On the maximum size of an antichain of linearly separable sets and convex pseudodiscs
Abstract
We answer a question raised by Walter Morris, and independently by Alon Efrat, about the maximum cardinality of an antichain composed of intersections of a given set of points in the plane with halfplanes. We approach this problem by establishing the equivalence with the problem of the maximum monotone path in an arrangement of lines. A related problem on convex pseudodiscs is also discussed in the paper.
1 Introduction
Let be a set of points in the plane, no three of which are collinear. A subset of is called linearly separable if it is the intersection of with a closed halfplane. A set of is a subset of points from which is linearly separable. Let denote the collection of all sets of . It is a wellknown open problem to determine , the maximum possible cardinality of , where varies over all possible sets of points in general position in the plane. The current records are by Dey ([D98]) and by Tóth ([T01]).
Let be the family of all linearly separable subsets of . The family is partially ordered by inclusion. Clearly, each is an antichain in . The following problem was raised by Walter Morris in 2003 in relation with the convex dimension of a point set (see [ES88]) and, as it turns out, it was independently raised by Alon Efrat 10 years before, in 1993:
Problem 1.
What is the maximum possible cardinality of an antichain in the poset , over all sets with points?
In Section 2 we show that in fact can be very large, and in particular much larger than .
Theorem 1.
, for some absolute constant .
In an attempt to bound from above the function one can view linearly separable sets as a special case of a slightly more general concept:
Definition 1.
Let be a set of points in general position in the plane. A Family of subsets of is called a family of convex pseudodiscs if the following two conditions are satisfied:

Every set in is the intersection of with a convex set.

