Chapter 9:

Geometric Selection Theorems

11/01/2013

9.1 A Point in Many Simplices: The 1st Selection Lemma

• Consider n points in the plane in general position, and draw

all the triangles with vertices at the given points. Then

there exists a point of the plane common to at least

of these triangles.

• Here is the optimal constant.

definition

2

Definition:

• If is a finite set, an X-simplex is the convex hull

of some (d+1)-tuple of points in X.

Convention: X-simplices are in bijective correspondence with

their vertex sets.

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• Let X be an n-point set in . Then there exists a point

contained in at least X-simplices,

where is a constant depending only on the

dimension d.

• For n very large, we may take

9.1.1 Theorem (1st Selection Lemma)

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9.1.1 The 1st Proof: from Tverberg and colorful Carathẽodory

• We may suppose that n is sufficiently large.

( by )

• Put . There exist r pairwise disjoints sets

whose convex hull have a point in common:

call this point .

Tverberg’s Theorem

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• Let be a set of (d+1)

indices. We apply for

the (d+1) “color” sets, which all contain

in their convex hull.

This yields a rainbow X-simplex containing

and having one vertex from each .

Colorful Carathẽodory’s Theorem

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• If are two (d+1)-tuples of indices, then .

Hence the number of X-simplices containing the point is:

• For n sufficiently large, say , this is at least

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9.1.1 The 2nd Proof: from Fractional Helly

• Let F denote the family of all X-simplices. Put

.

We want to apply to F.

• A (d+1)-tuple of sets of F is good if its d+1 sets have a

common point.

• It suffices to show that there are at least good

(d+1)-tuple for some independent of n (since then

the theorem provides a point common to at least members

of F).

Fractional Helly Theorem

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• Set and consider a t-point set .

Using we find that Y can be

partitioned into d+1 pairwise disjoint sets, of size d+1

each, whose convex hulls have a common point.

• Therefore, each t-point provides at least one good

(d+1)-tuple of members of F.

• Moreover, the members of this good (d+1)-tuple are

pairwise vertex-disjoint, and therefore the (d+1)-tuple

uniquely determines Y. It follows that the number of good

(d+1)-tuples is at least:

Tverberg’s Theorem

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9.1.2 A Point in the interior of many

X-simplices. Lemma:

• Let be a set of points in general

position, and let H be the set of the hyperplanes

determined by the points of X. Then no point is

contained in more than hyperplanes of H.

consequently, at most X-simplices have on their

boundary.

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9.1.2 Proof :

• For each d-tuple S whose hyperplane contains , we

choose an inclusion-minimal set whose affine

hull contains .

• We claim that if then either

or and share at most

points.

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• If and ,

, then the affine hulls of and are

distinct, for otherwise, we would have k+1 points in a

common (k-1)-flat, contradicting the general positions of X.

But then the affine hulls intersect in the (k-2)-flat

generated by and containing ,

and are not inclusion-minimal.

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• Therefore, the first k-1 points of determine the last

one uniquely, and the number of distinct sets of the form

of cardinality k is at most .

• The number of hyperplanes determined by X and

containing a given k-point set is at most

and the lemma follows by summing over k.

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9.2 The 2nd Selection Lemma

• Same as the previous section but instead of considering all

X-simplices, we consider some of all.

• it turns out that still many of them must have a point in

common.

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9.2.1 Theorem (2nd Selection Lemma)

• Let X be an n-point set in and let F be a family of

X-simplices, where is a parameter.

Then there exists a point contained in at least

X-simplices of F, where and are

constants (depending on d).

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Definitions:• Hypergraphs are a generalization of graphs where edges

can have more than 2 points. A hypergraph is a pair

, where is the vertex set and is a system

of subsets of , the edge set.

• A k-uniform hypergraph has

all edges of size k.

• A k-partite hypergraph is one

where the vertex set can be

partitioned into k subsets, such

that each edge contains at most

one point from each subset.16

9.2.1 Proof:

• We can view F as a (d+1)-uniform hypergraph: we regard

X as the vertex set and each X-simplex corresponds to an

edge.

• First, let us concentrate on the simpler task of exhibiting at

least one good (d+1)-tuple.

