Helly’s Type Theorems
av
Madeleine Leander
2008 - No 2
Madeleine Leander
Examensarbete i matematik 15 h¨ogskolepo¨ang, f¨ordjupningskurs Handledare: Paul Vaderlind
Abstract
The main topic of this paper is the theorem of Helly, some gener- alizations and applications of it. Eduard Helly (1884-1943) discovered the theorem during the first world war, while war prisoner in a Rus- sian camp, but published it first long after his release in 1923. The Helly theorem is one of the basic results of combinatorial geometry.
Helly was borne and educated in Vienna, where he was awarded his doctorate in 1907. Being a Jew he encountered some difficulties in receiving a permanent position in Austria. But finally, with some help of his friend Albert Einstein, he was appointed as a teacher in Paterson Junior College in New Jersey (1940). The central theorem of this paper, the Helly theorem states that that given r convex sub- set of Rn, r ≥ n + 1, such that every n + 1 of them has a nonempty intersection then the intersection of all r sets is nonempty.
Contents
1 Preliminary definitions and theorems 5
2 Helly’s theorem and elementary applications 8
3 Generalisations of Helly’s theorem 12
3.1 The ”Helly number” n + 1 . . . 12
3.2 Infinite collections of convex sets . . . 15
3.2.1 Countably many convex sets . . . 15
3.2.2 Uncountably many convex sets . . . 17
3.3 The Helly type theorem for star-shaped sets . . . 17
3.4 The transversal Helly Number . . . 20
References 24
1 Preliminary definitions and theorems
Definition 1.1. (Convex combination) Let α1. . . αk be k nonnegative real numbers such that the sum equals one. If ¯x1, . . . , ¯xk belongs to Rn then α1x¯1+ α2x¯2+ ... + αkx¯k is said to be a convex combination of ¯x1, ¯x2, ..., ¯xk. Definition 1.2. (Convex set) Let C be a subset of Rn. The set C is said to be a convex if for all ¯x, ¯y ∈ C.
(1 − t)¯x + t¯y ∈ C, 0 ≤ t ≤ 1.
Definition 1.3. (Convex hull) The convex hull of a subset S ⊂ Rn, denoted convS, is the intersection of all the convex sets which contain S.
Definition 1.4. (Convex hull) Alternatively convS can be defined as the set of all finite convex combinations of elements of S.
The equivalence of those two definitions is easy to prove, and it is done below.
Lemma 1.1. A set S is convex iff every convex combination of points in S lies in S.
Proof. Let ¯x = λ1x¯1 + . . . λmx¯m, where ¯xi ∈ S, Pm
i=1λi = 1 and λi ≥ 0. If m = 2 it follows by definition that ¯x ∈ S. Assume that ¯x ∈ S if m ≤ k.
Consider a combination of k + 1 points, ¯x = λ1x¯1+ . . . λk+1x¯k+1. If λk+1 = 1, then ¯x = ¯xk+1 ∈ S and there is nothing to proof. Suppose λk+1 < 1. Then
¯
x = (λ1+ . . . + λk)
λ1
λ1+ . . . + λkx¯1+ . . . + λk
λ1+ . . . + λkx¯k
+ λk+1xk+1 By the induction hypothesis ¯y ∈ S if ¯y can be expressed as a convex combi- nation of k points in S.
¯
y = λ1
λ1+ . . . + λkx¯1+ . . . + λk
λ1+ . . . + λkx¯k lies in S.
∴ ¯x = (1 − λk+1)¯y + λk+1x¯k+1 ∈ S
Theorem 1.1. Definition 1.3 ⇔ 1.4. In other words: For any set S, the convex hull of S consists precisely of all convex combinations of elements of S.
Helly’s Theorems Madeleine Leander
Proof. Let S0 be the set of all convex combinations of elements in S. Since convS is convex and S ⊂ convS, Lemma 1.1 implies that S0 ⊂ convS. Let
¯
x = α1x¯1 + . . . + αrx¯r and ¯y = β1y¯1+ . . . + βsy¯s be two elements in S0. For any 0 ≤ λ ≤ 1,
¯
z = λ¯x + (1 − λ)¯y
is an element in S0 since all coefficients (λαi, (1 − λ)βj) are positive and the sum
r
X
i=1
λαi+
s
X
j=1
(1 − λ)βj = λ
r
X
i=1
αi+ (1 − λ)
s
X
j=1
βj = λ · 1 + (1 − λ) · 1 = 1.
Thus S0 is a convex set containing S and convS ⊂ S0. Thus convS = S0.
Note: S is convex iff convS = S.
