MAT 388: Introduction to discrete geometry
Lim, Kyuson
October 24, 2021
MAT388
2
Lim Kyuson
Contents
1
2
Introduction to Geometric Combinatorics
5
1.1
Convex polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.1.1
Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.1.2
Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
Lecture on Discrete and Polyhedral Geometry
17
2.1
Basic discrete geometry . . . . . . . . . . . . . . . . . . . . . . . . . .
17
2.1.1
The Helly theorem . . . . . . . . . . . . . . . . . . . . . . . .
17
2.1.2
Application questions for the Helly theorem . . . . . . . . . . .
26
2.1.3
Relevant extra questions . . . . . . . . . . . . . . . . . . . . .
28
Radon’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
2.2.1
Application questions for the Radon’s theorem . . . . . . . . .
32
2.2.2
Extension of the Helly theorem . . . . . . . . . . . . . . . . .
34
The 𝑑-dimensional Helly theorem . . . . . . . . . . . . . . . . . . . .
35
2.3.1
Extension of the 𝑑-dimensional Helly theorem . . . . . . . . .
36
Some case studies of Topological Helly theorem . . . . . . . . . . . . .
41
2.2
2.3
2.4
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CONTENTS
Chapter 1
Introduction to Geometric
Combinatorics
1.1
Convex polytopes
In this chapter, we give definitions and some basic results about polyhedra following [1,
p.7~14].
1.1.1
Definitions
Definition 1.1.1. Unit 𝑑-cube: all cubes are bounded 𝐶𝑑 := {x ∈ R𝑑 : −1 ≤ 𝑥𝑖 ≤ 1} =
conv({+1, −1} 𝑑 ),



0 ≤ 𝑥1 ≤ 1 










0
≤
𝑥
≤
1
2
𝑑
𝐶𝑑 := (𝑥 1 , ..., 𝑥 𝑑 ) ∈ R :
.
..










0
≤
𝑥
≤
1
𝑑


Definition 1.1.2. A set 𝐶 ⊆ R𝑑 is convex if for any 2 points p and q in 𝐶, the entire line
segment joining them {𝜆p + (1 − 𝜆)q : 0 ≤ 𝜆 ≤ 1}, is contained in 𝐶.
Equivalently, set 𝐶 is convex if its intersection with every straight line is either empty,
or connected set [3, p.8].
Example 1.1.3. Convex set, open ball: B(0, 𝜀) ∈ R𝑛 for 0 < 𝜀.
Proof. Let for any x, y ∈ 𝐵(0, 𝜀).
Then, x + 𝜆(y − x) belong to the line segment joining x, y where 0 ≤ 𝜆 ≤ 1.
From the center 0 ∈ 𝐵(0, 𝜀) for a fixed radius 𝜀, |x +𝜆(y − x) − 0| = |x −𝜆x +𝜆0 − 0 +𝜆y −
𝜆0| = |(1−𝜆)(x−0) +𝜆(y−0)| ≤ |(1−𝜆)(x−0)| + |𝜆(y−0)| = (1−𝜆)|(x−0)| +𝜆|(y−0)|
< (1 − 𝜆)𝜀 + 𝜆𝜀 = 𝜀.
Equivalently, |x + 𝜆(y − x) − 0| < 𝜀. Hence, x + 𝜆(y − x) ∈ 𝐵(0, 𝜀).
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Example 1.1.4. The annulus = {1 ≤ 𝑥 2 + 𝑦 2 ≤ 2}, is not a convex set.
Figure 1. An Illustration of an annulus
Proposition 1. If 𝐴 and 𝐵 are convex the 𝐴 + 𝐵 and 𝐴 − 𝐵 are convex, and for any
𝜆 ∈ R, the set 𝜆𝐴 is convex [3, p.9].
Proof. The set 𝐴 is convex, contains {𝜆𝑎 1 + (1 − 𝜆)𝑎 2 } for any 𝑎 1 , 𝑎 2 ∈ 𝐴, where
𝜆 ∈ [0, 1]. Similarly, the set 𝐵 is convex, contains {𝜆𝑏 1 + (1 − 𝜆)𝑏 2 } for any 𝑏 1 , 𝑏 2 ∈ 𝐴,
where 𝜆 ∈ [0, 1].
Then, for the combined set of 𝐴 + 𝐵 := {𝑎 + 𝑏|𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵}, the convex set of line
segment is defined as {𝜆(𝑎 1 + 𝑏 1 ) + (1 − 𝜆)(𝑎 2 + 𝑏 2 )} for any 𝑎 1 , 𝑎 2 ∈ 𝐴 and 𝑏 1 , 𝑏 2 ∈ 𝐵,
where 𝜆 ∈ [0, 1].
Hence, the convex set of 𝐴 + 𝐵 is defined in terms of combined convex of each arbitrary 2 points from 𝐴 and each arbitrary 2 points from 𝐵. Thus, the sum of convex
sets 𝐴 and 𝐵 of {𝜆𝑎 1 + (1 − 𝜆)𝑎 2 } + {𝜆𝑏 1 + (1 − 𝜆)𝑏 2 } gives the convex set of 𝐴 + 𝐵 of
the {(𝜆𝑎 1 + 𝑎 2 − 𝜆𝑎 2 ) + (𝜆𝑏 1 + 𝑏 2 − 𝜆𝑏 2 )}.
Definition 1.1.5. A hyperplane in R𝑑 is a set 𝐻 := {x ∈ R𝑑 : 𝑎 1 𝑥 1 +𝑎 2 𝑥 2 +···+𝑎 𝑑 𝑥 𝑑 = 𝑏},
𝑎 1 , ..., 𝑎 𝑑 , 𝑏 ∈ R. The equivalent notation is {x | a𝑇 x = b}.
The hyperplane 𝐻 is defined 2 halfspaces in R𝑑 : 𝐻 = 𝐻 + ∩ 𝐻 − .
Proposition 2. All halfspaces can be considered to be of the form 𝐻 − .
Proof. 𝐻 − := {x ∈ R𝑑 : 𝑎 1 𝑥 1 + 𝑎 1 𝑥1 + · · · + 𝑎 𝑑 𝑥 𝑑 ≤ 𝑏} and 𝐻 + := {x ∈ R𝑑 : 𝑎 1 𝑥 1 + 𝑎 1 𝑥 1 +
· · · + 𝑎 𝑑 𝑥 𝑑 ≥ 𝑏} is multiplied by -1, then 𝐻 + := {x ∈ R𝑑 : −𝑎 1 𝑥 1 − 𝑎 1 𝑥 1 − · · · − 𝑎 𝑑 𝑥 𝑑 ≤
−𝑏}.
Example 1.1.6. Let 𝑇 is the halfspace, {(𝑥 1 , 𝑥2 , 𝑥3 ) ∈ R3 : 𝑥 1 ≤ 2} for 𝐴x ≥ b}, for
example [1, 0, 0]x ≤ [−2], then for any two points the 𝑇 is convex but not affine.
Definition 1.1.7. A polyhedron (or H -polyhedron) in R𝑑 is any set obtained as the
intersection of finitely many halfspaces in R𝑑 . This is 𝑃 = {x ∈ R𝑑 : 𝐴x ≤ b}, where
𝐴 ∈ R𝑚×𝑑 and 𝑏 ∈ R𝑚 .
(The prefix H stands for the fact, a polyhedron involves intersecting halfspaces)
The Unbounded polyhedron, 𝑃 = R𝑑 (any single halfspace in R𝑑 , 𝑑 ≥ 1).
The {𝑥 ∈ R : 𝑥 ≥ 0} and {𝑥 ∈ R : 𝑥 ≤ 1} are example of halfspaces in R, both are
unbounded polyhedra gotten by relaxing inequalities of 𝐶1 .
We define a polytope to be a bounded polyhedron
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Lemma 1.1.8. An H -polyhedron 𝑃 is a convex set.
Proof. Let 𝑃 = {x ∈ R𝑑 : Ax ≤ b}, and take p, q ∈ 𝑃, for all points 𝜆p + (1 − 𝜆)q on
the line segment joining them. Then, 𝑃 = {x ∈ R𝑑 : 𝐴x ≤ 𝑏} yields 𝐴(𝜆p + (1 − 𝜆)q) =
𝜆𝐴p + (1 − 𝜆) 𝐴q ≤ 𝜆b + (1 − 𝜆)b = b.
Definition 1.1.9. A convex combination of 2 points p,q ∈ R𝑑 is a point of the form
𝜆p + (1 − 𝜆)q, 0 ≤ 𝜆 ≤ 1.
The set of all convex combinations of p and q is called the convex hull of p and q.
Remark 1.1.10. The convex hull of p and q is the line segment joining them.
Example 1.1.11. Exercise 2.12
(1) Convex hull of points 0 and 1 in R: conv({0, 1}) = {(1 − 𝜆)0 + 𝜆1 : 0 ≤ 𝜆 ≤ 1} =
{𝜆 : 0 ≤ 𝜆 ≤ 1} = [0, 1]
0
1
(2) Convex hull of (0,0), (1,0), (1,1), (0,1):
0
1
1
0
Claim: conv
,
,
,
= [0, 1] 2 .
0
0
1
1
The convex hull is the set of all convex combinations
of points, which includes the square
𝑥
with an inclusion of its interior. Let
be the conv({(0, 0), (1, 0), (0, 1), (1, 1)}) so
𝑦
that
4
∑︁
𝑥
0
0
1
1
𝜆2 + 𝜆4
= 𝜆1
+ 𝜆2
+ 𝜆3
+ 𝜆4
=
, where
𝜆𝑖 = 1, 𝜆𝑖 ≥ 0.
𝑦
0
1
0
1
𝜆1 + 𝜆3
𝑖=1
Í4
By the non-negativity, 𝜆2 + 𝜆 4 ≥ 0, 𝜆3 + 𝜆 4 ≥ 0. Also, 𝜆 2 +
𝜆 4 ≤ 𝑖=1 𝜆𝑖 = 1 and
Í4
𝑥
𝜆3 + 𝜆 4 ≤ 𝑖=1
𝜆𝑖 = 1. Hence, 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1 such that
∈ [0, 1] 2 .
𝑦
0
1
1
0
This shows conv
,
,
,
⊆ [0, 1] 2 .
0
0
1
1
Thus, the consideration of 4 points for the convex hull exactly gives a square and
the interior parts. The max of the set is 𝑥 = 1 and 𝑦 = 1 such that the square enclose the
convex hull of given points. However, If we consider the convex combinations of two
points
at a time,
then
we
get
the union of six
line
segments.
0
1
𝜆
𝜆
i) (1 − 𝜆)
+𝜆
=
:0≤𝜆≤1 =
:0≤𝜆≤1 .
0
0
0
0
This is {𝑥 ∈ [0, 1]|(𝑥, 0)}, which is the bottom segment.
0
1
𝜆
𝜆
ii) (1 − 𝜆)
+𝜆
=
:0≤𝜆≤1 =
:0≤𝜆≤1 .
0
1
0
𝜆
This is {𝑥 ∈ [0, 1]|(𝑥, 𝑥)}, which is the square box.
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0
0
0
0
iii) (1 − 𝜆)
+𝜆
=
:0≤𝜆≤1 =
:0≤𝜆≤1 .
0
1
𝜆
𝜆
This is {𝑥 ∈ [0, 1]|(0, 𝑥)}, which is the right side.
1
1
1
1
iv) (1 − 𝜆)
+𝜆
=
:0≤𝜆≤1 =
:0≤𝜆≤1 .
0
1
𝜆
𝜆
This is {𝑥 = 1, 𝑦 ∈ [0, 1]|(1, 𝑦)}, which is the top.
1
0
1−𝜆
1−𝜆
v) (1 − 𝜆)
+𝜆
=
:0≤𝜆≤1 =
:0≤𝜆≤1 .
0
1
𝜆
𝜆
This is {𝑥 =∈ [0, 1]|(1 − 𝑥, 𝑥)}.
1
0
1−𝜆
1−𝜆
vi) (1 − 𝜆)
+𝜆
=
:0≤𝜆≤1 =
:0≤𝜆≤1 .
1
1
1
1
This is {𝑥 ∈ [0, 1]|(1 − 𝑥, 1)}.
The following shows a 6 line segments of consideration for 2 points.
(0, 1)
(1, 1)
(0, 0)
(1, 0)
Figure 2. The Square with 4 points
The conv({(0, 0), (1, 0), (1, 1)}), when the 𝜆 4 of points (0,1) is 0, gives an interior
right-angled triangle with vertices (0, 0), (1, 0), (1, 1). The conv({(0, 0), (0, 1), (1, 1)},
when the 𝜆2 of points (1,0) is 0, gives an interior right-angled triangle with vertices
(0, 0), (0, 1), (1, 1).
For example, the point (0.5,0.5) can be represented as 0.5 · (0, 0) + 0.5 · (1, 1), but
also as 0.5 · (0, 1) + 0.5 · (1, 0), 0.25 · (0, 0) + 0.25 · (1, 0) and 0.25 · (0, 1) + 0.25 · (1, 1).
Remark 1.1.12. We observe that there is always a representation of points in a square
with at most 3 nonzero coefficients, because we can triangulate the square.
Definition 1.1.13. A convex region [9, p.183] is a set of points, any two of which can
be joined by a segment consisting of points in the set, with extra condition that each of
the points is on at least 2 non-collinear segments consisting entire of points in the set.
Particularly, an angular region is the set of all points on rays within an angle, and the triangular region is the set of all points between pairs of points on distinct sides of a triangle.
Hence, the angular region is bounded by an angle, while the triangular region is bounded
by a triangle. The conjecture for general convex hulls of any points in the set 𝑆 ∈ R2 is
formulated for types of convex regions.
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Definition 1.1.14. Convex hull for 𝑡 number of points:
Í𝑡
A convex combination of p1 , ...., pt ∈ R𝑑 is any point of the form 𝑖=1
𝜆𝑖 p𝑖 where
Í𝑡
𝜆𝑖 ≥ 0, for all 𝑖=1,...t and 𝑖=1 𝜆𝑖 = 1.
The set of all convex combinations of p1 , ..., p𝑡 is called their convex hull, denoted as
conv({p1 , ...p𝑡 })
Taking the convex hull of a finite set of points is like ’shrink wrapping’ the points:
- conv({p1 , ...p𝑡 }) is the smallest convex set containing p1 , ..., p𝑡 .
- conv({p1 , ...p𝑡 }) is the intersection of all convex sets containing p1 , ..., p𝑡 , following
from the fact that the intersection of 2 convex sets is convex.
Definition 1.1.15. A V-polytope in R𝑑 is the convex hull of finite number of points in
R𝑑 . It is a set of the form 𝑃=conv({p1 , ...p𝑡 }) where p1 , ..., p𝑡 ∈ R𝑑 .
A V-polytope is a convex hull of finitely many points. [4, p.3]
An H -polytope is an intersection of finitely many closed half-spaces.
The two notions of a polytope as an H -polyhedron and as a V-polytope are the same.
Example 1.1.16. [4, p.3] Take 2 halfspaces in the plane whose borders are parallel to the
𝑥-axis and which intersect in a ’strip’. For example, {𝑦 ≥ 1, 𝑥 = 0} ∩ {𝑦 ≤ 3, 𝑥 = 0} in
R2 is {1 ≤ 𝑦 ≤ 3, 𝑥 = 0}.
The result is unbounded H -polyhedron, so it can not be a V-polytope. Notice, the
H -polytope is defined as bounded intersection of finitely many closed half-spaces.
Theorem 1.1.17. Main theorem of Polytope Theory
Every V-polytope has a description by inequalities as an H -polytope, and every H polytope is the convex hull of a minimal number of finitely many point called its vertices.
Definition 1.1.18. The cross-polytope in R𝑑 is the V-polytope.
𝐶𝑑4 := conv({±e1 , ±e2 ..., ±e𝑑 }),
where e1 , ...e𝑑 are the standard unit vectors in R𝑑 (ie. (1 00)𝑇 ), (0 10)𝑇 ), (0 01)𝑇 ∈ R3 ).
Example 1.1.19. Note that 𝐶14 is a line segment in R, 𝐶24 is a diamond in R2 and 𝐶34 is
an octahedron.
Figure 3. 𝐶14 : line segment, 𝐶24 : diamond, 𝐶34 : octahedron
CHAPTER 1. INTRODUCTION TO GEOMETRIC COMBINATORICS
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Expressed as an H -polytope,
𝑥1 + 𝑥2 + 𝑥3 ≤ 1 


−𝑥 1 + 𝑥 2 + 𝑥 3 ≤ 1 



𝑥1 − 𝑥2 + 𝑥3 ≤ 1 





𝑥
+
𝑥
−
𝑥
≤
1
1
2
3
4
3
𝐶3 = (𝑥1 , 𝑥2 , 𝑥3 ) ∈ R :
−𝑥1 − 𝑥 2 − 𝑥 3 ≤ 1 






𝑥1 − 𝑥2 − 𝑥3 ≤ 1 








−𝑥
+
𝑥
−
𝑥
≤
1


1
2
3




−𝑥1 − 𝑥 2 − 𝑥 3 ≤ 1 














Definition 1.1.20. The simplex is a a generalization of the notion of a triangle or tetrahedron to arbitrary dimensions, which is same as a generalization of a tetrahedral region
of space to n dimensions.
