Csaba Vincze (2013)

University of Debrecen

A recent trend in convex geometry is to investigate intersections of sets under a weaker condition than convexity. In 2001 N. A. Bobylev [9] provided a star-shaped set analogue of Helly's theorem. Especially Bobylev proved that if we have a family B of compact sets in the coordinate space of dimension n and every n+1 (not necessarily different) members of B have a star-shaped intersection then the intersection of all elements in B is star-shaped. The star- shaped set analogue of Klee's theorem 4.1.6 is true as well. In fact the result due to Marilyn Breen [15] states that the associated intersection is non-empty, star-shaped and its kernel is at least k-dimensional for an appropriate choice of k between 0 and n.

**Theorem 6.1.1 **
*Consider a collection B of sets in the coordinate space of dimension n and let *

$0\le k\le n$ |

be a fixed integer. If every countable subfamily containing not necessarily different members of B has a star-shaped intersection whose kernel is at least k-dimensional then the intersection of all the members in B is a star-shaped set whose kernel is at least k-dimensional.

The proof is actually an induction on the dimension of the embedding space. In case of the coordinate line of dimension n=1 any star-shaped set isconvex and, by Klee's theorem 4.1.6, the intersection of all the members in B is non-empty and convex (especially star-shaped). If the intersection contains at least two different points then it is of dimension 1 and the proof is finished independently of the values k=0 or 1. Suppose now that the intersection is a singleton:

${\cap}_{\gamma \in \mathrm{\Gamma}}{B}_{\gamma}=\left\{x\right\}$ |
(6.1) |

(for the sake of simplicity countable subfamilies will be labelled by natural numbers as usual to avoid double superscripts). We are going to prove that there exists a sequence of the members in B whose intersection reduces to a single point too. According to 6.1 for any natural number m there exists a set B(m) in B such that

$x+\frac{1}{m}\notin {B}_{m}.$ |

In a similar way let B(- m) be a set in B such that

$x-\frac{1}{m}\notin {B}_{-m}.$ |

By our assumption the intersection

$\bigcap _{m=1}^{\mathrm{\infty}}({B}_{-m}\cap {B}_{m})$ |
(6.2) |

is star-shaped (especially convex) which means that it must be reduced to a singleton. Otherwise we have a segment in 6.2 containing x together with points having arbitrarily small rational distance 1/m from x which is a contradiction. Suppose that the statement is true for the coordinate space of dimension at most n - 1. By the help of the inductive hypothesis one can prove the following lemma which is the key step in the proof of theorem 1.1.

**Lemma 6.1.2 **
*Under conditions of theorem 6.1.1 the intersection of all the members in B is non-empty and it contains a k-dimensional convex subset. *

Using 6.1.2 the proof of theorem 6.1.1 can be finished as follows. Let

$V=\bigcap _{\gamma \in \mathrm{\Gamma}}{B}_{\gamma}$ |

be the intersection of all the members in B. We should check that V is star-shaped and its kernel is at least k-dimensional. Consider a new collection

${M}_{\gamma}=\{p\in {B}_{\gamma}|s(p,v)\subset {B}_{\gamma}\text{}for\text{}all\text{}v\in V\}.$ |
(6.3) |

M(γ) contains the points of B(γ) from which any point in V can be visible. Consider a countable subfamily M(1), ..., M(m), ... together with the associated sets B(1), ..., B(m), ... and let

$K=\bigcap _{i=1}^{\mathrm{\infty}}{B}_{i}\text{}and\text{}M=\bigcap _{i=1}^{\mathrm{\infty}}{M}_{i}$ |

be the intersections of the corresponding countable family of sets. Choose a point p in Ker K. Since V is a subset in K

$s(p,v)\subset K\subset {B}_{i}$ |

for any element v in V and i=1, 2, ... Therefore p is in M. Especially M is a subset in K. This means that any element of M is visible from p, i.e.

