Multiple coverings with closed polygons

A planar set $P$ is said to be cover-decomposable if there is a constant $k=k(P)$ such that every $k$-fold covering of the plane with translates of $P$ can be decomposed into two coverings. It is known that open convex polygons are cover-decomposable. Here we show that closed, centrally symmetric convex polygons are also cover-decomposable. We also show that an infinite-fold covering of the plane with translates of $P$ can be decomposed into two infinite-fold coverings. Both results hold for coverings of any subset of the plane.

Splitting infinite-fold coverings can lead to very deep problems. Elekes, Mártai and Soukup [EMS11] constructed an infinite-fold covering of the line by translates of a closed set, whose decomposability is independent of ZFC. We believe that it is not the case for coverings of the plane with translates of a convex closed set.
It follows directly from Theorem 1.2 that an infinite-fold covering of the plane with translates of a closed, convex, centrally symmetric polygon is decomposable into two coverings. We prove the following stronger result.
Theorem 1.3. Let S be a closed, convex, centrally symmetric polygon. Then every infinite-fold covering of the plane with translates of S can be decomposed into two infinite-fold coverings.
Cover-decomposability has many other versions, instead of the plane, we can investigate, and decompose coverings of an arbitrary subset of the plane. We can consider only coverings with finite, countably many, or arbitrarily many translates. In the last section we review some of these versions of cover-decomposability. Our Theorems 1.2 and 1.3 hold for each of these versions, with the same proof.

Taking the dual, reduction to wedges
Just like in most of the papers about cover-decomposability, we formulate and solve the problem in its dual form. The idea is originally due to J. Pach [P86]. Suppose that S is an open or closed, centrally symmetric convex polygon, its vertices are v 1 , v 2 , . . ., v 2n , ordered clockwise. Indices are understood modulo 2n.
Definition 2.1. For any two points, a and b, let − → ab denote the halfline whose endpoint is a and goes through b. Let arg( − → ab) denote the clockwise angle from the positive x axis to − → ab.
Definition 2.2. For every i, 1 ≤ i ≤ 2n, let E i denote the convex wedge whose bounding halflines are the translates of −−−→ v i v i−1 and −−−→ v i v i+1 . If S is closed (resp. open), then let E i also be closed (resp. open). E i is called the wedge that belongs to vertex v i of S. We say that a wedge E belongs to S, or E is an S-wedge, if it belongs to one of its vertices. For any point p, let E i (p) denote the translate of E i such that its apex is in p.
Now we can state the dual version of Theorem 1.2 Theorem 2.3. Let S be a centrally symmetric closed convex polygon, with vertices v 1 , v 2 , . . . , v 2n , ordered clockwise. Then there is an m = m(S) > 0 with the following property. Any bounded point set H can be colored with red and blue such that any translate of an S-wedge, E i (p), if |E i (p) ∩ H| ≥ m, then E i (p) ∩ H contains points of both colors.
Proof of Theorem 1.2 from Theorem 2.3. Let x = x(S) be a number with the property that a square of side x intersects at most two consecutive sides of S. Divide the plane into squares of side x, by a square grid. There is a constant k such that any translate of S intersects at most k little squares.
For any point p, let S(p) denote the translate of S so that its center is at p. Let H = { S i | i ∈ I } be a collection of translates of S that form a k = k m-fold covering, where m = m(S) from Theorem 2.3. For every i ∈ I, let c i be the center of S i . Let H = { c i | i ∈ I } be the set of centers. For any point a, a ∈ S i if and only if c i ∈ S(a). Therefore, for every point a, S(a) contains at least k points of H .
The collection H can be decomposed into two coverings if and only if the set H can be colored with two colors, such that every translate of S contains a point of both colors.