Any 𝑑 + 1 points which does not lie in an (𝑑 − 1)-space are the vertices of an 𝑑dimensional simplex, whose elements are simplexes formed
by subsets
of the 𝑑 + 1
𝑑+1
𝑑+1
points, namely the vertices themselves, 𝑑+1
edges,
triangles,
tetrahedra
[5,
2
3
4
p.120]. Hence, the 𝑑-simplex is a polytope of dimension 𝑑 with 𝑑 + 1 vertices.
Simply, a line-segment is enclosed by 2 points, a triangle by 3 lines, a tetrahedron
by 4 planes, and so on. Thus, the general simplex is alternatively defined as finite region
of 𝑛-space enclosed by 𝑛 + 1 hyperplanes or (𝑛 − 1)-spaces [5, p.120].
Definition 1.1.21. An affine combination of vectors p1 , ..., p𝑡 ∈ R𝑑 is a combination of
Í𝑡
Í𝑡
the form 𝑖=1
𝜆𝑖 p𝑖 such that 𝑖=1
𝜆𝑖 = 1.
Definition 1.1.22. The affine hull of p1 , ..., p𝑡 is the set of all affine combinations of
p1 , ..., p𝑡 . It is denoted as aff({p1 , ...p𝑡 }). The affine hull of a set 𝑆 is, denoted as aff(𝑆),
the set of all affine combinations of finitely many points in 𝑆.
Example 1.1.23. The affine hull of two points p and q in R2 is the line through them.
This can be seen as a difference between aff({p, q}) and conv({p, q}). Notice, if the
condition 0 ≤ 𝜆 ≤ 1 is dropped from the conv({p, q}), then it becomes aff({p, q}.
Example 1.1.24. The affine hull of 3 points (-1,1),(0,1),(1,1)∈ R2 is the line 𝑥2 = 1
which is equivalent to the affine hull of the polytope conv({(−1, 1), (0, 1), (1, 1)}).
Notice, the aff({(−1, 1), (0, 1), (1, 1)}) = {(𝑥1 , 𝑥2 ) : 𝑥 2 = 1} is a translation of the vector
space {(𝑥 1 , 𝑥2 ) : 𝑥 2 = 0}
Example 1.1.25. The affine hull of the cube 𝐶3 ⊂ R3 is R3 . Notice, the aff(𝐶3 ) is already
the linear vector space R3 .
Proof. i) For any 𝑥 in R, (𝑥, 0, 0) = 𝑥(1, 0, 0) + (1 − 𝑥)(0, 0, 0), where other coefficients
are 0 for any p𝑖 ∈ 𝐶3 .
ii) For any 𝑥, 𝑦 in R, (𝑥, 𝑦, 0) = 𝑥(1, 0, 0) + 𝑦(0, 1, 0) + (1 − 𝑥 − 𝑦)(0, 0, 0), where other
coefficients are 0 for any p𝑖 ∈ 𝐶3 .
iii) For any 𝑥, 𝑦, 𝑧 in R, (𝑥, 𝑦, 𝑧) = 𝑥(1, 0, 0) + 𝑦(0, 1, 0) +𝑧(0, 0, 1) + (1−𝑥 − 𝑦 −𝑧)(0, 0, 0),
where other coefficients are 0 for any p𝑖 ∈ 𝐶3 .
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Definition 1.1.26. An affine space is any set of the form {𝑥 ∈ R𝑑 : 𝐴x = 𝑏}.
A non-empty affine space {x ∈ R𝑑 : 𝐴x = 𝑏} is a translate of the vector space {x ∈ R𝑑 :
𝐴x = 0} Further, every affine hull is an affine space.
We illustrate the differences among span, affine hull, and convex hull. Let v1 =
(1, 1), v2 = (2, 1) Then, span {v1 , v2 } = {𝑐 1 v1 + 𝑐 2 v2 : 𝑐 1 , 𝑐 2 ∈ R} = R2
The affine hull of v1 and v2 , aff({v1 , v2 }) = {𝑐 1 v1 + 𝑐 2 v2 : 𝑐 1 , 𝑐 2 ∈ R and 𝑐 1 + 𝑐 2 = 1}.
Simply, the affine set is a translates of a vector space
The convex hull of v1 and v2 , conv ({v1 , v2 }) = {𝑐 1 v1 + 𝑐 2 v2 : 𝑐 1 + 𝑐 2 = 1, 𝑐 1 , 𝑐 2 ∈ R≥0 }.
We illustrate the differences among linearly independence and affine independence
[3, p.9]. Let the points 𝑥 𝑘 ∈ R𝑛 , 𝑘 ∈ N.
Í
If 𝑘∈N 𝜆 𝑘 𝑥 𝑘 = 0 implies 𝜆 𝑘 = 0 for all 𝑘 ∈ N, then points are linearly independent.
Í
Í
If 𝑘 ∈N 𝜆 𝑘 𝑥 𝑘 = 0, 𝑘∈N 𝜆 𝑘 = 0 implies 𝜆 𝑘 = 0 for all 𝑘 ∈ N, then points 𝑥 𝑘 ∈ R𝑛 , 𝑘 ∈ N
are affinely independent.
The linear independence implies the affine independence, not vice versa.
In any linearly / affinely dependent set, some points are a linear / affine combination of
the remaining points. The 𝑑-dimensional space contains 𝑑-membered sets which are linearly independent, but every (𝑑 +1)-membered set in R𝑑 is linearly dependent [6, p.2~3].
Assume 𝐶 consists of 2 linearly independent points [3, p.10]:
𝑦
5
4
3
2
1
0
Conv(C)
0 1 2 3 4 5𝑥
𝑦
5
4
3
2
1
0
aff(C)
0 1 2 3 4 5𝑥
Figure 4. An Illustration of convex hull and affine hull of 2 points
Definition 1.1.27. The dimension of an affine space is the dimension of the linear vector
space that it is a translate of.
The dimension of a polytope 𝑃 is the dimension of its affine hull.
Example 1.1.28. [4, p.3]: The dimension of closed unit disk, denoted as 𝐵(x, 1) for
x ∈ R2 is 2. The dimension of a line segment (not a singleton) in any R𝑑 is 1.
Given the line segment being convex, the dimension of the affine hull is included in R1 .
Also, the 𝐵(x, 1) is convex and the dimension of the affine hull is included in R2
Example 1.1.29. Consider V-representation of 𝐶3 = 𝑐𝑜𝑛𝑣{v1 , v2 , ..., v8 } for 8 vertices.
Then, the H -representation for the 𝐶3 of dim-0 faces = {0 ≤ 𝑥1 , 0 ≤ 𝑥 2 , 0 ≤ 𝑥 3 , 𝑥1 ≤
1, 𝑥2 ≤ 1, 𝑥3 ≤ 1}, dim-1 faces are, edges, and dim- (𝑑 − 1) faces are, facets.
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1.1.2
Examples
Explain about the convex hull
The convex hull of a set 𝑆, of points is obtained by taking the collection of straight line
segments that join two arbitrary points in 𝑆.
Exercise 2.3
(1) Check each 𝐶𝑑 is convex:
Recall, to prove convexity of a set 𝑆, it needs to be shown that for every 2 points p, q ∈ 𝑆,
it is true that (1 − 𝜆)p + 𝜆q is also contained in 𝑆 for 𝜆 ∈ [0, 1].
Hence, to prove 𝐶𝑑 is convex, we choose any arbitrary two points p, q ∈ 𝐶𝑑 , and
show (1 − 𝜆)p + 𝜆q ∈ 𝐶𝑑 . Then, refer to individual components as 𝑝𝑖 and 𝑞𝑖 , so that
0 ≤ 𝑝𝑖 ≤ 1 and 0 ≤ 𝑞𝑖 ≤ 1 for all 𝑖 ∈ [𝑑].
Case 1. If 𝑝𝑖 < 𝑞𝑖 , then 0 ≤ 𝑝𝑖 = (1 − 𝜆) 𝑝𝑖 + 𝜆𝑝𝑖 < (1 − 𝜆) 𝑝𝑖 + 𝜆𝑞𝑖 < (1 − 𝜆)𝑞𝑖 + 𝜆𝑞𝑖 =
𝑞𝑖 ≤ 1. Therefore, 0 ≤ (1 − 𝜆) 𝑝𝑖 + 𝜆𝑞𝑖 ≤ 1.
Case 2. If 𝑞𝑖 ≤ 𝑝𝑖 , then the positions of 𝑝𝑖 and 𝑞𝑖 switch such that 0 ≤ 𝑞𝑖 =
(1 − 𝜆)𝑞𝑖 + 𝜆𝑞𝑖 ≤ (1 − 𝜆) 𝑝𝑖 + 𝜆𝑞𝑖 < (1 − 𝜆) 𝑝𝑖 + 𝜆𝑝𝑖 = 𝑝𝑖 ≤ 1 holds, which implies 0 ≤ (1 − 𝜆) 𝑝𝑖 + 𝜆𝑞𝑖 ≤ 1.
Thus, it is observed for any two points p, q ∈ 𝐶𝑑 that the 𝜆p + (1 − 𝜆)q for 𝜆 ∈ [0, 1] is
also in 𝐶𝑑 .
(2) Example of non-convex set: Any annular region of elliptic shape is not convex.
𝑦
𝑥𝑘
𝑥
𝑥𝑗
𝑦
(𝑐 2 , 𝑓 (𝑐 2 ))
𝑓 (𝑥) + ∇ 𝑓 𝑇 (𝑥)(𝑦 − 𝑥)
(𝑐 1 , 𝑓 (𝑐 1 ))
𝑥
𝑐 10
𝑐2
Figure 5. Non-convex set, Non-convex function and Convex function
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The line segment with endpoints is contained in the annulus but the the line segment
with endpoints 𝑥 𝑘 and 𝑥 𝑗 is not contained in the annulus.
Definition 1.1.30. A function 𝑓 : R𝑛 → R is convex if its domain is a convex set and
for all 𝑥, 𝑦 in its domain, and 𝑓 (𝜆𝑥 + (1−𝜆)𝑦) ≤ 𝜆 𝑓 (𝑥) + (1−𝜆) 𝑓 (𝑦), for all 𝜆 ∈ [0, 1] [8].
In words, if take any 2 points 𝑐 1 and 𝑐 2 , then 𝑓 evaluated at any convex combination of these 2 points should be no larger than the same convex combination of 𝑓 (𝑐 1 )
and 𝑓 (𝑐 2 ).
Geometrically, the line segment connecting (𝑐 1 , 𝑓 (𝑐 1 )) to (𝑐 2 , 𝑓 (𝑐 2 )) must sit above the
graph of 𝑓 , that is concave up which is 𝑓 00 > 0 everywhere.
Example 1.1.31. 𝑓 (𝑥) = 𝑒 𝑎𝑥 , −𝑙𝑜𝑔(𝑥), |𝑥| 𝑎 where 𝑎 ≥ 1
Theorem 1.1.32. Convexity along all lines[8]. A function 𝑓 : R𝑛 → R is convex if and
only if the function 𝑔 : R → R given by 𝑔(𝑡) = 𝑔(𝑥 + 𝑡𝑦) is convex for all 𝑥 in the domain
of 𝑓 and all 𝑦 ∈ R𝑛 .
Example 1.1.33. Multivariate convex functions [8] Given affine function 𝑓 (𝑥) = 𝑎𝑇 𝑥+𝑏
for any 𝑎 ∈ R𝑛 , 𝑏 ∈ R, the 𝑓 (𝑥) is convex and concave: 𝑓 (𝜆𝑥 + (1 − 𝜆)𝑦) = 𝑎𝑇 (𝜆𝑥 +
(1 − 𝜆)𝑦) + 𝑏 = 𝜆𝑎𝑇 𝑥 + (1 − 𝜆)𝑎𝑇 𝑦 + 𝜆𝑏 + (1 − 𝜆)𝑏 = 𝜆 𝑓 (𝑥) + (1 − 𝜆) 𝑓 (𝑦), where
𝑓 (𝑥) = 𝑎𝑇 𝑥 + 𝑏 and 𝑓 (𝑦) = 𝑎𝑇 𝑦 + 𝑏.
Exercise 2.4
The 𝑑-cube 𝐶𝑑 is the common intersection of 2𝑑 halfspaces in R𝑑
These halfspaces are {𝑥𝑖 ∈ R𝑑 , 𝑥𝑖 ≥ 0} and {𝑥𝑖 ∈ R𝑑 , 𝑥𝑖 ≤ 1}, where 𝑖 = 1, ..., 𝑑.
𝑑 ({𝑥 ∈ R𝑑 , 𝑥 ≥ 0} ∩ {𝑥 ∈ R𝑑 , 𝑥 ≤ 1}).
So, 𝐶𝑑 = ∩𝑖=1
𝑖
𝑖
𝑖
𝑖
(3) What is the the 𝐶3 the convex hull of and the 𝐶𝑑 the convex hull of :
0 1 1 0 0 1 1 0 !
© ª © ª © ª © ª © ª © ª © ª © ª
𝐶3 has vertices ­0® , ­0® , ­1® , ­1® , ­1® , ­1® , ­0® , ­0® is represented with
«0¬ «0¬ «0¬ «0¬ «1¬ «1¬ «1¬ «1¬
(
)
0
1
0
0 ∑︁
4
© ª
© ª
© ª
© ª
𝜆 1 ­0 ® + 𝜆 2 ­0 ® + 𝜆 3 ­1 ® + 𝜆 4 ­0 ® ,
𝜆𝑖 = 1, 𝜆𝑖 ≥ 0 =
«0 ¬
«0 ¬
«0 ¬
«1¬ 𝑖=1
conv ({𝜆 0 0 + 𝜆 1 e1 + 𝜆 2 e2 + 𝜆 3 e3 : 𝜆 1 , 𝜆2 , 𝜆3 ∈ {0, +1})
Similarly, the 𝐶𝑑 := conv({𝛼1 𝑒 1 + 𝛼2 𝑒 2 + · · · + 𝛼𝑑 𝑒 𝑑 | 𝛼1 , · · · , 𝛼𝑑 ∈ {0, +1}} for Vpolytope representation, and {(𝑥 1 , · · · , 𝑥 𝑑 ) ∈ R𝑑 | 0 ≤ 𝑥𝑖 ≤ 1, for 1 ≤ 𝑖 ≤ 𝑑} for
H -polytope representation [10].
Thus, the 𝐶3 is convex hull of e1 , e2 , e3 ∈ R3 , and the 𝐶𝑑 is convex hull of e1 , · · · , ed ∈ R𝑑 .
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Exercise 2.16
H -polygon representation of 𝐶𝑑4 [10].
- The H -polygon is the solution of finite system, 𝑃 = {x ∈ R𝑑 : 𝐴x ≤ b}, where 𝐴 is
𝑚 × 𝑑 matrix and the x consists of 𝑚 numbers of variables.
- The cross polytope in R𝑑 is the V-polytope of conv({+e1 , +e2 , · · · + ed , −e1 , −e2 , · · · −
Í𝑑
ed ) = {(𝑥 1 , · · · , 𝑥 𝑑 ) ∈ R𝑑 | 𝑖=1
|𝑥𝑖 | ≤ 1} for the H -polytope representation.
Comparison of representation
- V-polytope: for the convex hull of finite set 𝑋 = {x1 , · · · , xn } of points in R𝑑 ,
Í𝑛
Í𝑛
𝑃 = { 𝑖=1
𝜆𝑖 xi | 𝜆𝑖 ≥ 0, 𝑖=1
𝜆𝑖 = 1}.
- H -polytope: for bounded solution set of a finite system of linear inequalities,
𝑃 = 𝑃( 𝐴, 𝑏) = {x ∈ R𝑑 | 𝛼𝑖𝑇 x ≤ 𝑏𝑖 for 1 ≤ 𝑖 ≤ 𝑚} from 𝐴x ≤ 𝐵,
where the boundedness means there exists a constant 𝑁 such that ||x|| ≤ 𝑁 for all x ∈ 𝑃.
Exercise 2.19
Show if any inequality in the H -description of 4𝑑−1 is dropped, then it becomes unbounded polyhedron.
4𝑑−1 = {x ∈ R𝑑 : 𝑥 1 + 𝑥 2 + · · · + 𝑥 𝑑 = 1, 𝑥𝑖 ≥ 0} which is entirely hyperplane and also
has 𝑥𝑖 ≥ 0 such that this is a (𝑑 − 1)-dimensional polytope in R𝑑 .
In R2 , the 42−1 = {x ∈ R : 𝑥 1 + 𝑥 2 = 1, 𝑥1 and 𝑥 2 ≥ 0}. This defines 𝐻 − = {𝑥 1 + 𝑥2 ≤
1} ∩ 𝐻 + = {𝑥 1 + 𝑥 2 ≥ 1}, where 𝑥 1 , 𝑥2 ≥ 0.
