Open sets avoiding integral distances
Abstract
We study open point sets in Euclidean spaces without a pair of points an integral distance apart. By a result of Furstenberg, Katznelson, and Weiss such sets must be of Lebesgue upper density zero. We are interested in how large such sets can be in dimensional volume. We determine the exact values for the maximum volumes of the sets in terms of the number of their connected components and dimension. Here techniques from diophantine approximation, algebra and the theory of convex bodies come into play. Our problem can be viewed as a counterpart to known problems on sets with pairwise rational or integral distances. This reveals interesting links between discrete geometry, topology, and measure theory.
1 Introduction
Is there a dense set in the plane so that all pairwise Euclidean distances between the points are rational? This famous open problem was posed by Ulam in 1945, see e.g. [17, 18, 40]. Unlike this, a construction of a countable dense set in the plane avoiding rational distances is not hard to find, see e.g. [29, Problem 13.4, 13.9]. If all pairwise distances between the points in are integral and is noncollinear, i.e. not all points are located on a line, then is finite [2, 16]. Having heard of this result, Ulam guessed that the answer to his question would be in the negative. Of course the rational numbers form a dense subset of a coordinate line with pairwise rational distances; also, on a circle there are dense sets with pairwise rational distances, see e.g. [1, 2]. It was proved by Solymosi and De Zeeuw [38] that the line and the circle are the only two irreducible algebraic curves containing infinite subsets of points with pairwise rational distances. Point sets with rational coordinates on spheres have been considered in [35]. There is interest in a general construction of a planar point set of size with pairwise integral distances such that where is collinear, , , and has no three collinear points. The current record is [10]. And indeed, it is very hard to construct a planar point set, no three points on a line, no four points on a circle, with pairwise integral distances. Kreisel and Kurz [30] found such a set of size , but it is unknown if there exists one of size .
The present paper is concerned with a problem that may be considered as a counterpart to those just described, namely with large point sets in without a pair of points an integral distance apart. We write for the supremum of the volumes of open point sets with connected components without a pair of points whose distance apart is a positive integer. We determine the exact values of the function for all and .
This problem is related to the famous Hadwiger–Nelson open problem of determining the (measurable) chromatic number of , see e.g. [12, Problem G10]. Here one can also ask for the highest density of one color class in such a coloring, that is, we may ask for the densest set without a pair of points a distance apart. In [32] such a construction in has been given. In the plane the best known example, due to Croft [11], consists of the intersections of hexagons with circles and attains a density of . The upper bounds are computed in [5, 13]. Point sets avoiding a finite number of prescribed distances are considered e.g. in [9] and [12, Problem G4], so the point sets avoiding all distances that are positive integers correspond to the case with an infinite number of excluded distances. It is known [21] that for each subset of the plane with positive density, there is a constant such that all distances greater than occur between the points of . The same result is true in higher dimensions [34]. It follows that in every dimension , the Lebesgue measurable sets avoiding integral distances, which are of interest here, must be of upper density zero, so we consider the supremum of their volumes instead.
The paper is organized as follows: in Section 2 we introduce the basic notation and provide characterizations of arbitrary open point sets without pairs of points an integral distance apart. After stating first relationships between the upper bounds for the maximum volumes of those sets with different numbers of connected components we continue in Section 3 by considering a relaxed problem. We evaluate the maximum volumes of sets avoiding integral distances in the special case where the connected components of the sets are open balls. In our crucial constructions we make use of Weyl’s theorem from diophantine approximation and the fact, we derived from Mann’s theorem, that the lengths of the diagonals of a regular gon are linearly independent over whenever is a prime. In Section 4 we approach the main problem of evaluating the function in the general case. For twocomponent opens sets () we provide a complete solution in Subsection 4.1. Motivated by the necessary conditions for open point sets to avoid integral distances we consider dimensional open sets with connected components of diameter at most each whose intersection with every line has a total length of at most . At the end of Subsection 4.1 we state a conjecture on the exact values of that bears a strong resemblance to the problems of geometric tomography, see e.g. [22]. In Subsection 4.2 we provide some upper bounds for with . The main problem of evaluating generally is finally settled in Subsection 4.3. In Section 5 we give a summary of the results obtained and draw the appropriate conclusions.
2 General observations and basic notation
Denote by the Euclidean distance between two points and by the distance between two subsets and of . The minimum width of , i.e. the minimum distance between parallel support hyperplanes of the closed convex hull of , will be denoted by , and will stand for the Lebesgue measure in .
At first we observe that the diameter of any connected component of an open set avoiding integral distances, i.e. having no points an integral distance apart, is at most .
Lemma 1
Let be an open set avoiding integral distances. Then for every connected component of we have .

