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A New Topological Helly Theorem and Some Transversal Results

DOI:10.1007/s00454-014-9613-6
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L. Montejano at Universidad Nacional Autónoma de México
L. Montejano
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We prove that for a topological space (Formula presented.) with the property that (Formula presented.) for (Formula presented.) and every open subset (Formula presented.) of (Formula presented.), a finite family of open sets in (Formula presented.) has nonempty intersection if for any subfamily of size (Formula presented.) the (Formula presented.)-dimensional homology group of its intersection is zero. We use this theorem to prove new results concerning transversal affine planes to families of convex sets.
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A New Topological Helly Theorem and some
Transversal Results
L. Montejano
Instituto de Matemáticas, Unidad Juriquilla.
Universidad Nacional Autónoma de México, México
March 21, 2014
Abstract
We prove that for a topological space X with the property that H
(U) = 0 for
d and every open subset U of X, a …nite family of open sets in X has nonempty
intersection if for any subfamily of size j; 1 j d + 1; the (d j)-dimensional
homology group of its intersection is zero.
We use this theorem to prove new results concerning transversal a¢ ne planes to
families of convex sets
Key words. Helly theorem, homology group, transversal
MSC 2000. Primary 52A35,55N10
1 Introduction and Preliminaries
A prominent role in combinatorial geometry is played by the Helly Theorem [7], which
states that a …nite family of convex sets in R
d
has nonempty intersection if and only if any
subfamily of size at most d + 1 has nonempty intersection. Helly himself realized in 1930
(see [9]) that a …nite family of sets in R
d
has nonempty intersection if for any subfamily
of size at most d + 1; its intersection is homeomorphic to a ball in R
d
. In fact, the result
The author acknowledges support of CONACYT, proyect 166306
1
is true if we replace the notion of topological ball by the notion of acyclic set, see [3] and
[10]. In 1970, Debrunner [8] proved that a …nite family of open sets in R
d
has nonempty
intersection if for any subfamily of size j; 1 j d + 1; its intersection is (d j)-acyclic.
In fact, these hypothesis imply that each one the open sets are acyclic.
Unlike all previous topological Helly results in the literature, where the ambient space
is the euclidean space and the sets are acyclic, we only require as an ambient space a
topological space X in which H
(U) = 0 for d and for any subfamily of size j of our
sets, the (d j)-dimensional homology group of its intersection to be zero.
The fact that this is a non-expensive topological Helly theorem — in the sense that it
does not require the open sets to be simple— from the homotopy point of view (we only
require its (d 1)-dimensional homology group to be zero), allows us to prove interesting
new results concerning transversal planes to families of convex sets.
During this paper, we use reduced singular homology with any nonzero coe¢ cient group.
Let U be a topological space. We say that H
1
(U) = 0 if and only if U is nonempty
and we say that U is connected if and only if H
0
(U) = 0. For q = 1 and q = 0,
the exactness of the Mayer-Vietoris sequence should be understood as the statement that
if H
q
(U) occurs there between two vanishing terms, then H
q
(U) = 0: For an integer
n 1; we say that U is n-acyclic if H
(U) = 0 for 1 n: Furthermore, U is
acyclic if H
(U) = 0 for 1:
2 A Topological Helly-Type Theorem
We start with the following auxiliary proposition.
Proposition A
(m;)
: Let F = fA
1
; :::; A
m
g be a family of open subsets of a topological
space X and let 0 be an integer. Suppose that for any subfamily F
0
F of size j;
1 j m,
H
m1j+
(
T
F
0
) = 0.
Then
H
m2+
(
S
F ) = 0.
Proof. Proposition A
(2;0)
claims that the union of two connected sets with nonempty
intersection is a connected set, and Proposition A
(2;)
is just the statement of the exactness
2
of the Mayer-Vietoris sequence: 0 = H
(A
1
)H
(A
2
) ! H
(A
1
[A
2
) ! H
1
(A
1
\A
2
) =
0:
The proof is by induction on m. In fact, we shall prove that Proposition A
(m;)
together
with Proposition A
(m;+1)
implies Proposition A
(m+1;):
Suppose F = fA
0
; :::; A
m
g is a …nite collection of m + 1 open subsets of X such that
for 1 j m + 1 and any subfamily F
0
F of size j; H
mj+
(
T
F
0
) = 0: We
will prove that H
m1+
(
S
F ) = 0: Let us …rst prove, using Proposition A
(m;+1)
; that
H
m2+
(A
1
[ ::: [ A
m
) = 0: This is so because for any subfamily F
0
fA
1
; :::; A
m
g of size
j; 1 j m, we have H
m1j+(+1)
(
T
F
0
) = 0:
Let us consider the Mayer-Vietoris exact sequence of the pair ((A
1
[ ::: [ A
m
); A
0
) :
0 = H
m1+
(A
0
) H
m1+
(A
1
[ ::: [ A
m
) ! H
m1+
(
S
F ) !
