On the Sidon Ideal

#A91 INTEGERS 25 (2025) ON THE SIDON IDEAL Andrzej Nowik University of Gda´nsk, Institute of Mathematics, Gda´nsk, Poland [email protected] Received: 3/5/25, Accepted: 9/23/25, Published: 11/5/25 Abstract We investigate the ideals of the natural numbers generated by Sidon sets and weak Sidon sets, proving that they are, in fact, the same. We examine several properties of the resulting ideal and place this ideal in the Kowitz Diagram of ideals. 1. Introduction In recent years, we have witnessed the rapid development of the theory of ideals. Many examples of ideals have already been considered; some of them originate from number theory. Based on the well-known concept of a Sidon set, we define a new ideal — namely the Sidon ideal — and determine its place within the family of other known ideals. We denote by Nthe set of all natural numbers with the number 0 adjoined. For infinite sets A, B ⊆Nwe write A⊆∗Bif and only if A\Bis finite. If A⊆Nis an infinite set, then A(n) denotes the n-th element of A. If A⊆N and n∈N, then A+n={a+n:a∈A}. If A, B are nonempty subsets of the natural numbers, then we define dist(A, B) = min{|a−b|:a∈A, b ∈B}. Also, recall the following standard notation (see for example [3]): If A, B ⊆Nthen define A−B={a−b:a > b, a ∈A, b ∈B}and D(A) = A−A. Definition 1. Suppose that A⊆N. We define CenSym(A) by {n∈A: there exists h∈N, h > 0 such that n−h, n +h∈A}. An ideal is a nonempty family I ⊂ P(N) closed under taking subsets and finite unions, i.e., 1. ∅∈I; 2. for all A∈ I and all B⊂A, we have B∈ I; 3. for all A, B ∈ I, we have A∪B∈ I. DOI: 10.5281/zenodo.17535172 INTEGERS: 25 (2025) 2 An ideal is proper if N∈ I, or, equivalently, I =P(N). Unless stated otherwise, we assume all ideals are proper and we assume also that all singletons belong to the ideal. An ideal is dense if every infinite subset of Ncontains an infinite subset belonging to the ideal. An ideal Iis called a P-ideal if and only if, for each sequence A0⊆∗A1⊆∗··· of sets from I, there exists A∗∈ I such that An⊆A∗for all n. The following four definitions are taken from [6]. Definition 2. The difference ideal Dwe define by D={A⊆N: for all infinite Z⊆N, D(Z)⊆ A}. Definition 3. The Dfin ideal we define by Dfin ={A⊆N: there exists n∈Nsuch that for all Z∈[N]n, D(Z)⊆ A}. Definition 4. An infinite set A⊆Nis said to be a lacunary set if and only if limn→∞ A(n+ 1) −A(n) = ∞. Definition 5. The Lacunary ideal is the ideal generated by lacunary subsets and we denote it by Lac. Definition 6. An infinite set A⊆Nis a thin set if and only if limn→∞ A(n) A(n+1) = 0. Definition 7. An infinite set A⊆Nis an almost thin set if and only if lim sup n→∞ A(n) A(n+ 1) <1. Definition 8. The ideal generated by thin sets is called the thin sets ideal and we denote it by T. The ideal generated by almost thin sets is called the almost thin sets ideal and we denote it by A. Definition 9. Let Endenote the family of subsets of natural numbers which do not contain any arithmetic sequence of length n. Let us define W as the family of subsets of natural numbers which do not contain arithmetic sequences of arbitrary length n: W = {A⊆N: there exists n > 0 such that for all a∈N, r > 0 there exists j < n such that a+r·j∈ A}. This is an ideal and we call it the van der Waerden ideal. 