Combinatorial properties of convex sets.

Transcription

1 Winter School Combinatorial and algorithmic aspects of convexity Paris, Institut Henri Poincaré, January 9-23, 205. http ://wiki-math.univ-mlv.fr/gemecod/doku.php/winterschool205 Imre Bárány Combinatorial properties of convex sets. Scribes : J. Tkadlec and T. Tkocz S. Dann and M. B. Gomez Marta Strzelecka, Micha lstrzelecki G. Ambrus and L. Silverstein Evangelos Anagnostopoulos, Ioannis Psarros

2 Table of content. Lecture. Theorems of Carathéodory, Radon, Tverberg and Helly. Scribes : J. Tkadlec and T. Tkocz Lecture 2. More on Helly s theorem : applications, fractional and colorful versions. Scribes : S. Dann and M. B. Gomez Lecture 3. Extended colorful Carathéodory, Sarkaria s lemma and covering by simplices. Scribes : Marta Strzelecka, Micha lstrzelecki Lecture 4. Weak ε-nets, halving lines and lhalving planes. Scribes : G. Ambrus and L. Silverstein Lecture 5. Colored Tverberg, point selection theorem, (p, q)-property and order types. Scribes : Evangelos Anagnostopoulos, Ioannis Psarros 47

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4 I. Bárány, Lecture Scribers: J. Tkadlec and T. Tkocz In the first part of this lecture we shall recall some definitions as well as basic facts concerning convexity. Then we shall present a few classical theorems with combinatorial flavour, that is the theorems of Carathéodory, Radon and Tverberg respectively. For two points a, b R d we define a segment [a, b] joining a and b as the set [a, b] = {αa + βb, α, β 0, α + β = }. A set C in R d is called convex if for every two points a and b in C, the segment [a, b] is also contained in C. Plainly, an intersection of convex sets is a convex set. Given n points a,..., a n in R d and real coefficients α,..., α n, the point a = α a α n a n is called their positive combination if all α i s are nonnegative and if in addition α α n =, then the point a is called a convex combination (c.c. for short) of a i s. If the real coefficients α i s only satisfy the condition α α n =, then the point a is called an affine combination of a i s. Note the following easy observation. Fact. If C is a convex subset of R d and a,..., a n C, then all convex combinations of a,..., a n belong to C. For a subset S in R d we define conv S = {all convex combinations of elelements of S}, pos S = {all positive combinations of elelements of S}. These are clearly convex sets. Moreover, conv S is the smallest convex set containing S. It is called the convex hull of S. The set pos S is called the positive hull of S. For completeness, we also recall the definition of an affine hull, aff S = {all affine combinations of elelements of S}. Obviously, a segment in R d is a convex set. Another canonical example of a convex set is a (closed) half-space, H = {x R d, a x α}. The set of all positive semi-definite matrices of a fixed size is also convex. To see a nontrivial example, the interested reader is encouraged to look at the following exercise. Exercise. Let f be a polynomial with complex coefficients which is not constant. Show that the roots of f lie in the convex hull of the roots of f. Page 3

5 Our first theorem basically says that being in a convex hull is a very finite property. Theorem (Carathéodory). Let A be a subset in R d. Suppose that a conv A. Then there exists a subset B of A with B d + such that a conv B. ( ) x Proof. For a vector x in R d and a real number α by we mean a vector α in R d+ with the last coordinate α. The fact that a lies in the convex hull of A can be written shortly as ( ) a n ( ) ai = α i, i= for some nonnegative reals α i s (the last coordinate takes care of the condition that α i s add up to ). Without loss of generality we can assume that all α i s are positive. Moreover, let n be the smallest possible for which the ( above ) ai holds. We want to show that n d+. Suppose not; then the vectors, i =,..., n cannot be linearly independent as they lie in a d + dimensional space. Therefore we get that ( ) 0 0 = n i= ( ) ai β i for some β i s which are not all equal to zero. Hence, ( ) a n ( ) ai = (α i + tβ i ) i= for every t R. Since for t = 0 all the coefficients α i + tβ i s are positive, they remain positive for small t and there is a choice for t, say t 0 for which at least one of the coefficients becomes 0 with the rest remaining positive. This contradicts the minimality of n, as ( ) a n ( ) ai = (α i + t 0 β i ) i= ( ) ( ) a ai shows that the vector can be written as a positive combination of s with fewer than n nonzero coefficients. 2 Page 4

6 Geometrically speaking, Carathéodory s theorem says that a convex set A in R d can be covered by simplices of A (for us, a simplex in R d is a convex hull of at most d + points). With the same proof, we can also obtain an analogous result for cones (sets of all positive combinations). Theorem 2. Let A be a subset in R d. Suppose that a pos A. Then there exists a subset B of A with B d such that a pos B. Sometimes, the following derivatives of Carathédory s theorem can be useful in applications. Theorem 3. Let A be a subset in R d and b R d. Suppose that a conv A. Then there exists a subset B of A with B d such that a conv (B {b}). Theorem 4. Let A be a subset in R d. Suppose that a int conv A. Then there exists a subset B of A with B 2d such that a int conv B. Now we move on to Radon s theorem and its further generalisation, Tverberg s theorem, saying that there are some good partitions of sets having enough points. Theorem 5 (Radon). Let A be a subset in R d with A d + 2. Then there is a partition, A = X Y (X Y = ), such that conv X conv Y. Remark. The constant d + 2 is the best possible as shown by an example of a simplex in R d. Remark. When d = the theorem is clear as considering three points on a line, there is always one, say x between some two others, say y, z, so it suffices to take X = {x} and Y = A \ X {y, z}. Remark. When d = 2, considering 4 points in the plane, there are two possibilities. Either certain three of them are the vertices of a triangle containing the fourth point, or the points are the vertices of a convex quadrilateral. In any case, it is clear what to take for the partition. (See the pictures.) y x 2 y 3 y x z x y 2 y y 2 x 3 Page 5