After this definitions we are ready to state two basic theorems, of Caratheodory and Radon.
Before proceeding with the Radon theorem we will prove another useful result.
Theorem 1.2. (Caratheodory) Suppose S is a subset of Rn containing at least n + 1 points. Then every ¯x ∈ convS can be expressed as a convex combination of no more than n + 1 points ∈ S.
Example 1.1. If a point ¯p lies in a convex hull of a set P ⊂ R2 containing at least three points, then there are three points in P determining a triangle containing ¯p.
Proof. Let ¯x ∈ convS. Then ¯x is a convex combination of a finite number of points in S. Hence,
¯ x =
k
X
i=1
λix¯i where ¯xi ∈ S, λi ≥ 0 and Pk
i=1λi = 1. Suppose k ≥ n + 2. Then ¯x2 −
¯
x1, ¯x3 − ¯x1, . . . , ¯xk− ¯x1 a set of k − 1 ≥ n + 1 vectors in Rn, hence linearly dependent, thus ∃ µ2, µ3, . . . , µk ∈ R (not all zero) such that Pk
i=2µi(¯xi −
¯
x1) = Pk
i=2µix¯i−Pk
i=2µix¯1 = 0. By letting µ1 = −Pk
i=2µi we have
k
X
i=1
µix¯i =
k
X
i=2
µix¯i+ µ1x¯1 =
k
X
i=2
µix¯i−
k
X
i=2
µix¯1 = 0.
6
Figure 1: Here, (14,14) can be expressed as a convex combination of (0, 0), (0, 1), (1, 0) as follows: 12(0, 0) + 14(0, 1) +14(1, 0).
Furthermore, µ1+Pk
i=2µi = −Pk
i=2µi+Pk
i=2µi = 0. Since µi 6= 0 for some i ∈ {1, . . . , k} and the sum of all µi is equal to zero there exists µi > 0. Thus for any real α, x can be written as:
¯ x =
k
X
i=1
λix¯i− α
k
X
i=1
µix¯i =
k
X
i=1
(λi − αµi)¯xi
By letting α = min1≤i≤k{λµi
i : µi > 0} = λµj
j it follows that λi − αµi ≥ 0 and, by definition, λj − αµj = 0. We also have Pk
i=1λi − αµi = 1. In other words, ¯x is now expressed as a convex combination of at most k − 1 points.
By repeating this process ¯x can be represented as a convex combination of at most n + 1 points of S.
Now it is time to formulate and prove the key-theorem by Radon.
Theorem 1.3. (Radons theorem) Any set of at least n + 2 points in Rn can always be partitioned in two subsets S1 and S2 such that the convex hulls of S1 and S2 intersect.
Proof. Let S = {¯x1, ¯x2, ..., ¯xr}, r ≥ n + 2 be any finite set of r points in Rn. Since any set of at least n + 1 vectors in Rnis linearly dependent, there exists scalars α2, ..., αr not all zero such that Pr
i=2αi(¯xi− ¯x1) = 0. But
r
X
i=2
αi(¯xi− ¯x1) =
r
X
i=2
αix¯i− ¯x1
r
X
i=2
αi.
Helly’s Theorems Madeleine Leander
Let −Pr
i=2αi = α1. This gives us Pr
i=1αi = 0 andPr
i=1αix¯i = 0.
Fix one nonzero solution α1, α2, ..., αr. At least two of the αi’s must have opposite signs. Let I+ be the subset of {1, 2, ..., r} such that αi ≥ 0, ∀i ∈ I+ and let I− be the subset of {1, 2, ..., r} such that αi < 0, ∀i ∈ I−.
By letting S1 = {¯xi : i ∈ I+} and S2 = {¯xi : i ∈ I−} it follows that S1 6= ∅ and S2 6= ∅. Since the αi’s sum to zero, we may let
α =X
i∈I+
αi = −X
i∈I−
αi. Then α > 0 and we can let
¯ x =X
i∈I+
αi
αx¯i = −X
i∈I−
αi αx¯i.
It follows that ¯x is a convex combination of both S1 and S2. Then The- orem 1.1 implies that ¯x ∈ convS1T convS2 and hence convS1T convS2 6=
∅
2 Helly’s theorem and elementary applica- tions
Theorem 2.1. Suppose A1, A2, . . . , Ar ⊂ Rn is a family of r convex sets, with r ≥ n + 1 and that every n + 1 of them have a nonempty intersection.