H
m2+
(A
0
\ (A
1
[ ::: [ A
m
)) = 0:
Since by hypothesis, H
m1+
(A
0
) = 0; in order to conclude the proof of Proposition
A
(m+1;)
it is su¢ cient to prove that H
m2+
(A
0
\ (A
1
[ ::: [ A
m
)) = 0: For that purpose,
let G = fB
1
; :::; B
m
g be the family of open subsets of X given by B
i
= A
0
\ A
i
, 1
i m: Note that for any subfamily G
0
G of size j; 1 j m, H
m1j+
(
T
G
0
) = 0:
This is so because the homology group H
m1j+
(
T
G
0
) = H
m(j+1)+
(
T
F
0
) = 0; where
F
0
is the corresponding subfamily of F of size j + 1. Then by Proposition A
(m;)
; 0 =
H
m2+
(
S
G) = H
m2+
(A
0
\ (A
1
[ ::: [ A
m
)): This completes the proof.
We now give the Topological Berge’s Theorem. See [1].
Theorem B
(m;)
: Let F = fA
1
; :::; A
m
g be a family of open subsets of a topological space
X and let 0 be an integer. Suppose that
a) H
m2+
(
S
F ) = 0;
b) for 1 j m 1 and any subfamily F
0
F of size j;
H
m2j+
(
T
F
0
) = 0:
Then
H
1
(
T
F ) = 0:
3
Proof. The proof is by induction. Theorem B
(2;0)
claims that two nonempty open sets
whose union is connected must have a point in common and Theorem B
(2;)
is just the
statement of the exactness of the Mayer-Vietoris sequence: 0 = H
(A
1
[A
2
) ! H
1
(A
1
\
A
2
) ! H
1
(A
1
) H
1
(A
2
) = 0:
Let us prove that Theorem B
(m;)
implies Theorem B
(m+1;):
Let F = fA
0;
A
1
; :::; A
m
g as
in Theorem B
(m+1;)
: That is, H
m1+
(
S
F ) = 0 and for 1 j m and any subfamily
F
0
F of size j we have H
m1j+
(
T
F
0
) = 0. Let G = fB
1
; :::; B
m
g; where B
i
= A
0
\A
i
;
1 i m: In order to prove that H
1
(
T
F ) = H
1
(
T
G) = 0; it is enough to show
that the family G = fB
1
; :::; B
m
g satis…es properties a) and b) of Theorem B
(m;)
:
Proof of a). We need to prove that H
m2+
(A
0
\ (A
1
[ ::: [ A
m1
)) = H
m2+
(
S
G) = 0:
Note that H
m1+
(
S
F ) = 0 and H
m2+
(A
0
) = 0: Furthermore, by Proposition A
(m;)
;
for the family fA
1
; :::; A
m
g; we have that H
m2+
(A
1
[ ::: [ A
m
) = 0: Thus the conclusion
follows from the Mayer-Vietoris exact sequence of the pair (A
0
; (A
1
[ ::: [ A
m
));
0 = H
m1+
(A
0
[ ::: [ A
m
) ! H
m2+
(A
0
\ (A
1
[ ::: [ A
m
)) !
H
m2+
(A
0
) H
m2+
(A
1
[ ::: [ A
m
) = 0:
Proof of b). For 1 j m 1 and any subfamily G
0
G of size j;
T
G
0
=
T
F
0
; where
F
0
F has size j + 1: Thus H
m1(j+1)+
(
T
F
0
) = H
m2j+
(
T
G
0
) = 0: This completes
the proof of Theorem B
(m+1;)
:
We now state our main theorem.
Topological Helly Theorem. Let F be a …nite family of open subsets of a topological
space X. Let d > 0 be and integer such that H
i
(U) = 0 for i d and every open subset
U of X:
Suppose that
H
dj
(
T
F
0
) = 0
for any subfamily F
0
F of size j; 1 j d + 1: Then
T
F 6= ?:
Furthermore,
T
F is acyclic.