7 {( ) } a Proof. Since A d + 2, the set, a A of vectors in R d+ is not linearly independent. Therefore there are a i A and nonzero coefficients α i such that ( ) 0 = ( ) ai α 0 i. Because the sum of α i s is 0, some of them are positive, some are negative. Let I be the set of all the indices i for which α i > 0 and J for which α i < 0 (neither I nor J is empty). Breaking the sum into two pieces yields i I ( ) ai α i = i J ( α i ) ( ) ai. Dividing this by t = i I α i = i J ( α i) shows that we can take X = {a i, i I} and Y = A \ X, as then conv X i I α i t a i = i J α i a i conv Y. t It is left as an exercise to show a combinatorial analogue of Radon s theorem. We adopt the standard notation that [m] = {, 2,..., m}. Exercise 2. Let A,..., A n+ be nonempty subsets of [n]. Show that there are disjoint subsets I and J of [n + ] such that k I A k = k J A k. Then do the same for the union instead of the intersection. Now we would like to present a generalisation of Radon s theorem, Tverberg s theorem, along with a sketch of a proof. There are at least seven different proofs. The one we shall sketch comes from Roudneff. Theorem 6 (Tverberg). Let r 2 be an integer. Let X be a subset of R d with X = (r )(d + ) +. Then there is a partition X = X... X r (X i s are pairwise disjoint) such that r i= conv X i. Remark. Taking here r = 2 recovers Radon s theorem. Remark. The constant (r )(d + ) + is the best possible. To see this, consider (r )(d+) points in R d in general position, meaning that no d+ of them lie in the same hyperplane. 4 Page 6

8 Proof. We start off by stating a useful fact about intersections of affine subspaces. Claim. Suppose that a finite subset S of R d of points lying in general position is partitioned into r pairwise disjoint subsets S = S... S r with S i d+ for every i. If r i= aff S i = then S (r )(d + ). We shall not give a proof. Instead, we show a simple counting argument which we hope gives faith and indicates why the claim holds. Note that for a finite set Y of points in R d in general position we have that aff Y is the intersection of certain d + Y hyperplanes. Therefore, r i= aff S i is the intersection of at least d + hyperplanes, as r ( d + Si ) = r(d + ) X r(d + ) (r )(d + ) = d +. i= Hence, this intersection is empty. Now for the proof of Tverberg s theorem, we shall consider only partitions of X into sets with at most d + elements each. Fix such a partition, X = X... X r with X i d +, and consider the function R d x f r ( dist(x, conv Xi ) ) 2. i= As a sum of convex functions, it is convex, it goes to when x does, hence it attains its minimum. Choose a partition for which this minimum is the smallest possible and suppose that the minimum equals µ and is attained at x = z. If µ = 0 we are done, so suppose µ > 0 and we will arrive at a contradiction. Choose y i in conv X i such that z y i = dist(z, conv X i ). Now look at the function R d x g r x y i 2. i= Since its value at x = z is µ and it is bounded from below (pointwise) by the function f whose minimum is µ, the point x = z is also where g attains its minimum. Thus, taking the gradient of g at x = z yields r (z y i ) = 0 i= 5 Page 7

9 (i.e. z is the centre of mass of y i s). Notice that it is not possible that z = y i for all i s because otherwise µ = 0. Without loss of generality let us assume that z y, so z / conv X. For each i, let S i be a minimal subset of X i for which y i conv S i. We also have z / conv S. First, we show that z / r i=aff S i. If not, then in particular z aff S. But y is the point in the polytope conv S closest to z, which is outside the polytope as z y, so y has to lie on the boundary of the polytope which contradicts the minimality of S. Second, we show that it is not possible for any other point y z to belong to r i=aff S i. Indeed, if not, then for every i, by the minimality of y i, y y i S i z (y z) (y i z) 0, as y aff S i. Summing these inequalities yields 0 = (y z) r (y i z) 0, i= and, as a result, (y z) (y i z) = 0 for every i. Since y z, this implies that z = y i, for every i. We have shown that r i=aff S i =. By the claim we get r i= S i (r )(d + ). This means that we can choose a point x in X \ r i= S i. Observe that r r (z y i ) (x y i ) = (z y i ) ((z y i ) + (x z) ) r = (z y i ) 2 = µ > 0. i= i= Hence, there is j for which (z y j ) (x y j ) > 0. i= y j S j x z 6 Page 8

10 Moving x to the set X j leads to a partition with a smaller minimum of f than µ (see the picture). This contradiction finishes the proof. Now we shall discuss Helly s theorem which also concerns intersections of convex hulls. A family of sets in R d is said to have Helly s property if every d + of them have a nonempty intersection. Theorem 7 (Helly). Let C, C 2,..., C n, n d+, be convex sets in R d with Helly s property. Then n C. i= In other words, having empty intersection has a finite reason. Proof. By induction on n. Case n = d + is trivial. Now suppose that n d + 2 and the theorem holds for smaller n. For each j =,..., n set K j = n C i. i= i j By induction, K j. Take arbitrary z j K j. By Radon s Theorem, there is a partition of z,..., z n into two sets X, Y such that conv X conv Y. Take z in this intersection. We claim that z C i for all i =,..., n. Without loss of generality, focus on C and suppose that z X. Then all z j Y belong to C, hence z C too. Later on, we will see another proof related to Carathéodory s Theorem. In a special case d = (intervals on a line), a different proof is possible. Sketch: Take the rightmost left-endpoint x of these intervals. Then every interval starts to the left of x and ends to the right of x. Exercise 3. Extend the previous argument to trees, i.e. prove the following: If T,..., T n is a family of subtrees of a given tree T such that T i T j for each i, j =,..., n then n i= T i. Note that the theorem poses no restriction on the nature of the sets C i apart from being convex. However, it can be seen that the case with all sets compact already captures all of the complexity. Indeed, for I 7 Page 9

11 [n + ], I = d +, we can take Z I = i I C i and replace C i by K i = conv {Z I, i I} C i. Note that Helly s Theorem in general fails for infinite families. First example is the family of intervals of the form I n = [n, ), the other is the family of intervals of the form J n = (0, /n]. However, if all the sets are compact, the theorem holds. Theorem 8 (Helly s Theorem, infinite version). Let C, C 2,... be compact convex sets in R d with Helly s property. Then i= C i. Proof. Fix n and for i =,..., n set K i = C C i. Then the family of the sets K,..., K n has Helly s property, hence, by usual Helly s Theorem, there exists z n n i= K i. By compactness, take a convergent subsequence of {z n } and assume it tends to some z 0. Then z 0 C i for all i, thus i= C i. Helly s Theorem has many applications. Here we state six of them.. Let F be a finite family of convex sets in R d, F = n d + and let C R d be convex. Then there exists a translate of C intersecting every K F if and only if there exists a translate of C intersecting every (d + )-tuple from F. Proof. For K F, set K = {x R d, (x + C) K }. If we prove that K is convex, we are done by Helly s Theorem. And indeed, take x, y K. Then there exist x (x + C) K and y (y + C) K. Now for any α, β 0, α + β = we have αx αx + αc, βy βy + βc, and summing: αx + βy αx + βy + C. Thus αx + βy + C is a translate of C having nonempty intersection with K, i.e. αx + βy K and K is convex. 2. Let I,..., I n, n 3, be vertical intervals in R 2 such that every three of them have a line transversal. Then all of them have a common line transversal. 8 Page 0