Then r
\
i=1
Ai 6= ∅
Proof. The proof is by induction on r. If r = n + 1, then the theorem is trivially true. Suppose inductively that r > n + 1 and that the statement is true for r − 1 sets. The sets Bj = T
i6=jAi 6= ∅ by inductive hypothesis.
Pick a point ¯xj from each of Bj. Since r ≥ n + 2 then, by Radon’s Theorem, there is a partition of ¯x1, . . . , ¯xr into two sets S1 and S2 such that S = convS1T convS2 6= ∅. For every Aj either every point in S1 belongs to Aj or every point in S2 belongs to Aj. Therefore S ⊆ Aj∀j. Thus Tr
i=1Ai 6=
∅ ∀ r ≥ n + 1
Equivalently Helly’s theorem states that if
r
\
i=1
Ai 6= ∅
8
then there exist n + 1 sets Ai1, . . . , Ain+1 such that Ai1\
. . .\
Ain+1 6= ∅
In the following we give another proof of Helly’s theorem.
Lemma 2.1. Helly’s theorem is valid in the special case when A1, A2, . . . , Ar are closed half-spaces of Rn.
Proof. The proof is by induction on n. For n = 1 the lemma is obviously true.
Let A1, A2, . . . , Ar be closed half-spaces of Rn, defined by the hyperplanes π1, π2, . . . πr. Assume inductively that
(∗) A1 ∩ A2∩ . . . ∩ Ar 6= ∅.
Without loss of generality assume that no Ai may be omitted without making the intersection nonvoid. A1 is a closed half-space so A1 ⊃ π1. Furthermore
π1∩ A2∩ . . . ∩ Ar 6= ∅, and
(π1∩ A2) ∩ . . . ∩ (π1∩ Ar) 6= ∅.
Now π1∩Ai is a closed half-space of π1(considered as an (n−1)−dimensional space), or if π1and πi are parallel, then π1∩Aicoincides with π1or the empty set. By the induction hypothesis there are k, k ≤ n sets π1∩ Ai having no point in common. Without loss of generality we may assume
(π1∩ A2) ∩ . . . ∩ (π1∩ Ak+1) = π1∩ A2∩ . . . ∩ Ak+1 = ∅.
Letting B = A2 ∩ . . . ∩ Ak+1, note that B is convex. Denote also by Ac1 as the complement of A1 in Rn. Now either
(i) B ∩ Ac1 = ∅ or (ii) B ∩ A1 = ∅.
If none of them were true then there would exist two points P1 and P2 such that
P1 ∈ B ∩ Ac1, P2 ∈ B ∩ A1
and the segment [P1, P2] would have a nonempty intersection with π1. But [P1, P2] ⊂ B contradicts B ∩ π1 = ∅. Furthermore (ii) is impossible because it implies Ac1∩ A2∩ . . . ∩ Ar = ∅ which together with above implies
(Ac1∪ A1) ∩ A2∩ . . . ∩ Ar= ∅.
This contradicts the fact that none of the Ai could be omitted in (*). Thus case (ii) holds.
Helly’s Theorems Madeleine Leander
Proof. (Hellys theorem) Let A1, . . . , Ar be convex sets in Rn every n + 1 of them having a nonempty intersection. Let Ai1, . . . , Ain+1 be any n + 1 of them. Let further Pi1,...,in+1 be any point in the intersection of Ai1, . . . , Ain+1. Let P denote any finite set of such points. The sets P ∩ Ai are finite sets and all n + 1 of them have a nonempty intersection. Put Bi = conv(Ai∩ P ). The convex hull of a finite set can be represented as the intersection of a finite number of half-spaces, thus Bi = Di,1 ∩ . . . ∩ Di,ki say, where D1, . . . , Ds are all the half-spaces appearing for all the Bi. To every Dj corresponds a Bi where Dj ⊃ Bi ⊃ Ai ∩ P why every n + 1 of the Dj have a nonempty intersection. By lemma 2.1 D1∩. . .∩Ds6= ∅. Now D1∩. . .∩Ds=
1∩. . .∩Br. Furthermore
Ai ⊃ conv(Ai∩ P = Bi) and hence
r
\
i=1
Ai ⊃
r
\
i=1
Bi 6= ∅.
Example 2.1. Here is an illustration of Helly’s theorem. Let P, G, R, S be a family of four convex sets in the plane. Every three of them has a nonempty intersection. Here S1 = {p, q}, S2 = {r, s}. The four sets have a nonempty intersection, x.
Figure 2: An illustration of Helly’s theorem.
10
Exercise 2.1. Let n points be given in the plane such that each three of them can be enclosed in a circle of radius 1. Prove that all n of them can be enclosed in a circle of radius 1.