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• Hypergraphs with many edges need not contain complete

hypergraphs, but they have to contain complete

multipartite hypergraphs.

• Let denote the complete (d+1)-partite (d+1)-

uniform hypergraph with t vertices in each of its d+1

vertex classes.

Example for [only 3 edged are drawn as a sample]

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• If t is a constant and we have a (d+1)-uniform hypergraph

on n vertices with sufficiently many edges, then it has to

contain a copy of as a subhypergraph.

• In geometric language, given a family F of sufficiently many

X-simplices, we can color different sets of t points in (d+1)

colors in such a way that all the rainbow X-simplices on the

(d+1)t colored points are present in F.

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• In such situation, if t is sufficiently large constant, the

with r=d+1 claims that we

can find a (d+1)-tuple of vertex-disjoint rainbow X-

simplices whose convex hull intersected. And so there is a

good

(d+1)-tuple.

• For the we need not only one

but many good (d+1)-tuples. We use an appropriate

stronger hypergraph result, saying that if a hypergraph has

enough edges, then it contains many copies of :

Colored Tverberg’s Theorem

Factional Helly Theorem

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9.2.2 Theorem (The Erdös-Simonovits Theorem)

• Let d and t be positive numbers. Let H be a (d+1)-uniform

hypergraph on n vertices and with edges, where

for a certain sufficiently large constant C. Then

H contains at least

copies of , where is a constant.

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9.2.1… Proof• The given family F contains copies of

. Each such copy contributes at least one good

(d+1)-tuple of vertex-disjoint X-simplices of F.

• On the other hand, d+1 vertex disjoint X-simplices have

together vertices and hence their vertex set can

be extended to a vertex set of some (with t(d+1)

vertices) in at most ways.

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• is the maximum number of copies of

that can give rise to the same good (d+1)-tuple. Hence

there are at least good (d+1)-tuples of X-

simplices of F.

• By at least

X-simplices of F share a common point,

with .

• The best explicit value is

Factional Helly Theorem

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9.3 Order Types and the Same-Type Lemma -

Definitions:• The are infinitely many 4-point sets in the plane in general

positions, but there are only two “combinatorially distinct”

types of such sets.

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• What is “combinatorially the same”? Let’s see an explanation

for this notation for planar configuration in general position:

let and be two

sequences of points in , both in general positions. Then p

and q have the same order type if for any indices

we turn in the same direction when going from to

via and then going from to via .

• We say that the triples and have the same orientation.

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• The order type of a set. Let p and q be two sequences of

points in . We require that every (d+1)-element subsequence

of p have the same orientation as the corresponding

subsequence of q. Then p and q have the same-order type.

• notion of orientation: if are vectors in . Let

matrix A be the matrix that has vectors as the

columns. The orientation of is defined as the sign of

det(A) (it can be +1, -1, or 0).

• For a (d+1)-tuple of points , we define the

orientation of the d vectors .

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• i.e. In the planar: let ,

and be 3 points in . The

orientation of the 3 points is:

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• Back to the order type: let be a

point sequence in . The order type of p is defines as the

mapping assigning to each (d+1)-tuple of

indices, , the orientation of the

(d+1)-tuple . Thus, the order type of p

can be described as a sequence of +1’s, -1’s and 0’s with

terms.

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• Same-type transversals: let be an m-

tuple of finite sets in . By a transversal of this m-tuple

we mean any m-tuple such that

for all i. we say that has same-type

transversals if all of its transversals have the same order

type.

Example of 4 planar sets with same-type transversals

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• To see this color each transversal of

by its order type. Since the number of possible order types

of an m-point set in general position cannot be greater than

, we have a coloring of the edges of the complete

m-partite hypergaph on by r colors. By

Erdös-Simonovits theorem (9.2.2), there are sets ,

not too small, such that all edges induces by

have the same color, meaning – have same-type

transversals.

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9.3.1 Same-type Lemma (Theorem)

• For any integers , there exists

such that the following holds. Let be finite

sets in such that is in general

position*. Then there are such that

the m-tuple has same-type transversals

and for all .

• This is shorthand for saying that for all

and is in general position.