12 Proof. A set of lines y = αx + β intersecting some particular I j = x j [y j, y+ j ] corresponds to a set S j = {(α, β) R 2, y j αx j + β y + j }. Being the intersections of two halfspaces, these S j are convex, hence by Helly s Theorem they have a nonempty intersection. Any point in it determines a line intersecting all the intervals I j. y + j I j y j x j 3. Let H be a finite family of open (closed) half-spaces in R d and let C R d, C H H H be convex. Then there exists a subfamily H H, H = d + such that C H H H. Proof. For all H H, set H = C \ H. Then H is convex and H H H =. By Helly s Theorem, there exists a (d + )-tuple H,..., H d+ with empty intersection. Then C H H d+. 4. (Kirchberger s Theorem) Sets R and B of red and blue points in R d are given. Then R and B can be strictly separated by a hyperplane if and only if for all Y R B, Y d + 2 one can separate the sets Y R and Y B. (A hyperplane h strictly separates sets A and B if A lies in one open half-space determined by h and B lies in the opposite closed half-space.) Proof. With every r R we associate a half-space {( ) ( ) ( ) } a a r C r = R d+, > 0. α α 9 Page

13 Likewise, with every b B we associate {( ) ( ) ( ) } a a b D b = R d+, 0. α α By assumption, every d + 2 half-spaces have a point in common, hence by Helly s Theorem all the half-spaces have a point in common. This point determines a strictly separating hyperplane. 5. (Centrepoint Theorem) Let X R d, X d +. Then there exists a centrepoint of X, i.e. a point z R d such that any closed half-space H containing z satisfies X H d + X. Proof. Let { H = H, H is an open half-space such that X H < } d + X. By summing the sizes we see that no (d+)-tuple of sets in H covers the whole X. Hence the family of closed half-spaces satisfying X H X has Helly s property. The compactness issues in the infinite d+ Helly s Theorem are avoided by focusing on a sufficiently large ball instead of the whole R d. 6. Let X R d be finite with diameter diam(x) 2. Then there exists 2d a ball B of radius r = that contains X. Moreover, the cases requiring r = are precisely the regular simplices with side length d+ 2. 2d d+ Proof. For any x X, take the ball B(x, r) with centre x and radius 2d r =. We want to prove d+ x X B(x, r). By Helly s Theorem, we only need to prove it for a (d + )-tuple x 0, x,..., x d X. Let B(y, R) be the smallest ball containing X. By appropriate translation we may assume that y = 0. Suppose that x i = R for i = 0,,..., m d and x i < R for the rest. We will only deal with x 0, x,..., x m. 0 Page 2

14 If y conv {x 0, x,..., x m }, we could move y closer, thereby reducing the radius. Assume otherwise. Then there exist α 0, α,..., α m 0 with ( ) 0 = Now m i=0 ( ) xi α i. 4 x i x j 2 = x i 2 + x j 2 2x i x j = 2R 2 2x i x j. Multiplying by α i /2 and summing over i = 0,,..., m, i j, we get 2( a j ) R 2 ( a j ) + ( i j α i x i ) x j = R 2 ( a j ) + α j R 2 = R 2. Summing over all j = 0,,..., m finally gives the desired 2(m + ) (m + )R 2. Page 3

15 I. Bárány, Lecture 2 Scribes: S. Dann and M. B. Gomez We continue with two more applications of Helly s Theorem.. For any convex compact C R d there exists z C so that for every chord [u, v] of C with z [u, v] we have u z d + u v d d +. In other words, the ratio u z is separated from 0 and. u v Proof. For every x C, we define C x := x + d (C x). Note that d+ C x C. Claim: For any x,..., x d+ C, we have d+ C xi. Set y := d+ x d+ i. Let j {,..., d + } and rewrite y as Note that d i= d+ i=,i j y = x j + ( d d + d d+ i=,i j i= x i x j ) x i is a convex combination of points in the convex set C and hence belongs to C. We obtain that for any j {,..., d + }, y x j + d (C x d+ j) = C xj. This yields the claim. Helly s theorem now implies that C x. Let z C x C. x C Claim: The point z has the required property.. x C Let u, v C so that z [u, v]. By definition z C u. Hence z u + d ([u, v] u), d + i.e. z = u + d d (αu + ( α)v u) = u + ( α)(v u) for some d+ d+ α [0, ]. Rearranging this equality, we obtain z u = d ( α) v u d v u. d + d + Page 4

16 This immediately gives the upper bound and after a simple computation the lower bound as well. Indeed, since z C v as well, we also have z v u v. The lower bound now follows from d d+ u v = u z + z v u z + d u v. d + 2. Let a, b X R d. We say a sees b if [a, b] X. We say X is visible from a if every c X is seen from a, i.e. X is star-shaped. Krasnosel skii Theorem: Let X R d compact. If any d + points in X are seen from some x X, then X is star-shaped. Proof. For x X, define V x := {set of points visible from x} = {y : [x, y] X}. Observe that for any x,..., x d+ X, d+ convv xi d+ V xi, i.e. the collection of sets {convv x } x X enjoys the Helly i= property. Hence there is z convv x. x X Exercise. Show that z x X i= V x, i.e. z is visible from all x X. Sets C, D R d are separated if there is a hyperplane H = {x ax = b} with C H + and D H, where H + = {x ax b} and H = {x ax b}. The separation is strict if C inth + and D inth. The sets are separated by a slab (or skip) if there are parallel hyperplanes H, H 2 with H H 2, C H +, D H 2 and H + H 2 =. Theorem (Separation Theorem). Let C, D R d convex, C compact, D closed. Then C D = if and only if C, D are separated by a slab. Proof. Exercise. Lemma. Let K,..., K n be closed convex sets and let K be compact. Then n K i = if and only if there are closed halfspaces H i with K i H i and n H i =. 2 Page 5

