since y is not half-isolated in Y t , we can find two segments P, Q ∈ B • of type t that intersectboth N ε ( C ) and N ε ( D ), with A y B y being between P and Q . We can then find a closed polygonalcurve Γ ⊆ P ∪ Q ∪N ε ( C ) ∪N ε ( D ) whose interior region contains CD . Since S is p-componentwisesimply connected, we see that CD ⊆ S ◦ . Therefore, A y B y ∩ S ◦ is indeed an interval.Since ∂S is a closed set, we know that for every y ∈ Y ∗ t , the set A y B y ∩ ∂S is closed as well.Moreover, neither A y nor B y are isolated boundary points of A y B y , because then ( A y , B y ) wouldbelong to B (cid:67) or B (cid:66) . We conclude that A y B y ∩ ∂S is either equal to a single closed segment ofpositive length containing A y or B y , or it is equal to a disjoint union of two closed segments ofpositive length, one of which contains A y and the other contains B y .For an integer n ∈ N , define two sets Y (cid:67) t ( n ) and Y (cid:66) t ( n ) by Y (cid:67) t ( n ) := { y ∈ Y ∗ t : A y B y [(0 , n − )] ⊆ ∂S } ,Y (cid:66) t ( n ) := { y ∈ Y ∗ t : A y B y [(1 − n − , ⊆ ∂S } . Note that these sets are measurable: for instance, Y (cid:67) t ( n ) is equal to Y ∗ t (cid:114) ( S α Y t ( α )), wherewe take the union over all rational intervals α intersecting (0 , n − ). Moreover, we have Y ∗ t = S n ∈ N ( Y (cid:67) t ( n ) ∪ Y (cid:66) t ( n )). It follows that there is an n such that Y (cid:67) t ( n ) or Y (cid:66) t ( n ) has positivemeasure. Fix such an n and assume, without loss of generality, that λ ( Y (cid:67) t ( n )) is positive.Define the region of t , denoted by R t , by R t := [ y ∈ Y (cid:67) t ( n ) A y B y [(0 , n − )] . The set R t is a bijective affine image of Y (cid:67) t ( n ) × (0 , n − ), and in particular it is λ -measurablewith positive measure. Note that R t is a subset of ∂S . Consider now two types t, t ∈ R (cid:114) T with distinct slopes, such that both Y t and Y t havepositive measure. We will show that the regions R t and R t are disjoint.For contradiction, suppose there is a point C ∈ R t ∩ R t . Let AB and A B be the segmentscontaining C and having types t and t , respectively. Fix ε > A, B, A , B lies in N ε ( C ). Since Y ∗ t has no half-isolated points of Y t , we know that B • has segments of type t arbitrarily close to AB on both sides of AB , and similarly for segmentsof type t close to A B . We can therefore find four segments P, Q, P , Q ∈ B • (cid:114) { AB, A B } with these properties: • P and Q have type t , and P and Q have type t . • AB is between P and Q (i.e., AB ⊆ Conv( P ∪ Q )) and A B is between P and Q . • Both P and Q intersect both P and Q inside N ε ( C ).We see that the four points where P ∪ Q intersects P ∪ Q form the vertex set of a parallelogram W whose interior contains the point C . Moreover, the boundary of W is a closed polygonalcurve contained in S . Since S is p-componentwise simply connected, W is a subset of S and C belongs to S ◦ . This is a contradiction, since all points of R t (and R t ) belong to ∂S .We conclude that R t and R t are indeed disjoint. Since there cannot be uncountably manydisjoint sets of positive measure in R , there are at most countably many values a ∈ R for whichthere is a type t = ( a, b, c ) with λ ( Y t ) positive. Consequently, the right-hand side of ( ∗ ) is zero,and so λ ( B • ) = 0, as claimed. (cid:3) Proof of Lemma 2.1.
Observe that the inequalities b( S ) (cid:54) α c( S ) and λ (Seg( S )) (cid:54) α smc( S ) · λ ( S ) are equivalent. Call a set S bad if Seg( S ) is measurable and b( S ) > α c( S ) or equivalently λ (Seg( S )) > α smc( S ) λ ( S ). To prove the lemma, we suppose for the sake of contradictionthat there exists a bad p-componentwise simply connected set S ⊆ R of finite positive measure.By Lemma 2.9, for each ε >
0, there is a bounded p-componentwise simply connected set S ⊆ S such that λ ( S ) (cid:62) (1 − ε ) λ ( S ) and b( S ) (cid:62) b( S ) − ε . In particular, such a set S satisfies c( S ) (cid:54) c( S ) / (1 − ε ). Hence, for ε small enough, the set S is bad.Let S be the interior of S . By Lemma 2.10, λ (Seg( S )) = λ (Seg( S )). Clearly, λ ( S ) (cid:54) λ ( S ) and smc( S ) (cid:54) smc( S ), and therefore S is bad as well.Note that S is p-componentwise simply connected. Since S is an open set, all its p-components are open as well. In particular, S has at most countably many p-components. Let C be the set of p-components of S . Each T ∈ C is a bounded open simply connected set, andtherefore cannot be bad. Therefore, λ (Seg( S )) = X T ∈C λ (Seg( T )) (cid:54) X T ∈C α smc( T ) λ ( T ) (cid:54) α smc( S ) λ ( S ) , showing that S is not bad. This is a contradiction. (cid:3) Proof of Lemma 2.4.
Here we prove Lemma 2.4, which says that every bounded opensimply connected subset of R can can be split by a diagonal into two rooted sets. Lemma 2.11.
Let S be a bounded open simply connected subset of R , and let ‘ be a diagonalof S . Let h − and h + be the open half-planes defined by the supporting line of ‘ . It follows thatthe set S (cid:114) ‘ has exactly two p-components S and S . Moreover, for every point A ∈ ‘ andevery neighborhood N ε ( A ) ⊆ S , we have N ε ( A ) ∩ h − ⊆ S and N ε ( A ) ∩ h + ⊆ S .Proof. Notice first that any p-component of an open set is also open. This implies that anypath-connected open set is also p-connected, and therefore every open simply connected set isp-connected as well.
N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 17
Figure 7.