Notice, if the 4𝑑−1 = {𝑥1 + 𝑥2 ≤ 1, 𝑥1 + 𝑥2 ≥ 1, 𝑥1 ≥ 0 𝑥 2 ≥ 0} drops the 𝑥 1 ≥ 0,
then {𝑥 1 + 𝑥2 ≥ 1, 𝑥1 + 𝑥 2 ≤ 1, −𝑥 2 ≤ 0} in R2 , for 𝐴x ≤ b. This unbounded polyhedron contain directional vector ∇c ∈ R2 in the set of 𝐴c ≤ 0, which forms a ray, and
𝐴(x + c) = 𝐴x + 𝐴c ≤ b + 0 = b.
4𝑑−1 = {𝑥 1 ≥ 0, . . . , 𝑥 𝑑 ≥ 0, 𝑥1 + · · · + 𝑥 𝑑 ≤ 1, 𝑥1 + · · · + 𝑥 𝑑 ≥ 1}
= {−𝑥 1 ≤ 0, . . . , −𝑥 𝑑 ≤ 0, 𝑥1 + · · · + 𝑥 𝑑 ≤ 1, −𝑥 1 + · · · − 𝑥 𝑑 ≤ −1}
If any inequality of equations 𝑑 + 2 (ie. one of 𝑥𝑖 ≥ 0 is deleted) is dropped, then we get
the unbounded polyhedron 𝐴x with the directional vector ∇c ∈ R2 that is 𝐴c ≤ 0, and
𝐴(x + c) = 𝐴x + 𝐴c ≤ b + 0 = b.
Exercise 2.24
Calculate the dimension of the polytope, 𝑃=conv({(1, 0, 0), (1, 1, 0), (1, 2, 1), (1, 1, 2), (1, 0, 1)}).
How many equations are to be expected in its H -representation?
By the definition of convexity,
)
1
1
1
1
1
5
© ª
© ª
© ª
© ª
© ª ∑︁
𝑃 = 𝜆 1 ­0® + 𝜆2 ­1® + 𝜆3 ­2® + 𝜆 4 ­1® + 𝜆 5 ­0® :
𝜆𝑖 = 1, 𝜆𝑖 ≥ 0
𝑖=1
0
0
1
2
1
« ¬
« ¬
« ¬
« ¬
« ¬
(
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Example 1.1.34. The conv({(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 0, 1), (0, 1, 1)} has
H -representation of {𝑥1 ≥ 0, 𝑥2 ≥ 0, 1 ≥ 𝑥 3 ≥ 0, 𝑥1 + 𝑥 2 ≤ 1}. The inequality of
0 ≤ 𝑥 3 ≤ 1 includes the bottom and upper triangular face of the polytope. Others
include 3 faces, in total 6 inequalities.
Each sides are formed by separate hyperplanes, but the parallel sides of ||(1, 1, 0), (1, 2, 1)||
and ||(1, 0, 1), (1, 1, 2)|| can be grouped by 1 inequality. Other sides each need hyperplane and 𝑥 1 = 1 is fixed. Hence, 5 equations are needed, 3 inequalities ||(1, 0, 0), (1, 1, 0)||,
||(1, 0, 0), (1, 0, 1)||, ||(1, 2, 1), (1, 1, 2)||, 1 parallel sides and 2 inequalities for 𝑥 1 = 1
of 𝑥 1 ≤ 1, 𝑥 ≥ 1.
The equations include {−1 ≤ 𝑥 3 − 𝑥 2 ≤ 1, 0 ≤ 𝑥 2 , 0 ≤ 𝑥 3 , 𝑥2 + 𝑥 3 ≤ 3, 𝑥1 ≤ 1, 1 ≤ 𝑥 1 }
The dimension of this matrix is (3 × 5).
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CHAPTER 1. INTRODUCTION TO GEOMETRIC COMBINATORICS
Chapter 2
Lecture on Discrete and Polyhedral
Geometry
2.1
2.1.1
Basic discrete geometry
The Helly theorem
In this chapter, we give definitions and some basic results about the Helly’s theorem
following [11].
Helly theorem in 2-dimension
Let 𝑋1 , · · · , 𝑋𝑛 ⊂ R2 be convex sets such that 𝑋𝑖 ∩ 𝑋 𝑗 ∩ 𝑋 𝑘 ≠ ∅, for every 1 ≤ 𝑖 < 𝑗 <
𝑘 ≤ 𝑛, where 𝑛 ≥ 3. There there exists a point 𝑧 ∈ 𝑋1 , ..., 𝑋𝑛 .
In other words, if all triples of convex sets intersect, then all sets intersect. The convexity
condition in the Helly theorem is necessary.
- Restate: If any triple convex sets intersect, then there exists a common intersection point 𝑧, where all sets intersect.
- Analogy: The idea of 3 non-collinear circles intersect at most 1 point, as this point
fixes all other 3 points of centers with respect to a motion of an isometry.
- Restriction: This case is only restricted to R2 for every 𝑛 = 3 points to intersect,
compared with the 𝑑-dimensional Helly’s theorem on every 𝑑 + 1 points to intersect,
which is not generalized.
Proof. We will use induction. First, we consider the base case.
i) When 𝑛=3, from the assumption, triples of convex sets have nonempty intersection,
so the result follows trivially.
ii) When 𝑛 = 4, the argument is divided into 2 cases.
From the assumption, we can find points 𝑣 1 ∈ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 , 𝑣 2 ∈ 𝑋1 ∩ 𝑋3 ∩ 𝑋4 , 𝑣 3 ∈
𝑋1 ∩ 𝑋2 ∩ 𝑋4 , 𝑣 4 ∈ 𝑋1 ∩ 𝑋2 ∩ 𝑋3 .
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Case 1. One of the 𝑣 𝑖 where 𝑖 = 1, 2, 3, 4 is in the the convex hull of the other 3.
Without loss of generality, let 𝑣 4 ∈ conv({𝑣 1 , 𝑣 2 , 𝑣 3 }). We already know 𝑣 4 ∈ 𝑋1 ∩
𝑋2 ∩ 𝑋3 , so if 𝑣 4 ∈ 𝑋4 , we satisfy the desired property.
𝑣3
𝑣1
𝑣2
𝑣4
Figure 1. Illustration of Case 1.
Notice, 𝑣 1 , 𝑣 2 , 𝑣 3 ∈ 𝑋4 , where the 𝑋4 is convex. This gives conv({𝑣 1 , 𝑣 2 , 𝑣 3 }) ⊆ 𝑋4 , so
4 𝑋 , the
𝑣 4 ∈ con({𝑣 1 , 𝑣 2 , 𝑣 3 }) implies that 𝑣 4 ∈ 𝑋4 . Hence, this implies the 𝑣 4 ∈ ∩𝑖=1
𝑖
non-empty common intersection 𝑣 4 .
Case 2. If 𝑣 1 , 𝑣 2 , 𝑣 3 , 𝑣 4 are in convex position, then none of the points are contained in
the convex hull of the other 3 points. This means none of the triangles formed by any 3
points contain the other.
Without loss of generality, assume they are oriented 𝑣 1 , 𝑣 4 , 𝑣 2 , 𝑣 3 in a counterclockwise
manner. We claim that the intersection point 𝑧 = 𝑣 1 𝑣 2 ∩𝑣 3 𝑣 4 satisfies 𝑧 ∈ 𝑋1 ∩𝑋2 ∩𝑋3 ∩𝑋4 .
Since 𝑣 1 , 𝑣 2 ∈ 𝑋3 ∩ 𝑋4 , the line segment 𝑣 1 𝑣 2 ⊆ 𝑋3 ∩ 𝑋4 by convexity of 𝑋3 (are
result). Hence, 𝑧 ∈ 𝑋3 ∩ 𝑋4 . Similarly, since 𝑣 3 , 𝑣 4 ∈ 𝑋1 ∩ 𝑋2 , then 𝑣 3 𝑣 4 ⊆ 𝑋1 ∩ 𝑋2 .
Hence, similarly 𝑧 ∈ 𝑋1 ∩ 𝑋2 .
4 𝑋 ≠ ∅.
Thus, 𝑧 ∈ 𝑋1 ∩ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 , implying ∩𝑖=1
𝑖
𝑣3
𝑣1
𝑣2
𝑧
𝑣4
Figure 2. Illustration of Case 2.
iii) Induction step,
Define 𝑋 𝐼 for any subset 𝐼 ∈ [𝑛]. Assume for 𝑛 > 4, the subset 𝐼 is a set of arbitrary (𝑛 − 1) elements picked from the set of 𝑛-elements, {1, 2, ..., 𝑛}. Then, define the 𝑋 𝐼 = ∩𝑖∈𝐼 𝑋𝑖 to be the intersection for any set 𝐼 with cardinality (𝑛 − 1) (ie.
{1, 2, ..., 𝑛 − 1} ⊂ 𝐼, 𝑋𝐼 = ∩𝑖∈𝐼 𝑋𝑖 ).
Step 1. Assume for 𝑋 𝐼 ≠ ∅ for 𝐼 ⊂ [𝑛] = {1, 2, ..., 𝑛}, |𝐼 | = 𝑛 − 1. That is, any
collection of (𝑛 − 1) sets have a nonempty intersection.
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Step 2. For each 𝑖 ∈ [𝑛], write 𝑣 𝑖 ∈ 𝑋 [𝑛]−𝑖 to denote some point in the intersection
of all sets except 𝑋𝑖 (which is not arbitrary but for all 𝑖 = 1, · · · , 𝑛).
From the argument in ii), there are 2 possibilities: these 4 points are in convex position
or not.
Step 3. First, assume 4 points are in the convex position.
Suppose 𝑣 1 ∈ 𝑋 [𝑛]−1 , 𝑣 2 ∈ 𝑋 [𝑛]−2 , 𝑣 3 ∈ 𝑋 [𝑛]−3 and 𝑣 4 ∈ 𝑋 [𝑛]−4 . Similarly, let
𝑧 = 𝑣 1 𝑣 3 ∩ 𝑣 2 𝑣 4 . Then, 𝑧 ∈ 𝑋1 ∩ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 . Since 𝑧 ∈ 𝑣 1 𝑣 3 = conv({𝑣 1 , 𝑣 3 })
and 𝑣 1 , 𝑣 3 ∈ 𝑋2 ∩ 𝑋4 ∩ · · · ∩ 𝑋𝑛 , 𝑧 ∈ 𝑋2 ∩ 𝑋4 · · · ∩ 𝑋𝑛 . Also, 𝑧 ∈ 𝑣 2 𝑣 4 =conv({𝑣 2 , 𝑣 4 })
sucht that 𝑧 ∈ 𝑋1 ∩ 𝑋3 ∩ · · · ∩ 𝑋𝑛 .
𝑛 .
Notice, 𝑧 ∈ 𝑋2 ∩ 𝑋4 and 𝑧 ∈ 𝑋1 ∩ 𝑋3 such that 𝑧 ∈ 𝑋 [𝑛] = ∩𝑖=1
Otherwise, we already took 𝑣 4 ∈ 𝑋 [𝑛]−4 for fixed 𝑣 4 . Notice, 𝑣 1 , 𝑣 2 , 𝑣 3 ∈ 𝑋 [𝑛]−{1,2,3}
because 𝑣 1 ∈ 𝑋 [𝑛]−1 , 𝑣 2 ∈ 𝑋 [𝑛]−2 , 𝑣 3 ∈ 𝑋 [𝑛]−3 . Without a loss of generality, the 𝑣 4 ∈
𝑛 𝑋 = 𝑋 . conv({𝑣 1 , 𝑣 2 , 𝑣 3 }) ⊂ 𝑋4 of a convex set for missing 𝑋𝑖 such that 𝑣 4 ∈ ∩𝑖=1
𝑖
[𝑛]
Thus, we conclude that in each case there exists a point 𝑧 ∈ 𝑋 [𝑛] .
CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
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Problem 1.
If 𝑋 ⊆ R𝑑 is convex and x1 , x2 , ..., xn ∈ 𝑋, then conv({x1 , ..., xn }) ⊆ 𝑋. The goal is to
show that for all xi ∈ 𝑋 the convex hull of the set of points is a subset of 𝑋.
Let 𝑋 ⊆ R𝑑 be convex, and suppose x1 , x2 , ..., xn ∈ 𝑋. We must show all convex
Í𝑛
Í𝑛
combinations 𝑖=1
𝜆𝑖 x𝑖 for 𝑖=1
𝜆𝑖 = 1, 𝜆𝑖 ≥ 0 must be in 𝑋 as well.
For the proof, the approach is to use induction on the number of points x𝑖 in the
convex hull.
Step 1. For 𝑛 = 1, the convex hull of the vector x1 is just the set {x1 }, equivalently
conv({x1 }) = {x1 }, which is clearly a subset of 𝑋.
Next, we check the base case for two points, x1 and x2 in 𝑋.
Remark 2.1.1. The definition of convex set is, for any x1 and x2 in 𝑋, the entire line
segment joining the two points {𝜆 1 x1 + 𝜆 2 x2 , 𝜆1 + 𝜆 2 = 1, 𝜆1 + 𝜆 2 ≥ 0}, is contained in
the set 𝑋.
Remark 2.1.2. The convex hull, conv({x1 , x2 }), is the set of all convex combinations
of points in {x1 , x2 }.
Hence, this refers the results that all convex combinations must be in 𝑋.
The convex hull conv({x1 , x2 }) contains all points 𝜆 1 x1 +𝜆2 x2 where 𝜆 1 +𝜆 2 = 1, 𝜆1 , 𝜆2 ≥
0.
Notice, the convex combination x1 , x2 , equivalently let z = 𝜆x1 + (1−𝜆)x2 . By definition
of convexity, the line connecting x and y lies entirely within 𝑋 as an interval, [x, y] =
{𝜆x + 1 − 𝜆y : 0 ≤ 𝜆 ≤ 1} ⊂ 𝑋 . Hence, z are in 𝑋 where z ∈ [x1 , x2 ] [4, p.2].
Case study
Before the generalization using the induction, we must find how the convex combination
of points is obtained. This is to figure out how the last point is added, where the convex
combination holds.
Notice, the point is to generalize how the convex hull of 𝑛 points takes into account
of the convex hull of 𝑛 − 1, with a case example of any 2 points convex combination to
Í3
3 points, or conv({x1 , x2 , x3 }) = { 𝑖=1
𝜆𝑖 xi }.
When 𝜆 3 is 0, we just get the line segment connecting x1 , x2 and the x3 is derived to be
connected with other two points when 𝜆 ≠ 0 where it shares portion of parametrization
from existing line segment between x1 and x2 .
[strategy] By definition, the conv({x1 , x2 , x3 }) contains all vectors of 𝜆 1 x1 + 𝜆 2 x2 + 𝜆 2 x3 ,
from a combination of x1 , x2 such that third constant 𝜆 3 could be found with deficiency
of 1 − 𝜆 1 − 𝜆 2 = 𝜆 3 .
[idea] It is possible to position the third point with a manipulation of the equation
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from x1 and x2 to form the 𝜆 1 + 𝜆 2 + 𝜆 3 = 1 by the definition of the convex set as for the
parametrization of points.
x1
x3
𝜆 1 x1 + 𝜆 2 x2
x2
Figure 3. An Illustration of the third point x3 connected with x1 and x2 .
Therefore, the goal to connect the third point to the convex set of 2 points.
𝜆2
𝜆1
Notice, for the goal, rearrange the eexpression x1 + x2 .
1
1
𝜆1
𝜆2
Equivalently, 𝜆1 x1 + 𝜆 2 x2 = (𝜆1 + 𝜆 2 )
x1 +
x2 for 𝜆 1 and 𝜆 2
𝜆1 + 𝜆2
𝜆1 + 𝜆2
𝜆1
𝜆2
= (1 − 1 + 𝜆 1 + 𝜆 2 )
x1 +
x2
𝜆1 + 𝜆2
𝜆1 + 𝜆2
The point defined above represents the convex set, {𝜆 1 x1 + 𝜆 2 x2 }, but the rearrangement
delivers for 𝜆 3 to connect with the third point x3 Now, from the relocation of x3 for the
𝜆 3 takes the remaining portion to connect to the other x1 , x2 .
Start with 𝜆 1 x1 + 𝜆 2 x2 + 𝜆 3 x3 ,
3
∑︁
𝑖=1
𝜆𝑖 = 1, for all
3
∑︁
𝜆𝑖 x𝑖 is convex hull of 𝑋.
𝑖=1
𝜆1
𝜆2
x1 +
x2 + 𝜆 3 x3 , as 𝜆 3 = 1 − 𝜆 1 − 𝜆 2 .
= (1 − 𝜆 3 )
1 − 𝜆3
1 − 𝜆3
𝜆1
𝜆2
Notice that
x1 +
x2 is in conv ({x1 , x2 }) ⊂ 𝑋
1 − 𝜆3
1 − 𝜆3
𝜆2
𝜆1
x1 +
x2 + (1 − 𝜆 1 − 𝜆 2 )x3 , as 1 − 𝜆 1 + 𝜆2 ≥ 0
= {1 − (1 − 𝜆 1 − 𝜆 2 )}
𝜆1 + 𝜆2
𝜆1 + 𝜆2
𝜆1
𝜆2
= (𝜆 1 + 𝜆 2 )
x1 +
x2 , when 𝜆 3 = 1 − 𝜆 1 − 𝜆 2 = 0.
𝜆1 + 𝜆2
𝜆1 + 𝜆2
Define it to be y such that (1 − 𝜆 3 )y + 𝜆 3 x3 ∈ 𝑋.