Proof. Suppose there is a connected component with , then there exist such that . Since is locally connected, is open, so it is path connected. Hence there is a point on the image curve of a continuous path in joining and such that .
By the isodiametric inequality the open ball centered at the origin with unit diameter has the largest volume among measurable sets in of diameter at most , see e.g. [19], [6, chap. 2]. Thus we have
The first few values are given by , , , and . Note that the volume of the scaled ball with diameter in is .
Next we characterize dimensional open sets containing a pair of points an integral distance apart.
Lemma 2
A nonempty open set contains a pair of points with if and only if either or there is a pair of connected components (i.e. disjoint open intervals) , of such that and . If , then there exists a shift of such that .

Proof. The restriction of the canonical epimorphism , , to the interval is a continuous bijection of onto the dimensional torus , the inverse map being continuous at all points except . We consider the retraction , that is, for all (i.e. is the fractional part of ). We observe that the image under of any open interval of length is either the open interval of the same length , whenever both and are in , for some , or the union of two disjoint connected components
of the same total length , whenever , for some . If , then similarly either or whenever for some . Hence, in general, the total length of the connected components of is , whenever .
Let be the disjoint union of open intervals , say, with total length . Then by Lemma 1 and for all . We thus have from above that the total length of the connected components of all the images equals . Hence at least two images and must overlap, so there exists , that is, for some and . Thus , hence .
If for some connected components and of with , where , , so that , we can take a point in the leftmost interval, say and a point so that the length of is . Then
Conversely, suppose there are with . If and lie in the same connected component of , then because is open, hence . Suppose and lie in distinct connected components of , say and , , and let . Then as well whence the distance between the components is , because . Let where . Then
since because . Thus either or there is a pair of required connected components of .
If , then for all so the total length of the connected components of all the images equals , as shown previously. If , then clearly, . If , then again , whenever the images are not pairwise disjoint. Suppose all the images are pairwise disjoint and . Then there is exactly one that meets . Hence the complement is a nonopen set in that can not be covered by the images of the other connected components of , since they are all open intervals, so as well. Thus in all the cases we have . Take , , that is, . Then , i.e. so and the required shift is .
Applying Lemma 2 we establish a criterion for an open set to avoid integral distances in all dimensions.
Theorem 1
An open point set does not contain a pair of points an integral distance apart if and only if for every line

and

if hits a pair of distinct connected components , of in the intervals with , then .
Another criterion, which we will also be using is:
Lemma 3
Let be a dimensional disconnected open set all of whose connected components are of diameter at most . Then contains a pair of points with integral distance if and only if
for some of its connected components .

Proof. Since all the connected components of are open with diameter at most , any two distinct points of with integral distance must be in two different components, say and . Let , with . We then select two small closed balls and centered at and respectively with radii . The line through and meets the two balls in the intervals, say and , where and . With this notation we have
Conversely, if for an integer , then there exist and such that
Joining with in and with in by continuous paths, we can find and on the image curves of these paths with .
Sometimes it is helpful, if we can assume that the connected components of the point sets in question are not too close to each other. Specifically, we will be using the fact that in such cases the connected components of the sets have disjoint closures.
Lemma 4
Let , be distinct connected components of a dimensional open point set without a pair of points an integral distance apart. If , then .

