Kneser s theorem for upper Banach density

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1 Journal de Théorie des Nombres de Bordeaux 18 (2006), Kneser s theorem for upper Banach density par Prerna BIHANI et Renlin JIN Résumé. Supposons que A soit un ensemble d entiers non néatifs avec densité de Banach supérieure α (voir définition plus bas) et que la densité de Banach supérieure de A + A soit inférieure à 2α. Nous caractérisons la structure de A + A en démontrant la proposition suivante: il existe un entier positif et un ensemble W qui est l union des [2α 1] suites arithmétiques 1 avec la même différence tels que A + A W et si [a n, b n ] est, pour chaque n, un intervalle d entiers tel que b n a n et la densité relative de A dans [a n, b n ] approche α, il existe alors un intervalle [c n, d n ] [a n, b n ] pour chaque n tel que (d n c n )/(b n a n ) 1 et (A + A) [2c n, 2d n ] = W [2c n, 2d n ]. Abstract. Suppose A is a set of non-neative inteers with upper Banach density α (see definition below) and the upper Banach density of A + A is less than 2α. We characterize the structure of A+A by showin the followin: There is a positive inteer and a set W, which is the union of 2α 1 arithmetic sequences 1 with the same difference such that A + A W and if [a n, b n ] for each n is an interval of inteers such that b n a n and the relative density of A in [a n, b n ] approaches α, then there is an interval [c n, d n ] [a n, b n ] for each n such that (d n c n )/(b n a n ) 1 and (A + A) [2c n, 2d n ] = W [2c n, 2d n ]. 1. Introduction Let Z be the set of all inteers and let N be the set of all non-neative inteers. Capital letters A, B, C, and D are always used for sets of inteers and lower case letters a, b, c, d, e,, h, etc. are always used for inteers. Greek letters α, β, γ, and ɛ are reserved for standard real numbers. For any Manuscrit reçu le 22 septembre Mots clefs. Upper Banach density, inverse problem, nonstandard analysis. The research of both authors was supported in part by the 2003 summer underraduate research fund from the Collee of Charleston. The second author was also supported in part by the 2004 summer research fund from the Collee of Charleston. The first author is currently a raduate student. The second author is the main contributor of the paper. 1 We call a set of the form a + dn an arithmetic sequence of difference d and call a set of the form {a, a + d, a + 2d,..., a + kd} an arithmetic proression of difference d. So an arithmetic proression is finite and an arithmetic sequence is infinite.

2 324 Prerna Bihani, Renlin Jin a b we write [a, b] exclusively for the interval of all inteers between a and b includin a and b. We denote A ± B for the set {a ± b : a A & b B}. We write A ± a for A ± {a} and a ± A for {a} ± A. We denote ka for the set {ka : a A}. 2 Let A[a, b] denote the set A [a, b] and A(a, b) denote the cardinality of A[a, b]. For any set A the upper Banach density of A, denoted by BD(A), is defined by BD(A) = lim n sup k 0 A(k, k + n). n + 1 Clearly 0 BD(A) 1. It is easy to see that α = BD(A) iff α is the reatest real number satisfyin that there is a sequence of intervals {[a n, b n ] : n N} such that (I) lim (b n a n ) = and n lim n The followin is the main theorem of the paper. A(a n, b n ) b n a n + 1 = α. Theorem 1.1. Let A be a set of non-neative inteers such that BD(A) = α and BD(A + A) < 2α. Let {[a n, b n ] : n N} be a sequence of intervals satisfyin (I). Then there are N, G [0, 1], and [c n, d n ] [a n, b n ] for each n N such that (1) G = m = 2α 1, d (2) lim n c n n b n a n = 1, (3) A + A G + N, (4) (A + A) [2c n, 2d n ] = (G + N) [2c n, 2d n ] for all n N, (5) BD(A + A) = m 2α 1. Remark. (1) Since the upper Banach density of A is achieved in a sequence of intervals {[a n, b n ] : n N} satisfyin (I), it is natural to characterize the structure of (A + A) [2a n, 2b n ] because the part of A outside of those intervals [a n, b n ] may have nothin to do with the condition BD(A + A) < 2BD(A). (2) The interval [c n, d n ] in (4) of Theorem 1.1 cannot be replaced by [a n, b n ] because if we delete, for example, the first and the last k numbers for a fixed k from A[a n, b n ] for every n N, then is still true. lim n A(a n, b n ) b n a n + 1 = α 2 In some literature ka represents the k-fold sum of A. Since only the sum of two sets is considered in this paper, we would like to write A + A instead of 2A so that the term N can be reserved for the set of all multiples of without ambiuity.