Clearly, the set of points {(1 − 𝜆 3 )y + 𝜆 3 x3 } ⊆ 𝑋 for 0 ≤ 𝜆 3 ≤ 1.
- Obviously, in the case of 𝜆 3 = 0, we notice that the line segment, {𝜆 1 x1 + 𝜆 2 x2 :
𝜆1 + 𝜆 2 = 1, 𝜆1 , 𝜆2 ≥ 0} are included.
𝜆1
𝜆2
- In the (1 − 𝜆 3 ) 1−𝜆3 x1 + 1−𝜆3 x2 + 𝜆3 x3 , when 𝜆 3 < 1 represents the convex combination for convex hull, which leads to the subset of convex hull of 𝑋.
The process illustrates (1) generating all convex combinations of x1 , x2 , x3 , is equivalent
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to (2) generating all line segments with x3 at one endpoint and a point in conv({x1 , x2 })
at the others for fixed x3 .
As the base case 𝑛 = 2 holds, the induction hypothesis holds for 𝑛 − 1 of distinct
xi ∈ 𝑋.
Í𝑛
For any finite set of vectors x1 , ..., xn with 𝑖=1
𝜆𝑖 = 1 where 𝜆𝑖 ≥ 0 in 𝑋, the
conv({x1 , ..., xn }) contains all convex combinations,
𝜆 𝑛−1
𝜆1
x1 + · · ·
xn + 𝜆 𝑛 xn
𝜆 1 x1 + 𝜆 2 x2 + · · · 𝜆 𝑛 xn = (1 − 𝜆 𝑛 )
1 − 𝜆𝑛
1 − 𝜆𝑛
!
𝑛−1
𝑛−1
∑︁
∑︁
𝜆𝑖
𝜆𝑖
xi + 𝜆 𝑛 xn . and let z =
xi
= (1 − 𝜆 𝑛 )
1
−
𝜆
1
−
𝜆
𝑛
𝑛
𝑖=1
𝑖=1
This is clearly illustrate a convex combination of 𝑛−1 vectors from 𝑋 since the coefficients
satisfy the conditions for it to be a convex combination.
𝑛−1
∑︁
𝑖=1
𝜆𝑖
𝜆 1 + 𝜆 2 · · · + 𝜆 𝑛−1
=
= 1, where 1−𝜆 𝑛 = 𝜆 1 +· · ·+𝜆 𝑛−1 , for all 𝜆𝑖 ≥ 0 equals to 1.
1 − 𝜆𝑛
1 − 𝜆𝑛
By the induction hypothesis, z ∈ 𝑋.
Therefore, (1 − 𝜆 𝑛 )z + 𝜆 𝑛 xn ∈ 𝑋, since z and xn ∈ 𝑋 by the convexity of 𝑋.
Thus, conv{(x1 , ..., xn }) ⊆ 𝑋.
Problem 2. The finite intersection of convex sets is convex.
𝑛 𝑋 is convex.
The goal is to show that if 𝑋𝑖 is convex for 𝑖 = 1, ..., 𝑛 then ∩𝑖=1
𝑖
Proof. The approach for the proof is to use the induction where the base case is 𝑛 = 2.
For 2 sets 𝑋1 and 𝑋2 , we can assume that there exists at least 2 points in the intersection,
𝑋1 ∩ 𝑋2 .
Otherwise, if 𝑋𝑖 ∩ 𝑋2 is a point or the empty set, it is convex and the result follows.
Then, take any 2 points p and q in the intersection 𝑋1 ∩ 𝑋2 , and define 𝑋¯ to be the
line segment {𝜆𝑝 + (1 − 𝜆)𝑞 : 𝜆 ∈ [0, 1]}. The idea is that for any 2 points p and q
picked in the intersection, the line segment connecting them is also in the intersection.
By convexity, 𝑋¯ ⊆ 𝑋1 and 𝑋¯ ⊆ 𝑋2 , which together imply 𝑋¯ ⊆ 𝑋1 ∩ 𝑋2 . Thus,
since p and q were chosen arbitrary from 𝑋1 ∩ 𝑋2 , this shows that 𝑋1 ∩ 𝑋2 is convex.
𝑛 𝑋 = 𝑋 ∩ 𝑋 ∩ 𝑋 ∩ · · · 𝑋 = {[(𝑋 ∩ 𝑋 ) ∩ 𝑋 ] ∩ · · · 𝑋 }.
For finitely many 𝑋𝑖 , ∩𝑖=1
𝑖
1
2
3
𝑛
1
2
3
𝑛
We just showed that, if 𝑋 and 𝑌 are convex, then 𝑋 ∩ 𝑌 is convex.
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The inductive hypothesis is that the claim works for 𝑛 − 1: in particular, that the
intersection of 𝑛 − 1 convex sets is also convex.
Ñ𝑛
Now take 𝑛 convex sets 𝑋1 , ..., 𝑋𝑛 . We write their common intersection as 𝑖=1
𝑋𝑖 =
(𝑋1 ∩ 𝑋2 ∩ · · · ∩ 𝑋𝑛−1 ) ∩ 𝑋𝑛 . By the induction hypothesis, the set 𝑋1 ∩ 𝑋2 ∩ · · · ∩ 𝑋𝑛−1
is convex.
Now, apply the base case to the set 𝑌 = 𝑋1 ∩ 𝑋2 ∩ · · · ∩ 𝑋𝑛−1 (which is convex by
the previous statement) and 𝑋𝑛 (which is a convex set by the assumption), to get the
Ñ𝑛
desired result, as (𝑋1 ∩ 𝑋2 ∩ · · · ∩ 𝑋𝑛−1 ) ∩ 𝑋𝑛 = 𝑖=1
𝑋𝑖 is convex.
𝑛 𝑋 is contained in the convex set 𝑋 for 𝑖 = 1, ..., 𝑛, as the prior condiNotice, ∩𝑖=1
𝑖
𝑛
𝑛−1 𝑋 and 𝑋 is both convex where ∩𝑛−2 𝑋 and 𝑋
tion that ∩𝑖=1
𝑛−1 is both convex must be
𝑖
𝑛
𝑖=1 𝑖
𝑛
satisfied. Without loss of generality, the set of ∩𝑖=1 𝑋𝑖 contained in all 𝑋𝑖 from 1 to 𝑛
must be also convex.
Thus, the intersection of finitely many convex sets is convex.
Remark 2.1.3. The Helly’s theorem does not always work for an infinite collection of
sets. To show this, consider the collection of halfspaces.
Goal: Fix a point in any triple halfspaces that are in the intersection, but the point is
not contained in the intersection of all halfspaces for infinite collection.
Proof. Let 𝑋𝑖 = {(𝑥, 𝑦) : 𝑥 ≥ 𝑖}, 𝑋 𝑗 = {(𝑥, 𝑦) : 𝑥 ≥ 𝑗 }, 𝑋ℎ = {(𝑥, 𝑦) : 𝑥 ≥ ℎ} be
3 distinct halfspaces, chosen arbitrary with the inequality 𝑖 < 𝑗 < ℎ in R2 . Notice,
𝑋𝑖 ⊂ 𝑋 𝑗 ⊂ 𝑋ℎ .
Then, there exists some 𝑘 ∈ R where 𝑋𝑖 ∩ 𝑋 𝑗 ∩ 𝑋ℎ ⊇ {(𝑥, 𝑦) : 𝑥 ≥ 𝑘 } defined as
𝑋 𝑘 for 𝑘 ≤ ℎ. Now, fix a point in 𝑋 𝑘 . There exists some halfspaces that does not include
the fixed point in 𝑋 𝑘 .
Due to ordering of the metric space, such 𝑋𝑙 for 𝑙 < 𝑘 that exists and is contained
∞ 𝑋 = ∅, as it does not include
in 𝑋 𝑘 does not include the fixed point in 𝑋 𝑘 . Hence, ∩𝑖=1
𝑖
the fixed point in 𝑋 𝑘 = {(𝑥, 𝑦) : 𝑥 ≥ 𝑘 } (ie. for 𝑛 < 𝑘).
Hence, this is a counterexample for the Helly’s theorem on infinite sets.
However, the compact sets is sufficient, bounded and closed set, that satisfies the
condition for infinite collection of intersection to be non-empty.
Each of the following example illustrates a counterexample in which modified condition
is impossible to be hold.
False assertion 1. For any infinite collection of convex, closed sets 𝑋1 , 𝑋2 , · · · ∈ R𝑑 , if
the intersections of every 𝑑 + 1 of these sets is nonempty, then ∩∞
𝑗=1 𝑋 𝑗 ≠ ∅.
The counter-example is when 𝑑 = 1, the set of 𝑋𝑖 = R \ (−∞, 𝑖), 𝑖 ∈ N (complement
being closed).
Also, it is observed for any two 𝑋𝑖 ’s have a non-empty intersection because the 𝑋𝑖 ⊂ 𝑋 𝑗
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∞ 𝑋 = ∅.
if 𝑖 > 𝑗 such that ∩𝑖=1
𝑖
False assertion 2. For any infinite collection of convex, bounded subsets 𝑋1 , 𝑋2 , · · · ∈
R𝑑 , if the intersection of every 𝑑 + 1 of these sets is nonempty, then ∩∞
𝑗=1 𝑋 𝑗 ≠ ∅.
When 𝑑 = 1, 𝑋𝑖 = (0, 1𝑖 ), where ∈ N the 𝑋𝑖 are clearly bounded and any 2 𝑋𝑖 ’s have
nonempty intersection because the 𝑋𝑖 ⊂ 𝑋 𝑗 if 𝑖 > 𝑗 such that ∩∞
𝑗=1 𝑋 𝑗 ≠ ∅.
Excecise 1.1 Infinite Helly theorem
Claim: [13] For any infinite collection of convex, compact subsets 𝑋1 , 𝑋2 , · · · ∈ R𝑑 , if
the intersection of every 𝑑 + 1 is nonempty, then
Ù
𝑋𝑖 ≠ ∅.
𝑖→∞
In the 𝑑-dimensional Helly’s theorem, for any 𝑛 convex sets 𝑋1 , ..., 𝑋𝑛 with every 𝑑 + 1
of them having a nonempty intersection in R𝑑 , then there exists some point 𝑥 𝑛 such that
𝑥𝑛 ∈
𝑛
Ù
𝑋𝑖 .
𝑖=1
Assumption. For arbitrary family of compact sets, each of the finite families of has
non-empty intersection points, which is 𝑍𝑖 = {𝑥𝑖 , 𝑥𝑖+1 , ...} (among uncountably infinite
sets of 𝑋𝑖 ).
Step 1. For all 𝑗 ∈ N, 𝑗 > 𝑖, 𝑥 𝑗 ∈ 𝑋𝑖 because 𝑥 𝑗 is a point in the intersection of a
set of sets, one of which is 𝑋𝑖 . This means for all 𝑖 ∈ N, 𝑍𝑖 ⊂ 𝑋𝑖 . As 𝑋𝑖 is defined to be
compact (closed and bounded), so its subset 𝑍𝑖 also has to be bounded.
Simply, the compactness of arbitrary set family of compact set, delivers the property that the finite subfamilies to have a non-empty intersection (which is true by 𝑑 + 1
assumption), then the whole family of 𝑋𝑖 has a non-empty intersection [14], p.97.
Step 2. The 𝑍𝑖 is infinite and bounded, which implies that 𝑍𝑖 has at least one limit
point, due to the fact any infinite bounded set has a limit point (from Bolzano Weierstrass theorem, every bounded infinite set of real numbers has at least one limit point).
Step 3. Observe if 𝑖 > 𝑗 ∈ N, then 𝑍𝑖 ⊂ 𝑍 𝑗 which means that the set of limit
points of 𝑍𝑖 is a subset of the set of limit points of 𝑍 𝑗 (eg. Subsequences of a convergent
sequence converge to the same limit as the original sequence [15]).
Since every 𝑍𝑖 has at least 1 limit point, this means for every 𝑖 > 𝑗, 𝑍𝑖 shares at
least one limit points with 𝑍 𝑗 . Also, for all 𝑘 < 𝑗 ∈ N, 𝑍 𝑗 ⊂ 𝑍 𝑘 , and this implies that all
the limit points of 𝑍 𝑗 are also limit points of 𝑍 𝑘 , which means for every 𝑘 < 𝑗, 𝑍 𝑘 and
𝑍 𝑗 share a limit point (𝑘 < 𝑗 < 𝑖, 𝑍𝑖 ⊂ 𝑍 𝑗 ⊂ 𝑍 𝑘 ).
Step 4. Since this is all true for all 𝑗 ∈ N, all points in 𝑍𝑖 share a limit point, 𝑞,
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where at least one of 𝑍𝑖 is compact.
Simply, for uncountable families, the intersection of infinite numbers of closed sets, as
𝑍𝑖 ⊂ 𝑍 𝑗 ⊂ 𝑍 𝑘 , intersect for some countable sub-collection of 𝑋𝑖 s by the limit point of 𝑞.
Step 5. Now, as 𝑍𝑖 ⊂ 𝑋𝑖 , the limit points of 𝑍𝑖 are also limit points of 𝑋𝑖 . This
means that 𝑞 is the limit point of all the 𝑋𝑖 ’s as it is a limit point for all 𝑍𝑖 ’s.
Since all 𝑋𝑖 ’s are closed, for every 𝑖 ∈ N, 𝑞 ∈ 𝑋𝑖 ,
∞
Ù
𝑋𝑖 ≠ ∅
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Application questions for the Helly theorem
Application 1. Suppose 𝑛 distinct points, 𝑥 1 , 𝑥2 , ..., 𝑥 𝑛 are given, and any 3 points,
𝑥𝑖 , 𝑥 𝑗 , 𝑥 𝑘 can be contained in a disk of radius 𝑟, defined Γ𝑖 𝑗 𝑘 .
Then, there is the congruent disk of radius 𝑟, which contains all 𝑛 points.
The Goal is to use the Helly theorem to prove that there is a disk of radius 𝑟 centered at
𝑝, which is the common intersection, contains all points, 𝑥 1 , ..., 𝑥 𝑛 .
Notation: Suppose for 𝑛 finitely many points there is a disk of radius 𝑟 centered at
each 𝑥𝑖 points, denoted as 𝑌𝑖 = 𝐵(𝑥𝑖 , 𝑟). Then, these convex set of 𝑌𝑖 is contained in the
set of disks with each 𝑥𝑖 to be the center, 𝐵(𝑥𝑖 , 𝑟).
Now, by the assumption, for any 3 points each forms a disk of 𝐵(𝑥𝑖 , 𝑟), points 𝑥𝑖 , 𝑥 𝑗
and, 𝑥 𝑘 are contained in the disk with same radius 𝑟, Γ𝑖 𝑗 𝑘 . Hence, all 4 disks with same
radius 𝑟, all 𝐵(𝑥𝑖 , 𝑟), 𝐵(𝑥 𝑗 , 𝑟), 𝐵(𝑥 𝑘 , 𝑟) intersect inside the Γ𝑖 𝑗 𝑘 . This must hold, as the
distance of any 2 centers for 𝑥𝑖 , 𝑥 𝑗 and 𝑥 𝑘 is at most the length of the radius 𝑟.
Notice, due to the restriction of the radius 𝑟, 3 disks with radius 𝑟, 𝐵(𝑥𝑖 , 𝑟), 𝐵(𝑥 𝑗 , 𝑟), 𝐵(𝑥 𝑘 , 𝑟),
must at least intersect on a point 𝑂, which is the center of the Γ𝑖 𝑗 𝑘 .
𝑥𝑘
𝑥𝑗
𝑥𝑖
1
2𝑟
𝑂
Γ𝑖 𝑗 𝑘
(This is just a case study for one of the possibility of placement for 4 disks with same
radius)
The case of intersection for the center 𝑂 is shown to comprehend how 3 disks 𝐵(𝑥𝑖 , 𝑟), 𝐵(𝑥 𝑗 , 𝑟), 𝐵(𝑥 𝑘 , 𝑟)
and Γ𝑖 𝑗 𝑘 intersect, as 𝐵(𝑥𝑖 , 𝑟) ∩ 𝐵(𝑥 𝑗 , 𝑟) ∩ 𝐵(𝑥 𝑘 , 𝑟) ∩Γ𝑖 𝑗 𝑘 ≠ ∅ holds for the Helly theorem
to apply.
Now, the Helly theorem is applied to the set of 𝑌𝑖 for 𝑌𝑖 ∩ 𝑌 𝑗 ∩ 𝑌𝑘 ≠ ∅ such that
there exists a point of intersection 𝑝 ∈ 𝐵(𝑥𝑖 , 𝑟) for all 𝑖 = 1, ..., 𝑛. Then, the point 𝑝
belongs to every disk in B(𝑥𝑖 , 𝑟).
Therefore, by the existence of the 𝑝 at a distance less than 𝑟 from each centers 𝑥𝑖 ,
all the centers of the disks 𝑥𝑖 from all B(𝑥𝑖 , 𝑟) lie inside the disk of radius 𝑟 centered at
𝑝.
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Application 2. Jung’s theorem
2
Remark 2.1.4. Suppose
√ there are 𝑛 points in R , 𝑥 1 , ..., 𝑥 𝑛 , and the distance between any
2 points is at most 3. Then, there is a disk of radius 1, which contains all 𝑛 points.