Removing self-intersections and intersections between adjacent polyg-onal lines.Let A ∈ ‘ , and let N ε ( A ) be a neighborhood of A contained in S . We choose arbitrary points B ∈ N ε ( A ) ∩ h − and C ∈ N ε ( A ) ∩ h + . Suppose for a contradiction that S (cid:114) ‘ has a singlep-component. Then there exists a polygonal curve Γ in S (cid:114) ‘ with endpoints B and C . Let∆ ⊆ S be the closed polygonal curve Γ ∪ BC . We can assume that the curve ∆ is simple using alocal redrawing argument. See Figure 7.The curve ∆ separates R into two regions. The closure ‘ of the diagonal ‘ is a closed linesegment that intersects ∆ in exactly one point. It follows that one endpoint of ‘ is in the interiorregion of ∆. Since the endpoints of ‘ do not belong to S , this contradicts the assumption that S is simply connected.Now, we show that the set S (cid:114) ‘ has at most two p-components. For a point D ∈ ‘ , let N ε ( D )be a neighborhood of D in S . The set N ε ( D ) ∩ h − is contained in a unique p-component S of S (cid:114) ‘ , and N ε ( D ) ∩ h + is contained in a different p-component S . Choose another point E ∈ ‘ with a neighborhood N ε ( E ) ⊆ S . We claim that N ε ( E ) ∩ h − also belongs to S . To seethis, note that since DE is a compact subset of the open set S , it has a neighborhood N δ ( DE )which is contained in S . Clearly, N δ ( DE ) ∩ h − is p-connected and therefore belongs to S , hence N ε ( E ) ∩ h − belongs to S as well. An analogous argument can be made for the half-plane h + and the p-component S .Since for every p-component S of S (cid:114) ‘ , there is a point A ∈ ‘ and a neighborhood N ε ( A ) ⊆ S such that N ε ( A ) ∩ S = ∅ , we see that S and S are the only two p-components of S (cid:114) ‘ . (cid:3) Proof of Lemma 2.4.
By Lemma 2.11, the set S (cid:114) ‘ has of exactly two p-components S and S .It remains to show that S ∪ ‘ and S ∪ ‘ are rooted sets.Since S and S are p-connected, S ∪ ‘ and S ∪ ‘ are p-connected as well. To show that S ∪ ‘ and S ∪ ‘ are simply connected, choose a Jordan curve Γ in, say, S ∪ ‘ , and let Z be theinterior region of Γ. Suppose for a contradiction that Z is not a subset of S ∪ ‘ . Since S is simplyconnected, we have Z ⊆ S . Hence there is a point A ∈ Z ∩ S . Since both S and Z are open, wecan assume that A does not lie on the supporting line of ‘ . Let AB be the minimal closed segmentparallel to ‘ such that B ∈ Γ. Then B belongs to S , A belongs to S , and yet A and B are in thesame p-component of S (cid:114) ‘ . This contradiction shows that S ∪ ‘ and S ∪ ‘ are simply connected.As subsets of the bounded set S , the sets S ∪ ‘ and S ∪ ‘ are bounded. Lemma 2.11 andthe fact that S i is open imply that the set S i ∪ ‘ is half-open and S i ∩ ∂S i = ‘ for i ∈ { , } .Therefore, the sets S ∪ ‘ and S ∪ ‘ are rooted, and ‘ is their root. (cid:3) Proof of Lemma 2.5.
Here we prove Lemma 2.5, which explains the tree structure ofrooted sets. For this entire section, let R be a rooted set and ( L k ) k (cid:62) be the partition of R intolevels. We will need several auxiliary results in order to prove Lemma 2.5.For disjoint sets S, T ⊆ R , we say that the set S is T -half-open if every point A ∈ S has aneighborhood N ε ( A ) that satisfies one of the following two conditions:(1) N ε ( A ) ⊆ S , (2) N ε ( A ) ∩ ∂S is a diameter of N ε ( A ) splitting it into two subsets, one of which (including thediameter) is N ε ( A ) ∩ S and the other (excluding the diameter) is N ε ( A ) ∩ T .The only difference with the definition of S being half-open is that we additionally specify the“other side” of the neighborhoods N ε ( A ) for points A ∈ S ∩ ∂S in the condition (2). A rootedset R is T -half-open if and only if it is attached to T according to the definition of attachmentfrom Section 2. Lemma 2.12.
The set L is ( R (cid:114) R ) -half-open and L ∩ ∂L = R ∩ ∂R .Proof. We consider two cases for a point A ∈ L . First, suppose A ∈ L ∩ ∂R . It follows that A has a neighborhood N ε ( A ) that satisfies the condition (2) of the definition of a half-open set.By the definition of L , the same neighborhood N ε ( A ) satisfies the condition (2) for L beingan ( R (cid:114) R )-half-open set. In particular, A ∈ ∂L . Since R ∩ ∂R ⊆ L by the definition of L ,we have R ∩ ∂R ⊆ L ∩ ∂L .Now, suppose A ∈ L ∩ R ◦ . Let B be a point of the root of R such that AB ⊆ R . We have AB (cid:114) { B } ⊆ R ◦ , as otherwise the point t A + (1 − t ) B for t := sup { t ∈ [0 ,
1] : At + (1 − t ) B ∈ AB ∩ ∂R } would contradict the fact that R is half-open. There is a family of neighborhoods {N ε C ( C ) } C ∈ AB such that all N ε D ( D ) with D ∈ AB (cid:114) { B } satisfy the condition (1) and N ε B ( B )satisfies the condition (2) for R being half-open. Since AB is compact, there is a finite set X ⊆ AB such that AB ⊆ S C ∈ X N ε C / ( C ). Hence N ε ( AB ) ⊆ S C ∈ X N ε C ( C ), where ε := min C ∈ X ε C /
2. Itfollows that N ε ( AB ) ∩ ∂R is an open segment Q containing B but not A and splitting N ε ( AB )into two subsets, one of which (including Q ) is N ε ( AB ) ∩ R and the other (excluding Q ) is N ε ( AB ) (cid:114) R . Let ε be the minimum of ε and the distance of A to the line containing Q . Itfollows that N ε ( A ) ⊆ N ε ( AB ) ∩ R . Therefore, for every A ∈ N ε ( A ), we have A B ⊆ R , hence N ε ( A ) ⊆ L . It also follows that L ∩ ∂L ⊆ R ∩ ∂R . (cid:3) We say that a set P ⊆ R is R -convex when the following holds for any two points A, B ∈ P :if AB ⊆ R , then AB ⊆ P . Lemma 2.13.
The set L is R -convex.Proof. This follows directly from Lemma 2.3. (cid:3) A branch of R is a p-component of S k (cid:62) L k . Lemma 2.14.