3 Kneser s theorem for upper Banach density 325 (3) Due to the asymptotic nature of the upper Banach density the asymptotic characterization of the lenth of [c n, d n ] in (2) of Theorem 1.1 is the best we can do. Theorem 1.1 is motivated by Kneser s Theorem for lower asymptotic density. For a set A the lower asymptotic density of A, denoted by d(a), is defined by A(1, n) d(a) = lim inf. n n The lower asymptotic density is one of the densities that are of classical interest for many number theorists. Another classical density is Shnirel man density denoted by σ (see [4] or [11] for definition). It is often the case that a theorem about the Shnirel man density is obtained first and then a parallel theorem about the lower asymptotic density is explored. In early 1940 s, H. B. Mann proved a celebrated theorem, which says that if 0 A B, then σ(a + B) min{σ(a) + σ(b), 1}. Is the inequality true if the Shnirel man density is replaced by the lower asymptotic density? The answer is no and the followin is a trivial counterexample. We call that two sets A and B are eventually same, denoted by A B, if A [0, m] = B [0, m] for some m N. Example. Let k, k > 0 and k + k. B [0, k 1] + N be such that d(a) + d(b) > k+k 1 Let A [0, k 1] + N and. Then A + B [0, k + k 2] + N, A + B [0, k + k 2] + N, and d(a) + d(b) 1 d(a + B) < d(a) + d(b). However, Example 1 is basically the only reason that the inequality d(a + B) < d(a) + d(b) can be true. More precisely Kneser proved the followin theorem. Theorem 1.2 (M. Kneser, 1953). If d(a + B) < d(a) + d(b), then there are > 0 and G [0, 1] with G = m such that (1) A + B G + N, (2) A + B G + N, and (3) d(a + B) = m d(a) + d(b) 1. (II) For any A and B with d(a + B) < d(a) + d(b) we define A,B = min{ N : There is a G [0, 1] satisfyin (1) (3) of Theorem 1.2.} The oriinal version of Kneser s Theorem deals with the sum of multiple sets. For brevity and the purpose of this paper we stated only a version for the sum of two sets.

4 326 Prerna Bihani, Renlin Jin In [6] the second author discovered a eneral scheme, with the help of nonstandard analysis, how one can derive a theorem about the upper Banach density parallel to each existin theorem about the Shnirel man density or the lower asymptotic density. The results in [6] can also be obtained usin an erodic approach in symbolic dynamics (see [7] or [3]). In [7, Theorem 3.8] the second author derived a result about the upper Banach density, which is parallel to Kneser s Theorem. The result characterizes the structure of a very small portion of the sumset, i.e., in [7] one can only have lim n (d n c n ) = in the place of (2) of Theorem 1.1. We believe that the eneral scheme in [6] and the erodic methods in [7] are not sufficient for provin Theorem 1.1. In the followin few sections we develop stroner standard lemmas and extract more strenth from nonstandard methods so that we can have enouh tools for provin Theorem Standard lemmas In this section we first state some existin theorems as lemmas and then prove some new lemmas without involvin nonstandard analysis. Lemma 2.1 (Birkhoff Erodic Theorem). Suppose (Ω, Σ, µ) is a probability space and T is a measure-preservin transformation from Ω to Ω. For any function f L 1 (Ω), there exists a function f L 1 (Ω) such that n 1 1 µ({x Ω : lim f(t m (x)) = n n f(x)}) = 1. m=0 Birkhoff erodic theorem can be found in [3, p.59] or in [12, p.30]. Lemma 2.2 (M. Kneser, 1953). Let G be an Abelian roup and A, B be finite subsets of G. Let S = { G : + A + B = A + B} be the stabilizer of A + B. If A + B < A + B, then A + B = A + S + B + S S. Note that the stabilizer is always a subroup of G and if A + B < A + B 1, then the stabilizer of A + B is non-trivial, i.e., S > 1. Lemma 2.2 can be found in [11, p.115]. For a positive inteer let Z/Z be the additive roup of inteers modulo. Let π : Z Z/Z be the natural homomorphism from Z onto Z/Z. If d > 0 and d, then we write π,d : Z/Z Z/dZ for the natural homomorphism from Z/Z onto Z/dZ. Let d be the kernel of π,d, which is a subroup of Z/Z. Note that every non-trivial subroup of Z/Z has the form d for some proper factor d of. We now state another version of Theorem 1.2, which is convenient to use later. Lemma 2.3. Suppose d(a) = d(b) = α and d(a + B) < 2α. Let = A,B be as defined in (II) and let G [0, 1] with G = m be such that (1) (3) of Theorem 1.2 are true. Then there are F, F [0, 1] with F = F = k such that