17 Proof. : trivial. : The case n = 2 is clear (Separation theorem). Suppose n > 2. By assumption (K K 2 K n ) K n =. Thus there are closed halfspaces H n and H n with H n H n = and K K 2 K n H n, K n H n. Moreover, we have K K 2 K n H n =. Separate K n from K K 2 K n 2 H n by H n and so on. At the end we obtain H H 2 H n = with K i H i for i [n]. Lemma 2. Let H,..., H m be closed halfspaces in R d, H i = {x a i x α i }. Then ( ) {( ) ( )} m 0 H a am i = if and only if pos,...,. Proof. ( m H i ) = the system a i x α i, i =,..., m, has no solution. 0 : pos {... } means there are γ i 0, not all zero, so that ( ) 0 = m α ( ) ai γ i, α i i.e. 0 = m γ ia i and = m γ iα i. Suppose x is a solution of the system a i x α i, then multiplying by γ i and summing gives 0 = ( m γ ia i ) x m γ iα i =. A ( contradiction. ) 0 : Suppose / pos {... } =: D. Observe that D is closed and hence ( ) ( ) 0 b and D can be separated by a slab. It follows that there is R β d+ ( ) ( ) ( ) ( ) 0 b ai b so that > 0 and 0 for i [n]. This gives β < 0 β α i β and a i b + α i β 0. Thus b is a solution of the system a β i x α i. Remark. More is true. Let K,..., K n be closed convex sets in R d and let K be compact, n d +. Then n K i = if and only if there is I [n], I d + so that i I K i =. α m 3 Page 6

18 Another proof of Helly s Theorem First, let K,..., K n R d be convex sets, all closed and one compact. We have n K i = H i K i closed halfspaces with n i= H i = ( ) {( ) ( )} 0 a an pos,..., by the cone version of Carathéodory s Theorem in R d+ we have α α n ( ) {( ) ( )} 0 ai aid+ pos,..., α i α id+ d+ j= H i j =. This implies d+ j= K i j =. Now let C,..., C n R d be convex sets with Helly s property. For every J [n] with J = d + there exists z(j) R d, z(j) j J C j. For j [n], define K j := conv{z(j) j J}. Each K j is a polytope, K j C j and they satisfy the Helly condition. Hence n K j. This in turn implies that n C j. Remark. We will refer to the method in second part of the above proof as reduction to polytopes. Theorem 2 (Fractional Helly, Katchalski-Lin). Let α (0, ] and d 2 be fixed. If C is a finite family of convex sets, n = C, with at least α ( n d+) intersecting (d + )-tuples, then there exists an intersecting subfamily C C, with C βn, where β is a constant that depends only on d and α. Remark. The best known is β = ( α) d+ (Kalai). Proof. Let C = {C, C 2,..., C n }. First apply reduction so that we can consider the C i s to be polytopes. For each k [n], let I k := {I [n] : I = k, C i } be the set of indices of intersecting k-tuples. Given I I k, denote C(I) := i IC i. Since C(I) is a polytope, and there are finitely many such polytopes, we can choose a R d such that the functional f(x) = ax R takes different i I 4 Page 7

19 values in the vertices of C(I), for all possible I s. In particular, this means that the set argmin{ax : x C(I)} = {z C(I) : az ax, x C(I)} consists of one single point for each I. Let z(i) = argmin{ax : x C(I)}, then ax > az(i) for all x C(I) \ {z(i)}. Claim. For each I I d+, there exists J I d, J I, such that z(j) = z(i). Proof: Let H I := {x R d : ax < az(i)} the open halfspace (orthogonal to a) that has z(i) in the closed boundary. Then {C i : i I} {H I } is a family of d + 2 convex sets in R d such that ( i IC i ) H I =. By Helly, there exists a subfamily of d + sets with empty intersection. Since i IC i, this implies that there exists i 0 I such that ( C i ) H I =. Let J = I \ {i 0 }, since i I\{i 0} z(i) C(J) we get that z(i) = z(j) = argmin{ax : x C(J)}. There are at most ( ) n d possible sets J Id and exactly I d+ = α ( ) n d+ sets I I d+. Then there exists some J 0 I d such that at least α ( ) n d+ ( n d) distinct sets I I d+ are mapped to J 0. That is, such that z(i) = z(j 0 ). For each of those I, z(j 0 ) = z(i) C(I) C i0, where {i 0 } = I \ J 0. On the other hand, z(j 0 ) C(J 0 ) C j, for all j J 0. In total there are α ( ) n d+ n(n )... (n d) ( n + d = α d) (d + ) n(n )... (n d + ) + d = αn d d + + d sets C i that contain the point z(j 0 ). α d + n Theorem 3 (Colorful Carathéodory). Let A, A 2,..., A d+ R d and a d+ conv(a i ). Then there exist a A,..., a d+ A d+ such that a i= conv{a,..., a d+ }. 5 Page 8

20 Proof. By Carathéodory s Theorem, without loss of generality A i d +. We can also assume a = 0. Let {a,..., a d+ } be the set, with a i A i for each i =,..., d +, that minimizes the function dist(0, conv{a,..., a d+ }). Let µ be this minimum and z conv{a,..., a d+ } the point that achieves it. That is, µ = z = dist(0, z) = dist(0, conv{a,..., a d+ }) If µ = 0, then z = 0 conv{a,..., a d+ }. Suppose µ > 0. z must be a point in the boundary of conv{a,..., a d+ }. That is, it must lie in a facet, which is spanned by d vertices: without loss of generality, z conv{a,..., a d }. Let H be the hyperplane containing this facet, and let H + be the open halfspace defined by H and containing the origin. In order for 0 conv(a d+ ), there must exist a point a d+ A d+ H +. Then the segment (z, a d+ ] conv{a,..., a d, a d+ } H+, and there exists a point z (z, a d+ ] such that dist(0, z ) < dist(0, z) = µ, which is a contradiction. a d+ conv{a,..., a d+ } H H + µ a d+ o µ z z conv{a,..., a d, a d+ } Remark. General version: Carathéodory for one single set: take the A i to be d + copies of the same set. Cone version: Let A, A 2,..., A d R d and a d pos(a i ). Then there exist a A,..., a d A d such that a pos{a,..., a d }. i= 6 Page 9