$KerK\subset KerM.$ |
(6.4) |

Relation 6.4 says that we can apply lemma 6.1.2 to the collection 6.3:

$\bigcap _{\gamma \in \mathrm{\Gamma}}{M}_{\gamma}\ne \mathrm{\varnothing}$ |

and the intersection contains an at least k-dimensional convex subset. To finish the proof observe that

$\bigcap _{\gamma \in \mathrm{\Gamma}}{M}_{\gamma}=KerV$ |
(6.5) |

because the left hand side contains just the points of V from which any point in V can be visible, cf. 6.3. Therefore V is a star-shaped set and Ker V contains an at least k-dimensional convex subset.** The proof of the key lemma **6.1.2 is based on the inductive hypothesis. Two different cases should be considered.

I.** First case**. Suppose that for some countable subfamily B(1), ..., B(m), ... the intersection K of the sets is at most of dimension n - 1. Then the maximal dimension of a convex subset M in K is also less than the dimension of the embedding space. Let B(1), ..., B(m), ... be chosen in such a way thatthe following** minimax condition **is satisfied: the maximum of the dimension of convex subsets in the intersection is as small as possible. This means that if we can inscribe a τ-dimensional convex subset into K as that of maximal dimension, then the intersection of any countable subfamily contains an at least τ-dimensional convex subset.

**Lemma 6.1.3 **
*If the family *

${B}_{1},{B}_{2},\mathrm{\dots},{B}_{m},\mathrm{\dots}$ |

satisfies the minimax condition then

$KerK\subset affM,$ |

where M is a convex subset of maximal dimension in K.

**Proof **Suppose, in contrary, that we have a point p in Ker K which is not in the affine hull of M. Then the convex hull of the union of M and p would be a greater dimensional convex subset in K than M.

Conditions for M are

(M1) M is convex,

(M2) M is a subset in K,

(M3) M is of dimension τ.

Consider the intersection of the affine hulls of M as it runs through the subsets satisfying M1, M2 and M3. We have by lemma 6.1.3 that

$KerK\subset \bigcap _{M}affM=H,$ |

where H is an affine subspace of dimension at most τ. Suppose that the dimension of the affine subspace H associated with the countable family B(1), ..., B(m), ... satisfying the minimax condition is as great as possible. Choose an arbitrary countable subfamily B*(1), ..., B*(m), ... with intersection K* and let

${B}_{1},\mathrm{\dots},{B}_{m},\mathrm{\dots},{B}_{1}^{\mathrm{*}},\mathrm{\dots},{B}_{m}^{\mathrm{*}},\mathrm{\dots}$ |
(6.6) |

be the union of the subfamilies with intersection

$T=K\cap {K}^{\mathrm{*}}.$ |

Recall that T contains a convex subset of dimension at least τ but T is a subset in K. Therefore 6.6 also satisfies the minimax condition. Thus

$\mathrm{d}\mathrm{i}\mathrm{m}L\le \mathrm{d}\mathrm{i}\mathrm{m}H$ |

for the dimension of the associated affine subspace

$L=\bigcap _{M\mathrm{\text{'}}}affM\mathrm{\text{'}},$ |

where M' is a τ-dimensional convex subset in T. Especially, M' is a subset in K. Therefore

$H\subset L$ |

showing that H=L, i.e. the associated affine subspaces coincide because of the maximality condition for the dimension of H. On the other hand lemma 1.3 implies again that Ker T is a subset in L and, consequently,

$KerT\subset H.$ |

Finally we apply the inductive hypothesis to the family of sets

${B}_{\gamma}\cap K\cap H$ |
(6.7) |

in the coordinate space of dimension at most τ. The intersection of sets in 6.7 is contained in the intersection of all the members in B together with its k-dimensional kernel which is a convex set.