5 Kneser s theorem for upper Banach density 327 (1) A F + N and B F + N, (2) 2k 1 = m, (3) α > k 1 2. Furthermore, for any a [0, 1] F, there is a b F such that a + b G + N. Proof: Let F, F [0, 1] be the minimal sets such that A F +N and B F + N, respectively. Let k = F and k = F. By the minimality of F and F we have that F +F +N G+N and hence π [F +F ] = π [G]. If k + k m + 2, then π [F ] + π [F ] < k + k 1. By Lemma 2.2 the stabilizer S of π [F ] + π [F ] is non-trivial. Let S = d for some proper factor d of and S = /d = s > 1. Then This implies that π [A + B] = π [G] = π [F + F ] = π [G] + S. A + B G + N G + {0, d,..., (s 1)d} + N G + dn where G [0, d 1] and π d [G] = π d [G ]. We also have that d(a+b) = G = π [G] + S = G d d(a)+d(b) 1 > d(a)+d(b) 1 d. Hence (1) (3) of Theorem 1.2 are true when is replaced by d. This contradicts the minimality of = A,B. If k + k m, then d(a + B) = m k + k d(a) + d(b), which contradicts the assumption of the lemma. Hence we can assume that k + k 1 = m. If k k, say k < k, then 2α > d(a + B) = m = k + k 1 2k 2d(A) = 2α, which is absurd. Hence we have k = k, which implies 2α > d(a+b) = 2k 1 and α > k 1 2. Let a [0, 1] F. If a + F G + N, then π [F {a}] + π [F ] = m = 2k 1 < π [F {a}] + π [F ] 1. By Lemma 2.2 the stabilizer of π [G] is non-trivial. This contradicts the minimality of. (Lemma 2.3) Lemma 2.4. Let > 0 and F [0, 1] with F = k. If A F + N and d(a) = α > k 1 2, then for any A a = A (a + N) where a F we have d(a a ) > 1 2.

6 328 Prerna Bihani, Renlin Jin Proof: Let Ā = A A a. Then Ā (F {a}) + N. Hence d(a a ) d(a) k 1 = α k 1 Lemma 2.5. Let A, B N be such that d(a) > 1 2 A + B N. > k 1 2 k 1 = 1 2. Proof: Let m 0 N be such that for any n > m 0, B(0,n) n+ > 1 (Lemma 2.4) and d(b) > 1 2. Then A(0,n) n+ > Given any n > m 0, both A[0, n] and n B[0, n] are subsets of [0, n] N. Since [0, n] N = n+1 and A(0, n)+b(0, n) > n + 1, then A[0, n] (n B[0, n]), which implies n A[0, n] + B[0, n] A + B. (Lemma 2.5) and Hence A + B N. Suppose A, B N, d(a) = d(b) = α, > 0, and F, F [0, 1] with F = F = k such that (1) and (3) of Lemma 2.3 hold true, then A + B F + F + N by Lemma 2.4 and Lemma 2.5. Note that in Lemma 2.3, the sets F, F and the number k are uniquely determined by A,B. Lemma 2.6. Let d(a) = d(b) = α and d(a + B) < 2α. Let = A,B be as defined in (II) and let G [0, 1] with G = m be as defined in Theorem 1.2. Let b B and B b = B (b + N). Then d(b b ) + m 2α. Proof: Let k = m+1 2. By Lemma 2.3, there are F, F [0, 1] such that F = F = k, A F +N, and B F +N. Let d(b b ) = β. Clearly β 1. Since (B B b) is a subset of the union of k 1 arithmetic sequences of difference, then which implies Hence we have This now implies that α = d(b) = d(b b (B B b )) β + k 1, 2k 1 d(b b ) + m = β + 2k 1 k α β α 2β + 1. β + 2α 2β + 1 2α β + 1 2α. (Lemma 2.6)

7 Kneser s theorem for upper Banach density 329 Lemma 2.7. Let d(a) = d(b) = α. Suppose there is a positive inteer d such that (1) π d [A] π d [B] and (2) π d [A + B] 2 min{ π d [A], π d [B] }. Then d(a + B) 2α. Proof: Suppose d is the least positive inteer satisfyin the conditions of the lemma. Let k = π d [A] and k = π d [B]. Let π d [A] = {u 1, u 2,..., u k } and π d [B] = {v 1, v 2,..., v k }. For each i = 1, 2,..., k let a i = min(a π 1 d (u i)) and for each j = 1, 2,..., k let b j = min(b π 1 d (v j)). Without loss of enerality we assume k < k. Given any ɛ > 0, we want to show that d(a + B) 2α ɛ. Let t = max({a i : i = 1,..., k} {b j : j = 1,..., k }). Let m 0 > t be such that for every m > m 0, A(0, m) m t > α ɛ 2 and B(0, m) m t > α ɛ 2. Fix an m > m 0. For each i = 1,..., k let A i = A[0, m] π 1 d (u i). Without loss of enerality we can assume that A 1 A 2 A k. For each i = 1,..., k let U i = {u 1,..., u i }. Case There is an r [1, k] such that U r + π d [B] 2r 1. We want to show that Case is impossible. Let r be the least number satisfyin the case. Since U r + π d [B] 2r 1 < r + k 1, by Lemma 2.2 the stabilizer S of U r + π d [B] is non-trivial and U r + π d [B] = U r + S + π d [B] + S S. Let d < d and d d be such that S = d d. Then S = s = d/d > 1. Suppose π d [B] + S > U r + S. Since U r + S and π d [B] + S are all multiples of s, then π d [B] + S S U r + S. Hence we have U r + π d [B] 2 U r + S 2r, which contradicts the assumption of the case. From the aruments above we can assume that π d [B] + S U r + S. Now we have π d [B] = π d [B] + S /s U r + S /s = π d,d [U r ]. Suppose for each u {u r+1,..., u k } we have u + π d [B] U r + π d [B]. Then π d [A + B] = U r + π d [B] 2r 1 < 2k, which contradicts (2) of the lemma. Hence there is a u {u r+1,..., u k } and there is a v π d [B] such that u + v U r + π d [B] = U r + π d [B] + S.