Every branch of R is R -convex.Proof. Let P be a branch of R , and let A, B ∈ P be such that AB ⊆ R . Since R is half-open, itfollows that AB ⊆ R ◦ . Suppose AB P . It follows that AB ∩ L = ∅ . Since L is ( R (cid:114) R )-half-open (Lemma 2.12) and R -convex (Lemma 2.13), we see that AB ∩ L is an open segment A B for some A , B ∈ AB . It follows that A , B ∈ P .There is a simple polygonal line in P connecting A with B , which together with A B formsa Jordan curve Γ in R . Now, let C ∈ A B . Since C ∈ L , there is a point D on the root of R such that CD ⊆ R . Since A , B / ∈ L , D does not lie on the supporting line of A B . Extend thesegment DC beyond C until hitting ∂R at a point C . Here we use the fact that R is bounded.Since R is simply connected, the entire interior region of Γ is contained in R , so the points D and C both lie in the exterior region of Γ. However, since Γ ∩ L = A B , the line segment DC crosses Γ at exactly one point, which is C . This is a contradiction. (cid:3) Lemma 2.15.
The set L and every branch of R are p-connected and simply connected. N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 19
Proof.
Let P be the set L or a branch of R . It follows directly from the definitions of L and abranch of R that P is p-connected. To see that P is simply connected, let Γ be a Jordan curve in P , A be a point in the interior region of Γ, and BC be an inclusion-maximal open line segment inthe interior region of Γ such that A ∈ BC . It follows that B, C ∈ Γ and BC ⊆ R , as R is simplyconnected. Since B, C ∈ P and P is R -convex (Lemmas 2.13 and 2.14), we have A ∈ P . (cid:3) Lemma 2.16.
Every branch of R is L -half-open.Proof. Let P be a branch of R . It is enough to check the condition (2) for P being L -half-openfor points in ∂P ∩ P . Let A ∈ ∂P ∩ P . Since R is half-open, A has a neighborhood N ε ( A ) thatsatisfies the condition (1) or (2) for S being half-open. It cannot be (2), as then A would lie onthe root of R and thus in L . Hence N ε ( A ) ⊆ R .Since L is ( R (cid:114) R )-half-open (Lemma 2.12) and R -convex (Lemma 2.13) and A / ∈ L , theset N ε ( A ) ∩ L lies entirely in some open half-plane h whose boundary line passes through A .The set N ε ( A ) (cid:114) h is p-connected and contains A , so it lies entirely within P . The set N ε ( A ) ∩ h is disjoint from P . Indeed, if there was a point B ∈ N ε ( A ) ∩ h ∩ P , then by the R -convexity of P (Lemma 2.14), the convex hull of N ε ( A ) (cid:114) h and B would lie entirely within P and wouldcontain A in its interior, which would contradict the assumption that A ∈ ∂P . It follows that N ε ( A ) ∩ ∂P is an open segment that partitions N ε ( A ) into two half-discs, one of which (including N ε ( A ) ∩ ∂P ) is N ε ( A ) ∩ P .We show that N ε ( A ) ∩ ∂P ⊆ ∂L . Suppose to the contrary that there is a point A ∈N ε ( A ) ∩ ∂P (cid:114) L . It follows that A has a neighborhood N ε ( A ) ⊆ N ε ( A ) (cid:114) L . Since N ε ( A ) isp-connected and contains a point of P , it lies entirely within P . This contradicts the assumptionthat A ∈ ∂P .Since N ε ( A ) ∩ ∂P ⊆ ∂L , there is a point B ∈ N ε ( A ) ∩ L . Let A ∈ N ε ( A ) ∩ ∂P . Since L is( R (cid:114) R )-half-open and R -convex and A / ∈ L , there is a point C ∈ A B such that CB (cid:114) { C } ⊆ L while A C is disjoint from L . The latter implies that A C ⊆ P , as A ∈ P . Hence C = A .This shows the whole triangle T spanned by N ε ( A ) ∩ ∂P and B excluding the open segment N ε ( A ) ∩ ∂P is contained in L .Since A lies in the interior of N ε ( A ) ∩ ( P ∪ T ), it has a neighborhood N ε ( A ) that lies entirelywithin N ε ( A ) ∩ ( P ∪ T ). This neighborhood witnesses the condition (2) for P being L -half-open. (cid:3) Lemma 2.17.
Let P be a branch of R . If A , A ∈ P ∩ ∂P , then A A ⊆ R .Proof. Let A , A ∈ P ∩ ∂P . By Lemma 2.16, P is L -half-open, hence there are B , B ∈ L such that A B (cid:114) { A } ⊆ L and A B (cid:114) { A } ⊆ L . There is a polygonal line Γ in P connecting A with A , and a polygonal line Γ in L connecting B with B . These polygonallines together with the line segments A B and A B form a closed polygonal curve Γ in R . Wecan assume without loss of generality that Γ is simple (see Figure 7) and that the x -coordinatesof A and A are equal to 0. We also assume that no two vertices of Γ except A and A havethe same x -coordinates.We color the points of Γ ∩ L red and the points of Γ ∩ P blue. For convenience, we assumethat A and A have both colors. Let Z denote the interior region delimited by Γ including Γitself. Since R is simply connected, we have Z ⊆ R .Let x < · · · < x n be the x -coordinates of all vertices of Γ. We use [ n ] to denote the set ofindices { , . . . , n } . Since the x -coordinates of A and A are zero, there is j ∈ [ n ] such that x j = 0. For i ∈ [ n ], we let l i be the vertical line { x i } × R . Since the x -coordinates of the verticesof Γ (cid:114) { A , A } are distinct, there is at most one vertex of Γ on l i for every i ∈ [ n ] (cid:114) { j } . For x x x j = 0 x n · · · · · · A A Zx Γ CDAB El l l l n l j Figure 8.