From the consequence of the Application 1, for any 3 points 𝑥𝑖 , 𝑥 𝑗 , 𝑥 𝑘 if all 3 points are
contained in the Γ1 with a disk of radius 1, where 𝐵(𝑥𝑖 , 1) ∩ 𝐵(𝑥 𝑗 , 1) ∩ 𝐵(𝑥 𝑘 , 1) ≠ ∅,
then the Helly theorem is applied to the set of disks, B(𝑥𝑖 , 1) for all 𝑖 = 1, .., 𝑛.
𝑛 𝐵(𝑥 , 1) = 𝑝, all 𝑛 points is contained in the congruent disk of radius 1
Hence, as ∩𝑖=1
𝑖
with respect to the common intersection point, 𝑝.
Hence, it remains for
√ the goal to show, for any 3 points, 𝑥𝑖 , 𝑥 𝑗 , 𝑥 𝑘 if the distance between
any two is at most 3, then these 3 points is contained in the disk of radius 1, defined as
Γ1 .
The position of 3 points is best classified with the shape of the triangle, 𝑇 = (𝑥𝑖 , 𝑥 𝑗 , 𝑥 𝑘 ),
in terms of lengths between points, as 3 cases; acute, right and obtuse.
Case 1. Simply, for the obtuse
√ and right angle triangle, take the hypotenuse of the
𝑇, which has a length of max 3, and set the midpoint of hypotenuse√to be the center,
𝑂 of the Γ1 to include all 3 points. Clearly, the radius√of Γ1 is at most 3/2 < 1, as the
center takes the half of diameter of the max distance, 3.
To check the third point other than endpoints of diameter is inside the circle, denote
this third point as 𝑥 𝑗 and ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 to the be angle that is greater than or equal to 𝜋/2, but
never less.
If the ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 = 𝜋/2 then the point 𝑥 𝑗 on the arc of the circle, where the 𝑇 is inscribed in the circle. Otherwise, the 𝑥 𝑗 is inside the circle Γ1 when ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 > 𝜋/2. The
other 2 angles are less than 𝜋/2 such that two sides of 𝑥𝑖 𝑥 𝑗 and 𝑥 𝑗 𝑥 𝑘 intersect inside the
circle.
The hypotenuse of the 𝑇 is obtained as a diameter for obtuse angle such that the largest
side of the 𝑇 is taken for the circumcircle to contain 𝑇.
Case 2. The case is left with the acute angle case, has a circumscribed disk of radius 1, Γ1 with the center 𝑂.
The goal to show is, if the acute triangle that is circumscribed in a circle Γ1 (as shown
below), then the upper bound of the circle Γ1 is less than 1.
Notation: Hence, assume 𝑥𝑖 𝑥 𝑗 to be the largest side, followed from the largest angle ∠𝑥 𝑗 𝑂𝑥𝑖 in 𝑇. Let the point 𝑝 be the perpendicular bisector from the center 𝑂 to
the 𝑥 𝑗 𝑥𝑖 .
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𝑥𝑗
𝑝
𝑂
𝛽
𝑥𝑘
𝑥𝑖
Since the 2𝜋 is divided by 3 angles, ∠𝑥𝑖 𝑂𝑥 𝑗 , ∠𝑥𝑖 𝑂𝑥 𝑘 and ∠𝑥 𝑘 𝑂𝑥 𝑗 , and the largest
angles
√
is ∠𝑥 𝑗 𝑂𝑥𝑖 such that it is over
√
assumption |𝑥𝑖 𝑥 𝑗 | ≤ 3.
2𝜋 1
3 , 2 ∠𝑥 𝑗 𝑂𝑥𝑖
= 𝛽 ≥ 𝜋3 . Also, |𝑥𝑖 𝑝| = 21 |𝑥𝑖 𝑥 𝑗 | ≤
Hence, the radius of Γ1 is defined to be |𝑥𝑖 𝑂| =
√
nator is minimized, sin(𝛽) ≥ 23 =sin( 𝜋3 ),
√
√
3/2
3
√
2 such that the radius of Γ1 ≤ 3/2 = 1.
|𝑥 𝑖 𝑝|
𝑠𝑖𝑛(𝛽)
3
2 ,
by the
is maximized when denomi-
and numerator is maximized which is |𝑥𝑖 𝑝| =
Therefore, as the acute triangle is tight√on bound of the radius of 1 as desired, any triangle with its maximum side length of 3 is covered by the disk of radius 1, so called Γ1 .
For any 3 points 𝑥𝑖 , 𝑥 𝑗 and 𝑥 𝑘 , each point is a center of a disk of radius 1, contained in the
disk Γ1 such that all 3 disks has nonempty intersection, √
𝐵(𝑥𝑖 , 1)∩𝐵(𝑥 𝑗 , 1)∩𝐵(𝑥 𝑘 , 1) ≠ ∅,
as the maximum distance between 2 points is at most 3.
Now, the Helly theorem is applied to convex sets of disk, 𝐵(𝑥𝑖 , 1) for 𝑛 distinct points,
𝑛 𝐵(𝑥 , 1) = 𝑝. Then, the point 𝑝 is contained in all 𝐵(𝑥 , 1) with its centers
such that ∩𝑖=1
𝑖
𝑖
𝑥𝑖 distanced less than 1 such that the disk of radius 1 centered at 𝑝 contains all 𝑛 points.
2.1.3
Relevant extra questions
Application question 2: In case the triangle of 3 points is right or obtuse, the goal is
to show the opposite of the hypothesis is guaranteed to be inside the circle formed.
Proof. First, the endpoints are inscrbed in the circle Γ1 of radius
1 or less, as to take
√
the hypotenuse of the 𝑇 = (𝑥𝑖 , 𝑥 𝑗 , 𝑥 𝑘 ) the maximum length of 3 where the midpoint of
hypotenuse is a center, 𝑂 of the Γ1 to include the endpoints, which is 𝑥𝑖 and 𝑥 𝑘 .
Clearly, the radius of Γ1√is at most
eter of the max distance, 3.
√
3/2 < 1, as the center takes the half of diam-
Now, in the case where ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 is obtuse, the other two angles has less portion of
𝜋 − ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 . As the claim clearly states that the side 𝑥𝑖 𝑥 𝑘 is the largest side for ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘
to be obtuse, it is left to check when ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 is right angle.
Notice, if the third point is positioned in any place that forms ∠𝑥 𝑗 𝑥𝑖 𝑂 of a right angle, then given the 𝑥 𝑗 𝑥 𝑘 is a diameter the 𝑇 is circumscribed in the circle Γ𝑖 𝑗 𝑘 with the
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center being fixed.
With the opposite direction, there is Thales’s theorem that states that if two points
are the diameter, and the third point are distinct points on a circle, then the angle within
the third point is a right angle.
Even though the Thales’s theorem is not applicable, the assumption to circumscribe the
𝑇 in a circle with a diameter of 𝑥𝑖 𝑥 𝑘 lead to an implication that the point 𝑥 𝑗 which forms
an angle of 𝜋/2 to be inscribed on the circumference of the Γ𝑖 𝑗 𝑘 .
Hence, the ∠𝑥 𝑗 𝑥𝑖 𝑂 is right-angled.
𝑥𝑗
𝑥𝑖
𝑥𝑘
𝑂
Figure: Illustration of Γ1 and 𝑇 with 3 points
Thus, all 3 points 𝑥𝑖 , 𝑥 𝑗 and 𝑥 𝑘 is included in the Γ1 when the 𝑇 to be obtuse or right-angle
triangle, where the ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 is chosen to be larger than or equal to an angle of 𝜋/2.
Why is the empty set {∅} convex?
- The empty set is regarded as a convex set, since there is nothing to choose from the
empty domain. This mean there is nothing to produce for counter-example as to prove
why the empty set is not a convex set.
Hence, it is possible to assume for contradiction that the empty set is not convex. This
means for arbitrary two points from the set, it is possible to find a point 𝜆𝑎 + (1 − 𝜆)𝑏 is
not contained in the domain.
However, there is no point to choose to show the empty set is not convex. Thus,
the empty set is convex vacuously true by reasoning of the non-existence argument.
Why is any singleton set {x} convex?
- The convex set also includes singleton sets where 𝑎 and 𝑏 have to be the same point
such that the line between 𝑎 and 𝑏 is simply the same point.
In other words, for every one point set, the singleton set of {𝑎} is convex as {𝜆 ∈
[0, 1], 𝑎 ∈ R | 𝜆𝑎 + (1 − 𝜆)𝑎 = 𝑎} falls back into the same set.
Thus, as it is contradiction to find another point that is not in the singleton set where a
single point is the domain, it remains for a singleton set to be the convex set.
CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
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Lim Kyuson
Radon’s theorem
Let 𝑎 1 , 𝑎 2 , ...𝑎 𝑚 ∈ R𝑑 be any 𝑚 ≥ 𝑑 + 2 points. Then, there exists 2 distinct subsets
𝐼, 𝐽 ⊂ [𝑚], such that 𝐼 ∩ 𝐽 ≠ ∅ and conv{𝑎𝑖 | 𝑖 ∈ 𝐼}∩ conv{𝑎 𝑗 | 𝑗 ∈ 𝐽} ≠ ∅.
Remark 2.2.1. As given with enough number of points, at least with 𝑑 + 2, it is always
possible to find 2 disjoint subsets whose convex hull intersect.
The assumptions are
1. There exists 𝑚 ≥ 𝑑 + 2 points.
The bound of 𝑚 ≥ 𝑑 + 2 is tight. In case of 𝑚 = 3 < 4, 3 points are non-collinear, there
exists no choice of subset 𝐼 and 𝐽 to satisfy the theorem conclusion.
3. The subset 𝐼 and 𝐽 convex.
The goal is to show, two convex hull of 𝐼 and 𝐽 intersect. Notice, the subset 𝐼 and
𝐽 does not necessary partition the set of 𝑚 points, as there exists points that could be
counted for both 𝐼 and 𝐽 with 𝜏𝑖 = 0.
For vectors of a1 , ..., am ∈ R𝑑 , we need to find subsets of 𝐼 and 𝐽 so that we can
construct a point z that is contained in both convex hulls. Each ai is consists of the
coordinates (𝑎𝑖1 , 𝑎𝑖2 , ..., 𝑎𝑖𝑑 ) ∈ R𝑑 .
Proof. Step 1. Consider the following system of equations in 𝑚 variables 𝜏1 , ..., 𝜏𝑚 :
(Í
𝑚
𝑖=1 𝜏𝑖 =
Í𝑚
𝑟
𝑖=1 𝜏𝑖 a𝑖
0,
=0
⇔
1 1
a1 a2
! © 𝜏1 ª
· · · 1 ­­ 𝜏2 ®®
=0
· · · am ­­ ... ®®
«𝜏𝑚 ¬
with 𝑚 unknowns and 𝑑 + 1 equations. Since 𝑚 ≥ 𝑑 + 2 > 𝑑 + 1, this homogeneous
system has a nontrivial solution (𝜏1 , ...𝜏𝑚 ).
The nontrivial solutions are obtained via row reduction of (𝑑 + 1) × (𝑚 + 1) including
the column matrix of 0, and manually computed for any 𝜏𝑘 for convenience to derive the
solutions.
Í𝑛
Step 2. Since the system has a nonzero solution (𝜏1 , ..., 𝜏𝑚 ) but 𝑖=1
𝜏𝑖 = 0, some
of the 𝜏𝑖 are positive and some are negative. Set
∑︁
∑︁
𝐼 = {𝑖 : 𝜏𝑖 > 0}, 𝐽 = { 𝑗 : 𝜏 𝑗 < 0}, and 𝑐 =
𝜏𝑖 =
−𝜏 𝑗 .
𝑖∈𝐼
𝑗 ∈𝐽
Í
Clearly, 𝐼 ∩ 𝐽 = ∅. Then, by the assumption that 𝑖∈[𝑚] 𝜏𝑖 ai = 0,
∑︁
∑︁
𝜏𝑖 ai =
(−𝜏 𝑗 )aj
𝑖∈𝐼
𝑗 ∈𝐽
where 𝜏𝑖 > 0 for 𝑖 ∈ 𝐼 and −𝜏 𝑗 > 0 for 𝑗 ∈ 𝐽.
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Step 3. Finally, for the appropriate combinations of the a𝑖 we can scale the coefficients by dividing both sized by 𝑐,
∑︁ 𝜏𝑖
∑︁ −𝜏 𝑗
∑︁ 𝜏𝑖
∑︁ −𝜏 𝑗
z=
ai =
aj , where
= 1 and
=1
𝑐
𝑐
𝑐
𝑐
𝑖∈𝐼
𝑗 ∈𝐽
𝑖∈𝐼
𝑗 ∈𝐽
where 𝜏𝑖 /𝑐 > 0 for 𝑖 ∈ 𝐼, −𝜏 𝑗 /𝑐 > 0 for 𝑗 ∈ 𝐽. Since both sides are convex combinations
of points 𝛼𝑖 from disjoint sets 𝐼 and 𝐽, the z is contained in the convex hulls of both 𝐼
and 𝐽.
CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
31
MAT388
2.2.1
Lim Kyuson
Application questions for the Radon’s theorem
Q1. Given 6 points a1 = (1, 2) a2 = (0, 2) a3 = (−2, −4) a4 = (−3, 1) a5 =
(2, −1) a6 = (−1, 4), show for the two subsets that the convex hull intersect at a
point
The goal is to apply the Radon’s theorem to find the intersection points in both convex
hulls of 𝐼, 𝐽 ⊂ [6] in R2 .
𝜏1
© ª
𝜏®
­
1 1 1 1 1 1 ! ­ 2®
0!
1 1 1 1 1 1 0!
­𝜏3 ®
(a) 1 0 −2 −3 2 −1 · ­ ® = 0 . Then, solve 1 0 −2 −3 2 −1 0 ·
­𝜏 ®
2 2 −4 1 −1 4 ­ 4 ®
0
2 2 −4 1 −1 4 0
­𝜏5 ®
«𝜏6 ¬
(b) The homogenous system of equations are
5
8
𝜏1 = 𝜏4 − 3𝜏5 + 𝜏6
3
3
5
7
𝜏2 = − 𝜏4 + 𝜏5 − 3𝜏6
2
2
1
1
1
𝜏3 = − 𝜏4 − 𝜏5 + 𝜏6
6
2
3
i) When 𝜏4 = 6 and 𝜏5 and 𝜏6 is 0, the solution is (16, -21, -1, 6, 0, 0).
ii) When 𝜏5 = 2 and 𝜏4 and 𝜏6 is 0, the solution is (-6, 5, -1, 0, 2, 0).
iii) When 𝜏6 = 3 and 𝜏4 and 𝜏5 is 0, the solution is (5, -9, 1, 0, 0, 3).
(c) i) The 𝐼 = {𝜏1 , 𝜏4 , 𝜏5 , 𝜏6 } and 𝐽 = {𝜏2 , 𝜏3 }.
ii) The 𝐼 = {𝜏2 , 𝜏4 , 𝜏5 , 𝜏6 } and 𝐽 = {𝜏1 , 𝜏3 }.
iii) The 𝐼 = {𝜏1 , 𝜏2 , 𝜏3 , 𝜏4 , 𝜏6 } and 𝐽 = {𝜏2 }.
Notice, the 𝜏𝑖 = 0 does not necessary be contained in the set of 𝐼 or 𝐽, but for convenience
those points are included.
(d)
6
21
1
−2 38
16
,
Solution 1: 𝜏1 = 16, 𝜏4 = 6, 𝑐 = 22, −𝜏2 = 21, −𝜏3 = 1, a1 + a4 = a2 + a3 yields
22
22
22
22
22 22
5
2
6
1
4 8
Solution 2: 𝜏2 = 5, 𝜏5 = 2, 𝑐 = 7, −𝜏1 = 6, −𝜏3 = 1, a2 + a5 = a1 + a3 yields ,
7
7
7
7
7 7
5
1
3
Solution 3: 𝜏1 = 5, 𝜏3 = 1, 𝜏6 = 3, 𝑐 = 9, −𝜏2 = 9, a1 + a3 + a6 = a2 yields (0, 2)
9
9
9
Particularly, in the third case, the a2 is contained in the convex hull of a1 , a3 and a6 .
Thus, for
of 𝐼 and 𝐽 the conv({a𝑖 : 𝑖 ∈ 𝐼})∩ conv({a 𝑗 : 𝑗 ∈ 𝐽}) yields
43 8pairs
−2 38
,
,
,
and
(0,2).
22 22
7 7
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CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
Lim Kyuson
MAT388
−4
4
2
1
0
0
© ª®
© ª®
© ª®
© ª®
© ª®
© ª®
0
2
0
3
0
­
­
­
­
­
­0
Q2. a1 = ­ ®®, a2 = ­ ®®, a3 = ­ ®®, a4 = ­ ®®, a5 = ­ ®®, a6 = ­ ®®
­4®
­0 ®
­2®
­1 ®
­2®
­2®
«2¬
«0 ¬
«1¬
«1 ¬
«0¬
«2¬
𝜏1
0
© 1 1 1 1 1 1ª © ª © ª
©1
­
­
® ­𝜏2 ® ­0®
­−4 4 2 1 0 0® ­ ® ­ ®
­−4
­
­
® ­𝜏3 ® ­0®
(a) The equation is ­ 0 2 0 3 0 0®· ­ ® = ­ ®. Then, solve ­ 0
­
­
® ­𝜏 ® ­0®
­ 4 0 2 1 2 2® ­ 4 ® ­ ®
­4
­
­
® ­𝜏5 ® ­0®
2 0 0 1 0 2
2
«
«
¬ «𝜏6 ¬ «0¬
(b) The homogenous system of equations are
1
4
2
0
0
1
2
0
2
0
1
1
3
1
1
1
0
0
2
0
1
0
0
2
2
0ª
®
0®
®
0® ·
®
0®
®
0
¬
𝜏1 = −2𝜏6
𝜏2 = −3𝜏6
𝜏3 = 𝜏6
𝜏4 = 2𝜏6
𝜏5 = 𝜏6
When 𝜏6 = 1, the solution is (-2, -3, 1, 2, 1, 1).
