Abstract
A (homogeneous) d-interval is a union of d closed intervals in the line. Using topological methods, Tardos and Kaiser proved that for any finite collection of d-intervals that contains no k + 1 pairwise disjoint members, there is a set of O(dk) points that intersects each member of the collection. Here we give a short, elementary proof of this result. A (homogeneous) d-interval is a union of d closed intervals in the line. Let H be a finite collection of d-intervals. The transversal number τ(H) of H is the minimum number of points that intersect every member of H. The matching number ν(H) of H is the maximum number of pairwise disjoint members of H. Gyarfas and Lehel [3] proved that τ ≤ O(νd!) and Kaiser [4] proved that τ ≤ O(d2ν). His proof is topological, applies the Borsuk-Ulam theorem and extends and simplifies a result of Tardos [5]. Here we give a very short, elementary proof of a similar estimate, using the method of [2]. Theorem 1 Let H be a finite family of d-intervals containing no k + 1 pairwise disjoint members. Then τ(H) ≤ 2d2k. Proof. Let H′ be any family of d-intervals obtained from H by possibly duplicating some of its members, and let n denote the cardinality of H′. Note that H′ contains no k + 1 pairwise disjoint members. Therefore, by Turan’s Theorem, there are at least n(n − k)/(2k) unordered intersecting pairs of members of H′. Each such intersecting pair supplies at least 2 ordered pairs (p, I), where p is an end point of one of the intervals in a member of H′, I is a different member of H′, and p lies in I. Since there are altogether at most 2dn possible choices for p, there is such a point that lies in at least n(n−k) k2dn members of H ′ besides the one in which it is an endpoint of an interval, showing that there is a point that lies in at least n 2dk of the members of H ′. This implies that for any rational weights on the members of H there is a point that lies in at least a fraction 1 2dk of the total weight. By the min-max theorem it follows that there is a collection of m points so that each member of H contains at least m/(2dk) of them, and thus contains an interval that contains at least m/(2d2k) of the points. Order the points from left to right, and take the set of all points whose rank in this ordering is divisible by dm/(2d2k)e. This is a set of at most 2d2k points that intersects each member of H, completing the proof. 2 Remarks. It may be possible to improve the constant factor in the above proof. Kaiser’s estimate is indeed better by roughly a factor of 2; τ(H) ≤ (d2 − d+ 1)ν(H). It will be interesting to decide if the quadratic dependence on d is indeed best possible. Higher dimensional extensions are possible, using the techniques in [2], [1]. ∗Department of Mathematics, Raymond and Beverly Sackler Faculty of Exact Sciences, Tel Aviv University, Tel Aviv, Israel. Research supported in part by a USA-Israeli BSF grant and by the Hermann Minkowski Minerva Center for Geometry at Tel Aviv University. Email: noga@math.tau.ac.il.
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