Proof. Consider three circles c1, c2, c3 centered in a1, a2, a3. Since there is a circle c0, centered in a0, of radius 1 such that a1, a2, a3 ∈ c0 we have that a0 ∈ c1, c2, c3. Hence c1∩ c2 ∩ c3 6= ∅. By Helly’s theorem c1 ∩ . . . ∩ cn 6= ∅.
If x0 is in the intersection the distance from x0 to a1, . . . , an is not greater than 1. Thus all n points can be enclosed in the circle of radius 1 centered at x0.
Exercise 2.2. Prove that inside every compact convex figure F in R2 there is a point O such that every cord AB of F passing through O is divided into two segments each of which is not shorter than 13|AB|.
Proof. Let SX denote the image of F under the similitude centered at X and with the factor 23. For every point X of the border ∂F of F consider SX. Clearly SX is convex and compact. If such point O exists then apparently O ∈ SX. Let A, B and C be any three points on ∂F. Consider the triangle 4ABC, let A0 be the point dividing BC into two segments of the same length, B0 and C0 analogous. The medians AA0, BB0, CC0 have a point T in common. Thus T ∈ SA∩SB∩SC and AT = 23AA0, BT = 23BB0, CT = 23CC0. Since any three of the sets SX, for all X ∈ ∂F, have a nonempty intersec- tion then, by Hellys theorem
\
X∈∂F
SX 6= ∅.
Any point O of this intersection will do.
Exercise 2.3. Suppose that in the plane we have a set {Fα}α∈I, for some index set I, of polygons(convex or non-convex), any two of which have a point in common. Show that for any point a ∈ R2 there is a circle centered in a having a point in common with each polygon Fα.
Proof. Draw any halfline, l from a point a. Let x be a point in one of the polygons and d the distance between x and a. Let further f (x) be the point on l at the distance d to a. Now consider {f (Fα)}α∈I. Each set is contained in l. Since each Fα∩ Fα0 is nonempty, this implies
f (Fα)\
f (Fα0) 6= ∅.
Hence by Helly’s theorem
\f (Fα) 6= ∅.
Helly’s Theorems Madeleine Leander
Choose p ∈ T
α∈If (Fα). Let d∗ be the distance between p and a. Now, the circle centered in a with radius d∗ is a circle passing through all sets Fα. Exercise 2.4. Let F consist of k ≥ parallel line segments in R2. Suppose that for each three segments from F there exists a common transversal (a line passing through all three segments). Prove that there is a common transversal for all sets in F.
Proof. For every line segment x = a, b ≤ y ≤ c consider the mapping of all the lines y = kx + m, passing through this segment, in the km−plane. Every line in the xy−plane will represent a convex set in the km−plane, hence we have a family of convex sets all three of them with a nonempty intersection.
The statement follows by Helly’s Theorem.
3 Generalisations of Helly’s theorem
There are lots of generalisations of Helly’s theorem, some of them will be presented here. First generalisation tells that the ”Helly number” n + 1 can be reduced but under some additional restrictions of course.
Example 3.1. Consider a triangle as a set of three edges. Any two of them have a point in common (a vertex), but the three of them does not. However, given any point y ∈ R2 there exists a line through y that intersect all three sides of the triangle.
With other restrictions Helly’s theorem can also be expanded to an infinite collections of convex sets, while without any additional conditions the original Helly’s theorem is obviously not true for infinite collections. Consider for instance the system of all upper half-planes (regions consisting of all points on one side of a straight line parallel to the x−axis, and no points on the other side). It is obvious that any three of them have a non-empty intersection but all of them does not.
3.1 The ”Helly number” n + 1
The Helly number of a family F of sets is the minimal number n satisfying the following propperty: If every subfamily of n sets in F has a nonempty intesection then the whole F has a nonempty intersection.
With some additional restrictions the Helly number n + 1 can be further reduced, but it will of course result in some weaker conclusion. This was first observed by Alfred Horn in 1949.
12
Theorem 3.1. Let F be a a family of r compact, convex sets in Rn, F = {A1, . . . , Ar}, r > n. If every n of them have a nonempty intersection then, given any point y ∈ Rn there exists a line through ¯y that intersects all mem- bers of F .
To prove the theorem we need two more definitions.
Definition 3.1. (positive combination) A point ¯x is said to be a positive combination of ¯x1, . . . , ¯xk if ¯x = α1x¯1 + . . . + αkx¯k where αi ≥ 0 for all 1 ≤ i ≤ k.
Definition 3.2. (positive hull) The set of all positive combinations of points of a set S is said to be the positive hull of S, denoted posS.