Color the points of H in each square separately, satisfying the conditions of Theorem 2.3. Now return to the covering H and color each translate of S in H to the color of its center. We claim that both the red and the blue translates form a covering. Let p be an arbitrary point, we have to show that it is covered by a translate of both colors. Or equivalently, S(p) contains a point of H of both colors. Since S(p) contains at least k points of H , it contains at least k/k = m points in one of the little squares Q. But S(p) intersects Q "like a wedge" that is, S(p) ∩ Q = E(q) ∩ Q for some S-wedge E and point q. Therefore, by Theorem 2.3, S(p) ∩ Q contains a point of H of both colors. 2 Now we "only" have to prove Theorem 2.3. We need some preparation.

Some properties of boundary points
Theorem 2.3 has been proved by J. Pach [P86] in the special case when H is finite. Some parts of our proof are just modifications of his argument, but some other parts are completely new.
Let S be a centrally symmetric, open or closed convex polygon, its vertices v 1 , v 2 , . . . , v 2n in clockwise direction, the S-wedges are E 1 , E 2 , . . . , E 2n , respectively. Let H be a bounded point set.
It is easy to see that the map π is injective. If p 1 , p 2 ∈ B i and π(p 1 ) = π(p 2 ), then either p 2 ∈ E i (p 1 ) or p 1 ∈ E i (p 2 ), but both of them are impossible since both p 1 and p 2 are E i -boundary points.
Definition 2.5. Let 1 ≤ i ≤ 2n. For any two E i -boundary points p 1 and p 2 , let p 1 ≺ i p 2 if and only if π i (p 1 ) precedes π i (p 2 ) on − → i .
The relation π i is a linear ordering on B i . Based on π i , we can define intervals on B i , for example, We say that the first half of the interval [p 1 , p 2 ] is p ∈ B i : π i (p) ∈ π i (p 1 ), π(p 1 ) + π i (p 2 ) 2 .
We define (p 1 , p 2 ) similarly as [p 1 , p 2 ], and we define the second half of an interval similarly as the first half.
Claim 2.6. Suppose that p ∈ B i and p ∈ B i+1 , that is, p is a boundary point with respect to both E i and E i+1 . Let be the line through p, parallel to v i v i+1 . Then one of the closed halfplanes defined by contains all points of H.
Proof. If p is a boundary point with respect to both E i and E i+1 , then contains an open halfplane bounded by . This halfplane does contain any point of H, therefore, its complement satisfies the conditions.
It is easy now that if S is closed, then there is at most one point p ∈ H that is a boundary point with respect to both E i and E i+1 . If S is open and both p and q are such boundary points, then p ≺ i q if and only if p ≺ i+1 q. It also follows from Claim 2.6 that if p is a boundary point with respect to both E i and E i+1 and q is a boundary point with respect to E i but not E i+1 , then p ≺ i q.
There could be other types of boundary points with respect to more than one wedge.
Definition 2.7. A point p ∈ H is a singular boundary point if there are numbers 1 ≤ i 1 < n 1 < i 2 < n 2 ≤ 2n, or 1 ≤ n 1 < i 1 < n 2 < i 2 ≤ 2n such that p is a boundary point with respect to E i1 and E i2 , but not a boundary point with respect to E n1 and E n2 , see in Figure  Claim 2.8. If p is a singular E i -boundary point, then it is a boundary point with respect to E i and E i+n (the reflection of E i ) and no other wedge.
Proof. Suppose that 1 ≤ i 1 < n 1 < i 2 < n 2 ≤ 2n, p is a boundary point with respect to E i1 and E i2 , but it is not a boundary point with respect to E n1 and E n2 , and i 1 + n = i 1 . Assume wlog. that i 1 = 1, i 1 = k < n.
Then E 1 (p) and E k (p) do not contain any point of H, different from p. It follows from the convexity of S, , therefore, p is a boundary point with respect to E n1 , a contradiction, see in Figure 2a. The argument is the same if we have 1 ≤ n 1 < i 1 < n 2 < i 2 ≤ 2n.
Now we show the all singular boundary points are of the "same type".
Claim 2.9. If p is a singular boundary point with respect to E i and E i+n , then there is no singular boundary point with respect to some other pair of wedges.
