Pach, János

Tóth, Géza

Infoscience - École polytechnique fédérale de Lausanne

Comment on Fox NewsJanos Pahand Geza TothyRenyi Institute, Hungarian Aademy of SienesAbstratDoes there exist a onstant  > 0 suh that any familyof n ontinuous ars in the plane, any pair of whih intersetat most one, has two disjoint subfamilies A and B withjAj; jBj  n with the property that either every element ofA intersets all elements of B or no element of A intersetsany element of B? Based on a reent result of Fox, we showthat the answer is no if we drop the ondition that two arsan ross at most one.1 IntrodutionIt was shown in [4℄ that any family of n segments in the plane hastwo disjoint subfamilies A and B, eah of size at least onstant timesn, suh that either every element of A intersets all elements of Bor no element of A intersets any element of B. In [1℄, this resultwas extended to families of algebrai urves with bounded degree atmost D, where the orresponding onstant depends on D.More generally, letG be the intersetion graph of n d-dimensionalsemialgebrai sets of degree at most D. Then there exist two disjointsubsets A;B  V (G) suh that jAj; jBj  (d;D)n and one of thefollowing two onditions is satised:1. ab 2 E(G) for all a 2 A; b 2 B,Supported by NSF grant CCR-0514079 and grants from NSA, PSC-CUNY,Hungarian Researh Foundation, and BSF.ySupported by OTKA-T-038397 and OTKA-T-046246.12. ab 62 E(G) for all a 2 A; b 2 B.Here (d;D) is a positive onstant depending only on d and D.It is not ompletely lear whether the assumption that the setsare semialgebrai an be weakened. For example, a similar resultmay hold for intersetion graphs of plane onvex sets. Clearly, thesame theorem is false for intersetion graphs of three-dimensionalonvex bodies, beause any nite graph an be represented in suha way, and a random graph G with n verties almost surely does nothave A;B  V (G) satisfying onditions 1 or 2 with jAj; jBj   logn,if  is large enough.It would be interesting to analyze intersetion graphs of ontinu-ous ars in the plane. (These are often alled \string graphs" in theliterature [2℄.) We have been unable to answer the following questioneven for k = 1, that is, for pseudo-segments.Problem 1.1. Is it true that any family of n ontinuous ars inthe plane, any pair of whih interset at most k times, has twodisjoint subfamilies A and B with jAj; jBj  kn suh that eitherevery element of A intersets all elements of B or no element of Aintersets any element of B? (Here k> 0 is a suitable onstant.)It follows from a beautiful reent result of Jaob Fox [3℄ (seeTheorem 2.2 below) that the answer to the above question is negativeif we drop the ondition on pairwise intersetions.Proposition 1.2. Fix " 2 (0; 1). For every n, there is a family ofn ontinuous real funtions dened on [0; 1℄ suh that their inter-setion graph G has no omplete bipartite subgraph with at least(")nlognverties in eah of its vertex lasses, and every vertex of Gis onneted to all but at most n"other verties.Obviously, the last ondition implies that G has no two disjointnonempty sets of verties A and B with jA [ Bj > n"suh that novertex in A is onneted to any element of B by an edge.2 Proof of Proposition 1.2We need a simple representation lemma.2Lemma 2.1. The elements of every nite partially ordered set(fp1; p2; : : :g; <) an be represented by ontinuous real funtions f1; f2; : : :dened on the interval [0; 1℄ suh that fi(x) < fj(x) for every x ifand only if pi< pj(i 6= j).Moreover, we an assume that the graphs of any pair of funtionsfiand fjare either disjoint or have nitely many points in ommon,at whih they properly ross.Proof. Let P = fp1; p2; : : : p`g. We desribe a reursive onstrutionwith the additional property that for any extension of (P;<) to atotal order pk(1)< pk(2)< : : : < pk(`), there exists x 2 [0; 1℄ suh