Situation in the proof of Lemma 2.17. Here A is a left neighbor of C and B is a left neighbor of D . The points B , C , and D are two-sided, the points A and E are one-sided. i ∈ [ n ], the intersection of Z with l i is a family of closed line segments with endpoints from Γ ∩ l i .Some of the segments can be trivial , that is, consisting of a single point, and some segments cancontain a point of Γ in their interior.For i ∈ [ n ] and a point A ∈ Γ ∩ l i , we say that a point B is a left neighbor of A if B lies onΓ ∩ l i − and AB ⊆ Γ. Similarly, B is a right neighbor of A if B ∈ Γ ∩ l i +1 and AB ⊆ Γ. Notethat every point A ∈ Γ ∩ l i has exactly two neighbors and if A / ∈ { A , A } , then the neighbors of A have the same color as A . We distinguish two types of points of Γ ∩ l i . We say that a point A ∈ Γ ∩ l i is one-sided if it either has two right or two left neighbors. Otherwise, we say that A is two-sided . That is, A is two-sided if it has one left and one right neighbor. See Figure 8.Note that every one-sided point is a vertex of Γ and that one-sided points from Γ ∩ l i areexactly the points of Γ ∩ l i that either form a trivial line segment or that are contained in theinterior of some line segment of Z ∩ l i . Consequently, every line segment in Z ∩ l i contains atmost one point of Γ in its interior.For 2 (cid:54) i (cid:54) n and C, D ∈ l i ∩ Γ, let CD be a line segment in Z ∩ l i whose interior does notcontain a point of Γ with a left neighbor. Let A and B be left neighbors of C and D , respectively,such that there is no left neighbor of C and D between A and B on l i − . Since no point between A and B on l i − can have a right neighbor, we have AB ⊆ Z ∩ l i − and A, B, C, D are vertices ofa trapezoid whose interior is contained in Z . An analogous statement holds for right neighborsof C and D provided that the interior of CD does not contain a point of Γ with a right neighbor. Claim. Let i ∈ [ n ] (cid:114) { j } , and let C and D be points of Γ ∩ l i satisfying CD ⊆ Z ∩ l i . Then C and D have the same color. First, we will prove the claim by induction on i for all i < j . The claim clearly holds for i = 1,as Z ∩ l contains only a single vertex of Γ. Fix i with 1 < i < j and suppose that the claimholds for i −
1. Let
C, D ∈ Γ ∩ l i be points satisfying CD ⊆ Z ∩ l i . We show that C and D havethe same color. Obviously, we can assume that the line segment CD is non-trivial. Assume firstthat the points C and D are two-sided.Suppose the interior of CD does not contain a point of Γ with a left neighbor. Let A, B ∈ Γ ∩ l i − be the left neighbors of C and D , respectively. Then AB ⊆ Z ∩ l i − . Thus A and B have thesame color by the induction hypothesis. Since C, D / ∈ { A , A } , the points A and C have thesame color as well as the points B and D . This implies that C and D have the same color too. N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 21
If there is a point E of Γ in the interior of CD , then it follows from R -convexity of P (Lemma2.14) and L (Lemma 2.13) that E has the same color as C and D .Now, suppose the interior of CD contains a point E of Γ with a left neighbor. We have alreadyobserved that there is exactly one such point on CD . We also know that E has two left neighbors.The points C and E with their left neighbors A and B , respectively, where there is no left neighborof E between A and B on l i − , form a trapezoid in Z such that AB ⊆ Z ∩ l i − . From inductionhypothesis A and B have the same color which implies that C and E have the same color as well.Similarly, D and E have the same color which implies that C and D have the same color as well.The case where either C or D is one-sided is covered by the previous cases. The same inductiveargument but in the reverse direction shows the claim for all i with j < i (cid:54) n . This completesthe proof of the claim.Now, consider the inclusion-maximal line segment CD of Z ∩ l j that contains A . We canassume that either C and D are two-sided. Suppose for a contradiction that A is not containedin CD . If CD is trivial, that is, C = D = A , then A is one-sided and its neighbors A and B have different colors, as Γ changes color in A . This is impossible according to the claim, sincewe have AB ⊆ Z ∩ l i − or AB ⊆ Z ∩ l i +1 . Therefore CD is non-trivial.First, we assume that A is an endpoint of CD , say C = A . Then A is two-sided. Bysymmetry, we can assume that the left neighbor A of A and the left neighbor B of D havedifferent colors. If there is no point of Γ with a left neighbor in the interior of A D , then AB ⊆ Z ∩ l i − . This is impossible according to the claim. If there is a point E ∈ Γ with a leftneighbor in the interior of A D , then we can use a similar argument either for the line segment A E or for ED , as the neighbors of E have the same color. The last case is when A is aninterior point of CD . Since Γ does not change color in C nor in D , we apply the claim to oneof the line segments A C , A D , and CD and show, again, that none of the cases is possible.Altogether, we have derived a contradiction.Therefore, A and A are contained in the same line segment of Z ∩ l i . This completes theproof, as Z ⊆ R . (cid:3) Lemma 2.18.
For every branch P of R , the set P ∩ ∂P is an open segment.Proof. Let P be a branch of R . First, we show that the set P ∩ ∂P is convex. Let A , A ∈ P ∩ ∂P .By Lemma 2.17, we have A A ⊆ R . It follows that A A is disjoint from the root of R and thusis contained in R ◦ . By compactness, A A has a neighborhood N ε ( A A ) contained in R ◦ . Since P is L -half-open by Lemma 2.16, there are B , B ∈ N ε ( A A ) ∩ L such that A B (cid:114) { A } ⊆ L and A B (cid:114) { A } ⊆ L . For t ∈ [0 , A t = (1 − t ) A + tA and B t = (1 − t ) B + tB . Wehave A t ∈ A A and B t ∈ B B , hence A t , B t ∈ N ε ( A A ), for all t ∈ [0 , R -convexity of P (Lemma 2.14) and L (Lemma 2.13) that A t ∈ P and A t B t (cid:114) { A t } ⊆ L ,hence A t ∈ P ∩ ∂P , for all t ∈ [0 , P ∩ ∂P is convex.If P ∩ ∂P had three non-collinear points, then they would span a triangle with non-emptyinterior contained in P ∩ ∂P , which would be a contradiction. Since R is bounded, the set P ∩ ∂P is a line segment. That it is an open line segment follows directly from Lemma 2.16. (cid:3) Lemma 2.19.
