A HELLY NUMBER FOR UNIONS OF TWO BOXES IN R 2

Let S be a polygonal region in the plane with edges parallel to the coordinate axes. If every 5 or fewer boundary points of S can be partitioned into sets A and B so that cony AU conv B c S, then S is a union of two convex sets, each a rectangle. The number 5 is best possible. Without suitable hypothesis on edges of S, the theorem fails. Moreover, an example reveals that there is no finite Helly number which characterizes arbitrary unions of two convex sets, even for polygonal regions in the plane.

locally convex at some point q in S, then q is called a point of local noncon- vexity (inc point) of S. For points x and y in S, we say x sees y via S (x is visible from y via S) if and only if the corresponding segment Ix,y] lies in S. Points x l,...,xn in S are visually independent via S if and only if for I i < j n, x i does not see xj via S. Set S is said to be 3-convex if and only if for every 3 points in S, at least one of the corres- pondlng llne segments lies in S. Finally, S is called star-shaped if and only if there is some point p in S such that p sees each point of S via S. The set of all such points p is the (convex) kernel of S.
The following terminology will be used throughout the paper: Conv S, bdry S, and ker S will denote the convex hull, boundary, and kernel of set S, respectively, while Inc S will be the set of inc points of S. For distinct points x and y, R(x,y) will represent the ray from x emanating through y.The reader is referred to Valentine [9]  and to Imy [5] for a discussion of these concepts.
An interesting theorem by Lawrence, Hare, and Kenelly [4] provides the following characterization for unions of convex sets: For S a subset of a linear topological space, S is a union of k convex sets if and only if each finite subset F of S has a k-partition FI,...,F k with conv F.I --c S, I i k.A related problem concerns the existence of a finite Helly number to characterize such unions, and it is natural to ask when 'finite subset' in the hypothesis may be replaced by 'j-member subset' for an appropriate j. (See [3] for similar results.) Unfortunately, the Lawrence, Hare, Kenelly Theorem cannot be improved for k 2, even when S is a compact subset of the plane.(See Example 3.) However, the problem of obtaining a finite Helly number to characterize certain unions of convex sets remains open.Recent results suggest sets for which such a theorem might hold.In work by Danzer and Grnbaum [2], a kind of Helly number (called a piercing number) is found for certain families of boxes.Moreover, in [i] another kind of Helly number (a Krasnosel'skii number for unions of starshaped sets) is established for certain polygonal regions whose edges are parallel to the coordinate axes.Therefore, it seems reasonable to examine such polygonal regions.This is the situation studied here, and the following result is obtained: Let S be a polygonal region in the plane with edges parallel to the coordi- nate axes.If every 5 or fewer boundary points of S can be partitioned into sets A and B so that cony A U cony B c S, then S is a union of two convex sets, each a rectangle.The number 5 is best possible.Without suitable restrictions on the edges of S, the result fails and in fact no finite Helly number exists.
The following lemma will be useful.
LEMMA 1 Let S be a bounded subset of R d and let k be a fixed integer, k 3.If every k or fewer boundary points of S can be partitioned into sets A and B such that conv A U conv B c S, then every k or fewer points of S can be partitioned in this way as well.Moreover, S is 3-convex.
PROOF.Clearly S is closed.If S is not connected, then it is easy to show that S has exactly two components, each convex.Hence without loss of generality we may restrict our attention to the case in which S is connected.
Furthermore, we may assume that inc S # , for otherwise the closed, connected, locally convex set S will be convex by a theorem of Tietze [6].
We begin the proof by showing that inc S c ker S. Let q inc S # .We assert that for every z in bdry S, [q,z] c S: Otherwise, since S is closed, there would be a convex neighborhood N of q with no point of N N S seeing z via S. Since q Inc S, N S is not convex, and standard arguments produce points x,y in N bdry S with Ix,y] i S.However, then points x,y,z would be visually independent points in bdry S, contradicting our hypothesis.We conclude that [q,z] S, and the assertion is established.Now it is easy to show that q ker S: If not, then for some w in S, [q,w] S, and [q,w] S would contain an open interval (a,b) with a,b bdry S and q a < b < w.However, [q,b] c S. Again we have a contradiction, and q ker S, the desired result.
Finally, to establish the lemma, let x i S, i i k, and select q inc S # .Without loss of generality, assume q {x i i i k}, and let ray R(q,x i) meet bdry S at Yi with q < xi Yi" By our hypothesis, the points yl,...,y k can be partitioned into sets A and B such that conv A U conv B ! S.
Relabeling if necessary, assume that A {Yl Yj}' B {Yj+I Yk }" Since q inc S, conv ({q}UA) !S and conv {x I xj} !conv ({q} U A)! S. Similarly conv {xj+I ,x k} ! S. Since k > 3, it is easy to see that S is 3-convex, and the lemma is proved.
Without the requirement that set S be bounded, Lemma i fails, as Example i illustrates.EXAMPLE I. let S be the set of points in R 2 which lle either in the first quadrant or on one of the coordinate axes.(See Figure i.)The set bdry S may be partitioned into sets A and B satisfying the hypothesis of the lemma.However, the conclusion of the lemma fails for every k > 3.
We are ready to establish the following Helly-type theorem for unions of rectangles in the plane.THEOREM i. let S be a polygonal region in the plane with edges parallel to the coordinate axes.If every 5 or fewer boundary points of S can be partitioned into sets A and B so that conv A U conv B c S, then S is a union of two convex sets, each a rectangle.The number 5 is best posole.