Proof. Suppose the size of F is m. Take an integer 3 n m + 1: Using Theorem
B
(n1;)
, from = 0 up to = n 3; we can prove the following:
Claim C
n
: Suppose that for every 1 j n 1 and any subfamily F
0
F of size j;
4
H
n3
(
S
F
0
) = 0; and
for every 1 j n 2 and any subfamily F
0
F of size j;
H
nj3
(
T
F
0
) = 0:
Then for every 1 j n 1 and any subfamily F
0
F of size j;
H
nj2
(
T
F
0
) = 0:
Assume now H
(U) = 0 for every d and every open U X and suppose that
d n 3: By repeating the use of Claim C
n
; from n = d + 3 up to n = m + 1; we obtain
that
T
F 6= ?:
Arguing as above and using Theorem B
(m1;)
, from = 0 up to = m 3; we obtain
that H
0
(
T
F ) = 0: The conclusion of acyclicity can be achieved by repeating the use of
Theorem B
(n;)
; 2 n m 1; 1 m 3: Note that in the case m = d + 2; our
argument does not produce H
d1
(
T
F ) = 0, so we need to continue to = m 2: This
concludes the proof of our main theorem.
For completeness, we include here a Topological Breen’s Theorem.
Theorem
m
: Let F = fA
1
; :::; A
m
g be a family of open subsets of a topological space
X. Suppose that for 1 j m and any subfamily F
0
F of size j;
H
j2
(
S
F
0
) = 0:
Then
T
F 6= ?:
Proof. The proof is by induction. Theorem
2
claims that two nonempty open sets
whose union is connected must have a point in common. Suppose Theorem
m
is true
and let F = fA
0
; A
1
; :::; A
m
g be a family of open subsets of X such that for 1 j m
and any subfamily F
0
F of size j; H
j2
(
S
F
0
) = 0:
Let us prove …rst that for any subfamily F
0
fA
2
; :::; A
m
g of size j; 0 j m 1;
H
j1
((A
0
\ A
1
) [
S
F
0
) = 0:
5
To do so, simply consider the Mayer-Vietoris exact sequence of the pair (A
0
[
S
F
0
; A
1
[
S
F
0
) :
0 = H
j
(A
0
[A
1
[
S
F
0
) ! H
j1
((A
0
\A
1
)[
S
F
0
) ! H
j1
(A
0
[
S
F
0
)H
j1
(A
1
[
S
F
0
) =
0:
This implies that the family fA
0
\A
1
; A
2
; :::; A
m
g satis…es the hypothesis of Theorem
m
;
and therefore by induction that A
0
\ A
1
\ ::: \ A
m
6= ?: This completes the proof of this
theorem.
As an immediate consequence, we have the following theorem:
Topological Breen Theorem. Let F be a …nite family of open subsets of a topological
space X. Let d > 0 be and integer such that H
(U) = 0 for d and every open subset
U of X.
Suppose that
H
j2
(
S
F
0
) = 0
for 1 j d + 1 and any subfamily F
0
F of size j: Then
T
F 6= ?:
Remark. The corresponding theorems for
µ
Cech cohomology groups are also true. Fur-
thermore, the theorems in this section are true for a class of sets A
i
in which the Mayer-
Vietoris exact sequence of the pair (A
0
; (A
1
[ ::: [ A
m
)) is exact. For example, when X is
a polyhedron and every A
i
X is a subpolyhedron.
3 Transversal Theorems
3.1 Preliminary Lemmas
We start with some notation.
Let G(n; d); be the Grassmannian space of all n-planes in R
d
through the origin and let
M(n; d) be the space of all a¢ ne n-planes in R
d
as an open subset of G(n + 1; d + 1):
6
Let F be a collection of nonempty convex sets in euclidean d-space R
d
and let 0 n < d
be an integer. We denote by T
n
(F ) G(n; d) the topological space of all n-planes in R
d
transversals to F ; that is, the space of n-planes that intersect all members of F . We say
that F is separated if for every 2 n d and every subfamily F
0
F of size n, there is
no (n 2)-plane transversal to F
0
:
Let F = fA
1
; :::; A
n
g be a collection of closed subsets of a metric space X and let > 0
be a real number. We denote by F
= fA
1
; :::; A
n
g the collection of open subsets of X,
where A
denotes the open -neighborhood of A X:
Lemma 3.1.1. Let A be a nonempty convex set in R
d
and let 1 n < d: Then T
n
(fAg)
is homotopically equivalent to G(n; d); the Grassmannian space of all n-planes in R
d
through the origin.