For every j (cid:62) , every p-component P of S k (cid:62) j L k is a rooted set attachedto L j − . Moreover, for k (cid:62) , the k th level of P is equal to L j − k ∩ P .Proof. The proof proceeds by induction on j . For the base case, let P be a p-component of S i (cid:62) L i , that is, a branch of R . It follows from Lemmas 2.15, 2.16 and 2.18 that P is a rootedset attached to L . Let ‘ be the root of P , and let ( L k ) k (cid:62) be the partition of P into levels. Weprove that L k ⊆ S k +1 i =1 L i and L k +1 ∩ P ⊆ S ki =1 L i for every k (cid:62) Let A ∈ L k . It follows that there is a polygonal line Γ with k line segments connecting A to apoint B ∈ ‘ . Moreover, since there is no shorter polygonal line connecting A to ‘ , the last linesegment of Γ is not parallel to ‘ . Since P is L -half-open (Lemma 2.16), there is a neighborhood N ε ( B ) that is split by ‘ into two parts, one of which is a subset of L . Let C be a point in N ε ( B ) ∩ L such that BC is an extension of the last line segment of Γ. Since C ∈ L , there is apoint D on the root of R such that CD ⊆ L . The polygonal line Γ extended by BC and CD forms a polygonal line with k + 1 line segments connecting A to the root of R . This shows that L k ⊆ S k +1 i =1 L i .Now, let A ∈ L k +1 ∩ P . It follows that there is a polygonal line Γ with k + 1 line segmentsconnecting A to the root of R . Since L is an open subset of R (Lemma 2.12) and P is a p-component of R (cid:114) L , there is a point B ∈ Γ such that the part of Γ between A and B (inclusive)is contained in P and is maximal with this property. It follows that B ∈ ‘ . Since B / ∈ L , thepart of Γ between A and B consists of at most k segments. This shows that L k +1 ∩ P ⊆ S ki =1 L i .We have thus proved that L k ⊆ S k +1 i =1 L i and L k +1 ∩ P ⊆ S ki =1 L i for every k (cid:62)
1. To concludethe proof of the base case, we note that a straightforward induction shows that L k = L k +1 ∩ P for every k (cid:62) j (cid:62)
3, and let P be a p-component of S i (cid:62) j L i . Let Q be the branch of R containing P . Let ( L k ) k (cid:62) be the partition of Q into levels. As we have proved for the base case,we have L k = L k +1 ∩ Q for every k (cid:62)
1. Hence P is a p-component of ( S i (cid:62) j L i ) ∩ Q = S i (cid:62) j − L i .By the induction hypothesis, P is a rooted set attached to L j − ⊆ L j − . Moreover, for k (cid:62) k th level of P is equal to L j − k ∩ P = L j − k . This completes the induction step andproves the lemma. (cid:3) Proof of Lemma 2.5.
The statement (1) is a direct consequence of the definition of a rooted set,specifically, of the condition that a rooted set is p-connected.For the proof of the statement (2), let P be a j -body of R . If j = 1, then P = L = Vis( r, R ) =Vis( r, L ), where r is the root of R , and by Lemmas 2.12 and 2.15, L is rooted with the sameroot r . Now, suppose j (cid:62)
2. Let Q be the p-component of S k (cid:62) j L k containing P . By Lemma2.19, Q is a rooted set and L j ∩ Q is the first level of Q . Since P ⊆ L j ∩ Q , the definition of thefirst level yields P = L j ∩ Q = Vis( r, Q ) = Vis( r, P ), where r is the root of Q . By Lemma 2.12, P is a rooted set with the same root r .Lemma 2.12 and the fact that L is p-connected directly yield the statement (3).Finally, for the proof of the statement (4), let P be a j -body of R with j (cid:62)
2. Let Q bethe p-component of S k (cid:62) j L k containing P . As we have proved above, Q is a rooted set and P is the first level of Q and shares the root with Q . Moreover, by Lemma 2.19, Q (and hence P ) is attached to L j − . The definition of attachment implies that P is attached to a singlep-component of L j − , that is, a single ( j − R . (cid:3) General dimension
This section is devoted to the proofs of Theorem 1.5 and Theorem 1.6. In both proofs, we usethe operator Aff to denote the affine hull of a set of points.3.1.
Proof of Theorem 1.5.
Let T = ( B , B , . . . , B d ) be a ( d + 1)-tuple of distinct affinelyindependent points in R d . We say that a permutation B , B , . . . , B d of T is a regular permutation of T if the following two conditions hold:(1) the segment B B is the diameter of T , N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 23 (2) for i = 2 , . . . , d −
1, the point B i has the maximum distance to Aff( { B , . . . , B i − } ) amongthe points B i , B i +1 , . . . , B d .Obviously, T has at least two regular permutations due to the interchangeability of B and B .The regular permutation B i , B i , . . . , B i d with the lexicographically minimal vector ( i , i , . . . , i d )is called the canonical permutation of T .Let T be a ( d + 1)-tuple of distinct affinely independent points in R d , and let B , B , . . . , B d be the canonical permutation of T . For i = 1 , . . . , d −
1, we define Box i ( T ) inductively as follows:(1) Box ( T ) := B B ,(2) for i = 2 , . . . , d −
1, Box i ( T ) is the box containing all the points P ∈ Aff( { B , B , . . . , B i } )with the following two properties: • the orthogonal projection of P to Aff( { B , B , . . . , B i − } ) lies in Box i − ( T ), • the distance of P to Aff( { B , B , . . . , B i − } ) does not exceed the distance of B i toAff( { B , B , . . . , B i − } ),(3) Box d ( T ) is the box containing all the points P ∈ R d such that the orthogonal projection of P to Aff( { B , B , . . . , B d − } ) lies in Box d − ( T ) and λ d (Conv( { B , B , . . . , B d − , P } )) (cid:54) λ d ( S ) c( S ) . The definition of Box d ( T ) is independent of B d , so we can define Box d ( T (cid:114) { B d } ) byBox d ( T (cid:114) { B d } ) := Box d ( T ) . It is not hard to see that this gives us a proper definition of Box d ( T − ) for every d -tuple T − of d distinct affinely independent points in R d . Lemma 3.1. (1)
For i = 1 , . . . , d − , the box Box i ( T ) contains the orthogonal projection ofany point of T to Aff( { B , B , . . . , B i − } ) . (2) If Conv( T ) ⊆ S then Box d ( T ) contains the point B d .Proof. We prove the statement (1) by induction on i . First, let i = 1. Then the segment Box ( T )must contain every point A j ∈ T since otherwise one of the segments B A j and B A j wouldbe longer than the segment B B . Further, if a point A j ∈ T satisfies the statement (1) for aparameter i ∈ { , . . . , d − } then it also satisfies the statement (1) for the parameter i + 1 sinceotherwise A j = B i +1 and A j should have been chosen for B i +1 .The statement (2) follows from the fact that Conv( T ) ⊆ S implies λ d (Conv( T )) (cid:54) smc( S ) = λ d ( S ) c( S ). (cid:3) For i = 1 , . . . , d −
1, let d i be the distance of B i to Aff( { B , B , . . . , B i − } ). In particular, d is the diameter of T . The following observation follows from the definition of the canonicalpermutation of T and from the construction of the boxes Box i ( T ). Observation 3.2. (1)
The ( d − -dimensional measure of the simplex Conv( T (cid:114) { B d } ) is equalto d d · · · d d − / ( d − . (2) The sides of
Box d ( T ) have lengths d , d , . . . , d d − , and d ! d d ··· d d − λ d ( S ) c( S ) .Proof of Theorem 1.5. To estimate b d ( S ), we partition Simp d ( S ) into the following d + 2 subsets: X := { T ∈ Simp d ( S ) : T is affinely dependent } ,Y i := { T = ( A , . . . , A d ) ∈ Simp d ( S ) : T is affinely independent, and A i is the lastelement of the canonical permutation of T } ,for i = 0 , . . . , d . We point out that T is considered to be affinely dependent in the above definitions of X and Y i also in the degenerate case when some point of S appears more than once in T . We have λ d ( d +1) ( X ) = 0. Let i ∈ { , . . . , d } . The set Y i is a subset of the set Y i := { T = ( A , . . . , A d ) ∈ S d +1 : T (cid:114) { A i } is affinely independent and we have A i ∈ Box d ( T (cid:114) { A i } ) } .By Observation 3.2 (2), λ d (Box d ( T (cid:114) { A i } )) is equal to z := 2 d − d ! λ d ( S ) c( S ) for every set T (cid:114) { A i } appearing in the definition of Y i . Therefore, by Fubini’s Theorem, the set Y i is λ d ( d +1) -measurable and, moreover, λ d ( d +1) ( Y i ) = ( λ d ( S )) d z = ( λ d ( S )) d +1 d − d ! c( S ) . Thus, b d ( S ) = λ d ( d +1) (Simp d ( S )) λ d ( S ) d +1 (cid:54) λ d ( d +1) ( X ) + P di =0 λ d ( d +1) ( Y i ) λ d ( S ) d +1 = 2 d − ( d + 1)! c( S ) . This completes the proof of Theorem 1.5. (cid:3)
Proof of Theorem 1.6.