If and are two different sets in , then both sets and are connected (or empty).
One natural example for a family of convex pseudodiscs is the family , where is a set of points in general position in the plane. To see this, observe that every linearly separable set is the intersection of with a convex set, namely, a halfplane. It is therefore left to verify that if and , where and are two halfplanes, then both and are connected. Let . Since , we have . For any , we claim that there is a point such that the line segment is fully contained in . This will clearly show that is connected. Let be three points in such that is contained in the triangle . If each line segment , for , contains a point of , it follows that , contrary to our assumption. Thus there must be a line segment that is contained in , and we are done.
In Section 3 we bound from above the maximum size of a family of convex pseudodiscs of a set of points in the plane, assuming that this family of subsets of is by itself an antichain with respect to inclusion:
Theorem 2.
Let be a family of convex pseudodiscs of a set of points in general position in the plane. If no member of is contained in another, then consists of at most members.
2 Large antichains of linearly separable sets
Instead of considering Problem 1 directly, we will consider a related problem.
Definition 2.
For a pair of points and a pair of nonvertical lines, we say that strongly separate if lies strictly above and strictly below , and lies strictly above and strictly below .
We will also take the dual viewpoint and say that strongly separate . (In fact, this relation is invariant under the standard pointline duality.)
If we have a set of lines, we say that the point pair is strongly separated by , if contains two lines that strongly separate .
A pair of lines is said to be strongly separated by a set of points if there are two points that strongly separate and .
Using the above terminology one can reduce Problem 1 to the following problem:
Problem 2.
Let be a set of points in the plane. What is the maximum possible cardinality (taken over all possible sets of points) of a set of lines in the plane such that for every two lines , strongly separates and .
To see the equivalence of Problem 1 and Problem 2, let be a set of points and be a set of lines that answer Problem 2. We can assume that none of the points lie on a line of . Then with each of the lines we associate the subset of which is the intersection of with the halfplane below . We thus obtain subsets of each of which is a linearly separable subset of . Because of the condition on and , none of these linearly separable sets may contain another. Therefore we obtain elements from that form an antichain, hence .
Conversely, assume we have an antichain of size in for a set of points. Each linearly separable set is the intersection of with a halfplane, which is bounded by some line . We can assume without loss of generality that none of these lines is vertical, and at least half of the halfspaces lie below their bounding lines. These lines form a set of at least lines, and each pair of lines is separated by two points from the point set . Thus, .
Before reducing Problem 2 to another problem, we need the following simple lemma.
Lemma 1.
Let be nonvertical lines arranged in increasing order of slopes. Let be a set of points. Assume that for every , strongly separates and . Then for every , strongly separates and .
Proof.
We prove the lemma by induction on . For there is nothing to prove. Assume . We first show the existence of a point that lies above and below . Let denote the intersection point of and . Let denote the ray whose apex is , included in , and points to the right. Similarly, let denote the ray whose apex is , included in , and points to the right.
Since the slope of is between the slope of and the slope of , must intersect either or (or both, in case it goes through ).
Case 1. intersects . Then there is a point that lies above and below . This point must also lie below .
Case 2. intersects . Then, by the induction hypothesis, there is a point that lies above and below . This point must also lie above .
The existence of a point that lies above and below is symmetric. ∎
Problem 3.
What is the maximum cardinality of a collection of lines in the plane, indexed so that the slope of is smaller than the slope of whenever , such that there exists a set of points that strongly separates and , for every ?
We will consider the dual problem of Problem 3:
Problem 4.
What is the maximum cardinality of a set of points in the plane, indexed so that the coordinate of is smaller than the coordinate of , whenever , such that there exists a set of lines that strongly separates and , for every ?
We will relate Problem 4 to another wellknown problem: the question of the longest monotone path in an arrangement of lines.
Consider an monotone path in a line arrangement in the plane. The length of such a path is the number of different line segments that constitute the path, assuming that consecutive line segments on the path belong to different lines in the arrangement. (In other words, if the path passes through a vertex of the arrangement without making a turn, this does not count as a new edge.)
Problem 5.
What is the maximum possible length of an monotone path in an arrangement of lines?
A construction of [BRSSS04] gives a simple line arrangement in the plane which consists of lines and which contains an monotone path of length for some absolute constant . No upper bound that is asymptotically better than the trivial bound of is known.
Proposition 1.
(1) 
(2) 
Proof.
We first prove . Let be a simple arrangement of lines that admits an monotone path of length . Denote by the vertices of a monotone path arranged in increasing order of coordinates. In this notation are vertices of the line arrangement , while and are chosen arbitrarily on the corresponding two rays which constitute the first and last edges, respectively, of the path. For each let denote the line that contains the segment , and let denote the line through the segment .
For , we say that the path bends downward at the vertex if the slope of is greater than the slope of , and it bends upward if the slope of is smaller than the slope of . Without loss of generality we may assume that at least half of the vertices of the monotone path are downward bends.
Let be all indices such that is a downward bend, where . Observe that for every , the monotone path between and is an upwardbending convex polygonal path.
We will now define points such that for every the coordinate of is smaller than the coordinate of , and the line lies above and below while the line lies below and above . This construction will thus show that .
For every let and denote the left and respectively the right wedges delimited by and . That is, is the set of all points that lie below and above . Similarly, is the set of all points that lie above and below .
Claim 1.
For every , and have a nonempty intersection.
Proof.
We consider two possible cases:
Case 1. . In this case . Therefore any point above the line segment that is close enough to that segment lies both below and below and hence .
Case 2. . In this case, as we observed earlier, the monotone path between and is a convex polygonal path. Therefore, and are different lines that meet at a point whose coordinate is between the coordinates of and . Any point that lies vertically above and close enough to belongs to both and . ∎
Now it is very easy to construct , see Figure 2. Simply take to be any point in , and for every let be any point in . Finally, let be any point in . It follows from the definition of and that for every , lies above and below and the line lies below and above .
We now prove the opposite direction: .
Assume we are given points sorted by coordinate and a set of lines such that every pair is strongly separated by . By perturbing the lines if necessary, we can assume that none of the lines goes through a point, and no three lines are concurrent. For , let be the face of the arrangement that contains , and let and be, respectively, the leftmost and rightmost vertex in this face. (The faces are bounded, and therefore and are welldefined.) The monotone path will follow the upper boundary of each face from to .
We have to show that we can connect to by a monotone path. This follows from the separation property of . Let be a pair of lines that strongly separates and in such a way that lies above and below and lies below and above . Since lies on the boundary of the face that contains , lies also between and , including the possibility of lying on these lines. We can thus walk on the arrangement from to the right until we hit or , and from there we proceed straight to the intersection point of and . Similarly, there is a path in the arrangement from to the left that reaches . and these two paths together link with .
To count the number of edges of this path, we claim that there must be at least one bend between and (including the boundary points and ). If there is no bend at , the path must go straight through , say, on . But then the path must leave at some point when going to the right: if the path has not left by the time it reaches and lies on , then the path must bend upward at this point, since it proceeds on the upper boundary of the face that lies above .
Thus, the path makes at least bends (between and , for ) and contains at least edges. ∎
Now it is very easy to give a lower bound for , and prove Theorem 1. Indeed, this follows because and ,
The close relation between Problems 1 and 5 comes probably as no big surprise if one considers the close connection between sets and levels in arrangements of lines (see [E87, Section 3.2]). For a given set of points , the sets are in onetoone correspondence with the faces of the dual arrangements of lines which have lines passing below them and lines passing above them (or vice versa). The lower boundaries of these cells form the th level in the arrangement, and the upper boundaries form the st level.
Our chain of equivalence from Problem 1 to Problem 5 extends this relation between sets and levels in a way that is not entirely trivial: for example, establishing that we get sets that form an antichain requires some work, whereas for sets this property is fulfilled automatically.
3 Proof of Theorem 2
The heart of our argument uses a linear algebra approach first applied by Tverberg [T82] in his elegant proof for a theorem of Graham and Pollak [GP72] on decomposition of the complete graph into bipartite graphs.
Let be a collection of convex pseudodiscs of a set of points in general position in the plane. We wish to bound from above the size of assuming that no set in contains another. For every directed line passing through two points and in we denote by the collection of all sets that lie in the closed halfplane to the left of such that touches at the point only. Similarly, let be the collection of all sets that lie in the closed halfplane to the left of such that touches at the point only. Finally, let be those sets that lie in the closed halfplane to the left of such that supports at the edge .
Definition 3.
Let and be two sets in . Let be a directed line through two points and in . We say that is a common tangent of the first kind with respect the pair if and .
We say that is a common tangent of the second kind with respect to if and , or if and .
The crucial observation about any two sets and in is stated in the following lemma.
Lemma 2.
Let and be two sets in . Then exactly one of the following conditions is true.