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9.3.1 Proof:• It sufficient to prove the same-type lemma for . If

is the current m-tuple of sets, we go through all

(d+1)-tuple of indices, and we apply the same-type lemma to

the (d+1)-tuple . These sets are replaced by

smaller sets such that this (d+1)-tuple has

same-type transversals. After executing this for all

(d+1)-tuples of indices, the resulting current m-tuple of sets

has the same-type transversal. This method gives a smaller

bound:

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9.3.2 Lemma (handling the m=d+1 case):

• Let be convex sets. The following two

conditions are equivalent:

i) there is no hyperplane simultaneously intersecting all of

.

ii) for each nonempty index set , the sets

and can be by a

hyperplane.

• Moreover, if are finite sets such that

the sets have property (i) (and (ii)), then

has the same-type transversals. 33

strictly separated

9.3.1… Proof:• To prove the same-type lemma for the case , it

suffices to choose the sets in such a way that their

convex hulls are separated in the sense of (ii) in Lemma

9.3.2. this can be done by an iterative application of the

.

• Suppose that for some nonempty index set

the sets and cannot be

separated by a hyperplane. Lets assume that .

Let h be a hyperplane simultaneously bisecting

whose existence is guaranteed by

. Let be a closed half-space bounded by h and

containing at least half of the points of .

Ham-Sandwich Theorem

Ham-Sandwich Theorem

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• For all we discard the points of not lying in , and

for j we throw away the points of that lie in the interior of

.

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• We claim that union of the resulting sets with indices in I is

now strictly separated from the union of the remaining sets.

If h contains no points of the sets, then it is a separating

hyperplane. Otherwise, let the points contained in h be

. We have by the general position

assumption. For each , choose a point very near

to . If lies in some with , then is chosen in

the complement of . We let h’ be a hyparplane passing

through

and lying very close to h. Then h’ is the desired

separating hyperplane, provided that the are sufficiently

close to the corresponding .

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• The size of a set is reduces from to at least

. We can continue with the other index sets in the same

manner. After no more than halvings, we obtain sets

satisfying the separation condition and thus having

same-type transversals. The same-type lemma is proved.

• The lower bound for is doubly exponential,

roughly .

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9.3.3 Theorem (Positive-Fraction Erdös-Szekeres Theorem):

• For every integer there is a constant such

that every sufficiently large finite set in general

positions contains k disjoints subsets , of size

at least each, such that each transversal of

is in convex position.

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9.3.3 Proof:

• Let be the number as in the

. We partition X into n sets

of almost equal size, and we apply the same-

type lemma to them, obtaining sets , ,

with the same-type transversals.

Let be a transversal of . By the

, there are

such that are in convex positions.

Then are as required in the theorem.

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Erdös-Szekeres theorem

Erdös-Szekeres theorem

9.4 Hypergraph Regularity Lemma:

• Let be a k-partite hypergraph whose vertex set is in

the union of k pairwise disjoint n-element sets

, and whose edges are k-tuples containing precisely

one element from each . For subsets ,

, let denote the number of edges of H

contained in . In this notation, the total number of

edges of H is equal to .

• Let

denote the density of the graph induces by the .40

9.4.1 Theorem (Weak regularity lemma for hypergraphs):

• Let H be a k-partite hypergraph as defined before, and

suppose that for some . Let .

Suppose that n is sufficiently large in terms of k, and .

Then there exist subsets of equal size

, , such that:

(i) (High density) , and

(ii) (Edges on all large subsets) for any

with , .

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The following scheme illustrates the situation.

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9.4.1 proof

• We look at a modified density parameter that slightly favors

larger sets. Thus, we define the magical density

:

• We choose , , as sets of equal size that

have the maximum possible magical density .

We denote the common size by s.

• First we derive condition (i) in the theorem for this choice of

the . 43

• We have

and so , which verifies (i). Since obviously

, we have . Combining

with , we also obtain that .

• Since is a large number by assumption, rounding it up to

an integer doesn’t matter. We will assume is an integer,

and let be a -element sets. We want

to prove .

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• We have

• We want to show that the negative terms are not too large,

using the assumption that the magical density of

is maximum.