Proof. Let : T
n
(fAg) ! G(n; d) be given as follows: for every H 2 T
n
(fAg); let
(H) be the unique n-plane through the origin parallel to H. Then if 2 G(n; d);
1
() is homeomorphic to (A); where : R
d
!
?
is the orthogonal projection and
?
2 G(d n; d) is orthogonal to : Since has contractible …bers, it is a homotopy
equivalence.
Lemma 3.1.2. Let F = fA
1
; A
2
; :::; A
n
g be a separated family of nonempty convex sets
in R
d
; 2 n d; and let n m d be an integer. Then T
m1
(F ) is homotopically
equivalent to G(m n; d n + 1):
Proof. We start by proving that T
n1
(F ) is contractible. For this purpose let :
A
1
::: A
n
! T
n1
(F ) given by ((a
1
; :::; a
n
)) be equal to the unique (n 1)-plane
in R
d
through fa
1
; :::; a
n
g; for every (a
1
; :::; a
n
) 2 A
1
::: A
n
: Note that is well
de…ned because F is a separated family of sets. Furthermore, if H 2 T
n1
(F ); then
1
(H) = (H \ A
1
) ::: (H \ A
n
) is contractible. This implies that is a homotopy
equivalence and hence that T
n1
(F ) is contractible.
Let E = f(H; ) j H is a (n1)-plane of R
d
; is a (m1)-plane of R
d
and H g: Then
: E ! M(n 1; d); given by the projection in the …rst coordinate, is a classical …ber
bundle with …ber G(mn; dn+1). Now let Y = f(H; ) 2 T
n1
(F )T
m1
(F ) j H g:
Clearly, the restriction j: Y ! T
n1
(F ) is a …ber bundle with …ber G(m n; d n + 1)
and contractible base space T
n1
(F ): Therefore j: Y ! T
n1
(F ) is a trivial …ber bundle
and hence Y is homotopically equivalent to G(m n; d n + 1):
Consider now the projection : Y ! T
m1
(F ): Note that for every 2 T
m1
(F ); the
…ber
1
() is equal to T
n1
(fA
1
\; :::; A
n
\g): By the …rst part of this proof, the …bers
of are contractible, hence is a homotopy equivalence and T
m1
(F ) is homotopically
equivalent to G(m n; d n + 1):
7
Lemma 3.1.3. Let A; B, and C be three nonempty convex sets in R
d
such that A\B = ?:
Then
H
1
(T
1
(fA; B; Cg)) = 0:
Proof. Since A \ B \ C = ?; by Theorem 3 of [4], T
1
(fA; B; Cg) has the homotopy type
of the space C
1
(fA; B; Cg) C
1
2
of all a¢ ne con…gurations of three points in the line,
achieved by transversal lines to fA; B; Cg: Note now that the space of a¢ ne con…guration
of three points in a line, C
1
2
, is S
1
and note further that since A \ B = ?; the space of all
a¢ ne con…gurations of three points in the line achieved by transversal lines to fA; B; Cg
is a subset S
1
f1g; where 1 2 S
1
is the a¢ ne con…guration in which the …rst and the
second points coincide. This implies that H
1
(T
1
(fA; B; Cg)) = 0:
Lemma 3.1.4. Let F = fA
1
; :::; A
d+1
g be a separated family of closed convex sets in R
d
.
Suppose that H
0
(T
d1
(F )) = 0: Then there is
0
> 0 with the property that if 0 < <
0
;
then H
0
(T
d1
(F
)) = 0:
Proof. By Theorem 1 of [4], the space of transversals T
d1
(F ) of a separated family of
convex sets in R
d
has …nitely many components and each one of them is contractible.
In fact, each component corresponds precisely to a possible order type, of d 1 points
in a¢ ne (d 1)-space, achieved by the transversal hyperplanes when they intersect the
family F . In our case, since H
0
(T
d1
(F )) = 0; we have that T
d1
(F ) is contractible and
that the transversal hyperplanes intersect the family F consistently with a precise order
type :
Suppose now the lemma is not true, then there exist an order type
0
, di¤erent from ;
and a collection of hyperplanes H
i
that intersect F
i
consistently with the order type
0:
Since we may assume that f
i
g ! 0 and fH
i
g ! H; where H is a transversal hyperplane
to F consistently with the order type
0
; we have a contradiction.