21 Extra version: Let A, A 2,..., A d R d, b R d and a d conv(a i ). Then there exist a A,..., a d A d such that a conv{a,..., a d, b}. Proof. Without loss of generality a = 0 and, by Carathéodory, A i d +. If 0 d int(conv(a i )) = int( d conv(a i )), then b R d = d pos(a i ). i= i= By the cone version of Colorful Carathéodory, there exist α,..., α d 0, a i A i, such that b = d α i a i. Hence 0 = i= b+ d α i a i i= + d α i i= i= i= conv{a,..., a d, b}. If 0 int( d conv(a i )), then it lies in the interior of a lower dimensional i= face of the polytope d conv(a i ), and we can apply the previous for a i= lower dimension d < d. Theorem 4 (Application : Colorful Helly). Let C, C 2,..., C d+ be finite families of convex sets in R d. If d+ C i for all C C,..., C d+ C d+, i= then there exists i [d + ] such that C. C C i Proof : First apply reduction so that C is a polytope for all C C. Suppose that C = for all i =,..., d+. Since the C are polytopes, C C i they are compact. Then there exist H, H 2,..., H d+ finite families of closed halfspaces such that: H = for all i. H H i For each i and for each C C i, there exists H C H i such that C H C. Let a C R d and α C R such that H C = {x R d : a C x α C }. For each i, since H =, we have that H H i ( ) ( ) 0 ac pos{ : C C i } 7 α C Page 20

22 By the cone version of Colorful Carathéodory, there exist ( ac R d R, where C C,..., C d+ C d+, such that ( ) ) ( 0 acd+,..., α Cd+ pos{( ac α C ) } α C ) ( ) acd+,..., α Cd+ This implies that d+ H Ci =, thus d+ C i =, which is a contradiction. i= i= Proof 2: First apply reduction so that we can consider each C C i to be a polytope. Fix i 0 {,..., d+} and fix the set of indices I = {j,..., j i0, j i0+,..., j d } where each j i represents the set C ji C i. That is, I represents a d-tuple where each set comes from a different family and there is no element from family C i0. Denote the polytope C(I) := i IC i There are finitely many such polytopes (as many as choices for i 0, j,..., j d+ ) so there exists a R d such that the functional f(x) = ax R takes different values in the vertices of C(I), for all possible I s. In particular, this means that the set argmin{ax : x C(I)} consists of one single point for each I. Let z(i) = argmin{ax : x C(I)}, then ax > az(i) for all x C(I)\{z(I)}. Let now I 0 be the d-tuple such that the value az(i 0 ) is maximal and let i 0 be the index of the family C i0 that is not represented. Without loss of generality, i 0 = d +, and I 0 = {i,..., i d }. Claim. z(i 0 ) C, for all C C d+. Proof: Suppose that z(i 0 ) C, for some C = C id+ C d+. Then z(i 0 ) C(I 0 ) C. Let H := {x R d : ax az(i 0 )} the closed halfspace (orthogonal to a) that has z(i 0 ) in the boundary. Then {C i : i I 0 } {H} {C } is a family of d + 2 convex sets in R d such that ( i I 0 C i ) H C =. By Helly, there exists a subfamily of d + sets with empty intersection. Since ( i I 0 C i ) H = {z(i 0 )} =, this implies that there exists a (d )-tuple 8 Page 2

23 J I 0 such that ( C i ) H C =. Let the d-tuple I = J {i d+ }, i J then C(I ) = ( C i ) C R d \ H. That is, ax > az(i 0 ) for all x C(I ). i J In particular az(i ) > az(i 0 ), and so az(i 0 ) is not maximal, which is a contradiction. Hence C. C C d+ Theorem 5 (Application 2). Let G = (V, A) be a directed graph with n = V vertices, and let C,..., C n be directed cycles in G. Then there exist arcs a i C i such that {a, a 2,..., a n } contains a directed cycle. Proof. Label the vertices {v..., v n }, and represent each arch a A by a point p a = e j e i R n, where v i is the starting vertex of a and v j is the ending vertex. In particular, all the points p a lie in the hyperplane n x i = 0, which has dimension d = n. In this setting, a set of arcs C A contains a directed cycle if and only if a Cp a = 0 (there could be several pairwise disjoint cycles). C,..., C d+ are directed cycles in G. Then p a = 0 for each i [d + ] a C i = 0 = p n a conv{p a : a C i } for all i = 0 d+ conv{p a : a C i }. a C i i= By Colorful Carathéodory, there exist a C,..., a d+ C d+ such that 0 conv{p a,..., p ad+ }. Without loss of generality, 0 conv{p a,..., p ak }, k a i a j, for k minimal. Then there exist α,..., α k > 0, α i =, such that 0 = k α i p ai. i= It is easy to see that, under the previous conditions, all the (non-zero) coefficients must be equal: α = = α k =, and 0 = k k p k ai, which implies that {a,..., a k } contains a directed cycle in G. i= i= p k a i = k i= i= Exercise 2. Try to find a combinatorial proof of this last statement. 9 Page 22

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25 Imre Bárány: Combinatorial properties of convex sets lecture 3 Scribes: Marta Strzelecka, Micha l Strzelecki Wednesday, January 2, 205 Theorem (Extended Colorful Carathéodory). Let A,..., A d be nonempty sets in R d. Suppose that for any pair of distinct indices i, j we have 0 conv(a i A j ). Then there exist a i A i, i =,..., d +, such that 0 conv{a,..., a d+ }. Clearly, this is an extension of Colorful Carathéodory s Theorem, since the condition 0 d+ i= conv A i implies the above assumption. Remark. This theorem does not have a conic analog (but it has some consequences for the spherical colorful version of Helly s Theorem. However, we will not explain them here). Proof. We start as in the proof of Colorful Carathéodory s Theorem. Without loss of generality we may assume that all the sets A i are finite. Choose a i A i, i =,..., d +, so that the distance dist(0, conv{a,..., a d+ }) is minimal. If dist(0, conv{a,..., a d+ }) = 0 then we are done, so suppose further that dist(0, conv{a,..., a d+ }) = z > 0 (for some z conv{a,..., a d+ }). We may assume that the points a,..., a d+ are in general position and that z lies in the interior of the facet conv{a,..., a d }. Denote H = aff{a,..., a d }. Because of the optimality of z the set A d+ (and in particular the point a d+ ) has to lie above H (see Figure ). Consequently, because of the condition 0 conv(a i A d+ ), there exist points b A,..., b d A d, which lie below H. For i =,..., d we define f(e i ) = a i, f( e i ) = b i. Then we extend f to the mapping f : conv{±e,..., ±e d } R d simply by setting f to be affine on the facets. Note that the facets of conv{±e,..., ±e d } are mapped exactly to multicolor facets. The image of f divides R d into components, of which one is unbounded (we use topology here!). From the optimality of z we have Page 24