Example 3.2. Let S be the surface of the 2-dimensional ball in Figure 3, and let A be a convex set of points. Let A∗ = A ∩ posS. Then it is obvious that A∗ is the marked part of S in Figure 3.
Figure 3: A∗
Proof. Theorem 3.1 The proof is by induction on the number r of compact convex sets in the collection. If r = n there is nothing to proof. Suppose the theorem is true for every collection of r − 1 sets where r > n. Without loss of generality suppose ¯y is the origin. Either ¯y belongs to some of the sets in F or not. If ¯y belongs to some set in F then, by the induction hypothesis,
Helly’s Theorems Madeleine Leander
the theorem is true. If ¯y doesn’t belong to any of the sets in F we will need some more work. Let S be the surface of an n-dimensional ball centered at
¯
y. Define further
A∗i = S\
(posAi)
for i = 1, . . . , r, (see example 3.2). By the inductive hypothesis it follows that there exists
¯ wj ∈\
i6=j
A∗i, ∀ j = 1, 2, . . . , r Now we have two cases.
(1) The line through ¯w1 and ¯y meets A∗1.
(2) ¯w1 and A∗1 lies on an open hemisphere on S.
If (1) holds there exists a line through ¯y that intersects all A∗i and thus there exists a line through ¯y that intersects all the members of F. Thus the theorem is proved in this case.
If (2) holds then there exists a hyperplane H through ¯y such that ¯w1 and A∗1 lies on the same side of H. Hence ¯wilies on the same side of H for i = 1, . . . , r (because ¯wi ∈ A∗1, i = 2, . . . , r). Choose points ¯xj,i, in Ai, i 6= j such that ¯xj,i lies on the line trough ¯y and ¯wj. Let
Bi = conv[
j6=i
¯ xj,i.
It’s easy to realize that Bi ⊂ Ai. Let H0 be a hyperplane parallel to H. Let further π be the radial projection from ¯y onto H0. π(Bi) form a collection of r convex sets in H0 and each n of them has a nonempty intersection. H0 is a (n − 1)-dimensional hyperplane. By applying Helly’s Theorem 2.1 it follows that all these projections have a nonempty intersection, with at least one point in common. A line through ¯y and one of those points intersects all the sets Ai, i = 1, . . . , r.
This ”Helly number” can be reduced further, shown in the theorem below.
Theorem 3.2. Let F be a set containing r, r ≥ n, compact, convex sets in Rn, F = {A1, . . . , Ar}. If every k of them, 1 ≤ k ≤ n have a nonempty intersection then given any (n − k)-dimensional flat, a translate of a linear subspace of Rn, F1 ∈ Rn there exists a (n − k + 1)-dimensional flat, F2, such that F1 lies in F2 and F2 intersects all the members of F .
14
Proof. The case k = n follows from theorem 3.1, why we may assume 1 ≤ k <
n. Choose a (n − k)-dimensional flat F1. Let V be the orthogonal subspace to F1. Project, with (n − k)-dimensional flats parallel to F1, Rn onto V . Then in V we have a family of compact, convex sets where every k of them has a nonempty intersection. The dimension of V is obviously k. It follows, by applying theorem 3.1, that from any point ¯y in V there exists a line through
¯
y that intersects all the members of V. Choose ¯y to be in the projection of F1 onto V.
This line through ¯y is the projection of an (n − k + 1)-dimensional flat F2 that intersect all the members of F. It’s also easy to realize F1 ⊂ F2.
3.2 Infinite collections of convex sets
3.2.1 Countably many convex sets
Before we formulate and prove Helly’s theorem for families containing count- ably many sets we need some more definitions.
Definition 3.3. A subset E of a metric space X is said to be closed if every limit point of E is a point of E. Furthermore E is said to be open if every point of E is an interior point of E.
Definition 3.4. Let E be a subset of a metric space X. A collection {Gα} of open subsets of X such that E ⊂S
αGα is called an open cover of a E.
Definition 3.5. A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover.
A wellknown characterization of compact sets is the following: A subset K of a metric space X is compact iff K is closed and bounded.
Theorem 3.3. For any collection of closed sets {Fα}, α ∈ I, for some index set I, the set T
αFα is closed.
Proof. Let {Fα} be a collection of closed sets. Let further G =S
αGα where {Gα} is a collection of open sets. If ¯x ∈ G then ¯x ∈ Gα for some α. Thus ¯x is an interior point of Gα. It follows that ¯x is an interior point of G ⇒ G is open.