Proof. Suppose that p and q are singular boundary points with respect to different pairs of wedges, say, p with respect to E 1 and E n+1 , q with respect to E k and E n+k , 1 < k ≤ n. It follows that either or Suppose wlog. that (1) holds. Since q is a boundary point with respect to E k and E n+k , (Note, that if S is closed then the above inequalities are strict inequalities.) But (1) and (3) can simultaneously only if k = 2 and But in this case, q is also an E 1 -boundary point, so it is not singular, a contradiction see in Figure 2b.  From now on, suppose, without loss of generality, that all singular boundary points of H are E 1 -and E n+1 -boundary points. Observe, that if p and q are singular boundary points with respect to E 1 and E n+1 , then p ≺ 1 q ⇔ q ≺ n+1 p. The type of a boundary point p is the smallest i such that p is an E i -boundary point.
In the set B of boundary points, substitute each singular boundary point p by p and p , such that p is • Both p and q are of type i, and p ≺ i q.
Relation ≺ gives a linear ordering on B . We have the elements in the following order: • Boundary points with respect to both E 2n and E 1 , ordered according to ≺ 2n and ≺ 1 ; • E 1 -boundary points, ordered according to ≺ 1 ; • Boundary points with respect to both E 1 and E 2 , ordered according to ≺ 1 and ≺ 2 ; • E 2 -boundary points, ordered according to ≺ 2 ; • Boundary points with respect to both E 2 and E 3 , ordered according to ≺ 2 and ≺ 3 ; • E 3 -boundary points, ordered according to ≺ 3 ; • . . .
If we project the points of B on a circle, then there is a natural way to define intervals on B , and then also on B. No we define them precisely.
Definition 2.10. An I ⊂ B subset is called an interval of B , if one of the following two conditions hold.
(i) If p ≺ q ≺ r and p, r ∈ I, then q ∈ I.
(ii) If p ≺ r, p, r ∈ I, and either q ≺ p or r ≺ q, then q ∈ I.
An interval of B or B is called homogeneous if all its points are E i -boundary points, for some i.
Claim 2.11. A translate of an S-wedge E i intersects B in at most two intervals.
Proof. Consider a translate of an S-wedge, say, E 2 (z). Suppose that p is an E i -boundary point, q is an E j -boundary point, 3 ≤ i, j ≤ n + 1, p ∈ E 2 (z) and p q. Then so q ∈ E 2 (z). We can argue similarly if p and q are on the "other side", that is, i, j ∈ {n + 3, n + 4, . . . , 2n, 1}. Suppose now that p is an E i -boundary point, q is an E 2 -boundary point, 3 ≤ i ≤ n + 1, p, q ∈ E 2 (z) and p r q. Then again therefore, r ∈ E 2 (z). Again, we can argue similarly in the case i ∈ {n + 3, n + 4, . . . , 2n, 1}. Finally, suppose that p, q, and r are E 2 -boundary points, p, r ∈ E 2 (z) and p q r. Again, it is easy to check that r ∈ E 2 (z). The same holds if p, q, and r are E n+2 -boundary points.
It follows from these observations that E 2 (z) intersects B in at most two intervals. Definition 2.12. Two boundary points p, q ∈ B are neighbors if p ≺ q, and either (i) there is no r with p ≺ r ≺ q, or (ii) there is no r with r ≺ p or q ≺ r. Two boundary points p, q ∈ B are neighbors if the corresponding points in B are neighbors. Let p ∼ q denote that p and q are neighbors.
Let ≈ be the transitive closure of the relation ∼ on B, that is, p ≈ q if and only if there is a finite sequence of boundary points, starting with p, ending with q, such that the consecutive pairs are neighbors. The relation ≈ is an equivalence relation. Those boundary points p which belong to an equivalence class of size one, are called lonely boundary points. The others, which have a neighbor, are called social boundary points.
First we give a coloring procedure which colors the points black and white. Then we apply it several times to obtain our red-blue coloring.

Black-White-Boundary-Coloring(S, H)
Divide the boundary of H, B, into equivalence classes by relation ≈. First we color the social boundary points.