3.2 Transversal Lines in the Plane
A family of sets is called semipairwise disjoint if, given any three elements of F , two of
them are disjoint.
Theorem 3.2.1. Let F be a semipairwise disjoint family of at least 6 open convex sets
in R
2
: Suppose that for every subfamily F
0
F of size 5, T
1
(F
0
) 6= and for every
subfamily F
0
F of size 4, T
1
(F
0
) is connected. Then T
1
(F ) 6= ?.
8
Proof. Let X be the space of all lines in R
2
: Hence H
(U) = 0 for 2 and every
open subset U X. We are interested in applied the Topological Helly Theorem when
d = 4. Note …rst that H
3
(T
1
(fAg) = 0 for every A 2 F; and H
2
(T
1
(fA; Bg) = 0 for
A 6= B 2 F: By Lemma 3.1.3 and the fact that F is semipairwise disjoint, we have that
H
1
(T
1
(F
0
)) = 0; for any subfamily F
0
F of size 3. By hypothesis, H
0
(T
1
(fF
0
g) = 0; for
any subfamily F
0
F of size 4; and H
1
(T
1
(fF
0
g) = 0; for any subfamily F
0
F of size
5. This implies, by the Topological Helly Theorem, that T
1
(F ) is nonempty.
3.3 Transversal Lines in 3-Space
In this section we study transversal lines to families of convex sets in R
3
:
Theorem 3.3.1. Let F be a pairwise disjoint family of at least 6 open, convex sets in
R
3
. Suppose that for any subfamily F
0
F of size 5, T
1
(F
0
) 6= ?; and for any subfamily
F
0
F of size 4, T
1
(F
0
) is connected. Then, T
1
(F ) 6= ?.
Proof. Let X be the space of all lines in R
3
; hence X is an open 4-dimensional manifold
and therefore H
(U) = 0 for 4 and every open subset U X. We are interested
in applied the Topological Helly Theorem for d = 4. By Lemma 3.1.1, H
3
(T
1
(fAg) = 0;
for every A 2 F; since T
1
(fAg) has the homotopy type of G(1; 3) = RP
2
: By Lemma
3.1.2, H
2
(T
1
(fA; Bg)) = 0; for every A 6= B 2 F: By Lemma 3.1.3 and the fact that F is
pairwise disjoint, we have that H
1
(T
1
(F
0
)) = 0; for any subfamily F
0
F of size 3. By
hypothesis, H
0
(T
1
(fF
0
g) = 0; for any subfamily F
0
F of size 4; and H
1
(T
1
(fF
0
g) = 0;
for any subfamily F
0
F of size 5. This implies, by the Topological Helly Theorem, that
T
1
(F ) is nonempty.
3.4 Transversal Hyperplanes
This section is devoted to stating and proving a theorem concerning transversal hyper-
planes to families of separated convex sets in d-space.
Theorem 3.4.1. Let F be a separated family of at least d + 3 closed, convex sets in
R
d
: Suppose that for any subfamily F
0
F of size d + 2, T
d1
(F
0
) 6= ? and for any
subfamily F
0
F of size d + 1, T
d1
(F
0
) is connected. Then T
d1
(F ) 6= ?.
Proof. Let us …rst prove the theorem for a separated family of open convex sets. We are
going to use the Topological Helly Theorem. Let X be the space of all hyperplanes of R
d
:
9
Note that H
(U) = 0 for d and every open subset U X: In particular, H
(U) = 0
for every d +1:
By Lemma 3.1.2, for every subfamily F
0
F of size j, 1 j d; T
d1
(F
0
) is homotopi-
cally equivalent to G(dj; dj+1) and hence H
dj+1
(T
d1
(F
0
)) = H
dj+1
(
T
fT
d1
(fAg) j
A 2 F
0
g) = 0: Furthermore, by hypothesis, the same is true for j = d + 1 and j = d + 2:
Consequently, by our Topological Helly Theorem, T
d1
(F ) 6= ?.
By Lemma 3.1.4, there is > 0; such that F
is a separated family of open, convex sets in
R
d
and for any subfamily F
0
F
of size d + 1; T
d1
(F
0
) is connected. By the above,
this implies that T
d1
(F
)) 6= ?: Hence, by completeness of the Grassmannian spaces,
T
d1
(F ) 6= ?.
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11
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