Choose ¯x ∈ Fαc then ¯x /∈ Fα and x is not a limit point of Fα. Hence there exists a neighborhood N of ¯x such that Fα∩ N is empty.
∴ N ⊂ Fαc.
∴ ¯x is an interior point of Fαc and Fαc is open. According to above [
α∈I
(Fαc)
Helly’s Theorems Madeleine Leander
is open. But S
α∈I(Fαc) = (T
αFα)c, which means that (T
α∈IFα)c is open.
T
α∈IFα is closed.
Theorem 3.4. If F is closed and K is compact, then F ∩ K is compact.
Proof. Let F be a closed set, and K be a compact set of a metric space X.
According to Theorem 3.4 and Theorem 3.3 K is closed and K ∩ F is closed.
We have
K ∩ F ⊂ K
Let {Vα} be an open cover of F ∩K. If (F ∩K)c is adjoined to {Vα} we obtain an open cover Φ of K. But K is compact and there is a finite subcollection Φ0 of Φ which covers K and thus F ∩ K. If (F ∩ K)c ∈ Φ0 we remove it from Φ0 and we still have an open cover of F ∩ K.
∴ F ∩ K is compact.
Now let’s turn our attention to the Helly theorem for a countable family of convex sets.
Theorem 3.5. Let F = {B1, B2, . . .} be a family of closed convex sets in Rn, where at least one of the sets, say B1, is compact. If every n + 1 of the sets of F have a nonempty intersection then
∞
\
k=1
Bk 6= ∅.
Proof. Let Ci =Ti
k=1Bk. Then Ciis compact for every i = 1, 2, . . . , and since according to Helly’s Theorem 2.1 the intersection of every finite collection is nonempty, Ci 6= ∅ for finite i.
Fix a member K of {Ci} and put Gi = Cic. Assume that no point of K belongs to every Ci. The sets Gi form an open cover of K. Because K is compact there are finite many indices i1, i2, . . . , in such that
K ⊂ Gi1 ∪ . . . ∪ Gin But this means that
K ∩ Ci1 ∩ . . . ∩ Cin = ∅
We have a contradiction, hence there is a point in K belonging to every Ci and this implies that
∞
\
k=1
Bk 6= ∅.
16
3.2.2 Uncountably many convex sets
The Helly theorem can also be generalized to families of infinitely many convex sets. This time however we will demand that all sets of the family are compact.
Theorem 3.6. Suppose {Bα}α∈I is an infinite, possibly uncountable family of convex and compact sets in Rn, for some index set I. Suppose that every n + 1 of them have a nonempty intersection. Then
\
α∈I
Bα 6= ∅
Proof. According to Helly’s theorem 2.1 every collection of finitely many Bα has a nonempty intersection. Fix a member K of {Bα} and put Gi = Bαc
i. Assume, like in the proof of theorem 3.5, that no point of K belongs to every Bα. Then the sets Gi form an open cover of K. Because K is compact there are finite many indices i1, i2, . . . , in such that
K ⊂ Gi1 ∪ . . . ∪ Gin But this means that
K ∩ Bα1 ∩ . . . ∩ Cαn = ∅ We have a contradiction.
3.3 The Helly type theorem for star-shaped sets
The Helly theorem can be extended to star-shaped sets. A star-shaped set defines as a non-empty set M ⊂ Rn such that there exists a point x ∈ M from where all other points of M can be seen. In other words, for every point y ∈ M the segment
[y, x] = {z ∈ En : z = (1 − λ)x + λy, 0 ≤ λ ≤ 1}
lies in M .
The subset of M from where all other points of M can be seen is called the star, s(M ). The star of any star-shaped set is a convex sets.
Theorem 3.7. Let {Bα} be a family of star-shaped compact sets in Rn, such that every n + 1 of them has a nonempty and star-shaped intersection. Then
\
α
Bα 6= ∅
Helly’s Theorems Madeleine Leander
This theorem follows directly from the Helly theorem of homology-cells which states as below. A homology-cell is a nonempty set in Rn that is homologically trivial in all dimensions. This means this cell can be contracted down to a single point.
Theorem 3.8. [1] Let {Bα} be a family of homology-cells in Rn, such that the intersection of every n + 1 of them is a homology-cell. Then the intersection
\
α
Bα
is a homology-cell.
A star-shaped compact set is clearly a homology-cell why Theorem 3.7 follows by Theorem 3.8. We can develop Theorem 3.7 and without further restrictions find a more interesting result.
Theorem 3.9. Let {Bα}, (α ∈ I) be a family of star-shaped compact sets in Rn, such that the intersection of every n + 1 of them is nonempty and star-shaped. Then the intersection
\
α∈I
Bα
is nonempty and star-shaped.