Let C be an arbitrary equivalence class, |C| > 1. If C contains singular boundary points, then color them first, so that consecutive points receive different colors. Then, if there are regular boundary points between two consecutive singular boundary points, color them so that no two consecutive boundary points are black and no three consecutive are white. Do the same for each equivalence class |C| > 1. Now we color the lonely boundary points, denote their set by B lonely . It is the union of at most 2n homogeneous intervals, that is, B lonely = ∪ 2n i=1 I i where the elements of I i are all E i -boundary points. We color each interval separately. Recall that, based on projection π i , we defined the midpoint, the first and the second half of a homogeneous interval.
For each i, consider interval I i . If it contains infinitely many points, color one of them black. Then again, for each i, if I i contains infinitely many points, color an uncolored one white. If it contains finitely many points, color all of them white. Now half each interval which contained infinitely many points, and drop intervals with finitely many points. Let J 1 , J 2 , . . ., J m be the set of new intervals. Repeat the previous step, choose an uncolored point in each of the intervals with infinitely many points, and color them black, then the same with white, and then color all uncolored points in intervals with finitely many points white. Repeat this infinitely many times.
Then, if there is still an uncolored point, color it white.
Claim 2.13. If there are infinitely many points in an interval of B, then it contains infinitely many points of both colors.
Proof. We can assume that interval I is homogeneous, say, all of its points are E i -boundary points. Suppose first that I contains a lonely boundary point p in its interior. Then there is an accumulation point q of boundary points in the interior of I. (Note that q is not necessarily an element of H.) If q is an accumulation point of lonely boundary points, then our procedure Black-White-Boundary-Coloring(S, H) will arrive to an interval J ⊂ I which contains infinitely many lonely boundary points, and it colors one of them white, one black. Moreover, the procedure will find such an interval in infinitely many steps, so it colors infinitely many points white, and infinitely many black.
If q is an accumulation point of social boundary points, or if I does not contain a lonely boundary point p in its interior, then I contains infinitely many social boundary points. Then either I contains three consecutive such points, or contains two consecutive that form an equivalence class of size two. In both cases, at least one of them is white and at least one is black. We can proceed similarly to find infinitely many points of both colors.  Now we are ready to prove Theorem 2.3. Suppose that S is a closed, centrally symmetric convex polygon, its vertices are v 1 , v 2 , . . . , v 2n , ordered clockwise, the S-wedges are E 1 , E 2 , . . ., E 2n . Let S be S minus its boundary. Let H be a bounded set of points.
First we color the boundary, B, of H, then we color the boundary B of the interior points, and finally we color the remaining points. Very roughly speaking, the first level will be "responsible" for color blue in wedges which contain many, but finitely many points, the next level is responsible for color red, and coloring of the remaining points settles the wedges with infinitely many points. 4. Take a square that contain H . If it contains finitely many points of H (that is, H has finitely many points), color them red and stop. If it contains infinitely many points of H , then color one red and one blue. Divide the square into four smaller squares. In each of them, if there are finitely many points of H , then color all uncolored points red, and do not consider this square anymore. If there are infinitely many points in it, then color an uncolored point red and another one blue. Then divide it into four little squares. Repeat this infinitely many times. Finally color all points, which are still uncolored, red. Now we prove that this coloring satisfies the conditions. Suppose that E i (a) contains finitely many points of H, but at least 9. Then E i (a) contains at least one boundary point of H.
Case 1. E i (a) contains one point from the first level, that is, |E i (a) ∩ B| = 1. Then this point is rich, so it is blue, and E i (a) contains at least 8 interior points.
Case 2. |E i (a) ∩ B| = 2. By Claim 2.11, E i (a) intersects B in at most two intervals. If both contain one point, then at least one of them is rich, so it is blue, and E i (a) contains at least 7 interior points. Case 3. 3 ≤ |E i (a) ∩ B| ≤ 8. Since E i (a) intersects B in at most two intervals, it contains two consecutive boundary points, so one of them is blue, and E i (a) contains at least one interior point.