)#(

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• The problem is that maximize the magical

density only among the sets of equal size, while we have sets

of different sizes in the terms. To get back to equal size, we

use the following observation. If, say, is a randomly

chosen subset of of some given size r, we have

• For estimating the term , we use random

subsets of size of respectively.

Thus,

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• Now for any choice of , we have,

• Therefore,

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• To estimate the term , we

use random subsets and

this time all of size . A similar calculation as before yields

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• From we obtain that is at least

multiplied by the factor

)#(

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9.5 A Positive-Fraction Selection Lemma:

• Here we discuss a stronger version of the first selection

lemma. The theorem below shows that we can even get a

large collection of simplices with a quite special structure.

For example, in the plane, given n red points, n green points,

and n blue points, we can select red, green, and

blue points in such a way that all the red-green-blue triangles

for the resulting sets have a point in common.

• Here is the d-dimensional generalization

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9.5.1 theorem (Positive-Fraction Selection Lemma):

• For all natural numbers d, there exists with the

following property. Let be finite sets

of equal size, with in general positions.

Then there is a point and subsets

, with , such that the

convex hull of every transversal of contains

.

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9.5.1 Proof:• Let . We may suppose that all the

are large. Let be the set of all “rainbow” X-simplices,

i.e., of all the transversals of , where the

transversals are formally considered as sets for the moment.

The size of is, for d fixed, at least a constant fraction of

( are of equal size). Therefore, by the second

selection lemma there is a subset of at least

X-simplices containing a common point , where

.

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• For the subsequent argument we need to apply Lemma

9.1.2, which guarantees that lies on the boundary of at

most of the X-simplices of . So we let be the

X-simplices containing in the interior, and for a sufficiently

large n we still have .

• We consider the (d+1)-partite hypergraph H with vertex set

X and edge set . We let , where

is as in the same-type lemma, and we apply the weak

regularity theorem (theorem 9.4.1 ) to H. This yields sets

, whose size is at least a fixed

fraction of the size of , and such that any subsets

of size at least induce an edge. Meaning:

there is a rainbow X-simplices with vertices in the and

containing .53

• The argument is finished by applying the same-type lemma

with d+2 sets and . We obtain

sets and with same-

type transversals, and with for

. Now either all transversals of

contain the point in their convex hull or none does. But

the latter possibility is excluded by the choice of the (by

the weak regularity lemma).

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♦THE END ♦

Notation– General Position:

• For points in in general position, we assume that no

unnecessary affine dependencies exist: no

points lie in a common (k-2)-flat.

i.e.: for lines in the plane in general position we postulate,

that no 3 lines have a common point and no 2 are parallel.

Back to 9.1.1

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Tverberg’s Theorem:

Back to 9.1.1 (i)

• Let be points in , .

Then there is a partition of such

that

.

Back to 9.1.1 (ii)

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Colorful Carathẽodory’s Theorem :

• Let be d+1 sets in . Suppose that

.

Then there are such that

. Back to 9.1.1

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Fractional Helly Theorem:

• For every there exists with the

following property. Let be convex sets in ,

, and at least of the collection of sets

of size d+1 have nonempty intersection, so there exists a

point contained in at least sets.

Back to 9.1.1

Back to 9.2.1 (i)

Back to 9.2.1 (ii)

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Colored Tverberg’s Theorem :

• For every d and r there exists such that that for

every set of cardinality (d+1)t , partitioned into

t-point subsets , there exist r disjoint sets

that are rainbow, meaning that

for every i, j, and whose convex hulls all

have a common point. Back to 9.2.1

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Ham-Sandwich Theorem :

• Every d finite sets in can be simultaneously bisected

by hyperplane.

• A hyperplane h bisects a finite set A if each of the open

half-spaces defined by h contains at most points

of A.

Back to 9.3.2

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• Let be convex sets with . Then

there exists a hyperplane h such that C lies in one of the

closed half-spaces determined by h, and D lies in the

opposite closed half-space.

If C and D are closed and at least one of them is bounded,

they be separated strictly; in such a way that

Separation Theorem :

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Back to 9.3.2

Erdös-Szekeres Theorem :

• For any natural number k there is a natural number

such that any n-point set in the plane in general

position contains a subset of k points in convex position

(forming the vertices of a convex k-gon).

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Back to 9.3.3