(c) The 𝐼 = {𝜏3 , 𝜏4 , 𝜏5 , 𝜏6 } and 𝐽 = {𝜏1 , 𝜏2 }.
(d)
Solution: − 𝜏1 = 2, −𝜏2 = 3, 𝑐 = 22 and 𝜏3 = 1, 𝜏4 = 2, 𝜏5 = 1, 𝜏6 = 1.
3
1
2
1
1
4 6 8 4
2
Then, a1 + a2 = a3 + a4 + a5 + a5 yields , , ,
5
5
5
5
5
5
5 5 5 5
Thus, for the pair of 𝐼 and 𝐽 the conv({a𝑖 : 𝑖 ∈ 𝐼})∩ conv({a 𝑗 : 𝑗 ∈ 𝐽}) yields
4 6 8 4
5, 5, 5, 5 .
CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
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MAT388
2.2.2
Lim Kyuson
Extension of the Helly theorem
Corollary 1.6. explain statement related to Application question 1.
Remark 2.2.2. Let 𝐴 ⊂ R2 be a fixed convex set and let 𝑋1 , ..., 𝑋𝑛 ⊂ R2 be a finitely
many convex sets such that every three of them intersect a translation of 𝐴. Then, there
exists a translation of 𝐴 that intersects all sets 𝑋𝑖 .
The assumption includes,
i) The translation of 𝐴, so called 𝐴0 intersect with arbitrary sets, 𝑋𝑖 , 𝑋 𝑗 and 𝑋 𝑘 .
ii) The set 𝐴 and 𝑋 𝑗 for 𝑗 = 1, ..., 𝑛 is a convex sets.
Step 1. For every 𝑖, define 𝑌𝑖 . The 𝑌𝑖 contains a set of points 𝑦 such that 𝑦 is the
center of mass for translation of 𝐴0 of 𝐴 in which 𝐴0 ∩ 𝑋𝑖 ≠ ∅.
Remark 2.2.3. Define 𝑌𝑖 , 𝑋𝑖 ∩ 𝐴0 ≠ ∅ ⇔ cm(𝐴0) ∈ 𝑌𝑖 .
Equivalently,
i) If the translation of 𝐴0 does not intersect with 𝑋𝑖 , then the corresponding center 𝑦 𝑗 ∉ 𝑌𝑖 .
ii) If 𝑦 ∈ 𝑋𝑖 , then the translations of 𝐴 centered with point 𝑦 definitely intersect 𝑋𝑖 , as
𝑦 ∈ 𝑌𝑖 .
Step 2. Given 3 arbitrary convex sets 𝑋𝑖 , 𝑋 𝑗 and 𝑋 𝑘 suppose there is a translation
𝐴0 intersect with 𝑋𝑖 , 𝑋 𝑗 and 𝑋 𝑘 by the assumption, 𝑋𝑖 , 𝑋 𝑗 , 𝑋𝑘 ∩ 𝐴0 ≠ ∅.
Equivalently, cm(𝐴0) ∈ 𝑌𝑖 , 𝑌 𝑗 , 𝑌𝑘 , where each set contains corresponding 𝑋𝑖 , 𝑋 𝑗 , 𝑋𝑘 .
Hence, the convex sets of 𝑌𝑖 ∩ 𝑌 𝑗 ∩ 𝑌𝑘 ≠ ∅.
Three arbitrary sets of 𝑌𝑖 , 𝑌 𝑗 , 𝑌𝑘 intersect at least for a point cm(𝐴0) and the Helly’s
theorem is applied.
𝑛 𝑌 ≠ ∅ for cm( 𝐴0) such that 𝐴0 to intersect for all sets of 𝑋 ,
Step 3-1. Now, ∩𝑖=1
𝑖
𝑖
by the defined sets of 𝑌𝑖 .
Notice, if the 𝐴 is just a point, then 𝐴0 intersect with arbitrary 𝑋𝑖 , 𝑋𝑘 and 𝑋 𝑘 such
that 𝑋𝑖 ∩ 𝑋 𝑗 ∩ 𝑋 𝑘 ≠ ∅.
Step 3-2. Related to Application 1, it is assumed that an arbitrary 𝑌𝑖 , 𝑌 𝑗 and 𝑌𝑘 is
substituted for 𝐵(𝑥𝑖 , 𝑟), 𝐵(𝑥 𝑗 , 𝑟) and 𝐵(𝑥 𝑘 , 𝑟). Analogously, the set of 𝑋𝑖 , 𝑋 𝑗 , and 𝑋 𝑘
can be defined as points for centers of 𝑌𝑖 , 𝑌 𝑗 and 𝑌𝑘 .
Also, the 𝐴0 is a disk with radius 𝑟, which is Γ𝑖 𝑗 𝑘 Then, this gives the intersected
point 𝑝 to be cm(𝐴0).
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MAT388
The 𝑑-dimensional Helly theorem
Let 𝑋1 , ..., 𝑋𝑛 ⊂ R𝑑 be 𝑛 ≥ 𝑑 + 1 convex sets such that 𝑋 𝐼 ≠ ∅ for every subset
𝐼 ⊂ [𝑛], |𝐼 | = 𝑑 + 1.
Then, there exists a point 𝑧 ∈ 𝑋1 , ..., 𝑋𝑛 .
The assumption includes,
i) For every subset 𝐼 ∈ [𝑛] of 𝑑 + 1 number of them, the convex sets of 𝑋 𝐼 ≠ ∅.
This is that for any 𝑑 + 1 number of convex sets chosen from [𝑛], the ∩𝑖∈𝐼 𝑋𝑖 at least
intersect for a point 𝑧.
ii) The number of convex sets 𝑛 ≥ 𝑑 + 1 which lives in R𝑑 .
The goal is to find a point 𝑧 as a common intersection for all 𝑋𝑖 , 𝑖 ∈ [𝑛].
Proof. i) When 𝑛 = 𝑑 + 1,the induction used in the Helly theorem holds for 𝑑-dimension
with 𝑛 + 1 numbers of distinct convex sets.
Since the induction hypothesis holds for the base case of 𝑛 − 1 of at least 𝑑 + 1 number
of distinct 𝑋 𝐼 has nonempty intersection, it is left to show for 𝑛 ≥ 𝑑 + 2 to show for
non-empty intersection.
ii) Suppose for 𝑛 ≥ 𝑑+2 and every (𝑛−1)-element subset of convex sets 𝑋 𝑘 has a common
points 𝑣 𝑘 ∈ 𝑋 [𝑛]−𝑘 , which the notation means it is equivalent to take off an arbitrary 𝑘
ranges from 1 to 𝑛 for exact 𝑛−1 elements of subset. (ie. 𝑣 1 ∈ ∩∀𝑛 𝑋𝑛−1 , 𝑣 2 ∈ ∩∀𝑛 𝑋𝑛−2 , ...)
Step 1. Then, the point 𝑣 𝑘 is found for every distinct 𝑛−1 elements of subset that is different combination from each other. Now, the distinct intersection points of every different
(𝑛 − 1)-element subsets, 𝑣 𝑘 for 𝑘 = 1 to 𝑛, forms a family points [𝑛], |[𝑛]| = 𝑛 ≥ 𝑑 + 2.
Step 2. Followed from the Radon’s theorem, there exists 2 disjoint subsets 𝐼, 𝐽 ⊂ [𝑛]
and a point 𝑧 ∈ R𝑑 such that 𝑧 ∈ conv({𝑣 𝑖 |𝑖 ∈ 𝐼}∩ conv({𝑣 𝑗 | 𝑗 ∈ 𝐽} (eg. [𝑛] =
{𝑖1 , 𝑖2 , 𝑖3 , · · · , 𝑖 𝐼 , 𝑗1 , 𝑗2 , · · · , 𝑗 𝐽 }). Notice, for all 𝑖 ∈ 𝐼, 𝑣 𝑖 ∈ 𝑋 [𝑛]−𝑖 such that for all
𝑣 𝑖 ∈ 𝑋 [𝑛]\𝐼 , and 𝑣 𝑗 ∈ 𝑋 [𝑛]\𝐽 for defined subsets of corresponding 𝑖 and 𝑗.
Step 3. Hence, 𝑧 ∈ conv({𝑣 𝑖 |𝑖 ∈ 𝐼} ⊂ 𝑋 [𝑛]\𝐼 = 𝑋𝐽 = ∩ 𝑗 ∈𝐽 𝑋 𝑗 and 𝑧 ∈ conv({𝑣 𝑖 |𝑖 ∈
𝑛 𝑋.
𝐼} ⊂ 𝑋 [𝑛]\𝐽 = 𝑋 𝐼 = ∩𝑖∈𝐼 𝑋𝑖 where both are subset of 𝑋 [𝑛] of 𝐼 ∩ 𝐽 = [𝑛], 𝑧 ∈ ∩𝑖=1
𝑖
Specifically, 𝑧 ∈ 𝑋 𝐼 = ∩𝑖∈𝐼 𝑋𝑖 = 𝑋𝑖 , for all 𝑖 ∈ 𝐼 and 𝑧 ∈ 𝑋𝐽 = ∩ 𝑗 ∈𝐽 𝑋 𝑗 = 𝑋 𝑗 , for
all 𝑗 ∈ 𝐽. This gives 𝑧 ∈ (∩ 𝑗 ∈𝐽 𝑋 𝑗 ) ∩ (∩𝑖∈𝐼 𝑋𝑖 ) = 𝑋 𝐼 ∩ 𝑋𝐽 = 𝑋 [𝑛] such that 𝑧 ∈ 𝑋 [𝑛] ,
which is 𝑧 ∈ {𝑋1 , · · · , 𝑋𝑛 }. (Notice, it is simple to divide subsets as ones belong to 𝐽
and not in 𝐽 to 𝑛 for 𝑣 𝑗 ∈ 𝑋 [𝑛]\𝐽 and 𝑣 𝑛− 𝑗 ∈ 𝑋𝐽 )
Thus, the Radon’s theorem is directly applied to sets of 𝑛 point to derive that the
common intersection is contained in both subsets of 𝐼 and 𝐽 for all 𝑛 convex sets.
CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
35
MAT388
2.3.1
Lim Kyuson
Extension of the 𝑑-dimensional Helly theorem
Exercise 1.3 Let 𝑄 ⊂ R2 be a polygon with perimeter 𝐿. Prove that 𝑄 can be covered
by a disk of radius 𝐿/4
Proof. Method 1. In case the polygon is in the disk without being circumscribed,
find 2 points 𝐴 and 𝐵 apart from each other that is the longest distanced, on the boundary of the polygon 𝑃. Then, | 𝐴𝐵| < 𝐿/2 as assumed the 𝑃 is inside the disk of radius 𝐿/4.
Notice that the point 𝐴 and 𝐵 is fixed to be arbitrary chosen for the sum of edges
from 𝐴 to 𝐵 to be 𝐿/2 as two points are the farthest points from each other.
Step 1. For each point 𝑀 on the 𝑃, | 𝐴𝑀 | + |𝐵𝑀 | ≤ 𝐿/2 as the perimeter is 𝐿.
Now, let 𝑆 be the midpoint of the line 𝐴𝐵 such that the disk of 𝐵(𝑆, 𝐿/4) covers the
boundary of 𝑃, which also include the whole 𝑃.
Step 2. For an arbitrary chosen point 𝑀, construct a point 𝑀 0 symmetric to 𝑀 with
respect to a midpoint 𝑆, which is a reflection about a point, 𝑟 𝑠 . The 𝑀 0 could be located
outside 𝑃 but still stay inside the 𝐵(𝑆, 𝐿/4). The use of 𝑀 and 𝑀 0 is to achieve the
𝐵(𝑆, 𝐿/4) covers the 𝑃.
Step 3. From the triangle inequality, |𝑀 𝑀 0 | ≤ |𝑀 𝐴| + | 𝐴𝑀 0 | for 4𝐴𝑀 𝑀 0 as well
as |𝑆𝑀 | ≤ 21 (|𝑀 𝐴| + | 𝐴𝑀 0 |) for the 1/ 2 of 𝑀 𝑀 0. Due to the symmetry of 𝑀 for 𝑀 0,
| 𝐴𝑀 0 | = |𝐵𝑀 | and | 𝐴𝑀 | + |𝐵𝑀 | ≤ 𝐿/2 such that |𝑆𝑀 | ≤ 21 (|𝑀 𝐴| + |𝐵𝑀 |) ≤ 𝐿/4.
Hence, for any chosen 𝑀 the 𝑃 is inside 𝐵(𝑆, 𝐿/4).
𝑀0
𝐵
𝐴
𝑆
𝑀
Figure: Possible illustration of 𝑃
Thus, any polygon with perimeter to be 𝐿 is covered by the disk of radius 𝐿/4.
Method 2. The points on the edges contained in the circle does not mean vertices
are contained on the circle.
The aim of the problem is to find the disk of radius 𝐿/4 as a given value (among all
congruent disks) that covers the planar graph (simply connected with interior angles) of
the 𝑛-polygon 𝑃 for 𝑛 vertices.
Note that once vertices are inscribed inside the circle, as a result the points on the edges
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CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
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MAT388
are contained in the circle (or as a consequence there exists no circle).
Claim: If any 3 vertices of the 𝑛-polygon to be inside the disk of radius 𝐿/4 for
𝑛 ≥ 3, then all vertices of the 𝑛-polygon is inside the disk of radius 𝐿/4 for perimeter 𝐿.
Denote 3 arbitrary vertices as 𝑥𝑖 , 𝑥 𝑗 , and 𝑥 𝑘 and the simply connected triangle as
𝑇 = (𝑥𝑖 , 𝑥 𝑗 , 𝑥 𝑘 ).
The goal is to show the 𝑇 is inscribed inside the circle for 3 cases, obtuse, acute, and
right-angle triangle.
Case 1. Simply, for the obtuse ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 the mid point of the hypotenuse 𝑥𝑖 𝑥 𝑘 of the
𝑇, which has a length less than 𝐿/2, is set to be the center, the point 𝑂 of the 𝐵(𝑂, 𝐿/4)
to include all 3 points. The hypotenuse is the longest side is taken from the obtuse 𝑇
such that 𝑇 is contained in the 𝐵(𝑂, 𝐿/4) naturally.
Note that from method 1 any 2 points 𝐴 and 𝐵 is distanced less than 𝐿/2 and | 𝐴𝑀 | +
|𝑀 𝐵| ≤ 21 𝐿for arbitrary point 𝑀.
Case 2. The goal is to show the existence of the circumcircle that inscribe the rightangled 𝑇 such that the circumcircle is covered by a larger disk of 𝐵(𝑂, 𝐿/4).
Claim: There exists a circumcircle that inscribes the 𝑇 where ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 is 𝜋/2, using
Thale’s theorem conversely.
Theorem 2.3.1. (Thale’s theorem) If 𝑥𝑖 , 𝑥 𝑗 , and 𝑥 𝑘 are distinct points on a circle where
the line 𝑥𝑖 𝑥 𝑘 is a diameter, then the ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 is a right angle.
When the 𝑇 has right-angle which is ∠𝑥𝑖 𝑥 𝑗 𝑥 𝑘 , take the circle Γ1 whose diameter is
𝑥𝑖 𝑥 𝑘 and the mid point of the hypotenuse of |𝑥𝑖 𝑥 𝑘 | < 𝐿/2 is the center 𝑂.
Let 𝑃 be the intersection of Γ1 and the ray 𝑂𝑥 𝑗 . By the Thale’s theorem, ∠𝑥𝑖 𝑃𝑥 𝑘 is
right-angle. But then 𝑃 must equal to 𝑥 𝑗 because if 𝑃 lies outside of the 4𝑥𝑖 𝑥 𝑗 𝑥 𝑘 then
∠𝑥𝑖 𝑃𝑥 𝑘 must be acute, and if 𝑃 lies inside of the 4𝑥𝑖 𝑥 𝑗 𝑥 𝑘 then ∠𝑥𝑖 𝑃𝑥 𝑘 must be obtuse
for contradiction.
Thus, wherever the point 𝑥 𝑗 is on the arc of the Γ1 there exists a circle that inscribe
the right-angled 𝑇 (|𝑥𝑖 𝑂| = |𝑥 𝑘 𝑂| = |𝑥 𝑗 𝑂|) such that this Γ1 with radius less than 𝐿/4 is
contained in the 𝐵(𝑂, 𝐿/4).