26 a d+ a 2 z a a d H 0 Figure : The point a d+ lies above H. [0, z) Im f =. Therefore 0 lies in a bounded component of R d \ Im f (since the image of f lies below H and contains z). On the other hand the point a d+ lies in the unbounded component of R d \Im f, since it lies above H. Therefore the half-ray starting from a d+ (in the unbounded component) after passing through 0 (in a bounded component) has to pierce a multicolor facet conv{c,..., c d }, for some c i {a i, b i } (see Figure 2). a d+ c d H 0 c 2 c Figure 2: The half-ray starting at a d+ pierces a multicolor facet. Page 25 2

27 The simplex conv{c,..., c d, a d+ } is better than the originally chosen simplex conv{a,..., a d+ } (note that we cannot have c i = a i for all i =,..., d since our half-ray goes first through H, then passes through 0 and only then pierces the facet conv{c,..., c d }). This contradiction with the assumption dist(0, conv{a,..., a d+ }) > 0 and ends the proof. Remark (homework). Let Γ be a curve in R d (a continuous image of the interval [0, ]). Then every point in conv Γ can be written as a convex combination of only d points from Γ (this is one point less than in Carathéodory s Theorem, but our set is of a special form). We will give now a second proof of Tverberg s Theorem. Sarkaria s proof of Tverberg s Theorem. Let X := {x 0, x,..., x n } R d, where n = (r )(d + ). We want to prove that there exists a partition X = X... X r such that r i= conv X i. We will use an artificial tool. Namely, let v,..., v r be vectors in R r such that every (r )-tuple of them is linearly independent and v +...+v r = 0. For any i = 0,... n we define the set A i := { v j Page 26 ( xi ) } : j =,..., r R n. Since 0 = r i= v r i, the origin belongs to n i=0 conv A i. By Colorful Carathéodory s Theorem there exist a i A i such that 0 conv{a 0,..., a n }. Hence there exist weights α i 0, n i=0 α i =, such that 0 = n α i a i = i=0 n α i v j(i) i=0 ( xi ). () Let I j := {i {0,..., n} : j(i) = j}. These sets are a partition of {0,..., n} and thus the sets X j := {x i : i I j } are a partition of X. We will find a common point to all the sets conv X i, which will end the proof. Note that for any distinct k, l {,... r} there exists u R r such that u v k =, u v l = and u v i = 0 for all other i. We multiply the equation () by u from the left side (in the sense of scalar product of vectors in R r ) and get 0 = ( ) xi α i ( ) xi α i. i I k i I l In particular i I k α i = i I l α i for every pair of k, l. Thus the sum i I k α i does not depend on k and since the sum of all α i is, all the sums i I k α i are equal to r. 3

28 Similarly we get that i I k α i x i does not depend on k. Denote this sum by z. Then rz = i I (rα i )x i = i I 2 (rα i )x i =... = i I r (rα i )x i, which (together with i I k (rα i ) = ) means that rz conv X i for all i =,..., r. Remark. We can prove Tverberg s Theorem in a different way. We will give only the main idea of the third proof: take n + points in R d (n is as in the above proof), build a simplex in R d+ by lifting the given points through fibres of an affine projection. Then it is enough to prove that there exist r disjoint faces of this simplex such that their images under this projection on R d have a common point. Remark. The question what happens if we use a continuous map (rather than an affine one) in the formulation of Tverberg s Theorem described in the previous Remark was long open. It is now known that the version with the continuous maps is true for r = p k, where p is a prime number (in the proof actions of groups are used). The second proof of Tverberg s Theorem is actually a copycat of the proof of Radon s Theorem from the first lecture. We can formulate this method, invented by Sarkaria, as an easy lemma: Lemma (Sarkaria). Assume X is a finite set in R d and X = X... X r is its partition. Let n = (r )(d + ). With every x X, if x X j, we associate a vector x := v j ( x ) R n, where (v i ) r i= is a sequence introduced in the previous proof. By X be denote a set of all x. Then r j= conv X j = if and only if 0 / conv X. Remark. In Sarkaria s Lemma we can skip the assumption that X i are pairwise disjoint. Then we have to consider multiset X. Remark. This gives an interior condition equivalent to r j= conv X j = (which means that X is separated along the colors,..., r). Note that we found an exterior condition before, in two lemmas from the Tuesday lecture. Proof of Sarkaria s Lemma. We will show that r j= conv X j if and only if 0 conv X. We will proceed in the same way as in the second proof of Tverberg s Theorem. Page 27 4

29 Assume first that 0 conv X. Thus there exist weights α(x) such that 0 = α(x) x = ( ) x r α(x)v j(x) = v j ( ) x α(x). x X x X j= x Xj Take u such that uv k =, uv l = and uv i = 0 for all other i. We multiply the equation by u and get, as in the previous proof, that both x X k α(x) and z := x X k α(x)x do not depend on k. Thus rz is a convex combination of points of X k for all k =,..., r, which ends the proof of this implication. Assume now that z r j= conv X j. Then there exist nonegative weights α(x) such that z = x X α(x)x =... = x X r α(x)x, thus ( ) ( ) z z 0 = 0 = (v +...+v r ) = r v j ( ) x α(x) = α(x) x, x Xj x X j= which means (since α(x) are nonegative), that 0 conv X and proves the second implication. We will also use Sarkaria s Lemma to prove another theorem: Theorem (Kirchberger). Assume X = X... X r R d is a partition of a finite set X. Then X is separated along the colors (i.e. r j= conv X j = ) if and only if all Y X of cardinality at most (r )(d+)+ are separated along the colors (i.e. r j= conv(y X j) = ). This theorem will follow by a more general fact. Consider a partition of X R d as above and let n = (r )(d + ). Let X i,0,..., X i,n be such that for any i =,... r we have For j = 0,... n we introduce sets which we will call groups. Page 28 X i = G j := n X i,k. k=0 r X l,j, l= 5