2. The Sidon Ideal Let us recall the main tool of our paper. INTEGERS: 25 (2025) 3 Definition 10. A subset (sequence) A⊆Nis called a Sidon set if and only if for n, m ∈A,n≤m, all sums n+mare distinct, i.e., for all n, m, k, l ∈A, if n≤m, k ≤l, and n+m=k+l, then n=k. The following definition is taken from [5]. Definition 11. A subset (sequence) A⊆Nis called a weak Sidon set (or wellspread) if for n, m ∈A,n<m, all sums n+mare distinct, i.e., for all n, m, k, l ∈A, if n < m, k < l, and n+m=k+l, then n=k. Recall here that the notion of Sidon sets appeared also in [3], page 6, and in [6] under the name D-sparse sets. Definition 12. A set A⊆Nis D-sparse if for every k∈D(A) there is only one pair n, m ∈Awith k=m−n. Notice that this notion is equivalent to the so-called “Golomb ruler” (in fact, Golomb ruler is a finite version of D-sparse sets), see for example [2]. A simple computation shows that the notions of D-sparse sets and Sidon sets are in fact identical. Let us define the main notion of this paper. Definition 13. The Sidon ideal, denoted by Si, is the ideal generated by the family of all Sidon sets, i.e., Si={A⊆N: there exists Sidon sets A1, . . . , Ansuch that A⊆ n [ j=1 Aj}. It is also tempting to define “a weak Sidon ideal” as an ideal generated by all weak Sidon sets, but this ideal coincides with Si. Theorem 1. Every weak Sidon set can be partitioned into two Sidon sets. Proof. It is well known that, if Ais a weak Sidon set, then A\CenSym(A) is a Sidon set. This is because the only equality in the definition of a Sidon set that shows that a weak Sidon set Ais not a Sidon set is the equation n+m= 2 ·k,n < m, where n=k, and this is the case when k∈CenSym(A). Now, assume that Ais a finite weak Sidon set. Define a sequence by setting A0:= A;An+1 := CenSym(An). This sequence is decreasing and since Ais a finite set there exists n0such that, for all n>n0, we have An=∅(for a finite Bwe cannot have CenSym(B) = Bunless B=∅). Let us define E(A) = S∞ k=0 A2k+1 \A2k+2 and O(A) = S∞ k=0 A2k\A2k+1. Of course, {E(A), O(A)}is a partition of A. We claim that the sets E(A) and O(A) are Sidon sets. To prove the claim, suppose by way of contradiction that there exist nand h > 0 such that n−h, n, n+ h∈E(A). Then n∈A2k0+1 \A2k0+2 for some k0. Since n∈CenSym(A2k0), there INTEGERS: 25 (2025) 4 exists h1>0 such that n−h1, n, n +h1∈A2k0. It is clear that, since Ais a weak Sidon set, we have h=h1. Hence n−h, n +h∈A2k0. Now, suppose that we have n−h, n +h∈A2k0+1. Then we would have n, n −h, n +h∈A2k0+1 and this would imply that n∈CenSym(A2k0+1), so n∈A2k0+2, which would be impossible. Thus, either n−h∈ A2k0+1 or n+h∈ A2k0+1. Suppose for example that n−h∈ A2k0+1. Then n−h∈A2k0\A2k0+1, so n−h∈O(A) which is a contradiction. This proves the claim. Thus, we have proved that each finite weak Sidon set can be partitioned into two Sidon sets. Using a standard compactness argument, we conclude that any infinite weak Sidon set can be partitioned into two Sidon sets. Corollary 1. The ideal generated by weak Sidon sets is the same as the ideal generated by Sidon sets. Note that the collection Sof all Sidon sets A⊆Nis a perfect subset of the Cantor space 2N. Corollary 2. The Sidon ideal is an Fσideal. Proof. Observe that the function ∪n: (2ω)n→2ω, defined by ∪n(A1, . . . , An) = Sn j=1 Aj, is continuous, and Si=S∞ n=1 ∪n[Sn]. 