Measurable sets with excluded distances^{1}^{1}1This work is a part of a Ph. D. thesis under the supervision of Benjamin Sudakov.
Abstract
For a set of distances a set is called avoiding if no pair of points of is at distance for some . We show that the density of is exponentially small in provided the ratios , , …, are all small enough. This resolves a question of Székely, and generalizes a theorem of FurstenbergKatznelsonWeiss, FalconerMarstrand, and Bourgain. Several more results on avoiding sets are presented.
1 Introduction
The problem of determining the least number of colors required to color the points of the plane so that no pair of points at distance is colored in the same color was first investigated by Nelson and Hadwiger in 1940s. This number, which we denote by , is called the chromatic number of because it is the chromatic number of the graph whose vertices are the points of and the edges are pairs of points that are distance apart. We denote this graph by .
In the dimension two, there has been no improvement on the bounds in the past fortyfive years [Had61, MM61]. In higher dimensions, however, Frankl and Wilson [FW81] showed that the chromatic number grows exponentially in the dimension, , confirming an earlier conjecture of Erdős. The paper of Frankl and Wilson in conjunction with the earlier work of RayChaudhuri and Wilson[RCW75] laid down the theory of set families with restricted intersection, which led to many other results including the disproof of Borsuk’s conjecture by Kahn and Kalai [KK93].
It was first shown by Erdős and de Bruijn [dBE51] that the chromatic number of any infinite graph, and in particular, is the maximum of the chromatic numbers of its finite subgraphs, provided the maximum is finite. The proof relied on the axiom of choice, which suggested that the chromatic number might depend on the underlying axiom system. This was partially confirmed by Falconer [Fal81] who showed that there is no coloring of into four colors such that each color class is a Lebesgue measurable set and no pairs of points at distance have the same color. Since as shown by Solovay [Sol70] the axiom that all subsets of are Lebesgue measurable is consistent with the usual ZermeloFraenkel set theory without the axiom of choice, is unprovable in the set theory without the axiom of choice.
Thus, we denote by the least number of colors required to color so that no points at distance are assigned the same color, and each color class is a measurable set. A set with no pairs of points at distance is going to be called avoiding. The most natural way to show that is large is by showing that no color class can be large. Denote by the upper limit density of (which is formally defined in section 3). Let be the supremum over all measurable avoiding sets. Then . Unfortunately, Falconer’s proof that does not show that . The best known bounds are (see [SU97, p. 61] and [Szé85] respectively), and it is a conjecture of Erdős that [Szé02].
The problem of forbidding more than one distance was first studied by Székely in his thesis [Szé85]. There he established the first bounds on and which denote the analogues of and , respectively, where a finite set of distances is forbidden. Székely conjectured that in dimension for any set with there is a such that all the distances greater than occur among the points of . The conjecture was proved by Furstenberg, Katznelson and Weiss [FKW90]. Their proof was ergodictheoretic. Later Bourgain found a harmonicanalytic proof [Bou86], and Falconer and Marstrand gave a direct geometric proof [FM86]. Székely also conjectured that if is a sequence converging to , then as . This was proved by Falconer [Fal86] and Bourgain [Bou86].
It is not known how large can be for a set of given size. It has been long known that [CFG94, p. 180]. The only known upper bound comes from the observation that the coloring, which is a product of colorings that avoid and , avoids both and . Croft, Falconer and Guy asked whether is exponential in [CFG94, Prob. G11]. Erdős conjectured that is polynomial in [Erd81].
In this paper we answer the question of Croft, Falconer and Guy in the measurable setting by showing that in the dimension as the ratios all tend to infinity tends to , and thus . We will also show that for every set of distances , answering question of Székely [Szé02, p. 657], who asked for the value of . This also generalizes the abovementioned theorems of FurstenbergKatznelsonWeiss and Falconer. Indeed, to deduce FurstenbergKatznelsonWeiss theorem suppose there is a set with and a sequence going to infinity such that the distance does not occur between points of . Then there is a subsequence such that tends to infinity, implying for any positive integer . In fact our result is stronger:
Theorem 1.
Suppose and let be arbitrary finite sets. If the ratios tend to infinity, then
It is conceivable that there might be denser and denser avoiding sets whose density approaches without there being a avoiding set of density . However, that is not the case. We show that there is a set which not just achieves this density, but whose measure cannot be increased by an alteration on a bounded subset. Moreover, we show that the constants can in principle be computed for any finite set . However, the high time complexity of our algorithm prohibits us from settling the question whether .
The principal tool of the paper is the socalled zoomingout lemma stating that under the appropriate conditions we can ignore the smallscale details of the measurable sets in question. In this sense, it is similar to the celebrated Szemerédi regularity lemma. The Szemerédi regularity lemma implies that for the purpose of counting subgraphs every graph can be replaced by a much smaller “reduced graph” [KS96]. The zoomingout lemma states that every measurable set can be replaced by a “zoomedout set” which captures some of information about counting (by an appropriate integral) pairs of points that are at a given distance away.
2 The dimensional case and the main idea