thatfk(1)(x) < fk(2)(x) < : : : < fk(`)(x).The proof is by indution on the number of elements of P . For` = 1, there is nothing to prove. For ` = 2, there are two possibilities.If p1< p2, then the funtions f1 1, f2 2 meet the requirements.If p1and p2are inomparable, then let f1(x) = x, f2(x) = 1   x.Now (P;<) an be extended to a total order in two dierent ways.Aordingly, f1(x) < f2(x) at x = 0 and f2(x) < f1(x) at x = 1.Let `  3, and suppose without loss of generality that p`is aminimal element of P . Assume reursively that we have alreadyonstruted ontinuous real funtions f1; f2; : : : ; f` 1with the re-quired properties representing the elements of the partially orderedset (P n fp`g; <). Consider now an extension of (P;<) to a totalorder pk(1)< pk(2)< : : : < pk(`). Clearly, p`appears in this sequene,i.e., ` = k(m) for some 1  m  `. By our assumption, there existsx 2 [0; 1℄ suh thatfk(1)(x) < : : : < fk(m 1)(x) < fk(m+1)(x) < : : : < fk(`):In fat, there exists a whole interval I  [0; 1℄ suh that the aboveinequalities hold for all x 2 I. Now pik a point x2 I and a numberysuh that fk(m 1)(x) < y< fk(m+1)(x), and denef`(x) := y:Repeating this proedure for every permutation (k(1); k(2); : : : ; k(`))for whih pk(1)< pk(2)< : : : < pk(`)is an extension of (P;<) to atotal order, we dene the funtion f`at nitely many points. (Toavoid inonsistenies, we an make sure that we pik a dierent pointxfor eah permutation.)3It remains to verify that this partially dened funtion an beextended to a ontinuous funtion f`: [0; 1℄ ! R meeting the re-quirements. The following two onditions must be satised:1. if p`< pjin (P;<) for some j 6= `, then f`(x) < fj(x) for allx 2 [0; 1℄;2. if p`and pjare inomparable in (P;<) for some j 6= `, thenthe graphs of f`and fjross eah other.Notie that eah point (x; y) onstruted during the above pro-edure lies below the lower envelope (pointwise minimum) of thefuntions fj(x) over all j for whih pj> p`in (P;<). Pik a pointx02 [0; 1℄ distint from all previously seleted points x2 [0; 1℄, andlet f`(x0) := y0for somey0< min1j<`fj(x0):Extend f`to a ontinuous funtion on [0; 1℄ whose graph lies stritlybelowminffj(x) : for all j suh that pj> p`g:Obviously, f`satises ondition 1. To see that ondition 2 isalso satised, x an index j suh that p`and pjare inomparablein (P;<). Consider an extension of (P;<) to a total order in whihpj< p`. It follows from our onstrution that there exists a pointx 2 [0; 1℄ at whih the values fi(x) are in the same total order as theelements pi(1  i  `). In partiular, we have fj(x) < f`(x). Onthe other hand, by denition, f`(x0) = y0< fj(x0). Therefore, thegraphs of f`and fjmust ross eah other, ompleting the proof. 2Theorem 2.2. (Fox) Fix " 2 (0; 1). For every n, there is a partiallyordered set (P;<) of size n with the following two properties. (i)There are no two disjoint subsets A;B  P suh that jAj; jBj (")nlognand no element of A is omparable to any element of B. (ii)Every element of P is omparable to at most n"other elements. 2To dedue Proposition 1.2, apply Lemma 2.1 to the partially or-dered set whose existene is guaranteed by Theorem 2.2. To seethat the intersetion graph G of the resulting funtions meets therequirements, it is enough to notie that two verties of G are on-neted by an edge if and only if the orresponding elements of P areinomparable.4Referenes[1℄ N. Alon, J. Pah, R. Pinhasi, R. Radoii, and M. Sharir:Crossing patterns of semi-algebrai sets, Journal of Combina-torial Theory Ser. A, aepted.[2℄ P. Brass, W. Moser, and J. Pah: Researh Problems in Dis-rete Geometry, Springer-Verlag, New York, 200.[3℄ J. Fox: A bipartite analogue of Dilworth's theorem,manusript, 2005.[4℄ J. Pah and J. Solymosi: Crossing patterns of segments, Jour-nal of Combinatorial Theory Ser. A 96 (2001), 316{325.5

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