3. The Place of the Sidon Ideal in the Kowitz Diagram It is proved in [3] that (see Proposition 4.3) for each infinite A⊆Nthere exists an infinite D-sparse B⊆A. This shows that the Sidon ideal is dense. Moreover, the author proves (see Proposition 4.3 (1)) that if Ais D-sparse and n∈Nthen A+n is in the difference ideal. In particular, this shows that the Sidon ideal is contained in the difference ideal. Let us prove a little more. Theorem 2. The Sidon ideal is contained in the lacunary ideal. Proof. Suppose that S⊆Nis a Sidon set, and let M > 0. For any m≤Mthere exists at most one pair {S(n), S(n+ 1)}such that S(n+ 1) −S(n) = m. If there exists such a pair, define nm=n. If there is no such pair, define nm= 0. Define N= max{nm:m≤M}+ 1. Then, for n>N we have S(n+ 1) −S(n)> M, so S is a lacunary set. It is straightforward that the Sidon ideal is included in the Van der Waerden ideal. This is because any Sidon set cannot contain a 3-element increasing arithmetic sequence. Recall the following results concerning the existence of so-called “fat” Sidon set. INTEGERS: 25 (2025) 5 Theorem 3. There exists an infinite Sidon set Asuch that 1. |[1, n]∩A|> c 3 √nfor some c > 0(see [7]); 2. |[1, n]∩A|> c 3 pnlog(n)(see [1]); 3. |[1, n]∩A|> n√2−1−o(1) (see [8]). Even the first example of Theorem 3 suffices to prove the following proposition, which is a construction of a special Sidon set. Proposition 1. There exists a Sidon set Awhich is not in the ideal A. Proof. Let us begin with the following (probably well known) asymptotic behavior of any almost thin set. Suppose that A⊆Nis an almost thin set. Then there are C, D > 0 such that |[0, M)∩A| ≤ Clog M+D. Let us sketch the proof. There exist N0>0 and α > 1 such that A(n) A(n+1) <1 αfor n≥N0, so A(N0)·αn≤A(N0+n). A simple computation gives us the estimation |(A(N0), M)∩A| ≤ logα(M); hence, there exist C, D > 0 such that |[0, M)∩A| ≤ Clog M+D. Obviously, the same estimation holds for any element of the ideal A. By [7] there exists a Sidon set A such that |[1, M]∩A|> C1 3 √Mfor some C1>0. If this set were an almost thin set then we would have C1 3 √M≤Clog(M) + Dfor any M, which is impossible. Let us start with an easy lemma. Lemma 1. Suppose that A⊆[1,∞)∩Nis a set such that 2·A(n)≤A(n+ 1).(1) Then Ais a Sidon set. Proof. Suppose that A(n) + A(m) = A(k) + A(l) for some n≤mand k≤l. If l < m then we would have A(k) + A(l)≤2A(l)≤A(l+ 1) ≤A(m)< A(m) + A(n), which is impossible. Hence m=land therefore n=k. Theorem 4. Every Aset is in the ideal Si. Proof. It suffices to observe that since any almost thin set satisfies the estimate A(N0)·αn≤A(N0+n), we can easily divide it into a finite amount of sets which fulfills (1). It is straightforward to see that the Sidon ideal is not a P-ideal. We can use the same example of sets (Ak={n! + k:n∈N}) from the Lemma 1.2.8 from [4] since the sets Akbelong to Si. The Figure 1 is a part of the Kowitz Diagram (see [6], page 20) with the new ideal Siin the appropriate place. We show that all inclusions in the Figure 1 concerning the Sidon ideal are irreversible. By virtue of Example 1 it suffices to show the following theorem. INTEGERS: 25 (2025) 6 T A Dfin SiW Lac Figure 1: A part of the Kowitz diagram with the Sidon ideal Theorem 5. There exists a set A∈W∩Lac ∩Dfin which is not in the ideal Si. Definition 14. A set A⊆Nis said to be 3-artihmetic free if there are no a, b, c ∈A, where a<b<c, such that 2 ·b=a+c. Definition 15. A set