30 We call a set Y = {y 0,..., y n } a transversal, if y j G j for j = 0,..., n. We say that a group G j is separated along the colors if r i= conv X i,j =. Kirchberger s Theorem is a special case of the following one (to recover it we have to set X i,j := X i ). Theorem. Under above notation and conditions, if every transversal is separated along the colors then there exists a group separated along the colors. Proof. By Sarkaria s Lemma if a transversal Y is separated along the colors then 0 / conv Ȳ, where in the construction of Ȳ we use a partition of Y induced by the partition X,..., X r of X, namely Y = (Y X )... (Y X r ). On the other hand, if every group G j was not separated along the colors, the origin would belong to all the sets conv Ḡj, where again in the construction of Ḡ j we use the partition given by the partition X,... X r of X. Then, by Colorful Carathéodory s Theorem, there would exist a transversal Y, such that conv Ȳ would contain the origin, which would be a contradiction. We will consider now another problem: by how many simplices can a single point be covered? The following theorem gives an exact answer in the case of the plane. Theorem (Boros-Füredi). Assume X R 2 is a set of n points. Then there exists a point covered by ( ) n 3 ( 2 o()) triangles with vertices in X. 9 Remark. We will prove later that the constant 2 9 Page 29 is optimal. Proof. Consider a nice finite measure on the plane R 2, say absolutely continuous with respect to the Lebesgue measure with a positive continuous density. By M we denote the measure of the plane. Fix any direction v S. Note that there exists a line l = l(v) in this direction dividing the plane into two half-planes H = H (v) and H 2 = H 2 (v) of the same measure. We assume that H is on the left-hand side of v (see Figure 3). Then fix any point x on this line. It divides the line l into two half-lines l and l 2 (assume that l goes in the direction of v and l 2 in the direction of v). Note that there exist two half-lines k H and k 2 H 2 originating at x, such that the sector between l and k has measure M and 6 so does the sector between l 2 and k 2. When we move the point x along l from to, the angle between k and k 2 changes continuously from 0 to 2π and hence there exists a point x = x(v), such that these two half-lines form a line (see Figure 3). Moreover there exist two lines m H and m 2 H 2 originating at x such that the sector between k and m is of measure M and so is the 6 6

31 k H l x v l 2 k 2 H 2 Figure 3: We move x so that k and k 2 form a line. sector between k 2 and m 2 (see Figure 4). Let α = α(v) be the angle between m and m 2. Note that this angle is a continuous function of v S (we do not assume this angle belongs to [0, 2π)). Moreover H (v) = H 2 ( v), l (v) = l 2 ( v), m (v) = m 2 ( v), etc. and thus α(v) = α( v) (mod 2π), hence there exists a direction v 0 such that α(v 0 ) = π (mod 2π), which means that m (v 0 ) and m 2 (v 0 ) are collinear (see Figure 4). k m x l v l 2 k 2 m 2 Figure 4: We rotate v so that m and m 2 form a line. We have constructed three lines with a common point, dividing the plane into three sectors each of measure M. If we considered a counting measure 6 (which is not absolutely continuous) on the set X, we could find (if n = X is 7 Page 30

32 large) three lines with a common point dividing the plane into three sectors each containing almost n points of X. We skip this technical argument. 6 We will now prove that the point z common to these three lines is covered by at least ( ) n 3 ( 2 n3 o()) 9 27 o(n3 ) triangles. Note that if we pick three points, each from a different sector in such a way, that any two points do not lie in neighbouring sectors, we get vertices of a triangle covering z (see Figure 5). We can pick these points in 2( n 6 )3 ways. z z Figure 5: Triangles covering z. Moreover, if we choose two nonneighbouring groups of neighbouring sectors and pick a point from each of them, we get vertices of possible four triangles of which at least two cover point the z (see Figure 5). We can pick these triangles in 3 2 ( n 6 )4 n ways (we divide by n because we count 6 6 every triangle n times). This gives us a total number of triangles covering z 6 approximately equal to at least n3. 27 Page 3 Theorem. For any d 2 there exists a positive constant c(d) with the following property: for any set X R d of n points in general position there is a point in at least c(d) ( n d+) simplices with vertices in X. First proof. Let r be the largest possible number such that n = (r )(d + ) + k for some k. Then X has a Tverberg partition into r pieces, i.e. X = X... X r and there exists a point z r i= conv X i. By Colorful Carathédory s Theorem for any i... i d+ r there exist x i X i,..., x id+ X id+ such that z conv{x i,..., x id+ }. Clearly the simplices conv{x i,..., x id+ } are pairwise distinct for different choices of the indices i... i d+ r. Therefore the point z is covered by at least ( ) r + d = d + ( n k + + d ) d+ d + different simplices with vertices from X. ( n ) d+ d + 8 ( ) n (d + ) d+ d +

33 Remark. We proved the theorem with c(d) = (d + ) d. We could improve this constant slightly by using the stronger version of Colorful Carathéodory s Theorem, where one point is fixed. Second proof. Let F = {conv S : S X, S = d + }. Clearly F = ( n d+). We want to show that many of the elements of F have a nonempty intersection. We will use Fractional Helly s Theorem for this purpose. It is enough to show that a positive fraction of all (( n d+ d + (d + )-tuples of elements from F (or in other words, a positive fraction of all (d + )-tuples of convex hulls of d + elements from X) has a nonempty intersection. Suppose we have a subset Y X, Y = (d + ) 2. Any d 2 + d + = ((d + ) )(d + ) + points from Y have a Tverberg s partition, i.e. they can be splited into mutually disjoint Y,..., Y d+ such that d+ i= conv Y i. By Carathédory s Theorem we may assume that each of the sets Y i has no more than d + elements. Then we distribute the points from Y \ d+ i= Y i among the sets Y i so that each of the d + resulting sets Ỹi has exactly d + elements. Summing up, we obtain a partition Y = d+ i= Ỹi, Ỹi = d+, d+ i= conv Ỹi. There are ( ) n (d+) possible choices of the subset Y, each resulting in a 2 different (d + )-tuple of d + elements from X such that the convex hulls of those elements intersect. Since ( ) (( n n )) d+ c(d) (d + ) 2 d + for some positive c(d) (both sides of the above inequality are polynomials of the variable n, each of degree (d + ) 2 ), we can use Fractional Helly s Theorem and the assertion of the theorem follows. Remark. The constant form the first proof has not been improved for a long time. One can get a better constant considering a lifting of X into R n. Gromov has shown that getting back we can find a point covered many times. His proof uses topology and gives the constant ed. It is still an open d d question if this is best possible. )) Page 32 9