II.** Second case**. Suppose that for every countable subfamily has an n-dimensional intersection and let

$P=\left\{G\right|G={\cap}_{m=1}^{\mathrm{\infty}}{B}_{m}\}$ |

be the collection of intersections of countable subfamilies. It can be easily seen that

$\bigcap _{G\in P}G=\bigcap _{\gamma \in \mathrm{\Gamma}}{B}_{\gamma}$ |
(6.8) |

and for any countable subfamily

${G}^{1},...,{G}^{m},...$ |
(6.9) |

we have a corresponding countable subfamily

${B}_{1}^{1},\mathrm{\dots},{B}_{m}^{1},\mathrm{\dots},{B}_{1}^{2},\mathrm{\dots},{B}_{m}^{2},\mathrm{\dots},{B}_{1}^{j},\mathrm{\dots},{B}_{m}^{j},\mathrm{\dots},$ |
(6.10) |

where

${G}^{j}=\bigcap _{m=1}^{\mathrm{\infty}}{B}_{m}^{j}$ |

for any index j. The intersection of 6.9 is just that of 6.10. Therefore the intersection of 6.9 has a non-empty interior and the same is true for the intersection of the closures of sets in 6.9. Let T be the intersection of the closures of sets in P (cf. theorem 4.1.5). Since the complement of T is expressed as the union of open subsets we can choose, by Lindelöf's theorem, a countable subcover. Taking the complement again T can be expressed as the intersection of a countable subfamily (say 6.9) of the closures:

$T=\bigcap _{j=1}^{m}the\text{}closure\text{}of\text{}{G}^{j}=\bigcap _{G\in P}the\text{}closure\text{}of\text{}G.$ |

Since the interior of T is non-empty it contains an open ball D in its interior. If D is a subset of the intersection of sets without closure operator the proof is finished. Otherwise suppose that p is a point in D such that p is not in G(1) (but p is in the closure of G(1)). Since G(1) is represented as a countable intersection of sets from B we can suppose thatp is not in B(1) for some set in B. Consider a countable subfamily B(1), ..., B(m), ... of subsets in B. The intersection

$K=\bigcap _{i=1}^{\mathrm{\infty}}{B}_{i}$ |
(6.11) |

is a star-shaped set. Let z(1) be a point of Ker K. Since p is not in B(1) we have that p is not in K and the ray emanating from p into the opposite direction relative to z(1) does not contain points from K because they would be visible from z(1) together with p.

Recall that p is an interior point of the intersection of the closures of sets in P and thus it is an interior point of the closure of K (belonging to P). So we have a segment S(1) on the line of p and z(1) such that it has no common points with K but it is in the interior of the closure of K. If the kernel of K is a subset of aff S(1) we are ready. Otherwise we can construct a two-dimensional triangle S(2) such that it is in the interior of the closure of K but there are no common points with K.

If the kernel of K is a subset of aff S(2) we are ready. Otherwise repeat the process to construct a tetrahedron S(3) such that it has no common points with K but it is in the interior of the closure of K.

Repeat the algorithm as far as possible we can construct a j-dimensional simplex S(j) such that it has no common points with K but it is in the interior of the closure of K. By the constructing process,

$\mathrm{d}\mathrm{i}\mathrm{m}{S}_{j}=\mathrm{d}\mathrm{i}\mathrm{m}KerK\ge k.$ |

Let τ be the number of the maximal dimension of sets in D which are disjoint from K. Then τ is greater than or equal to k but it must be less than n. It is clear because in case of τ=n there would be an n-dimensional simplex S which is disjoint from K (i.e. the interior points of S could not be limits of sequences in K) but S is a subset ofthe closure of K. Therefore

$k\le \tau \le n-1.$ |

Suppose that the collection B(1), ..., B(m), ... has the greatest possible value of τ and consider the family of sets

${B}_{\gamma}\cap K\cap H,$ |
(6.12) |

where H is the affine hull of S(τ). Using the inductive hypothesis, the intersection

$\bigcap _{\gamma \in \mathrm{\Gamma}}{B}_{\gamma}\cap K\cap H$ |

is a star-shaped set with kernel of dimension at least k and the proof is finished.