Case 4. |E i (a) ∩ B| ≤ 9. Then E i (a) contains at least 5 consecutive boundary points, say, p 1 , p 2 , . . ., p 5 . At least three of them are blue. Suppose that all of them are blue. Procedure Black-White-Boundary-Coloring(S, H) did not color three consecutive points white, therefore, at least one of p 2 , p 3 and p 4 got color blue, because it is rich. It is not hard to see that E i (a) contains the interior points corresponding to this rich boundary point.
Summarizing, if E i (a) contains at least 9 but finitely many points, then either it contains a point of both colors, or it contains a blue point on the boundary, and at least one interior point. But in this case it contains a point of B , the boundary of the interior points, which is red, so we are done in the case when E i (a) contains finitely many but at least 9 points of H. Now suppose that E i (a) contains infinitely many points of H, and suppose for contradiction that it does not contain a point of both colors.
Case 1. E i (a) contains infinitely many points from the boundary of H. By Claim 2.11, E i (a) ∩ B consists of of at most two intervals, one of the intervals, say I, is infinite. Procedure Black-White-Boundary-Coloring(S, H) colors infinitely many points of I to both colors. It follows immediately, that there are infinitely many blue points in E i (a). Therefore, by our assumption, all points of I got color blue. Then infinitely many of them are rich. But then the infinitely many interior points that correspond to these rich points, are also in E i (a).
Case 2. E i (a) contains finitely many points from the boundary of H, but at least one. Then, just like in the finite case, it is not hard to see that E i (a) contains at least one blue point from the boundary, and infinitely many interior points.
Case 3. E i (a) does not contain boundary points. Obviously, it contains infinitely many interior points, and by the definition it doesn't contain boundary points of H .
So we can conclude that E i (a) contains infinitely many interior points, and either it contains a blue boundary point, or no boundary points at all. Since we colored the boundary of the interior points, B red, we obtain that E i (a) contains a red point from its boundary, or no boundary points at all, and infinitely many interior points of H .
We assumed that E i (a) does not contain a point of both colors, therefore, either E i (a) ∩ B = ∅ (and therefore E i (a) ∩ B = ∅), or E i (a) ∩ B = ∅. Assume the first, the argument in the second case is the same.
We know that E i (a) contains infinitely many points from H , the set of interior points of H . We distinguish two cases.
Case 1. E i (a) contains infinitely many points from B , the boundary of H with respect to S . The set E i (a) ∩ B is again the union of at most two intervals, therefore, one of the intervals contain infinitely many points, so by Claim 2.13 it contains infinitely many points of both colors.
Case 2. E i (a) contains finitely many points from B . Then it contains infinitely many points from the set H , the interior points of H , with respect to S . We claim that in this case E i (a) contains a point in its interior. Suppose not. Then all points in E i (a) ∩ H are on the boundary of E i (a), so they all belong to B , a contradiction. Therefore, there is a point a 0 ∈ H in the interior of E i (a). Clearly, E i (a 0 ) is also in the interior of E i (a). E i (a) ∩ B = ∅, hence a 0 is not a boundary point of H, so there is a point a 1 in E i (a 0 ). Since a 1 is not a boundary point either, there is an a 2 in E i (a 1 ). This way we get an infinite sequence a 0 , a 1 , . . . of points in E i (a 0 ). With the exception of finitely many, they belong to H . They have an accumulation point x. The point x ∈ E i (a 0 ) since E i (a 0 ) is closed, so x is in the interior of E i (a). Therefore, when we colored H , in step 4 of procedure Red-Blue-Coloring(S, H), once we arrived to a little square which is in E i (a), contains x, and contains infinitely many points. So we colored one of the blue and one of them red. This concludes the proof of Theorem 2.3.
Just like in the proof of Theorem 1.2, we can take the dual of the problem, and divide the plane into small squares. Therefore, it is enough to prove the following result.