8 330 Prerna Bihani, Renlin Jin This implies π d,d (u + v) π d,d [U r ] + π d [B] and π d,d (u) π d,d [U r ]. Hence π d [A] > π d,d [U r ] π d [B] and π d [A + B] π d,d [U r {u}] + π d [B] 1 + π d,d [U r ] + π d [B] = π d,d [U r ] + π d [B] 2 π d [B]. Hence d < d is a positive inteer, which satisfies (1) and (2) of the lemma. This contradicts the minimality of d. (Case 2.7.1) Case For every r [0, k], U r + π d [B] 2r. By the assumption of the case, we can select, inductively on i, the elements w i,j U i and z i,j π d [B] for j = 1, 2 such that {w i,j + z i,j : i = 1,..., k & j = 1, 2} in Z/dZ has the cardinality 2k. For each w i,j and z i,j we have π 1 d (w i,j + z i,j ) (A + B)[0, m + t] A i by the enumeration of U k, the definition of t, and the choices of w i,j s. Hence k (A + B)(0, m + t) 2 A i = 2A(0, m). This implies (A + B)(0, m + t) m t i=1 2A(0, m) m t 2(α ɛ ) = 2α ɛ. 2 (Case 2.7.2) Since Case is impossible, then by Case we have d(a + B) 2α ɛ. Now the lemma follows because ɛ > 0 is arbitrary. (Lemma 2.7) The followin lemma miht be considered as trivial. Lemma 2.8. Let, be positive and d = cd(, ). Let G [0, 1] and G [0, 1] be such that X = G + Z = G + Z. Then there is an F [0, d 1] such that X = F + dz. Proof: Let F [0, d 1] be minimal such that X F + dz. It suffices to show F + dz X. Let s, t Z be such that d = s + t. Given any c F and n Z, by the minimality of F we can find e G and k Z such that c + kd = e. Hence c + nd = e + (n k)d = e + (n k)s + (n k)t. Clearly e + (n k)s G + Z. Hence e + (n k)s G + Z. This implies e + (n k)s + (n k)t G + Z. Therefore, c + nd X. (Lemma 2.8) Corollary 2.1. Let, be positive and d = cd(, ). Let G [0, 1] and G [0, 1]. If A G + N and A G + N, then there is an F [0, d 1] such that A F + dn.

9 Kneser s theorem for upper Banach density 331 Proof: Since G + N G + N, then G + Z = G + Z. Hence by Lemma 2.8 we have G + Z = G + Z = F + dz for some F [0, d 1]. This implies A F + dn. (Corollary 2.1) We now prove a lemma, which may be interestin for its own sake. Lemma 2.9. Suppose d(a) = d(b) = α, d(a + A) < 2α, and d(a + B) < 2α. Let 0 = A,A and 1 = A,B be as defined in (II). Then 0 = 1. Proof: We proceed the proof in two claims. We show 0 1 in the first claim and show 0 = 1 in the second claim. Claim Proof of Claim 2.9.1: By Lemma 2.3 we have k = π 1 [A] = π 1 [B] and α > k By Lemma 2.4 and Lemma 2.5 we can find a G [0, 1 1] such that A+A G+ 1 N. On the other hand there is a G [0, 0 1] such that A+A G + 0 N by the definition of 0. Hence there is a G [0, d 1] where d = cd( 0, 1 ) such that A + A G + dn by Corollary 2.1. By the minimality of 0, d = 0. This ends the proof. (Claim 2.9.1) Claim = 1. Proof of Claim 2.9.2: Suppose π 0 [A] = k and π 0 [B] = k. By Lemma 2.3, we have α > k If k k, then α > k For each v π 0 [B] let B v = B π 1 0 (v). Then d(b v ) d(b) k 1 0 > by Lemma 2.4. Hence by Lemma 2.5 there is a G [0, 0 1] such that A + B G + 0 N. By the minimality of 1 we have 0 = 1. So we can assume k < k. If π 0 [A + B] 2k, then by Lemma 2.7 we have d(a + B) 2α, a contradiction. Hence we can assume π 0 [A + B] 2k 1 < k + k 1. By Lemma 2.2 the stabilizer S Z/ 0 Z of π 0 [A] + π 0 [B] is non-trivial and π 0 [A + B] = π 0 [A] + S + π 0 [B] + S S. Let d 0 and d < 0 be such that S = d 0 and let s = S = 0 /d. If π d [A] π d [B], say π d [A] < π d [B], then π d [A + B] = π 0 [A + B] /s = π 0 [A] + S /s + π 0 [B] + S /s S /s = π d [A] + π d [B] 1 2 π d [A]. By Lemma 2.7 we have d(a + B) 2α, which contradicts the assumption of the lemma. So we can assume π d [A] = π d [B] or equivalently π 0 [A] + S = π 0 [B] + S. This implies π 0 [A + B] = 2 π 0 [A] + S s. On the other hand, we have π 0 [A + B] 2k 1. Hence π 0 [A] + S k + s 1 2. This