In the following, we make no serious effort to optimize theconstants. As the first step towards the proof of Theorem 1.6, we show that if we remove anarbitrary n -tuple of points from the open d -dimensional box (0 , d , then the d -index of convexityof the resulting set is of order Ω( n ). Lemma 3.3.
For every positive integer n and every n -tuple N of points from (0 , d , the set S := (0 , d (cid:114) N satisfies b d ( S ) (cid:62) / n .Proof. Let S and N = { B , . . . , B n } be the sets from the statement and let 0 be the origin. Weuse S d − ∗ to denote the set of ( d − A , . . . , A d − ) ∈ S d − that satisfy the following: forevery B ∈ N the points A , . . . , A d − , B are affinely independent and Aff( { A , . . . , A d − , B } ) ∩ ( N ∪ { } ) = { B } . Note that the set S d − ∗ is measurable and λ d ( d − ( S d − ∗ ) = 1. If h is ahyperplane in R d that does not contain the origin, we use h − and h + to denote the open half-spaces defined by h such that 0 ∈ h − .Let ( A , . . . , A d − ) ∈ S d − ∗ . For a point B ∈ N , we let h A ,...,A d − ,B be the hyperplane deter-mined by the d -tuple ( A , . . . , A d − , B ). Since ( A , . . . , A d − ) ∈ S d − ∗ , we see that h A ,...,A d − ,B satisfies h A ,...,A d − ,B ∩ N = { B } and that it does not contain the origin. Therefore the half-spaces h − A ,...,A d − ,B and h + A ,...,A d − ,B are well defined.For every ( d − A , . . . , A d − ) ∈ S d − ∗ , we split the set S into 2 n pairwise disjointopen convex sets that are determined by the hyperplanes h A ,...,A d − ,B for B ∈ N . Thisis done by induction on n . For n = 1, we set P ( A , . . . , A d − ) := S ∩ h − A ,...,A d − ,B and P ( A , . . . , A d − ) := S ∩ h + A ,...,A d − ,B . Suppose we have split the set S into sets P i ( A , . . . , A d − )for 1 (cid:54) i (cid:54) n −
1) and n (cid:62)
2. Consider the hyperplane h A ,...,A d − ,B n . Since for every k ∈ { , . . . , n − } the intersection h A ,...,A d − ,B k ∩ h A ,...,A d − ,B n is the affine hull of A , . . . , A d − ,we see that h A ,...,A d − ,B n (cid:114) Aff( { A , . . . , A d − } ) is contained in two sets P i ( A , . . . , A d − ) and P j ( A , . . . , A d − ) for some 1 (cid:54) i < j (cid:54) n − h − A ,...,A d − ,B n and set P n − ( A , . . . , A d − ) and P n ( A , . . . , A d − ) as theintersection of h + A ,...,A d − ,B n with P i ( A , . . . , A d − ) and P j ( A , . . . , A d − ), respectively. SeeFigure 9 for an illustration.Since none of the sets P i ( A , . . . , A d − ) contains a point from N , it can be regarded asan intersection of (0 , d with open half-spaces. Therefore every set P i ( A , . . . , A d − ) is anopen convex subset of S . Let P ( A , . . . , A d − ) be the set S (cid:114) (cid:0)S B ∈ N h A ,...,A d − ,B (cid:1) . Clearly, N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 25 A P P
01 1 P P P P P P B B B P P P P Figure 9.