34 We will end this lecture with a proof that the leading term n3 in the 27 Boros-Füredi Theorem is optimal. Our example will be the stretched grid. The n n stretched grid consists of n 2 points in the plane. They form n parallel rows (n points in each), in each row the distance between two neighbouring points is one, and the points are aligned so that they also form n parallel columns. The difference between the stretched grid and a normal n n grid is the spacing between rows. The second row lies above the first (bottom) row, in distance from it (no difference so far). The third row lies above the second row, on such a height that the segment connecting the rightmost point from the first row and the leftmost point from the third row passes between the two rightmost points in the second row (so the third row has to be in distance strictly greater than n from the bottom row; it is convenient to think of it as lying in distance n from the bottom row). This construcion is continued: the j-th row lies in such a distance above the first row, that the segment connecting the rightmost point from the first row and the leftmost point from the j-th row passes between the two rightmost points in the (j )-th row (see Figure 6 for an example of the 4 4 stretched grid). Figure 6: The 4 4 stretched grid (stretched on the left, compressed on the right). We will be interested in the following question: how many points from the stretched grid lie in the interior of a triangle with vertices from stretched grid? The answer to this question will not change if we identify the n n stretched grid with a normal n n grid and the segments connecting two points with two perpendicular segments: we first go vertically from the lower Page 33 0

35 of the two points to the height of the higher of these points and then right or left to the second point (see Figure 6; more examples of segments and triangles are depicted in Figure 7). Figure 7: Segments and triangles. We can now proceed to the proof of the optimality of the Boros-Füredi Theorem. Proof of the optimality of the Boros-Füredi Theorem. Recall that the theorem states that for every set X R 2, X = n, there exists a point covered by ( ) n 3 ( 2 o()) triangles with vertices in X. We need to show that (for 9 every n N) there exists a set X R 2, X = n, such that every point in the plane is covered by at most ( ) n 3 ( 2 + o()) triangles with vertices in X. 9 n n 2 n 3 Figure 8: The points on the diagonal are divided into three groups. Let X be the set of points from diagonal of the n n stretched grid (going from the bottom left-hand side corner to the top right-hand side corner). Using the above explications we can identify X with the set {(, ),..., (n, n)} of R 2. If a point (x, y) R 2 is covered by some triangles with vertices in Page 34

36 X, then we must have x y n. A horizontal and a vertical line through the point (x, y) divide the points in the set X into three groups, of cardinality n, n 2, n 3 respectively (see Figure 8). Notice that if a triangle with vertices from X coveres the point (x, y) then each of its vertices comes from a different group (and the covering triangle looks like the bottom right-hand side triangle in Figure 7). Therefore the point (x, y) is covered by at most ( ) 3 n + n 2 + n 3 n n 2 n 3 3 triangles with vertices from the set X. (n + 2)3 27 = n o(n3 ) Remark. A similar example can be constructed also in higher dimensions. Page 35 2

37 Imre Bárány, Lecture 4 Scribes: G. Ambrus and L. Silverstein Weak ε-nets. Let X R d be a set of n points in general position (i.e. no d + points of X lie on the same hyperplane.) For a positive 0 < ε <, we define F ε = {conv Y : Y X, Y εn}. A finite set S R d is called a (weak) ε-net of X if S F for any F F ε. We note that S is called a strong ε-net if, in addition, it also satisfies S X. Since we are only going to discuss weak ε-nets, we will simply call these ε-nets. Our goal is to find an ε-net of X with small cardinality. Theorem. For any X R d and any 0 < ε <, there exists an ε-net S of X with cardinality S c d ε d+, where c d is a constant depending only on the dimension d. Note that this is a striking result: the cardinality of S does not depend on the set X! Proof. We may assume that ε > 2 d+ n since, otherwise, 2 d+ (d + ) ε d+ 2d+ (d + ) 2 d+ n. (d + ) d+ n d+ Therefore, X is a suitable ε-net. We construct S by an iterative algorithm. We start with S 0 =, and H 0 = ( X d+). At the ith step, for i, we check if there exists Y X with Y εn and S i Y =. If there is no such Y, then the algorithm terminates - we have found an ε-net. Otherwise, take such a Y = Y i. By Page 36

38 the results of the previous lecture, there exists a highly covered point z i of Y i ; a point z, which is covered by at least ( ) Y (d + ) d d + many different { simplices with vertices from Y i. Let S i = S i {z i }, and set H i = H i \ T ( ) } X d+ : z conv (T ). Since X is finite, the algorithm terminates. Therefore, we have to estimate the number of steps until this happens this is the same as the cardinality of the constructed ε-net. Note that, for every i, the convex hull of every element of ( X d+) \ Hi contains a point from S i. On the other hand, conv Y i S i =, which implies that every (d+)-element subset of Y i is contained in H i. Therefore, since z i is a highly covered point of Y i, we obtain that ( ) Y H i \ H i (d + ) d d + Note that the condition ε > 2(d+) n implies that n (d + ) εn (d + ) < 2 ε. (d + ) d ( ) εn. d + Therefore, the number of iterations until termination is at most ( n ) d+ ) (d + ) d 2 d+ ε d+. ( εn (d+) d d+ We note that a stronger upper bound of c d (/ε d )(log(/ε)) d has been proven. What about lower bounds for the cardinality of an ε-net? The simplest construction is the following: take /ε + parallel lines on the plane, and let X have n ε points in each of the disjoint regions between the consecutive lines. An ε-net must contain a point in each of these components, therefore, its cardinality is at least /ε. A more elaborate example comes from the stretched grid and gives the stronger bound (/ε) (log(/ε)) d. One has to investigate the sets that have a negative staircase form (see below). Still, there is a huge gap between the lower and upper bounds: apart from logarithmic factors, it is /ε versus /ε d. The general belief is that the truth may be around /ε. 2 Page 37