Proof. Making use of the isodiametric inequality we deduce from that . By Lemma 1 we have and . So we can choose , with . If , then there exist and such that . Since and are open, they are path connected, hence we can join and by a continuous path in and similarly and in and on the image curves of these paths we then find and such that , but avoids integral distances, a contradiction. Thus we have .
As Lemma 1 and Theorem 1(i) will be our main tools in estimating upper bounds for , we denote by the supremum of the volumes of open point sets with connected components of diameter at most each (condition (a)), and with total length of the intersection with every line at most (condition (b)). Clearly and for all and . Note that omitting condition (b) trivializes the problem of estimating the extreme volumes, the extreme configurations obviously consist of disjoint open dimensional balls of diameter . Dropping condition (a) makes the problem more challenging. It turns out that there are open connected dimensional point sets with infinite volume and diameter even though the length of the intersection of with every line is at most , i.e. for all .
Example 1
For integers and denote by the dimensional open spherical shell, or annulus, centered at the origin with inner radius and outer radius , i.e. are bounded by concentric dimensional spheres centered at the origin. These shells will guarantee that the volume of their union is unbounded as increases. So far the constructed point set is disconnected. To obtain a connected point set, we denote by the dimensional open spherical shell centered on the axis at with inner radius and outer radius . With this is open and connected with infinite volume and diameter even though the length of its intersection with every line is smaller than 1. In Figure 1 we depicted such a configuration in dimension with first few annuli getting thinner and thinner and being blue and green respectively. The detailed computations demonstrating the assertions claimed are provided in the Appendix, see Subsection A.1.
In order to make the problem of evaluating the functions and more tractable, we consider both problems in the special case, where the connected components are restricted to dimensional open balls. We denote the corresponding maximum volumes by and respectively. Clearly we have and . In Section 3 we determine the exact values of both functions and for all and .
Based on a simple averaging argument, any given upper bound on one of the four introduced maximum volumes for connected components yields an upper bound for connected components in the same dimension.
Lemma 5
For each we have whenever and whenever .

Proof. Let be a dimensional open set with corresponding property in either case and connected components. The volume of each of the different unions of connected components inheriting these properties is at most . Since each connected component occurs exactly times in those unions, the stated inequalities hold.
3 Unions of open dimensional balls
Here we consider open point sets that are unions of disjoint dimensional open balls of diameter at most each such that they either do not contain a pair of points with integral distance or intersect each line in the intervals with total length at most . As introduced in the previous section, we denote the supremum of possible volumes of such by and respectively.
In dimension we can consider one open interval of length and open intervals of length , where , arranged in a unit interval so that they are pairwise disjoint. Clearly the set does not have a pair of points an integral distance apart and the total length of the intervals tends to , as approaches . It follows from Theorem 1 that for all . For , by the isodiametric inequality, only the volumes of dimensional open balls of diameter attain the maximum value .
Lemma 6
for all .

Proof. Consider disjoint open dimensional balls with diameters , where we can assume w.l.o.g. that . Clearly in dimension we have for all , and for all dimensions , we have , so in the cases where either or is the stated inequality holds. Hence we can assume that and . Then by Theorem 1(i) we have for all . If , then , so the required inequality holds. Otherwise we have and it remains to maximize the function with domain . Since , every inner local extremum of is a minimum, so the global maximum of is attained at the boundary of the domain. Finally, we compute , , so the lemma follows.
Remark. The special case of balls of diameter is directly related to point sets with pairwise integral distances. Let be the union of dimensional open balls of diameter each without a pair of points an integral distance apart. Then the distances between the centers of the balls and in some enumeration must be of the form for some integers . By dilation with a factor of two we obtain the set of size of the centers of the balls with pairwise odd integral distances. However, it has been shown in [23] that for such sets , where equality holds if and only if . The exact maximum number of odd integral distances between points in the plane has been determined in [33].
Theorem 2
for all and .

Proof. By Lemma 6 it suffices to provide configurations whose volumes (asymptotically) attain the upper bound.
For , we consider the union of one dimensional open ball of diameter and disjoint open balls of diameter arranged in the interior of an open ball of diameter . As approaches , the volume of the union tends to .
For the remaining part we consider the union of open dimensional balls with diameter centered at the vertices of a regular gon with circumradius . Clearly, for large enough, every line hits at most two balls.
An alternative construction would be the union of open dimensional balls of diameter centered at for . If is large enough, then again there is no line intersecting three or more balls.
Corollary 1
for all and .
It turns out that, in fact, the equalities hold in all dimensions . To explain the underlying idea, we first consider the special case where and , i.e. the first case that is not covered by Corollary 1.
Lemma 7