𝑥𝑗 = 𝑃
𝑥𝑖
𝑥𝑘
𝑂
𝐵(𝑂, 𝐿/4)
Γ1
Figure: Illustration of Γ1 and 𝐵(𝑂, 𝐿/4) in the right-angle 𝑇
CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
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Case 3. Claim: The acute 𝑇 is covered by the circumcircle 𝐵(𝑂, 𝐿/4).
Step 1. First, note that there exists not all 3 sides of 𝑇 have side-lengths of
that add up to be over 𝐿 (which is
lengths greater than
√
the bound of
3
4 𝐿
√
3
4 𝐿
√
3 3
4 𝐿
√
3
4 𝐿
> 𝐿) such that the equilateral triangle side-
is not possible but isosceles triangle is only possible. Note that
is attained when ∠𝑥𝑖 𝑂𝑥 𝑗 ≥ 𝜋/3.
Step 2. We can pick 2 points say 𝐴 and 𝐵 in 𝑇 that has the longest path from the
point 𝐴 to 𝐵, which is less than 𝐿/2. Then, for any point 𝑀 on the boundary of 𝑇
between the point 𝐴 and 𝐵, we have | 𝐴𝑀 | + |𝐵𝑀 | ≤ 𝐿/2 as the 𝑃 has a perimeter equal
to 𝐿.
Let 𝑆 be defined as the midpoint of the point 𝐴 and 𝐵 that is the center of the circle
𝐵(𝑆, 𝐿/4). The goal is to figure out with the arbitrary point 𝑀 that 𝑇 is inside the
𝐵(𝑆, 𝐿/4).
Step 3. For any 𝑀, construct a point 𝑀 0 that is symmetric from the point 𝑆, reflected by the 𝑟 𝑠 . The restriction also applies to 𝑀 0, | 𝐴𝑀 0 | + |𝐵𝑀 0 | ≤ 𝐿/2, such that 𝑀 0
is inside the 𝐵(𝑆, 𝐿/4).
Now, from the 4𝐴𝑀 𝑀 0 the side |𝑀 𝑀 0 | ≤ |𝑀 𝐴| + | 𝐴𝑀 0 |. Then, dividing by 1/2
of |𝑀 𝑀 0 |, |𝑆𝑀 | ≤ 21 (|𝑀 𝐴| + | 𝐴𝑀 0 |).
Due to the symmetry of point 𝑀, | 𝐴𝑀 0 | = |𝐵𝑀 | and | 𝐴𝑀 | + |𝐵𝑀 | ≤ 𝐿/2 such
that |𝑆𝑀 | ≤ 21 (|𝑀 𝐴| + |𝐵𝑀 |) ≤ 𝐿/4, implying the 𝑇 to be inside the 𝐵(𝑆, 𝐿/4).
Thus, the circumcircle of the 𝑇 has a radius less than or equal to 𝐿/2.
𝑀
𝐵
𝑆
𝐴
𝑀0
Figure: An illustration of 𝑃 and the 𝐵(𝑆, 𝐿/4)
Failed method of acute triangle case.
In case 𝑇 is acute, the circumcircle 𝐵(𝑂, 𝐿/4)
√ is able to cover 𝑇 is able to cover 𝑇
which all pairwise distance of vertices at most 43 𝐿 followed from the Jung’s theorem.
However, the pairwise vertices chosen from the 𝑛-polygon is at most 𝐿/2 >
√
3
4 𝐿.
i) In other words,√ the Jung’s theorem assures for any pairwise vertices with maximum distance of 43 𝐿 to be covered by the 𝐵(𝑂, 𝐿/4).
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CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
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MAT388
√
3
4 𝐿
ii) It remains to show the side-length of 𝑇 are distanced at most between 𝐿/2 and
is circumscribed in 𝐵(𝑂, 𝐿/4).
√
iii) Note that the the upper bound of radius 63 𝐿, when 𝑇 has a side that is at most 𝐿/2,
is only possible when 𝑇 is regular triangle.
Simply, the radius of the circumcircle is attained when the ∠𝑥 𝑗 𝑂𝑥𝑖 = ∠𝑥𝑖 𝑂𝑥 𝑘 = ∠𝑥 𝑘 𝑂𝑥 𝑗 =
2𝜋
3 which is minimized with maximum side-lengths of 𝐿/2.
𝜋
Notice that this minimized angle of 2𝜋
3 gives an interior angle of the 𝑇 with 3 .
𝑥𝑗
𝐵
𝑂
𝑝
𝑀
𝑥𝑘
𝛽
𝑥𝑖
𝐴
Figure: Maximum radius, Jung’s theorem
The ∠𝑥 𝑗 𝑂𝑥𝑖 = 2𝜋 − (∠𝑥𝑖 𝑂𝑥 𝑘 + ∠𝑥 𝑘 𝑂𝑥 𝑗 ), where the ∠𝑥𝑖 𝑂𝑥 𝑘 = 𝜋 − 2∠𝑂𝑥 𝑘 𝑥𝑖 and
∠𝑥 𝑗 𝑂𝑥 𝑘 = 𝜋 − 2∠𝑂𝑥 𝑘 𝑥 𝑗 because the 4𝑥𝑖 𝑂𝑥 𝑘 and 4𝑥 𝑗 𝑂𝑥 𝑘 is an isosceles triangle.
Equivalently, the 2𝜋
3 = ∠𝑥 𝑗 𝑂𝑥𝑖 = 2𝜋 − (𝜋 − 2∠𝑂𝑥 𝑘 𝑥𝑖 + 𝜋 − 2∠𝑂𝑥 𝑘 𝑥 𝑗 ) = 2(∠𝑂𝑥 𝑘 𝑥𝑖 +
∠𝑂𝑥 𝑘 𝑥 𝑗 ) which is same as 𝜋3 = ∠𝑂𝑥 𝑘 𝑥𝑖 + ∠𝑂𝑥 𝑘 𝑥 𝑗 .
√
Hence, the radius of the circumcircle of 63 𝐿 is attained when the 𝑇 is an equilateral
triangle with maximum side-lengths of 𝐿/2.
But we know that such equilateral triangle with side-length of 𝐿/2 is not possible where
the perimeter is 𝐿.
√
The maximum possible equilateral triangle is 31 𝐿 which is clearly less than 43 𝐿 that is
possible to be inscribed inside the 𝐵(𝑂, 𝐿/4). In other words, the 𝑇 is able to covered
by the 𝐵(𝑂, 𝐿/4) with restriction of the perimeter equal to 𝐿.
However, consider the case ii) where at least one of the 3 sides is greater than
√
√
3
4 𝐿.
Assume that the greatest side-length 𝑥𝑖 𝑥 𝑗 > 43 𝐿 (greater than L/3) with ∠𝑥𝑖 𝑥 𝑘 𝑥 𝑗 > 𝜋/3
as the ∠𝑥𝑖 𝑂𝑥 𝑗 ≤ 2𝜋/3 is fixed inside the disk (close to the center) and the other two
sides 𝑥𝑖 𝑥 𝑘 and 𝑥 𝑗 𝑥 𝑘 is circumscribed inside the 𝐵(𝑂, 𝐿/4) .
Step 1. First, note that there exists not all 3 sides of 𝑇 have side-lengths of
that add up to be over 𝐿 (which is
lengths greater than
the bound of
√
3
4 𝐿
√
3
4 𝐿
√
3 3
4 𝐿
√
3
4 𝐿
> 𝐿) such that the equilateral triangle side-
is not possible but isosceles triangle is only possible. Note that
is attained when ∠𝑥𝑖 𝑂𝑥 𝑗 ≥ 𝜋/3.
Step 2. We can pick 2 points say 𝐴 and 𝐵 in 𝑇 that has the longest path from the
point 𝐴 to 𝐵, which is less than 𝐿/2. Then, for any point 𝑀 on the boundary of 𝑇
between the point 𝐴 and 𝐵, we have | 𝐴𝑀 | + |𝐵𝑀 | ≤ 𝐿/2 as the 𝑃 has a perimeter equal
to 𝐿.
CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
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Lim Kyuson
Let 𝑆 be defined as the midpoint of the point 𝐴 and 𝐵 that is the center of the circle
𝐵(𝑆, 𝐿/4). The goal is to figure out with the arbitrary point 𝑀 that 𝑇 is inside the
𝐵(𝑆, 𝐿/4).
Step 3. For any 𝑀, construct a point 𝑀 0 that is symmetric from the point 𝑆, reflected by the 𝑟 𝑆 . The restriction also applies to 𝑀 0, | 𝐴𝑀 0 | + |𝐵𝑀 0 | ≤ 𝐿/2, such that 𝑀 0
is inside the 𝐵(𝑆, 𝐿/4).
Now, from the 4𝐴𝑀 𝑀 0 the side |𝑀 𝑀 0 | ≤ |𝑀 𝐴| + | 𝐴𝑀 0 |. Then, dividing by 1/2, |𝑆𝑀 | ≤
1
0
2 (|𝑀 𝐴| + | 𝐴𝑀 |).
Due to the symmetry of point 𝑀, | 𝐴𝑀 0 | = |𝐵𝑀 | and | 𝐴𝑀 | + |𝐵𝑀 | ≤ 𝐿/2 such that
|𝑆𝑀 | ≤ 21 (|𝑀 𝐴| + |𝐵𝑀 |) ≤ 𝐿/4, implying the 𝑇 to be inside the 𝐵(𝑆, 𝐿/4).
So, the circumcircle of the 𝑇 has a radius less than or equal to 𝐿/4.
Thus, i), ii) iii) and 3 cases altogether prove for the 𝑛-polygon to be contained in
the 𝐵(𝑂, 𝐿/4) followed from the Jung’s theorem.
Method 2.
Step 1. First, note that there exists not all 3 sides of 𝑇 have side-lengths of
that add up to be over 𝐿 (which is
lengths greater than
√
the bound of
3
4 𝐿
√
3
4 𝐿
√
3 3
4 𝐿
√
3
4 𝐿
> 𝐿) such that the equilateral triangle side-
is not possible but isosceles triangle is only possible. Note that
is attained when ∠𝑥𝑖 𝑂𝑥 𝑗 ≥ 𝜋/3.
𝑥𝑗
𝑂
𝑝
𝑟
𝛽
𝑥𝑘
𝑥𝑖
𝑞
Figure: Maximum radius, Jung’s theorem
Step 2. Second, 𝜋 < ∠𝑥𝑖 𝑂𝑥 𝑘 + ∠𝑥 𝑘 𝑂𝑥 𝑗 ≤ 4𝜋/3(−∠𝑥𝑖 𝑂𝑥 𝑗 ) and the points 𝑞 and 𝑟 is a
perpendicular bisector such that |𝑥𝑖 𝑥 𝑘 | = 2× radius × sin(∠𝑥 𝑘 𝑂𝑞) and |𝑥 𝑗 𝑥 𝑘 | = 2× radius
× sin(∠𝑥 𝑘 𝑂𝑟).
But the domain is defined as 0 < ∠𝑥 𝑘 𝑂𝑞 < 𝜋2 and 0 < ∠𝑥 𝑘 𝑂𝑟 < 𝜋2 depended on the
second point, 0 < |𝑥𝑖 𝑥 𝑘 | < 𝐿4 × 2 = 𝐿2 and 0 < |𝑥 𝑗 𝑥 𝑘 | < 𝐿4 × 2 × 1 = 𝐿2 .
Note that the |𝑥𝑖 𝑥 𝑘 | and |𝑥 𝑗 𝑥 𝑘 | is flexible on the change of values for |𝑥𝑖 𝑥 𝑗 | with ∠𝑥 𝑗 𝑂𝑥𝑖 .
For example from the first point,√in case 𝑇 is isosceles where 2 sides, defined
as 𝑥𝑖 𝑥 𝑘 and
√
3
3
𝑥𝑖 𝑥 𝑗 , have lengths greater than 4 𝐿, the side-length of 𝑥 𝑗 𝑥 𝑘 is less than 4 𝐿 where the
angle of 𝑥 𝑗 𝑂𝑥 𝑘 less than 2𝜋/3.
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Likewise,
the range of |𝑥 𝑗 𝑥 𝑘 | + |𝑥𝑖 𝑥 𝑘 | < 𝐿 depend on the size of the ∠𝑥𝑖 𝑥 𝑘 𝑥 𝑗 where
√
3
4 𝐿
≤ 𝑥𝑖 𝑥 𝑗 ≤ 𝐿/2 ranges such that any side-length is covered by the 𝐵(𝑂, 𝐿/4). Being
flexible on the restriction among the range of side-lengths, the 𝑇 only need to find the
lower bound for the disk to be verified.
Step 3. Third, the radius of the circumcircle is determined by the ratio of angle and
|𝑥 𝑥 |
|𝑥 𝑥 |
|𝑥𝑖 𝑥 𝑘 |
= 𝑠𝑖𝑛(∠𝑥𝑗 𝑗 𝑘𝑥𝑖 𝑥 𝑘 ) , as well.
side-length which is the diameter = 𝑠𝑖𝑛(∠𝑥𝑖 𝑗 𝑥𝑗 𝑘 𝑥𝑖 ) = 𝑠𝑖𝑛(∠𝑥
𝑗 𝑥 𝑗 𝑥𝑘 )
This law of sines gives the lower bound of the diameter to be
for ∠𝑥 𝑗 𝑥 𝑗 𝑥 𝑘 < 𝜋/2 and |𝑥𝑖 𝑥 𝑘 | ≥
√
(𝐿/4 >
√
3
4 𝐿,
√
3𝐿/4
1
<
|𝑥𝑖 𝑥 𝑘 |
𝑠𝑖𝑛(∠𝑥 𝑗 𝑥 𝑗 𝑥 𝑘 )
<
𝐿/2
1
where clearly the 𝐵(𝑂, 𝐿/4) is in the range
3
8 𝐿).
In summary, the first point shows the reason how the upper bound of greatest sidelength is not fixed and gives a general range for other side-lengths in the second point
which lead to the third point for lower bound of the radius only, which is the only added
necessary restriction.
Reason: First of all, this way fails because of the constructive argument made in
the step 2. While the goal to prove the 𝑇 circumscribed has radius less than or equal
to 𝐿/4, the method assumes the radius of the circle to be less than or equal to 𝐿/4 and
show sides have restricted lengths (which is clearly double argument).
|𝑥 𝑥 |
|𝑥 𝑥 |
|𝑥 𝑖 𝑥 𝑘 |
= 𝑠𝑖𝑛(∠𝑥𝑗 𝑗 𝑘𝑥𝑖 𝑥 𝑘 ) ) is only used to deAlso, the law of sine (= 𝑠𝑖𝑛(∠𝑥𝑖 𝑗 𝑥𝑗 𝑘 𝑥𝑖 ) = 𝑠𝑖𝑛(∠𝑥
𝑗 𝑥 𝑗 𝑥𝑘 )
rive the upper bound of the circumcircle, which does not clearly find the lower bound
the radius of the circumcircle.
2.4
Some case studies of Topological Helly theorem
Exercise 1.21. Let 𝑋1 , · · · , 𝑋𝑛 ⊂ R2 be simple polygons (simply connected finite union
of convex polygons) in the plane, such that all double and triple intersections of 𝑋𝑖 are
also (nonempty) simple polygons. Prove that the intersection of all 𝑋𝑖 is also a simple
polygon.
Given conditions:
- Finitely many 𝑋1 , ..., 𝑋𝑛 ∈ R2 polygons are simply connected, meaning each 𝑋𝑖 has no
holes.
- Each polygon is a connected, simply connected finite union of convex polygons.
- All double intersections of 𝑋𝑖 are simple polygons.
- All triple intersections of 𝑋𝑖 are also simple polygons.
𝑛 𝑋 to
Claim: Given 𝑋𝑖 ∩ 𝑋 𝑗 ≠ ∅ and 𝑋𝑖 ∩ 𝑋 𝑗 ∩ 𝑋 𝑘 ≠ ∅, the induction yields ∩𝑖=1
𝑖
be a simply polygon.
1. Intersecting two simple polygons give a simple polygon (if nonempty).
If 𝐴, 𝐵 are simple polygons and 𝐴 ∩ 𝐵 ≠ ∅, then 𝐴 ∩ 𝐵 is a simple polygon.
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The simple polygon is simply connected finite union of convex polygons. Generally, for
example given two convex set 𝐴 and 𝐵, 𝐴 = 𝑌1 ∩ · · · ∩ 𝑌𝑚 and 𝐵 = 𝑍1 ∩ · · · ∩ 𝑍𝑛 , where
𝑚 𝑌 )∩(∪𝑛 𝑍 ) = ∪
each 𝑌𝑖 and 𝑍 𝑗 is a convex set. Then, 𝐴∩𝐵 = (∪𝑖=1
𝑖
1≤𝑖≤𝑚,1≤ 𝑗 ≤𝑛 (𝑌𝑖 ∩𝑍 𝑗 )
𝑗=1 𝑗
As the intersection of (𝑌𝑖 ∩ 𝑍 𝑗 ) is convex, the intersection of two simple polygon is
a finite union of convex polygon, where 𝑌𝑖 and 𝑍 𝑗 is nonempty.
Note that the intersection must be a simple polygon, not a disconnected collection of
convex sets. The goal is to inductive argument for 𝑛-case with the base case of 𝑛 = 4.