This section is devoted to the proof of the theorem above.
Lemma 3.1. Let B be a homology cell in Rncontaining at least n + 1 points.
Suppose that for each n + 1 points of B there is a point in B from which all n + 1 pints can be seen. Then B is star-shaped.
Proof. First let for all ¯x ∈ BVx¯ be the set of points in B which can be seen from ¯x. By the hypothesis every n + 1 of the sets Vx¯ have a nonempty intersection. By Helly’s theorem there exists a point
¯ y ∈ \
¯ x∈B
convVx¯.
To prove the lemma we need to show that ¯y ∈ T
¯
x∈BVx¯. Assume, to get a contradiction, that ¯y /∈T
¯
x∈BV¯x. Then there exists ¯x ∈ B and ¯u ∈ [¯x, ¯y] such that ¯u /∈ B. Let ¯x0 be a point of the relative boundary of B in [¯x, ¯u]. There exists a point ¯w in ¯x0u such that¯
ρ( ¯w, ¯x0) = 1
2ρ(¯u, B)
18
because of the fact that B is compact. Furthermore there is a point ¯v in [ ¯w, ¯u] and ¯z ∈ B such that
ρ(¯v, ¯z) = ρ([ ¯w, ¯u], B).
By construction ¯z is the point in B nearest to ¯v. Further V¯z lies in the closed half-space P that is bounded by the hyperplane H through ¯z orthogonal to [¯v, ¯z]. Bus since ¯y in conv ¯z ⊂ P, the angle ¯y ¯z¯v ≥ π/2, and so the angle
¯
z¯v ¯y < π/2. Furthermore since
ρ(¯v, B) ≤ ρ( ¯w, B) < ρ(¯u, B)
it follows that ¯v 6= ¯u. Thus some point of [¯v, ¯u] is closer to ¯z than ¯v is. This contradicts our choice of ¯v and the lemma is proved.
Figure 4: An illustration of the proof of lemma 3.1.
Proof. (Theorem 3.9) Set
B∗ = \
α∈I
Bα.
By Helly’s theorem for homology cells B∗ is a nonempty homology cell. We show that B∗ is star-shaped. Let ¯x1, . . . , ¯xn+1 be some points in B∗. Let Mα be the set of points ¯x in Bα from which all the points ¯x1, . . . , ¯xn+1can be seen in Bα, for all α ∈ I. The sets Mα is nonempty because the star of Bα lies in Mα. Fix a positive integer k ≤ n + 1. Consider arbitrary sets Mα1, . . . , Mαk and set
Mα1···αk =
k
\
j=1
Mαk. Consider further the set
Bα1···αk =
k
\
j=1
Bαj.
Helly’s Theorems Madeleine Leander
From the hypothesis of the theorem it follows that Bα1···αk is nonempty and star-shaped. Let ¯x0 ∈ s(Bα1···αk). Then every point ¯xj can be seen from ¯x0. Hence ¯x0 ∈ Mα1···αk and therefore
Mα1···αk 6= ∅.
Now let ¯x ∈ s(Bα1···αk), ¯y ∈ Mα1···αk, and consider some point ¯xj, 1 ≤ j ≤ n + 1. From the choice of ¯x and ¯y we can see that the triangle ¯x¯y ¯xj lies in Bα1···αk. Hence, the point ¯xj can be seen, in Bα1···αk, from every point of the segment [¯x, ¯y]. This implies
[¯x, ¯y] ∈ Mα1···αk
and therefore Mα1···αk is star-shaped and
s(Bα1···αk) ⊂ s(Mα1···αk).
By Helly’s theorem for homology cells M∗ = \
α∈I
Mα 6= ∅.
Thus for every n+1 points in B∗ there exists a point, in M∗, from where these n + 1 points can be seen. But then the set B∗ is stare-shaped by lemma 3.1.
The theorem is proved.
3.4 The transversal Helly Number
A family of disks, closed circles of diameter 1, where every k of the members have a common line transversal, a straight line intersecting all the members, is said to be a family with property T (k). We say that the family is t−disjoint if the distance between every pair of of centers is larger than t > 0. We consider the infimum tk of the distances t for which any finite family of t−disjoint unit diameter disks with the property T (k) has a line transversal.
If property T (k) implies property T but property T (k − 1) does not, we say that the family has transversal Helly Number kt. Our concern is to investigate the relation between t−disjointness and the property T (k). The function kt is completely described by the sequence of its points of discontinuity tk = inf{t|kt≤ k − 1}, k ≥ 4.