Proof. For each integer and , we consider a regular pentagon with side length and the union of five open round discs of diameter centered at the vertices of , see Figure 2. Since each connected component of has diameter less than , there is no pair of points an integral distance apart inside a connected component. For every two points and from different components, we either have
whenever the discs are adjacent with their centers located on an edge of , or
otherwise.
Let stand for the positive fractional part of a real number , i.e. . If, given , one can find an integer such that , then the set with parameters and does not contain a pair of points with integral distance.
Since is irrational, we can apply the equidistribution theorem, see e.g. [39, 41], to ensure that is dense (even uniformly distributed) in . The same holds true if we shift the set by the fixed real number . Thus we can find a suitable integer for each . As approaches , the total area of tends to , which is best possible by Lemma 6.
We illustrate this by a short list of suitable values of :
,
,
,
, and
.
We shall generalize Lemma 7 to an arbitrary dimension and arbitrary number of connected components. The idea is to locate the centers of small dimensional open balls of diameter slightly less than at some points in a twodimensional subplane so that the set of different pairwise distances between their centers are linearly independent over the rational numbers, that is, the distances are either confluent or rationally independent. The appropriate candidates for the center points would be the vertices of a regular gon, where is an odd prime. We use a theorem of Mann, see [31], to prove the desired property of the set of distances. The condition that the point set in question avoids integral distances can be translated into a system of inequalities of the form , where , and we are looking for an integer such that the above fractional parts of the scaled pairwise distances are arbitrarily small. By a theorem of Weyl, see e.g. [41, Satz 3] or a textbook on Diophantine Approximation like e.g. [26], such systems have solutions whenever the are irrational and linearly independent over . (Weyl actually proves equidistribution while we only need denseness, a weaker result that Weyl himself attributes to Kronecker.)
Note that the same construction, using the vertices of a regular hexagon, does not work. Indeed, there would be only three disinct values for the lengths of the diagonals, namely , , and . The required inequalities
would trivially hold for , but fail for and small enough. We note in passing that quite recently Mann’s theorem was used in another problem from Discrete Geometry see [14, 36].
Theorem 3
(Mann, 1965, [31]) Suppose we have
with , roots of unity, and no subrelations , where . Then
for all , where .
The vertices of a regular gon with a circumcircle of radius centered at the origin are given by
for . In standard complex number notation with they coincide with the th roots of unity . The distance between the vertices and is equal to . Since , there are only distinct distances in a regular gon, attained for . We note in passing that this number is not far away from the minimum number of distinct distances in the plane, which is bounded below by for a suitable constant , see [24]. We can express these distances in terms of th roots of unity via
for all .
Lemma 8
Given an odd prime , let for , where the are th roots of unity. Then the are irrational and linearly independent over .

Proof. A folklore result, see e.g. [27], states that , where , is a rational number if and only if . Since is odd, this cannot occur in our context. It remains to show that the irrational numbers are linearly independent over . Suppose to the contrary that there are rational numbers for such that . We then have
Now let be a subset of those indices , such that , where , and no vanishing subcombination. We have . Hence by Mann’s Theorem for all , since
This yields for . Since is a subset of
this is impossible, so the numbers have to be linearly independent over .
Theorem 4
for all and .

Proof. Since , by Theorem 2 . By Corollary 1 we can assume that . For the construction we fix an odd prime with . For each integer and each we consider a regular gon with side lengths , i.e. with circumradius . At arbitrarily chosen vertices of the gon we place the centers of dimensional open balls with diameter and consider their union. Since each of the connected components of the union has a diameter less than , there is no pair of points with integral distance inside the components. Now consider arbitrary points and from two different connected components (open balls). Let stand for the distance between their centers. The triangle inequality yields
Since all possible distances are given by for , we look for a solution of the system of inequalities
where . By Lemma 8 the factors are irrational and linearly independent over , so by Weyl’s Theorem [41] such systems admit solutions for all .
Therefore, for every we can choose a suitable value of and construct an open component set without pairs of points an integral distance apart with volume . As approaches 0, this volume tends to , completing the proof.
Thus, in the case of round connected components the values of and are completely determined. In the general case of arbitrary connected components the problem is more challenging for and will be addressed in the following section.
4 Bounds for and the exact value of
In dimension we can consider the disjoint union of open intervals of length inside an open interval of length 1. Obviously avoids integral distances and the line intersection property holds trivially for the total length of being 1 which is the largest possible by Theorem 1. Thus for all . For , we have and only dimensional open balls of diameter can have that large volume. For the evaluation of and gets more involved. In Subsection 4.1 we treat the 2component case . As to the general case, we only could find some bounds for in Subsection 4.2 and succeeded in determining the exact values of the function in Subsection