10 332 Prerna Bihani, Renlin Jin implies that for each u π 0 [A] we have π 0 [A] (u + S) k ( π 0 [A] + S S ) s s = s Hence each coset of S is either disjoint from π 0 [A] or contains more than s/2 elements from π 0 [A]. So π 0 [A + A] is the union of S cosets, i.e., there exists an F [0, d 1] such that π 0 [F ] + S = π 0 [A + A]. This implies A + A G + 0 N = π 1 0 [π 0 [A + A]] = π 1 0 [π 0 [F ] + S] = F + dn, which contradicts the minimality of 0. (Lemma 2.9) 3. Nonstandard Analysis In this section, we introduce needed notation, basic lemmas, and principles in nonstandard analysis. The reader is recommended to consult [6] for the detailed proofs of the lemmas and principles if they are omitted here. The reader who is familiar with nonstandard analysis should skip this section. The detailed introduction for nonstandard analysis can also be found in [5, 10] or many other nonstandard analysis books. [5] is written for the reader who has no prior knowlede in mathematical loic. Let (R; +,,, 0, 1) be the (standard) real ordered field. We often write R for this field as well as its base set. Let (R) be the collection of all subsets of R. We call the structure V = (R (R); +,,, 0, 1,,, F) the standard model, where is the membership relation between R and (R), is the cardinality function from the collection F in(r) of all finite subsets A of R to N such that A is the number of elements in A, and F is a set of n f -ary functions or partial functions f from R n f to R. For example, a sequence of real numbers x n : n N can be viewed as a partial function from R to R. We will also write V for the set R (R). We use an ultrapower construction to construct a nonstandard model V as the followin. Let U be a nonprincipal ultrafilter on N (see [6] for definition) and let V N be the set of all sequences x n : n N in V. Two sequences x n and y n are equivalent (modulo U) if {n N : x n = y n } U. For a sequence x n in V let [ x n ] denote the equivalence class containin x n. Let V N /U be the set of all equivalence classes of the sequences in V N. One can embed V into V N /U by identifyin each x V with x = [ x ], the equivalence class of the constant sequence x. We consider R as a subset of R N /U and denote X, where X R, for the version of X in V N /U, i.e., X = [ X ]. A number is called standard if it is in R and a set Y R is called standard if there is an X R with Y = X. There is a natural way to extend relations and functions on V such as +,,,, and to V N /U. The nonstandard model V for the purpose of this paper is the set V N /U toether with the extensions of all relevant relations and functions

11 Kneser s theorem for upper Banach density 333 such as +,,,,, and each function in F to V N /U. 3 The details of the construction can be found in [6]. In V there are non-zero infinitesimals, which are numbers whose absolute values are less than all standard positive real numbers. There are many positive inteers in N but not in N. All inteers in N N are called hyperfinite inteers that are finite from nonstandard point of view but reater than every standard inteer. We reserve the letters H, K, N for hyperfinite inteers. For two real numbers x, y R, x and y are called infinitesimally close, denoted by x y, if x y is an infinitesimal. We write x y (x y) if x < y (x > y) and x y. By the least upper bound axiom one can show that for any real number r R with α < r < β for some α, β R there is a unique standard real number γ such that r γ. The number γ is called the standard part of r denoted by st(r) = γ where st is called the standard part map. A subset Y of R is called internal if Y = [ X n ] where X n is a subset of R for each n. So a standard set is an internal set but an internal set may not be a standard set. A subset of R, which is not internal, is called an external set. For example N and R are external sets. An internal set [ X n ] is called hyperfinite if there exists a sequence of finite sets X n such that X = [ X n ], hence X = [ X n ] is a hyperfinite inteer. Any hyperfinite set [ X n ] has a reatest element [ max X n ] and a least element [ min X n ]. By a formula in this paper we mean a first-order formula (see [6] for definition). The followin lemma [6, Proposition 2] is called the transfer principle. Lemma 3.1. Let ϕ(α 1,..., α m, X 1,..., X n ) be a formula such that α 1,..., α m R and X 1,..., X n R are the only constants in ϕ and ϕ contains no free variables. Then if and only if ϕ(α 1,..., α m, X 1,..., X n ) is true in V ϕ(α 1,..., α m, X 1,..., X n ) is true in V. The next lemma can be found in [6, Proposition 2] Lemma 3.2. Let ϕ(x, v 1,..., v k ) be a formula such that v 1,..., v k V and x is the only free variable. Then the set is internal. {r R : ϕ(r, v 1,..., v k ) is true in V} 3 We consider that the standard model V contains only R and its power set (R) because we want to introduce nonstandard model as simple as possible for the reader who does not have prior knowlede of nonstandard analysis.