2. From Helly’s theorem, if all triples of convex set intersect in a nonempty set,
then all of them intersect in a nonempty set.
i) For the base case, using the condition that all double and triple intersections of 𝑋𝑖 from
𝑋1 , · · · , 𝑋𝑛 , we need to show that 𝑋1 ∩ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 is a simple polygon. Equivalently,
∪1≤𝑖≤𝑚,1≤ 𝑗 ≤𝑛 (𝑌𝑖 ∩ 𝑍 𝑗 ) ∩ ∪1≤𝑘 ≤𝑜,1≤ℎ≤𝑙 ( 𝐴 𝑘 ∩ 𝐵 ℎ ) = ∪𝑖, 𝑗,𝑘,ℎ (𝑌𝑖 ∩ 𝑍 𝑗 ∩ 𝐴 𝑘 ∩ 𝐵 ℎ ), where
(𝑌𝑖 ∩ 𝑍 𝑗 ∩ 𝐴 𝑘 ∩ 𝐵 ℎ ) is a convex polygon such that the intersection of 4 simple polygons
is a finite union of convex polygons.
ii) Let any point 𝑣 1 ∈ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 , 𝑣 2 ∈ 𝑋1 ∩ 𝑋3 ∩ 𝑋4 , 𝑣 3 ∈ 𝑋1 ∩ 𝑋2 ∩ 𝑋4 and
𝑣 4 ∈ 𝑋1 ∩ 𝑋2 ∩ 𝑋3 , in the form of 𝑣 𝑖 ∈ 𝑋 [𝑛]−𝑖 .
Definition 2.4.1. A subset 𝐴 of a metric space 𝑋 is not connected (disconnected) if there
are disjoint open sets 𝑈 and 𝑉 such that 𝐴 ⊂ 𝑈 ∪ 𝑉, 𝐴 ∩ 𝑈 ≠ ∅ and 𝐴 ∩ 𝑉 ≠ ∅. If no
such disjoint open sets 𝑈 and 𝑉 exist then 𝐴 is connected [16, p.1].
In other words, a path-connected domain is said to be simply connected, if any simple
closed curve can be shrunk to a point continuously in the set.
Case 1. (Figure 1) Among randomly scattered 𝑣 1 to 𝑣 4 , if the connected path of
Γ𝑣 1 𝑣 2 , Γ𝑣 2 𝑣 3 and Γ𝑣 3 𝑣 1 is connected to contain 𝑣 4 , then simply connected path of Γ𝑣 1 𝑣 2 , Γ𝑣 2 𝑣 3 , Γ𝑣 3 𝑣 1 ⊆
4 𝑋 ≠ ∅, as there is a loop without
𝑋4 contains 𝑣 4 ∈ 𝑋1 ∩ 𝑋2 ∩ 𝑋3 ⊆ 𝑋4 such that ∩𝑖=1
𝑛
any hole to be contractible.
𝑣1
𝑣2
𝑣1
𝑣4
𝑣1
𝑣2 𝑣4
𝑣1
𝑣4
𝑣3
𝑣2
𝑣4
𝑣2
𝑣3
𝑣3
𝑣1
𝑣2
𝑣3
𝑣1
𝑣4
𝑣2
𝑣3
𝑣3
𝑣1
𝑣4
𝑣2
𝑣4
𝑣3
Figure 1-6: possible path connected for 𝑣 1 , 𝑣 2 , 𝑣 3 , 𝑣 4 in topological R2 .
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Case 2. (Figure 2) Each path is specified according to the intersection of simple polygons
belongs to. If Γ𝑣 1 𝑣 3 is a path connected trough the intersection between 𝑋1 ∩ 𝑋3 ∩ 𝑋4
and 𝑋1 ∩ 𝑋2 ∩ 𝑋3 of Γ𝑣 2 𝑣 4 , then simply Γ𝑣 1 𝑣 3 ∩ Γ𝑣 2 𝑣 4 ⊆ (𝑋1 ∩ 𝑋3 ) ∩ (𝑋2 ∩ 𝑋4 ). Let
𝑧 ∈ Γ𝑣 1 𝑣 3 ∩ Γ𝑣 4 𝑣 2 ⊆ 𝑋1 ∩ 𝑋3 ∩ 𝑋2 ∩ 𝑋4 be defined such that 𝑋1 ∩ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 ≠ ∅.
Hence, if any two distinct connected paths connected, then Γ𝑣 𝑖 𝑣 𝑗 ∩ Γ𝑣 𝑘 𝑣 ℎ ⊆ (𝑋 [4]−𝑖, 𝑗 ) ∩
(𝑋 [4]−𝑘,ℎ ) = ∩4𝑛=1 𝑋𝑛 ≠ ∅ for 𝑖, 𝑗, 𝑘, ℎ ∈ {1, 2, 3, 4}.
Now, suppose Γ𝑣 1 𝑣 3 ∩ Γ𝑣 2 𝑣 4 = ∅, and no two paths connected intersects.
Case 3. (Figure 3) If Γ𝑣 3 𝑣 4 , Γ𝑣 1 𝑣 4 , Γ𝑣 1 𝑣 3 ⊆ 𝑋2 is simply connected and the intersection of 𝑣 2 ∈ 𝑋1 ∩ 𝑋3 ∩ 𝑋4 is contained in connected paths of Γ𝑣 3 𝑣 4 , Γ𝑣 1 𝑣 4 , Γ𝑣 1 𝑣 3 , then
𝑋1 ∩ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 ≠ ∅.
Case 4. (Figure 4) If Γ𝑣 3 𝑣 4 surrounds Γ𝑣 1 𝑣 3 and Γ𝑣 1 𝑣 4 without an intersection and
Γ𝑣 2 𝑣 3 , Γ𝑣 2 𝑣 4 , Γ𝑣 3 𝑣 4 ⊆ 𝑋1 is path connected where 𝑣 1 ∈ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 is contained, then
𝑋1 ∩ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 ≠ ∅.
Case 5. (Figure 5) If Γ𝑣 1 𝑣 4 surrounds Γ𝑣 1 𝑣 3 and Γ𝑣 3 𝑣 4 without an intersection and
Γ𝑣 1 𝑣 2 , Γ𝑣 2 𝑣 4 , Γ𝑣 1 𝑣 4 ⊆ 𝑋3 is path connected where 𝑣 3 ∈ 𝑋1 ∩ 𝑋2 ∩ 𝑋4 is contained, then
𝑋1 ∩ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 ≠ ∅.
Note that combinatorially in the first case Γ𝑣 3 𝑣 4 , Γ𝑣 4 𝑣 1 , Γ𝑣 1 𝑣 3 contains 𝑣 2 , in the second case Γ𝑣 3 𝑣 4 , Γ𝑣 2 𝑣 4 , Γ𝑣 2 𝑣 3 contains 𝑣 1 , in the third case Γ𝑣 2 𝑣 4 , Γ𝑣 1 𝑣 4 , Γ𝑣 1 𝑣 2 contains 𝑣 3 ,
with respect to 𝑣 4 .
Notice that in case there is two distinct paths that does not intersect, Γ𝑣 1 𝑣 3 and Γ𝑣 2 𝑣 3 ,
three points of intersection are path connected to contain the other point such that
Γ𝑣 𝑖 𝑣 𝑗 , Γ𝑣 𝑗 𝑣 𝑘 , Γ𝑣 𝑘 𝑣 𝑖 ⊆ 𝑋 [𝑛]−𝑖, 𝑗,𝑘 ∩ 𝑋𝑙 ≠ ∅ for 𝑖, 𝑗, 𝑘, 𝑙 ∈ {1, 2, 3, 4}, 𝑋1 ∩ 𝑋2 ∩ 𝑋3 ∩ 𝑋4 ≠ ∅.
Case 6. (Figure 6) Otherwise, if Γ𝑣 1 𝑣 2 , Γ𝑣 2 𝑣 3 is connected with Γ𝑣 1 𝑣 3 to surround the 𝑣 4
4 𝑋 ≠ ∅.
of Γ𝑣 2 𝑣 4 , then 𝑣 4 ∈ 𝑋1 ∩ 𝑋2 ∩ 𝑋3 ⊆ 𝑋4 such that ∩𝑖=1
𝑖
Notice that no three paths connected is disconnected from each other, as a path connected
needs at least 2 points among 4 possible intersections of simple polygons.
When the planar graph by three intersection is fully connected with each other, the
last point is placed to path connect with other three intersection containing the former
planar graph within three connected paths.
Case 7. (Figure 7) If 4 intersection has each connected path to nearby intersection
as a 4-gon, Γ𝑣 1 𝑣 3 , Γ𝑣 3 𝑣 4 , Γ𝑣 4 𝑣 2 , Γ𝑣 2 𝑣 1 , then simply two diagonal intersection is path connected, Γ𝑣 1 𝑣 4 or Γ𝑣 2 𝑣 3 , to contain the intersection between them, Γ𝑣 1 𝑣 4 or Γ𝑣 2 𝑣 3 and
Γ𝑣 4 𝑣 2 , Γ𝑣 2 𝑣 1 .
Hence, no matter the orientation of paths, one of the points inside paths formed by
the other three by the simply connected condition such that three connected paths contain the remaining point by paths connected.
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Planar graph
Given 4 points in R2 , if the paths between each pair of points does not intersect with
each other, then one of the four points is contained in paths by the other three points with
complete planar graph.
However, there is a case where two paths from an intersection of simple polygons
intersect, Γ𝑣 𝑖 𝑣 𝑗 and Γ𝑣 𝑖 𝑣 𝑘 . In this case, 𝑣 𝑖 ∈ 𝑋 [𝑛]\𝑖 is in the intersection of a same simple polygon of Γ𝑣 𝑖 𝑣 𝑗 ∩Γ𝑣 𝑖 𝑣 𝑘 ⊆ (𝑋 [𝑛]\𝑖, 𝑗 ∩ (𝑋 [𝑛]\𝑖,𝑘 = 𝑋 𝑗 ∩ 𝑋 𝑘 ∩ 𝑋𝑙 for 𝑖, 𝑗, 𝑘, 𝑙 ∈ {1, 2, 3, 4}.
Self-intersecting path case
Apart from the previous case, there is a self-intersecting continuous loop in R2 , non
Jordan curve, which is also simple closed curve.
Theorem 2.4.2. (Jordan Curve Theorem) Any continuous simple closed curve in the
plane, separates the plane into two disjoint regions, the inside and the outside. [17]
Every Jordan curve divides into an interior region bounded by the curve and an
exterior region R2 , where the intersection of 𝑣 𝑖 is contained in the path connected
, so that every continuous path connecting a point of one region to a point of the
other intersects with that loop somewhere. If the path connected self-intersect and the
intersection is contained in the interior of the path connected, then the region bounded
by the Jordan curve contains the intersection resulting in a common intersection.
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Claim: 𝑋1 ∩ · · · ∩ 𝑋𝑛 ≠ ∅.
By the inductive hypothesis, define 𝑋 𝐼 for any (𝑛 − 1) simple polygon picked from the
𝑛-sets. Then, define 𝑋 𝐼 = ∩𝑖∈𝐼 𝑋𝑖 and the point 𝑣 𝑖 ∈ 𝑋 [𝑛]−𝑖 in the simple polygon of the
intersection for all 𝑛 − 1-sets except 𝑋𝑖 .
Case 1. Among randomly scattered 𝑣 1 to 𝑣 4 , if the connected path of Γ𝑣 1 𝑣 2 , Γ𝑣 2 𝑣 3 and
Γ𝑣 3 𝑣 1 is connected to contain 𝑣 4 , then simply connected path of Γ𝑣 1 𝑣 2 ⊆ 𝑋 [𝑛]−1,2 , Γ𝑣 2 𝑣 3 ⊆
𝑋 [𝑛]−2,3 , Γ𝑣 3 𝑣 1 ⊆ 𝑋 [𝑛]−1,3 , in which Γ𝑣 1 𝑣 2 , Γ𝑣 2 𝑣 3 , Γ𝑣 1 𝑣 3 ⊆ 𝑋 [𝑛]−1,2,3 , contains 𝑣 4 ∈ 𝑋 [𝑛]−4 ⊆
𝑛 𝑋 ≠ ∅, as there is a loop without any hole to be contractible.
𝑋 [𝑛]−1,2,3 such that ∩𝑖=1
𝑖
Case 2. Each path is specified according to the intersection of simple polygons belongs to. If Γ𝑣 1 𝑣 3 is a path connected trough the intersection of Γ𝑣 2 𝑣 4 ⊆ 𝑋 [𝑛]−2,4 , then
simply Γ𝑣 1 𝑣 3 ∩ Γ𝑣 2 𝑣 4 ⊆ (𝑋 [𝑛]−1,3 ) ∩ (𝑋 [𝑛]−2,4 ). Let 𝑧 ∈ 𝑋 [𝑛]−1,3 ∩ 𝑋 [𝑛]−2,4 be defined,
𝑛 𝑋 ≠ ∅.
where ∩𝑖=1
𝑖
Hence, if any two distinct connected paths connected, then Γ𝑣 𝑖 𝑣 𝑗 ∩ Γ𝑣 𝑘 𝑣 ℎ ⊆ (𝑋 [𝑛]−𝑖, 𝑗 ) ∩
𝑛 𝑋 ≠ ∅ for 𝑖, 𝑗, 𝑘, ℎ ∈ {1, .., 𝑛}.
(𝑋 [𝑛]−𝑘,ℎ ) = ∩𝑙=1
𝑙
Now, suppose Γ𝑣 1 𝑣 3 ∩ Γ𝑣 2 𝑣 4 = ∅, and no two paths connected intersects.
Case 3. If Γ𝑣 3 𝑣 4 , Γ𝑣 1 𝑣 4 , Γ𝑣 1 𝑣 3 ⊆ 𝑋 [𝑛]−1,3,4 is simply connected and the intersection of
𝑣 2 ∈ 𝑋 [𝑛]−2 is contained in connected paths of Γ𝑣 3 𝑣 4 , Γ𝑣 1 𝑣 4 , Γ𝑣 1 𝑣 3 , then 𝑋 [𝑛]−2 ∩𝑋 [𝑛]−1,3,4 =
𝑛 𝑋 ≠ ∅.
∩𝑖=1
𝑖
Case 4. If Γ𝑣 3 𝑣 4 surrounds Γ𝑣 1 𝑣 3 and Γ𝑣 1 𝑣 4 without an intersection and Γ𝑣 2 𝑣 3 , Γ𝑣 2 𝑣 4 , Γ𝑣 3 𝑣 4 ⊆
𝑋 [𝑛]−2,3,4 is path connected where 𝑣 1 ∈ 𝑋 [𝑛]−1 is contained, then 𝑋 [𝑛]−1 ∩ 𝑋 [𝑛]−2,3,4 =
𝑛 𝑋 ≠ ∅.
∩𝑖=1
𝑖
Case 5. If Γ𝑣 1 𝑣 4 surrounds Γ𝑣 1 𝑣 3 and Γ𝑣 3 𝑣 4 without an intersection and Γ𝑣 1 𝑣 2 , Γ𝑣 2 𝑣 4 , Γ𝑣 1 𝑣 4 ⊆
𝑋 [𝑛]−1,2,4 is path connected where 𝑣 3 ∈ 𝑋 [𝑛]−3 is contained, then 𝑋 [𝑛]−3 ∩ 𝑋 [𝑛]−1,2,4 =
𝑛 𝑋 ≠ ∅.
∩𝑖=1
𝑖
Notice that in case there is two distinct paths that does not intersect, Γ𝑣 1 𝑣 3 and Γ𝑣 2 𝑣 3 ,
three points of intersection are path connected to contain the other point such that
Γ𝑣 𝑖 𝑣 𝑗 , Γ𝑣 𝑗 𝑣 𝑘 , Γ𝑣 𝑘 𝑣 𝑖 ⊆ 𝑋 [𝑛]−𝑖, 𝑗,𝑘 ∩ 𝑋 [𝑛]−𝑙 = ∩𝑛𝑚=1 𝑋𝑚 ≠ ∅ for 𝑖, 𝑗, 𝑘, 𝑙 ∈ {1, ..., 𝑛}.
Case 6. Otherwise, if Γ𝑣 1 𝑣 2 , Γ𝑣 2 𝑣 3 is connected with Γ𝑣 1 𝑣 3 to surround the 𝑣 4 of Γ𝑣 2 𝑣 4 ,
𝑛 𝑋 ≠ ∅.
then 𝑣 4 ∈ 𝑋 [𝑛]−4 ⊆ 𝑋 [𝑛]−1,2,3 such that ∩𝑖=1
𝑖
Case 7. In the case where no point is contained by the other 4 points, the the diagonals of 4-gon formed by Γ𝑣 1 𝑣 3 , Γ𝑣 3 𝑣 4 , Γ𝑣 4 𝑣 2 , Γ𝑣 2 𝑣 1 which Γ𝑣 1 𝑣 4 , Γ𝑣 2 𝑣 3 intersects resulting
𝑛 𝑋 ≠ ∅.
in 𝑋 [𝑛]−2,3 ∩ 𝑋 [𝑛]−1,4 = ∩𝑖=1
𝑖
Hence, no matter the orientation of paths, one of the points inside paths formed by
the other three by the simply connected condition such that three connected paths contain the remaining point by paths connected.
Thus, the inductive argument and composition of simple polygon support the intersection
of finite simple polygons to be defined as a simple polygon.
CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
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CHAPTER 2. LECTURE ON DISCRETE AND POLYHEDRAL GEOMETRY
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