There is precisely one common tangent of the first kind with respect to and no common tangent of the second kind with respect to , or

there is no common tangent of the first kind with respect to , and there are precisely two common tangents of the second kind with respect .
Proof.
The idea is that because and are two pseudodiscs and none of and contains the other, then as we roll a tangent around , there is precisely one transition between and , and this is where the situation described in the lemma occurs (see Figure 3).
Formally, by our assumption on , none of and contains the other. Any directed line that is a common tangent of the first or second kind with respect to and must be a line supporting at an edge.
Let denote the vertices of arranged in counterclockwise order on the boundary of . In what follows, arithmetic on indices is done modulo .
There must be an index such that , for otherwise every belongs to and therefore and therefore (because both and are intersections of with convex sets) in contrast to our assumption. Similarly, there must be an index such that .
Let be the set of all indices such that , and let be the set of all indices such that .
We claim that (and similarly ) is a set of consecutive indices. To see this, assume to the contrary that there are indices arranged in a cyclic order modulo such that and . Then it is easy to see that is not a connected set because and are in different connected components of this set.
We have therefore two disjoint intervals and . It is possible that or .
Observe that are arranged in this counterclockwise cyclic order on the boundary of , and for every index , . The only candidates for common tangents of the first kind or of the second kind with respect to and are of the form , that is, they must pass through two consecutive vertices of .
We distinguish two possible cases:

. In this case the line through and is the only common tangent of the first kind with respect to and there are no common tangents of the second kind with respect to .

. In this case, there is no common tangent of the first kind with respect to . The line through and and the line through and are the only common tangents of the second kind with respect to .
This completes the proof of the lemma. ∎
Let be all the sets in , and for every let be an indeterminate associated with . For each directed line , define the following polynomial :
This polynomial contains a term whenever is a tangent line for the pair or for the pair (of the first or of the second kind, and with coefficient 1 or , accordingly). If we sum this equation over all directed lines , it follows by Lemma 2 that every term with appears with coefficient 2:
(3) 
Consider the system of linear equations and , where varies over all directed lines determined by . Add to this system the equation . There are equations in this system and if , there must be a nontrivial solution. However, it is easily seen that a nontrivial solution will result in a contradiction to (3). This is because the lefthand side of (3) vanishes, while the righthand side equals . We conclude that . ∎
We now show by a simple construction that Theorem 2 is tight apart from the multiplicative constant factor of . Fix three rays , and emanating from the origin such that the angle between two rays is degrees. For each , let be points on , indexed according to their increasing distance from the origin. Slightly perturb the points to get a set of points in general position in the plane. For every define
It can easily be checked that the collection of all such that and is an antichain of convex pseudodiscs of . This collection consists of sets.
Footnotes
 footnotetext: This research was supported by a Grant from the G.I.F., the GermanIsraeli Foundation for Scientific Research and Development.
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