12 334 Prerna Bihani, Renlin Jin The next lemma [6, Proposition 4] is called countable saturation. Lemma 3.3. Suppose {X (k) : k N} is a collection of non-empty internal subsets of R such that X (1) X (2)... X (k).... Then there is a v R such that v X (k) for every k N. Note that if H is hyperfinite, then there are always hyperfinite inteers in [0, H] (N (H N)). To see this let A n = [n, H n]. Then A n A n+1 for every n N. Hence there exists N, which belons to A n for every n N. It is easy to check that we have N + Z [0, H]. The next lemma [6, Lemma 1] establishes a nonstandard equivalence of BD(A) α. Lemma 3.4. Given α, for any set A N, BD(A) α iff there is an infinitesimal ι 0 and an interval I = [n, n + H] N of hyperfinite lenth such that A(n, n + H) α ι. H + 1 Loeb spaces: Given a hyperfinite inteer H, Ω = [0, H 1] is a hyperfinite set. Let A Ω be an internal set. Then A is an inteer between 0 and H. Hence A /H is a real number in R between 0 and 1. Let Σ 0 be the collection of all internal subsets of Ω and let µ(a) = st( A /H) for every A Σ 0. Then (Ω, Σ 0, µ) is a finitely-additive probability space from the standard point of view. For any subset S of Ω, internal or external, define µ(s) = inf{µ(a) : A Σ 0 and A S}, µ(s) = sup{µ(a) : A Σ 0 and A S}, and Σ = {S Ω : µ(s) = µ(s)}. It is easy to see that Σ 0 Σ. For each S Σ, define µ L (S) = µ(s) = µ(s). Then (Ω, Σ, µ L ) is a standard, countably-additive, atomless, complete probability space, which is called a hyperfinite Loeb space enerated by a normalized uniform countin measure /H. Let s call it simply a Loeb space on Ω. Note that the Loeb space construction can be carried out on any hyperfinite set. Note also that the verification of the countable-additivity requires usin countable saturation. The followin notation is non-traditional and will be frequently used in the proof of the main theorems of the paper. Let A Z and x Z. Define A x+n = x + (A x) N, A x N = x (x A) N, A x+z = x + (A x) Z,

13 Kneser s theorem for upper Banach density 335 d x+n (A) = d((a x) N), and d x N (A) = d((x A) N). In words, A x+n is the part of A in x + N where x + N is a copy of N lyin upward from x. Likewise, A x N is the part of A in x N and A x+z is the part of A in x + Z. The followin lemma is a variation of [6, Lemma 2]. Althouh the style of the lemma is slihtly different from [6, Lemma 2], the ideas of the proofs are similar. Lemma 3.5. Suppose A N, BD(A) = α, and {[a n, b n ] : n N} is a sequence of intervals of standard inteers satisfyin (I). Let N be any hyperfinite inteer. Then for almost all x [a N, b N ] in terms of the Loeb measure µ L on [a N, b N ], we have d x+n ( A) = d x N ( A) = α. On the other hand, if A N and there is an x N such that d x+n ( A) α or d x N ( A) α, then BD(A) α. Proof Suppose BD(A) = α and {[a n, b n ] : n N} is the sequence satisfyin (I). By Lemma 3.1 A(a N, b N ) b N a N + 1 α. Hence the Loeb measure of the set A[a N, b N ] in [a N, b N ] is α. Let T be the map from [a N, b N ] to [a N, b N ] such that T (b N ) = a N and T (x) = x + 1 for every x [a N, b N 1]. Let T be the map from [a N, b N ] to [a N, b N ] such that T (a N ) = b N and T (x) = x 1 for every x [a N +1, b N ]. Then T and T are Loeb measure-preservin transformation. Let f be the characteristic function of the set A[a N, b N ]. By Lemma 2.1 there is an L 1 function f such that for almost all x [a N, b N ], (III) n 1 1 lim f(t m (x)) = n n f(x). m=0 Since the fdµ interation over [a N, b N ] of the left-hand side of (III) is α, then [a N,b N ] L = α. We need to show that f(x) = α almost surely. Note that the set n=0 [a N, b N n] has Loeb measure 1. Suppose there is an x n=0 [a N, b N n] such that the left-hand side of (III) is β > α. Then A(x,x+n) n+1 > β+α 2 for sufficiently lare n N. Let D be the set of all n N such that A(x,x+n). By Lemma 3.2, D is an n+1 > β+α 2 internal subset of N, which contains all sufficiently lare n in N. Then D must contain a hyperfinite inteer N. Hence A(x,x+N ) N +1 > β+α 2. By Lemma 3.4, BD(A) β+α 2 > α. This contradicts BD(A) = α.