Before delving into the proof of the results in it is instructive to examine the situation in , for it is much simpler, of interest on its own right, and illustrates some of the ideas used in the main results.
Throughout the paper we identify sets with their characteristic functions, i.e., for a set we define if and if . In this section we use the notation to denote the interval of the integers from to , i.e., .
For a set define upper and lower densities by
The set is avoiding if , where is the difference set of . Define where the supremum is over all avoiding sets.
The simpleminded analogue of theorem 1 is false. If is avoiding set, then and thus showing that . On the other hand, the set of even integers shows that . However, for every odd integer the set of even integers shows that . This example also shows why the theorem 1 is itself false in . In the integration of the inequality yields . The set shows that , and the same set shows that for every odd integer .
The version of theorem 1 that works in one dimension involves excluding thickened sets, in order to avoid this kind of congruential obstacles. For a set we denote by the neighborhood of , i.e., .
Theorem 2.
For every finite set there is a such that for every finite nonempty set we have
for every positive integer .
Proof.
Denote . Let be any even integer so that .
Suppose is avoiding. Then the set is avoiding. To see that suppose is a pair of elements such that . By the definition of there are with and . By the triangle inequality , which is a contradiction.
Write the set as a union of disjoint intervals where for no we have . Each of these intervals has length at least . If is the smallest element of , then none of these intervals has length exceeding , for is avoiding. The density of does not exceed . The set is contained , so it suffices to bound the density of on each of the intervals . By translating the interval it suffices to consider the case .
So, suppose is avoiding and . Then is avoiding because the copies of a avoiding set are too far from each other for there to be elements in different copies such that . Since has density , we infer .
Now let us turn back to the proof of the theorem. For each interval the subintervals and do not meet . Thus each interval in of length contains no more than elements of .
Similarly no more than elements belong to in any interval of length . Let be an arbitrary positive integer. Consider . Since at most two intervals contain elements in , but not contained in , we have . Hence,
Letting we conclude that . ∎
As remarked above the reason why theorem 1 fails in the dimension one is because the largest avoiding set can be periodic (in fact there is always a set of density which is periodic as shown by Cantor and Gordon [CG73]), and thus avoid many more distances than required of it. By the theorem of Furstenberg, Katznelson and Weiss [FKW90] this cannot happen in higher dimensions because any periodic set has positive density, and all sufficiently large distances occur in sets of positive density. So, it is not surprising that in the higher dimensions it becomes possible to carry out a proof very similar in spirit to the proof of theorem 2 above, but technically more complicated.
The approach employed in this paper is rooted in the proof of Bourgain [Bou86] of the FurstenbergKatznelsonWeiss theorem.
3 Notation
Throughout the rest of the paper the dimension is going to be fixed, so we will often omit the dependency on from our notation.
For a measurable set the notation denotes the measure of . The notation denotes the open axisparallel cube of side length centered at the point .
For a set and a bounded domain the density of on is
The upper and lower limit densities of are
Whenever we write . Note that we measure the densities with respect to cubes, and not balls as it is usually done. Whereas, in general these densities might be different, corollary 13 below implies that our results do not depend on the kind of density chosen, and the proofs are cleaner for the density measured on cubes since there are fewer edge effects one needs to worry about. The advantage of using cubes centered at the origin lies in less cluttered notation. However, since the properties we consider in this paper are translationinvariant, we incur no loss of generality.
Being interested in the largest avoiding sets, we define
More generally, we will be looking at the properties of sets that are more general than the property of being avoiding. So, we let denote the family of all the measurable subsets of and call a function a property. If , we say that has the property , and if , we say that does not have it. We define
For a property and a real number the property is the property that holds for precisely when the property holds for . This is in agreement with the definition of avoiding set as a set such that is avoiding. Note that the function is scaleinvariant: for every we have .
If and are two properties, then denotes the property asserting that both and hold, i.e., . In particular, if and are the properties of being  and avoiding respectively, then is the property of being avoiding.
4 Supersaturable properties
In this section we prove basic theorems about a class of properties for which the analogue of theorem 1 holds.
As explained in the introduction, the crucial tool is the ability to ignore the fine details of the sets. The intuition here is that given a set and a large real number in order to understand whether the set has points which are at distance apart we should zoomout away from the set and look at a scale comparable to . If we think of the set as colored black on the otherwise white background, then the very fine details of will blur into some shade of gray. The zoomingout lemma says that for our purposes if the shade is not too light, then we can treat gray points as if they were black.
More formally, for each and we define a zoomingout operator acting on by
One can think of the zoomingout operator as the replacement for the operation of thickening sets in the integers. In the sequel we use the following easy properties of the zoomingout operator which we now state.
Lemma 3.