A⊆Nis said to be sum-free if there are no a, b ∈A, where a < b, such that a+b∈A. Note that this definition differs from the standard definition of a sum-free set in Schur’s theorem, because we assume that a<bnot a≤b. Notice that if a set A⊆N is 3-arithmetic free then A∈W. Also, if a set A⊆Nis sum-free, then A∈ Dfin — this is a standard argument which mimics the argument from [3] that a D-sparse set is in the ideal D(see Proposition 4.3 (1) in [3]). Suppose that n, C, M, K are natural numbers. Define Block(n, C, M, K) = {Ml+C·2k:k= 0,1, . . . , n −1; l= 1,2, . . . , K}. A simple computation shows that if we assume that M > C ·2nthen Block(n, C, M, K) is both 3-arithmetic and sum-free. Let us formulate the following lemma. Lemma 2. For any sequence (Nk)of natural numbers and for any sequence (Ak)of finite sets, each of which are both 3-arithmetic and sum-free sets of natural numbers, we can find a sequence (mk)of natural numbers such that 1. the set A∗=S∞ k=1 Ak+mkis both 3-arithmetic and sum-free; 2. for all k, dist(Ak+mk, Ak+1 +mk+1)≥Nk. Proof. The proof is a straightforward inductive construction. Lemma 3. The set Block(n, C, C ·2n+ 1,(n−1) ·n 2+ 1) is not a sum of n−1 Sidon sets. INTEGERS: 25 (2025) 7 Proof. By way of contradiction, suppose that Block(n, C, C·2n+1,(n−1)·n 2+1) = Sn−1 j=1 Sj, where Sjare Sidon sets. For each l= 1, . . . (n−1) ·n 2+ 1 by the pigeonhole principle we can find k1(l), k2(l)∈ {0, . . . , n −1},k1(l)< k2(l), and j(l)∈ {1, . . . , n −1}such that Ml+C·2k1(l), Ml+C·2k2(l)∈Sj(l). Since there are at most (n−1) ·n 2triplets (k1, k2, j)∈ {0, . . . , n −1}2×{1, . . . , n −1}, such that k1< k2, again by the pigeonhole principle we can find l1< l2such that (k1(l1), k2(l1), j(l1)) = (k1(l2), k2(l2), j(l2)). Let ˆ k1=k1(l1) = k1(l2), ˆ k2=k2(l1) = k2(l2), and ˆ j=j(l1) = j(l2). Then we have Ml1+C·2ˆ k1, Ml1+C·2ˆ k2, Ml2+C· 2ˆ k1, Ml2+C·2ˆ k2∈Sˆ jand this is impossible since Sˆ jis a Sidon set. Proof of Theorem 5. Now apply Lemma 2 to the sequences Nn=nand An=Block(n, n, n ·2n+ 1,(n−1) ·n 2+ 1) and we obtain a suitable sequence (mn) and define a set A∗=[ n=2 Block(n, n, n ·2n+ 1,(n−1) ·n 2+ 1) + mn. By Lemma 2 this set is both 3-arithmetic and sum-free. Hence A∗∈W and A∗∈ Dfin . It is easy to see that also A∗∈ Lac. By Lemma 3, A∗is not a union of finitely many Sidon sets; therefore, A∗∈ Si. Acknowledgement. I would like to express my gratitude to Professor Krzysztof Kowitz and Professor Jacek Tryba for inspiring and fruitful discussions. References [1] M. Ajtai, J. Koml´os, and E. Szemer´edi, A dense infinite Sidon sequence, European J. Combin. 2(1) (1981), 1-11. [2] K. Drakakis, A review of the available construction methods for Golomb rulers, Adv. Math. Commun. 3(3) (2009), 235-250. [3] R. Filip´ow, On Hindman spaces and the Bolzano–Weierstrass property, Topology Appl. 160 (15) (2013), 2003-2011. [4] J.Flaˇskov´a, Ultrafilters and Small Sets, Ph.D. thesis, Univerzita Karlova , 2006. [5] P.M. Kayll, A note on weak Sidon sequences, Discrete Math. 299 (1) (2005), 141-144. [6] K. Kowitz, The use of Katˇetov order in the study of topological spaces and ultrafilters, Ph.D. thesis, University of Gda´nsk, 2023. [7] A.M.Mian and S.Chowla, On the B2sequences of Sidon, Proc. Nat. Acad. Sci. India Sect. A 14 (1944), 3-4. [8] I.Z.Ruzsa, An infinite Sidon sequence, J. Number Theory 68 (1) (1998), 63-71.