The inductive splitting of S = (0 , (cid:114) N with respect to the point A .The points B , B , B of N are denoted as empty circles and we use a shorthand P i for P i ( A ). λ d ( P ( A , . . . , A d − )) = 1. Since the sets P i ( A , . . . , A d − ) form a partitioning of P ( A , . . . , A d − ),we also have P ni =1 λ d ( P i ( A , . . . , A d − )) = 1.For i = 1 , . . . , n , we let R i be the subset of S d − ∗ × S defined as R i := { ( A , . . . , A d +1 ) ∈ S d − ∗ × S : A d , A d +1 ∈ P i ( A , . . . , A d − ) } , and we let R := S ni =1 R i . The sets R i are pairwise disjoint and it is not difficult to argue that thesesets are measurable. If a ( d + 1)-tuple ( A , . . . , A d +1 ) is contained in R i for some i ∈ { , . . . , n } ,then ( A , . . . , A d +1 ) is contained in Simp d ( S ), as P i ( A , . . . , A d − ) ∪ (Aff( { A , . . . , A d − } ) ∩ S )is a convex subset of S . Therefore, to find a lower bound for b d ( S ) = λ d ( d +1) (Simp d ( S )), itsuffices to find a lower bound for λ d ( d +1) ( R ), because R is a subset of Simp d ( S ). By Fubini’sTheorem, we obtain λ d ( d +1) ( R i ) = Z ( A ,...,A d +1 ) ∈ S d − ∗ × S [( A , . . . , A d +1 ) ∈ R i ]= Z ( A ,...,A d − ) ∈ S d − ∗ Z ( A d ,A d +1 ) ∈ S [ A d , A d +1 ∈ P i ( A , . . . , A d − )] ! = Z ( A ,...,A d − ) ∈ S d − ∗ λ d ( P i ( A , . . . , A d − )) , where [ φ ] is the characteristic function of a logical expression φ , that is, [ φ ] equals 1 if thecondition φ holds and 0 otherwise. For the measure of R we then derive λ d ( d +1) ( R ) = n X i =1 λ d ( d +1) ( R i ) = Z ( A ,...,A d − ) ∈ S d − ∗ n X i =1 ( λ d ( P i ( A , . . . , A d − )) . Since the function x x is convex, we can apply Jensen’s inequality and bound the last termfrom below by Z ( A ,...,A d − ) ∈ S d − ∗ n P ni =1 λ d ( P i ( A , . . . , A d − ))2 n ! = 12 n . (cid:3) The next step in the proof of Theorem 1.6 is to find a convenient n -tuple N of points from(0 , d whose removal produces a set with sufficiently small convexity ratio. We are going to find N using a continuous version of the well-known Epsilon Net Theorem [10]. Before stating thisresult, we need some definitions.Let X be a subset of R d and let U be a set system on X . We say that a set T ⊆ X is shattered by U if every subset of T can be obtained as the intersection of some U ∈ U with T . The Vapnik-Chervonenkis dimension (or
VC-dimension ) of U , denoted by dim( U ), is the maximum n (or ∞ if no such maximum exists) for which some subset of X of cardinality n is shattered by U . Let U be a system of measurable subsets of a set X ⊆ R d with λ d ( X ) = 1, and let ε ∈ (0 , N ⊆ X is called an ε -net for ( X, U ) if N ∩ U = ∅ for every U ∈ U with λ d ( U ) (cid:62) ε . Theorem 3.4 ([14, Theorem 10.2.4]) . Let X be a subset of R d with λ d ( X ) = 1 . Then for everysystem U of measurable subsets of X with dim( U ) (cid:54) v , v (cid:62) , there is a r -net for ( X, U ) ofsize at most vr log r for r sufficiently large with respect to v . To apply Theorem 3.4, the VC-dimension of the set system U has to be finite. However, itis known that the VC-dimension of all convex sets in R d is infinite (see e.g. [14, page 238]).Therefore, instead of considering convex sets directly, we approximate them by ellipsoids.A d -dimensional ellipsoid in R d is an image of the closed d -dimensional unit ball under anonsingular affine map. A convex body in R d is a compact convex subset of R d with non-emptyinterior. The following result, known as John’s Lemma [11], shows that every convex body canbe approximated by an inscribed ellipsoid. Lemma 3.5 ([14, Theorem 13.4.1]) . For every d -dimensional convex body K ⊆ R d , there is a d -dimensional ellipsoid E with the center C that satisfies E ⊆ K ⊆ C + d ( E − C ) . In particular, we have λ d ( K ) /d d (cid:54) λ d ( E ) . As the last step before the proof of Theorem 1.6, we mention the following fact, which impliesthat the VC-dimension of the system E of d -dimensional ellipsoids in R d is at most (cid:0) d +2 d (cid:1) . Lemma 3.6 ([14, Proposition 10.3.2]) . Let R [ x , . . . , x d ] (cid:54) t denote the set of real polynomials in d variables of degree at most t , and let P d,t = (cid:8) { x ∈ R d : p ( x ) (cid:62) } : p ∈ R [ x , . . . , x d ] (cid:54) t (cid:9) . Then dim( P d,t ) (cid:54) (cid:0) d + td (cid:1) .Proof of Theorem 1.6. Suppose we are given ε > d .We show how to construct a set S ⊆ R d with λ d ( S ) = 1 satisfying c( S ) (cid:54) ε andb d ( S ) (cid:62) (cid:0) d +2 d (cid:1) d d · ε log /ε . Without loss of generality we assume that ε = d d /r for some integer r (cid:62) d d .Consider the open d -dimensional box (0 , d and the system E (cid:22) (0 , d of d -dimensionalellipsoids in (0 , d . Since the restriction of E to (0 , d does not increase the VC-dimension,Lemma 3.6 implies dim( E (cid:22) (0 , d ) (cid:54) (cid:0) d +2 d (cid:1) .If we set n := 2 (cid:0) d +2 d (cid:1) r d log r e , then, by Theorem 3.4, there is a r -net N for the system((0 , d , E (cid:22) (0 , d ) of size n , having r sufficiently large with respect to d . Let S be the set(0 , d (cid:114) N . Clearly, we have λ d ( S ) = 1.Suppose K is a convex subset of (0 , d with λ d ( K ) > ε . Since the measure of K is positive,we can assume that K is a convex body of measure at least ε . By Lemma 3.5, the convex body K contains a d -dimensional ellipsoid E with λ d ( E ) (cid:62) ε/d d = r . Therefore E ∩ N = ∅ . Since wehave E ⊆ K and N ∩ S = ∅ , we see that K is not a subset of S . In other words, we have c( S ) (cid:54) ε .By Lemma 3.3, we have b d ( S ) (cid:62) n . According to the choice of n and r , the term n isbounded from below by14 (cid:0) d +2 d (cid:1) r log r = ε (cid:0) d +2 d (cid:1) d d log (2 d d /ε ) (cid:62) (cid:0) d +2 d (cid:1) d d · ε log /ε , N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 27 where the last inequality follows from the estimate 2 d d (cid:54) /ε . This completes the proof ofTheorem 1.6. (cid:3) It is a natural question whether the bound for b d ( S ) in Theorem 1.6 can be improved tob d ( S ) = Ω(c( S )). In the plane, this is related to the famous problem of Danzer and Rogers (see[5, 15] and Problem E14 in [8]) which asks whether for given ε > N ⊆ (0 , ofsize O ( ε ) with the property that every convex set of area ε within the unit square contains atleast one point from N .If this problem was to be answered affirmatively, then we could use such a set N to stab(0 , in our proof of Theorem 1.6 which would yield the desired bound for b ( S ). However it isgenerally believed that the answer is likely to be nonlinear in ε .3.3. A set with large k -index of convexity and small convexity ratio.Proposition 3.7. For every integer d (cid:62) , the set S := [0 , d (cid:114) Q d satisfies c( S ) = 0 and b k ( S ) = 1 for every positive integer k < d .Proof. Since Q d is countable and λ d ([0 , d ) = 1, we have λ d ( S ) = 1. Every convex subset K of[0 , d with positive d -dimensional measure contains an open d -dimensional ball B with positivediameter, as there is a ( d + 1)-tuple of affinely independent points of K . Since Q d is a densesubset of R d , we see that B ∩ Q d = ∅ and thus c( S ) = 0.It remains to estimate b k ( S ). By Fubini’s Theorem, we haveb k ( S ) = Z ( B ,...,B k ) ∈ S k λ d (cid:0) { A ∈ S : Conv( { B , . . . , B k , A } ) ⊆ S } (cid:1) λ d ( S ) k +1 . If A is a point of S such that Conv( { B , . . . , B k , A } ) is not contained in S , then A is a point ofthe affine hull Aff( { B , . . . , B k , Q } ) of B , . . . , B k and some Q ∈ Q d . Therefore, b k ( S ) is at least Z ( B ,...,B k ) ∈ S k λ d ([0 , d ) − λ d (cid:0)S Q ∈ Q d Aff( { B , . . . , B k , Q } ) (cid:1) λ d ( S ) k +1 . A countable union of affine subspaces of dimension less than d has d -dimensional measure zeroand we already know that λ d ( S ) = 1 = λ d ([0 , d ), hence b k ( S ) = 1. (cid:3) Other variants and open problems
We have seen in Theorem 1.3 that a p-componentwise simply connected set S ⊆ R whoseb( S ) is defined satisfies b( S ) (cid:54) α c( S ), for an absolute constant α (cid:54) S satisfies smc( S ) (cid:62) b( S ) λ ( S ) / K ⊆ R contains a triangle ofmeasure at least √ π λ ( K ). In view of this, Theorem 1.3 yields the following consequence. Corollary 4.1.