for any .

.
Proof.
We say that a property is supersaturable if there is a function such that the following seven conditions are satisfied:

.

is monotone nondecreasing and is monotone, i.e., and if .

implies that does not have the property .

Both and are translationinvariant: and for every .

There is a real number, which we denote by , such that if and are sets which are at distance at least away from each other, then and has the property iff both and have the property .

There is an and a strictly positive function , such that if does not have the property , then .

(Zoomingout lemma) If then , where is positive and as for any fixed .
We call a saturation function for the property . An example of supersaturable property to keep in mind is the property of being avoiding, for which the saturation function can be chosen to be where is the uniform measure on the unit circle, and here for the second time we use the convention that a set is identified with the characteristic function of . In this example, with the exception of the zoomingout lemma all the conditions are not hard to check, and the zoomingout lemma will be proved in section 5. More generally in theorem 17 we will show that the property of being avoiding is an example of a supersaturable property. The proof of theorem is independent of the results in this section, and might be read before this section.
The motivation for the definition of the supersaturable properties is that not only implies that does not have the property , but also implies that . The latter statement is the content of the following lemma.
Supersaturation lemma.
Let be a supersaturable property. For every there is a constant such that for any there is such that the following holds. For any and any measurable set if
then
In particular, does not have the property .
Moreover, is a monotone nondecreasing function of for any fixed .
Before proving the supersaturation lemma, we need two lemmas. The first lemma shows that the rate of convergence in the definition of cannot be too slow, whereas the second lemma assures us that we need not to worry about small values of . \piccaptionTiling \parpic[r]
Proof.
Set . Then the tiling has the property because the distance between the translates of is and has the property . Since , the lemma follows. ∎
Proof.
Assume the contrary. We will show there is no set of positive measure with property , contradicting condition I. Suppose there is a set of positive measure with property . By the Lebesgue density theorem there is a point such that tends to as tends to . By condition IV we may assume that . Then the set is a subset of having property . Since the density of this set tends to as tends to zero we have reached a contradiction. ∎
Proof of supersaturation lemma.
Since the condition of the lemma refers only to the set we can assume without any loss of generality that . By lemma 5 for every there is such that if , then . Thus if , then the premise of the supersaturation lemma cannot hold since no set can have density . Hence we can assume that throughout the proof.
In the course of the proof of the supersaturation lemma we will prove following three statements:

is the statement that the supersaturation lemma holds for some specific and .

is the statement that if with , then the inequality holds with .

is the statement that for the conditions of the supersaturation lemma imply the weaker conclusion in which the constant is allowed to depend not only on but also on . Here is a positive number which depends only on the property .
First, we will establish for every . Then we will show that implies for any . Finally, we will demonstrate that and together imply . Since for the is vacuously true, all of these imply for all by induction on . Then the proof will be complete.
: We let to be the whose existence is postulated in the condition VI. We set . Choose so large that
If then since does not have the property , the condition VI tells us provided is chosen small enough. So, assume . Let . Let be a collection of disjoint cubes inside of side length each. Let . Then
where in the last inequality we used that . Thus
From lemma 4, and the choice of we get  
Let . Set . Then
(1) 
Since by lemma 4
the inequality (1) implies
Since we have . Therefore if , then the condition VI implies . Since and are at distance at least for distinct , we can apply the condition V to deduce
The monotonicity condition II allows us to conclude that .
implies : Suppose a set satisfies conditions of . Then the zoomingout lemma and tell us that
If small enough, we obtain that .
and imply : With hindsight we set . Condition I asserts that ensuring that . Recall that and let
Suppose we have a set satisfying the conditions of . If also satisfies the conditions of , then is as large as it should be, and we are done. Hence, the conditions of do not hold. Since , and in is equal to , the only way in which the conditions of can fail is
Since the average density of is at least and the inequality above says that the density of points that are centers of cubes of large density is no more than , there should be many points that are centers of cubes with medium density . For this we need to first relate to . For that we need to allow for the edge effects due to averaging over the cube of edge length rather than . Since ,