Theorem 3.1. Let S be a closed, convex, centrally symmetric polygon, its vertices are v 1 , v 2 , . . . , v 2n , oriented clockwise. Then any bounded point set H can be colored with red and blue such that for any translate of an S-wedge E i (p), if |E i (p) ∩ H| = ∞, then E i (p) ∩ H contains infinitely many red and infinitely many blue points.
Let S be S minus its boundary and for 1 ≤ i ≤ 2n, let E i be E i minus its boundary. That is, E 1 , E 2 , . . ., E 2n are the S -wedges.
From now on, boundary of a point set is understood according to S , and not S.
i . Now we are ready to give the coloring algorithm.
Multiple-Red-Blue-Coloring(S, H) Step 1. We color a subset of B * so that we color at most one point from each four consecutive levels. For each p ∈ B * let h(p) = n if and only if p ∈ B (n) . That is, each p is on the h(p)-th level. Take a square Q 1 which contains B * . Divide it into four little squares, these are Q 2 , Q 3 , Q 4 and Q 5 . Then divide Q 2 into four little squares, these are Q 6 , Q 7 , Q 8 , Q 9 . Similarly, divide Q 3 to get Q 10 , . . . , Q 13 , and continue similarly. Eventually we divide each square in the list into four little squares, and put them in the list. This way we obtain an infinite list Q 1 , Q 2 , . . . of squares.

In
Step 1.1, if Q 1 contains infinitely many points of B * , then color one of them, p 1 , red. Otherwise, we stop. In Step 1.2, if Q 1 contains infinitely many points of B * \ l<h(p1)+3 B (l) , then color one of them, p 2 , blue. Otherwise, we stop.
In general, in Step 1.(2k − 1), if Q k contains infinitely many points of the set B * \ l<h(p 2k−1 )+3 B (l) , then color one of them, p 2k−1 , red. Otherwise, we stop. Then, in Step 1.2k, if Q k contains infinitely many points of the set B * \ l<h(p 2k−2 )+3 B (l) , then color one of them, p 2k , blue. Otherwise, we stop.
After countably many steps, we are done with Step 1.
In the following steps we color the uncolored points.
Step 2. For each even n, color B (n) with procedure Black-White-Boundary-Coloring(S , H (n) ). Now a boundary point p ∈ B (n) will be -blue, if it is rich of white, -red otherwise.
Step 3. For each odd n, color B (n) with procedure Black-White-Boundary-Coloring(S , H (n) ). Now a boundary point p ∈ B (n) will be -red, if it is rich of white, -blue otherwise.
That is, we change the roles of the colors.
Step 4. Take a square which contains H * . If it contains infinitely many points from H * , (that is, H * has infinitely many points) then color one of them blue and one of them red. Divide the square into four little squares. In each of them, which contains infinitely many points from H * , color one of the uncolored points blue and one of them red, and divide it into four smaller squares. Continue recursively. Once we obtain a square which contains only finitely many points from H * , color all uncolored points red, and do not divide it into smaller squares.
Suppose that H is colored by procedure Multiple-Red-Blue-Coloring(S, H). We show that if a translate of an S-wedge contains infinitely many points of H, then it contains infinitely many points of both colors. First we show that a wedge contains an accumulation point in its interior, then it contains infinitely many points of both colors.
Lemma 3.2. Suppose that E i (a) ∩ H is infinite and this set has an accumulation point in the interior of E i (a). Then E i (a) contains infinitely many points of both colors.
Every covering in the sequel is a family of translates of a planar set S. Our Theorem 1.2 states that every centrally symmetric closed convex polygon is plane-arbitrary-coverdecomposable. It is not hard to see, that our proof works also for the other versions of cover-decomposability. It was known only for those versions which could be reduced to a finite problem. The next Our proof, with hardly any modification, implies the same results for open centrally symmetric convex polygons. In this case it is easier to reduce the problem to the finite case, therefore, cover-decomposability was proved for more versions. The next 2. In the proof of Theorems 1.2 and 1.3 we used different colorings. In fact, there is a single coloring algorithm which can be used in both proofs, but we found it too technical to present it.