There is a constant α > such that every p-componentwise simply connectedset S ⊆ R whose b( S ) is defined contains a triangle T ⊆ S of measure at least α b( S ) λ ( S ) . A similar argument works in higher dimensions as well. For every d (cid:62)
2, there is a constant β = β ( d ) such that every convex set K ⊆ R d contains a simplex of measure at least βλ d ( K ) (seee.g. Lassak [13]). Therefore, Theorem 1.5 can be rephrased in the following equivalent form. Corollary 4.2.
For every d (cid:62) , there is a constant α = α ( d ) > such that every set S ⊆ R d whose b d ( S ) is defined contains a simplex T of measure at least α b d ( S ) λ d ( S ) . What can we say about sets S ⊆ R that are not p-componentwise simply connected? Firstof all, we can consider a weaker form of simple connectivity: we call a set S p-componentwisesimply -connected if for every triangle T such that ∂T ⊆ S we have T ⊆ S . We conjecturethat Theorem 1.3 can be extended to p-componentwise simply -connected sets. Conjecture 4.3.
There is an absolute constant α > such that every p-componentwise simply -connected set S ⊆ R whose b( S ) is defined satisfies b( S ) (cid:54) α c( S ) . What does the value of b( S ) say about a planar set S that does not satisfy even a weak formof simple connectivity? As Proposition 3.7 shows, such a set may not contain any convex subsetof positive measure, even when b( S ) is equal to 1. However, we conjecture that a large b( S )implies the existence of a large convex set whose boundary belongs to S . Conjecture 4.4.
For every ε > , there is a δ > such that if S ⊆ R is a set with b( S ) (cid:62) ε ,then there is a bounded convex set C ⊆ R with λ ( C ) (cid:62) δλ ( S ) and ∂C ⊆ S . Theorem 1.3 shows that Conjecture 4.4 holds for p-componentwise simply connected sets,with δ being a constant multiple of ε . It is possible that even in the general setting of Conjecture4.4, δ can be taken as a constant multiple of ε .Motivated by Corollary 4.1, we propose a stronger version of Conjecture 4.4, where the convexset C is required to be a triangle. Conjecture 4.5.
For every ε > , there is a δ > such that if S ⊆ R is a set with b( S ) (cid:62) ε ,then there is a triangle T ⊆ R with λ ( T ) (cid:62) δλ ( S ) and ∂T ⊆ S . Note that Conjecture 4.5 holds when restricted to p-componentwise simply connected sets, asimplied by Corollary 4.1.We can generalize Conjecture 4.5 to higher dimensions and to higher-order indices of convexity.To state the general conjecture, we introduce the following notation: for a set X ⊆ R d , let (cid:0) Xk (cid:1) be the set of k -element subsets of X , and let the set Skel k ( X ) be defined bySkel k ( X ) := [ Y ∈ ( Xk +1 ) Conv( Y ) . If X is the vertex set of a d -dimensional simplex T = Conv( X ), then Skel k ( X ) is often calledthe k -dimensional skeleton of T . Our general conjecture states, roughly speaking, that sets withlarge k -index of convexity should contain the k -dimensional skeleton of a large simplex. Here isthe precise statement. Conjecture 4.6.
For every k, d ∈ N such that (cid:54) k (cid:54) d and every ε > , there is a δ > such that if S ⊆ R d is a set with b k ( S ) (cid:62) ε , then there is a simplex T with vertex set X suchthat λ d ( T ) (cid:62) δλ d ( S ) and Skel k ( X ) ⊆ S . Corollary 4.2 asserts that this conjecture holds in the special case of k = d (cid:62)
2, sinceSkel d ( X ) = Conv( X ) = T . Corollary 4.1 shows that the conjecture holds for k = 1 and d = 2if S is further assumed to be p-componentwise simply connected. In all these cases, δ can betaken as a constant multiple of ε , with the constant depending on k and d .Finally, we can ask whether there is a way to generalize Theorem 1.3 to higher dimensions,by replacing simple connectivity with another topological property. Here is an example of onesuch possible generalization. Conjecture 4.7.
For every d (cid:62) , there is a constant α = α ( d ) > such that if S ⊆ R d is aset with b d − ( S ) defined whose every p-component is contractible, then b d − ( S ) (cid:54) α c( S ) . N THE BEER INDEX OF CONVEXITY AND ITS VARIANTS 29
A modification of the proof of Theorem 1.5 implies that Conjecture 4.7 is true for star-shapedsets S . Acknowledgment
The authors would like to thank Marek Eliáš for interesting discussions about the problemand participation in our meetings during the early stages of the research.
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E-mail : [email protected]ff.cuni.cz, [email protected]ff.cuni.cz(Vít Jelínek) Computer Science Institute, Faculty of Mathematics and Physics, Charles University, Prague, CzechRepublic
E-mail : [email protected]ff.cuni.cz(Bartosz Walczak) Department of Theoretical Computer Science, Faculty of Mathematics and Computer Science,Jagiellonian University, Kraków, Poland