First we will concentrate on the general lower bound for tk. Theorem 3.10. If k ≥ 4 then tk ≥ sk. Where
sk= 2 sinπk 1 + cos2πk ,
20
if k is odd and if k is even then
sk = 2 sinπk cos2πk + cosπk.
Proof. Let sk be the largest side length of a regular k−gon, R(k), such that a strip of width 1 covers all the vertices of R(k) except one (se Figure 5).
Figure 5: An illustration of R(5) and R(6).
The family of disks centered at the k vertices of R(k) has property T (k−1) but does not have a line transversal. This family is clearly sk− −disjoint, for any positive which implies tk ≥ sk. Some simple geometry yields the indicated values of sk.
For two distinct points X, Y one defines the center sheaf P(X, Y ) to be the locus of the center of a disk D, sush that D and the disks have a common line transversal. A center sheaf is bounded of at most six lines, four of them pass through X or Y and are the tangent to the other disk of radius one and two common tangents to the two disks.
Helly’s Theorems Madeleine Leander
Figure 6: The center sheaf of X and Y.
Lemma 3.2. Let P1(0, −1), P2(x2, 0), x2 ≥ 1. Denote by A(x2) the inter- section of the second quadrant ({(x, y)|x ≤ 0 ≤ y}) and P(P1, P2). Then A1 is covered by the disk of radius 212 centered at P1 and x2 > 1 implies A(x2) ⊂ A(1).
Proof. The lemma easily folows from the fact that only the common upper tangent, of the boundary lines, of the disks of radius 1 centered at P1 and P2 has points in common with the open second quadrant.
Figure 7: A special case of Lemma 3.2.
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Lemma 3.3. Let P1(0, −1), P3(x3, −1), x3 ≥ 2, and denote by B(x3), the intersection of the half-strip {(x, y)|0 ≤ x ≤ x3/2, y ≥ 0} and P(P1, P3).
Then B(2) is covered by the closed disk of radius 212 centered at P1, and x3 > 2 implies B(x3) ⊂ B(2).
Proof. The claim easily follows from the fact that of the boundary lines of P(P1, P3) only the upper tangent of the disk of radius 1 centered at P1 and through P3 cuts into the half-strip {(x, y)|0 ≤ x ≤ x3/2, y ≥ 0}.
Theorem 3.11. A √
2−disjoint family of unit disks having property T (3) also have property T.
Proof. For each triple of centers, consider the narrowest covering strip and let w be maximum of the width, the minimum of the distances between pairs of its parallel support lines, of these strips. Without loss of generality we may assume that w = 1, by the T (3) property we have w ≤ 1. If w < 1 then replace all disks by a disk of radius w/2 and rescale to get a family of disks of diameter 1. Clearly the new family of disks has property T (3), and if it has a transversal then so has the original family. Let S has maximal width 1 and be the narrowest covering strip of the three centers X1, X2, Y. Suppose that X1 and X2 are on one boundary line of S and Y is on the other. The orthogonal projection of Y on the line [X1, X2] is between X1 and X2, since S is the narrowest strip covering the triangle X1X2Y. Each of the other disks share a transversal with any two of the disks around X1, X2 and Y, and thus, their centers lie in the intersection of P(X1, X2),P(X1, Y ),P(X2, Y ). The family is 212-disjoint, and hence all other centers lie outside of the disks of radius 212 centered at X1, X2 and Y. By applying lemma 3.4 and lemma 3.5 at the points of the intersection of P(X1, X2),P(X1, Y ),P(X2, Y ) outside S are covered by the disks of radius 212 centered at X1, X2 and Y, hence none of these points can be the center of a disk in the family. It follows that the axis of S is a transversal of the family. The theorem is thereby proved.
The theorem of Helly was a start point of a rapid development of com- binatorial geometry. Today eighty years later, this particular theorem still attracts interest of many mathematicians. Each year brings new generaliza- tions in different contexts and new applications of this result.
The author is grateful to Paul Vaderlind for helpful discussions and re- marks.
Helly’s Theorems Madeleine Leander
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[2] Lay, Steven R. (1982), Convex sets and their applications. John Wiley
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[8] Philip W. Bean, Helly and Radon-type theorems in interval convexity spaces Pacific Journal of mathematics Vil. 51, No 2, 1974.
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[13] K. Bezdek, T. Bisztriczky, B. Scikos and A. Heppes On the transversal Helly numbers of disjoint and overlapping disksArchiv der Mathematik, Birkh¨auser Basel, Volume 87, Number 1 / July, 2006.
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