14 336 Prerna Bihani, Renlin Jin Since f(x) α for almost all x [a N, b N ] and [a N,b N ] f(x)dµ L = α, then f = α almost surely. Since n 1 1 lim f(x + m) = α n n m=0 implies d x+n ( A) = α, then for almost all x [a N, b N ], we have d x+n ( A) = α. If T is replaced by T in the aruments above, then d x N ( A) = α for almost all x [a N, b N ]. Hence for almost all x [a N, b N ] we have d x+n ( A) = d x N ( A) = α. Suppose d x+n ( A) α (d x N ( A) α). Then there is a hyperfinite inteer N such that A(x,x+N) N+1 α or > α ( A(x N,x) N+1 α or > α). By Lemma 3.4 we have BD(A) α. (Lemma 3.5) 4. Proof of theorem 1.1 In this section we state Theorem 4.1, which can be viewed as a nonstandard version of Theorem 1.1. We first prove Theorem 1.1 assumin Theorem 4.1 and then prove Theorem 4.1. Theorem 4.1. Let B be an internal subset of an interval [0, H] of hyperfinite lenth. Suppose (a) for any x [0, H], d x+n (B) α and d x N (B) α, (b) there is an S [0, H] with Loeb measure one such that for any x S, d x+n (B) = d x N (B) = α, (c) for any x [0, 2H], d x+n (B + B) < 2α and d x N (B + B) < 2α. Then there is a N such that for any x S, = C,C as defined in (II) where C = (B x) N, and there are G [0, 1] and [c, d] [0, H] such that (1) d c H+1 1, (2) B + B G + N, (3) (B + B) [2c, 2d] = (G + N) [2c, 2d], (4) G 2α 1. Proof of Theorem 1.1 assumin Theorem 4.1: Let A N be such that BD(A) = α and BD(A+A) < 2α. Let {[a n, b n ] : n N} be a sequence of intervals satisfyin (I). Let N be an arbitrary hyperfinite inteer. There is a subset S [a N, b N ] of Loeb measure one (on [a N, b N ]) such that for any x S we have d x+n ( A) = d x N ( A) = α by Lemma 3.5. For any x [a N, b N ] we have d x+n ( A) α and d x N ( A) α, and for any x [2a N, 2b N ] we have d x+n ( A + A) < 2α and d x N ( A + A) < 2α, aain by Lemma 3.5. Applyin Theorem 4.1 with [0, H] = [a N, b N ] a N and B = A[a N, b N ] a N we can find N, G N, [c N, d N ] such that

15 Kneser s theorem for upper Banach density 337 (1) d N c N b N a N 1, (2) A[a N, b N ] + A[a N, b N ] G N + N N, (3) ( A[a N, b N ] + A[a N, b N ]) [2c N, 2d N ] = (G N + N N) [2c N, 2d N ], (4) G N N 2α 1 N. We show first that N and G N do not depend on N and then show that G N does not depend on N. Let N be another hyperfinite inteer. Let x [a N, b N ] and y [a N, b N ] be such that d x+n ( A[a N, b N ]) = d y+n ( A[a N, b N ]) = α. Then d x+y+n ( A x+n + A y+n ) d x+y+n ( A + A) < 2α by Lemma 3.5. Let = C,D be as defined in (II) where (C, D) = (( A x) N, ( A y) N). Then by Lemma 2.9 we have N = = N. Clearly m = G N = 2α 1 depends only on α and. We now show that G N does not depend on N. Given a hyperfinite inteer N, let F N [0, 1] be minimal such that A[a N, b N ] F N + N. Note that F N +F N = G in Z/Z. By Lemma 2.2 and the minimality of = C,C as defined in Theorem 4.1 we have F N = m+1. Claim A F N + N. Proof of Claim 1.1.1: Suppose not and let z A (F N + N). By Lemma 3.1 we can assume z A (F N + N). Choose another hyperfinite inteer N such that b N a N > 2b N. Such N exists by Lemma 3.1. Let x [a N, b N ] be such that x + N [a N, b N ] and d x+n ( A) = α. Note that A x+n F N + N. Since (z + b N ) (x + a N ) b N a N 2, then by Lemma 3.5 one can find y 1, y 2 [a N, b N ] such that d y1 +N( A) = d y2 +N( A) = α and x + y 1 = z + y 2 = u. Since d u+n ( A x+n + A y1 +N) < 2α, then there is a G [0, 1] such that A x+n + A y1 +N x + y 1 + G + N. Let F, F [0, 1] be minimal such that A x+n F + N and A y1 +N F + N. By Lemma 2.3 we have F = F = m+1. Hence F = F N. Since A y1 +Z A[a N, b N ] F N + N, then F = F N. Since A y2 +N F N + N, then A y2 +N F + N. Note that z F + N. Hence there is a v A y2 +N such that z + v F + F + N by Lemma 2.3. Now we have z + A y2 +N x + y 1 + N and z + A y2 +N (v + N) A x+n + A y1 +N. Since d x+y1 +N( A + A) d x+y1 +N( A x+n + A y1 +N) + d z+y2 +N(z + ( A y2 +N (v + N))), then by Lemma 2.6 we have d x+y1 +N( A + A) 2α, which implies BD(A+A) 2α by 3.5, a contradiction to the assumption BD(A+A) < 2α. (Claim 1.1.1)