PROOF.
We begin with some preliminary observations.By Lemma i, set S is 3-convex.Moreover, using [8, Theorem i] S is starshaped and inc S c ker S. Now in case S has O, i, or 2 inc points, we may use [8, Theorem 3] to conclude that S is a union of two or fewer closed convex sets, the desired result.There- fore, throughout the argument it suffices to assume that S has at least 3 inc points.
As in [7] and [i], it is helpful to order bdry S in a clockwise direction.
This in turn induces a natural order on the edges of S, and we may classify each edge as 'right', 'left', 'up', or 'down' according to the order it inherits from bdry S. For convenience of notation, we label the inc points of S by ql,q2,...,qn, again according to the clockwise order on bdry S. Recall that n 3 by a previous assumption.The proof of the theorem will require some results concerning these inc points.
It is easy to see that for each inc point of S, exactly one of the following must occur relative to our order: The Inc point Is either preceded by a 'right' edge and followed by an 'up' edge, preceded by a 'left' edge and followed by a 'down' edge, preceded by a 'down' edge and followed by a 'right' edge, or preceded by an 'up' edge and followed by a 'left' edge.Moreover, using this classifi- cation together with the 3-convexlty of S, it is not hard to show that no three  consecutive Inc points of S (in our order) can be colllnear: Otherwise, for an appropriate orientation of S, this would yield a 'right' edge whose predecessor is a 'down' edge and whose successor is an 'up' edge, producing three visually independent points of S, impossible.
Furthermore, we assert that every two consecutive inc points of S determine a segment parallel to one of the coordinate axes: Suppose on the contrary that the condition fails for ql and q2" Without loss of generality, assume that S is oriented in the plane so that ql is to the left of q2 and above q2" We examine possible classifications for ql and q2" Using the fact that q2 follows ql in our clockwise order, it is not hard to see that ql cannot be preceded by a 'left' edge and followed by a 'down' edge.For convenience of terminology, we say that ql cannot be classified as 'left-down'.Similarly, q2 'right-up'.
cannot be classified as Examine the remaining 9 possibilities. (See Figure 2, Cases 1-9, for corres- ponding illustrations.) i. ql is 'right-up' and q2 is 'left-down' 2. ql is 'right-up' and q2 is 'down-rlght' 3. ql is 'right-up' and q2 is 'up-left' 4. ql is 'down-rlght' and q2 is 'left-down' 5. ql is 'down-right' and q2 is 'down-right' 6. ql is 'down-rlght' and q2 is 'up-left' 7. ql is 'up-left' and q2 is 'left-down' 8. ql is 'up-left' and q2 is 'down-right' 9. ql is 'up-left' and q2 is 'up-left' It is not hard to prove that each of Cases 2 through 9 above violates the condition that ql,q 2 ker S, so none of these can occur.Thus Case 1 is the only possibility.
Moreover, by a similar argument involving the classification of Inc point q3' q3 must be 'up-left': Classification 'down-right' is impossible in our clock- wise order.Classifications l)'right-up' and 2)'left-down' violate the facts that ql ker S and q2 ker S, respectively. (See Figure 3, Cases 1-2.)However, this forces S to have 5 boundary points Pi' 1 i 5, for which there is no partition satisfying our hypothesis: Select Pl near ql on the edge preceding ql' P2 above ql and right of q2' P3 near q2 on the edge P4 q2 Case I.  Figure 5.
following q2' P4 near q3 on the edge preceding q3' and P5 near q3 on the edge following q3" (see Figure 4.) We have a contradiction, our supposition is false, and every two consecutive inc points of S must determine a segment parallel to one of the coordinate axes.Thus the assertion is established.
The rest of the argument is easy.Using the assertion above together with the fact that no three consecutive inc points of S are collinear, it is not hard to show that S has at most 4 inc points.Similarly, S cannot have exactly 3 inc points, and since we are assuming that n 3, we conclude that n 4. Hence by [8, Theorem 3], S is a union of two closed convex sets.Finally, using decomposition techniques from [8, Theorem 3], it is easy to show that S may be expressed as a union of two rectangles, and the proof of Theorem i is complete.
To see that the number 5 in Theorem i is best possible, consider the following example.EXAMPLE 2. Let S be the set in Figure 5. Every 4 points of S may be partitioned into sets A and B such that conv A U conv B ! S.However, S is not a union of fewer than three convex sets.
In conclusion, it is interesting to observe that the result in Theorem 1 fails without the restriction on edges of S. In fact, there is no finite Helly number which characterizes unions of two convex sets, even for polygonal regions in the plane, and the Lawrence, Hare, Kenelly Theorem is best possible in this case.This is illustrated in Example 3.  Finally let e i such that s i meets si_ 2 at a i and s i i+2 i" S(n) be the simply connected region determined by U{s i i < i < 2n +l}. (The case for n 2 is illustrated in Figure 6.)Every 2n points of S(n) may be partitioned into sets A and B so that conv A U cony B c__ S.However, the 2n + i points a i i < i < 2n + 1 have no such partition, and S(n) is not a union of fewer than three cot:vex sets.

Figure i.
Figure i.

EXAMPLE 3 .
For n > 2, let P(n) be a regular (2n + l)-gon whose edges are labeled in a clockwise direction by el,e2,...,e2n+l.Adjusting our subscripts modulo 2n + i, for i i < 2n +I, Let s i [al,bi] be a segment containing meets s at b