16 338 Prerna Bihani, Renlin Jin By Claim we conclude that A F N + N for any hyperfinite inteer N. Hence there is an F [0, 1] such that F N = F for any hyperfinite inteer N by the minimality of F N. Let G [0, 1] be such that π [F + F ] = π [G]. Then G N = G for any hyperfinite inteer N and m = G = 2α 1. Clearly A+ A G+ N and hence A+A G+N. Note that I = {[a n, b n ] : n N} is an internal sequence of intervals 4. Given k N let ϕ k (x,, G, A, I) be the formula sayin that x is a positive inteer and there is a [c x, d x ] [a x, b x ] with dx cx b x a x > 1 1 k such that ( A + A) [2c x, 2d x ] = (G + N) [2c x, 2d x ]. We will omit the parameters of the formula ϕ k and simply write ϕ k (x) instead. Let X k be the set of all positive inteers x N such that ϕ k (x) is true in V. Then X k is internal by Lemma 3.2 and contains all hyperfinite inteers. Let n k = min{x N : N [0, x 1] X k }. Then n k N. One can now choose a strictly increasin sequence n k n k in N. For each n [n k, n k+1 1] for k > 0 choose [c n, d n ] uaranteed by the truth of ϕ k (n) in V. For n < n 1 choose [c d n, d n ] =. Then lim n c n n b n a n = 1. Finally m (A + A) [2c n, 2d n ] = BD(G + N) BD(A + A) lim = m n 2d n 2c n + 1 implies BD(A + A) = m 2α 1. (Theorem 1.1) Proof of Theorem 4.1: Without loss of enerality let S = {x [0, H] : d x+n (B) = d x N (B) = α}. Given x, y S, let x,y = C,D be as defined in (II) where and let h x,y = C,D where (C, D) = ((B x) N, (B y) N) (C, D) = ((x B) N, (y B) N). Note that for any k Z, x S iff x + k S. We also have x,y = x+k,y = x,y+k and h x,y = h x+k,y = h x,y+k. For any C, D x + N we write C D if C x D x. For any C, D x N we write C D if x C x D. Claim For any x, y S, x,x = x,y and h x,x = h x,y. Proof of Claim 4.1.1: The claim follows from Lemma 2.9. (Claim 4.1.1) 4 I can be viewed as a pair of sequences in F as we define the nonstandard universe.

17 Kneser s theorem for upper Banach density 339 By Claim we can fix and h such that = x,y and h = h x,y for any x, y S. Claim = h. Proof of Claim 4.1.2: Assume the claim is not true. By symmetry we can assume h <. Let x S and let G [0, h 1] with G = m be such that B x N + B x N 2x + G hn. Let F [0, h 1] with F = k be minimal such that B x N x + F hn. Then m = 2k 1. Note that α > k h 1 2h. Since d x N (B) = α, then there is a hyperfinite inteer N such that B(x N,x) N+1 α and B[x N, x] x + F h N by Lemma 3.1 and Lemma 3.2. By the proof of Lemma 3.5 there is a y [x N, x] such that y + Z [x N, x] and d y+n (B) = d y N (B) = α. So y S. Since B y+n x+f h N, then for any z B y+n and B z = B y+n (z +h Z), we have d y+n (B z ) > 1 2h by Lemma 2.4. This implies that there is G [0, h 1] such that B y+n + B y+n G + hn. Hence = y,y h by Claim and by the definition of. This contradicts the assumption that h <. (Claim 4.1.2) Claim For any x, y S, there is G [0, 1] such that G = 2α 1 and B x+z + B y+z = (x + y) + G + Z. Proof of Claim 4.1.3: Let G, G [0, 1] be such that B x+n + B y+n (x + y) + G + N, B x+n + B y+n (x + y) + G + N, B x N + B y N (x + y) + G N, and B x N + B y N (x + y) + G N. Note that G = G = 2α 1. Suppose G G. There are b B x N and c B y N such that b + c (x + y) + G + Z. Clearly b, c S. Let G [0, 1] be such that B b+n + B c+n (b + c) + G + N. Then G m + 1, which contradicts that G = 2α 1. This shows G = G. Suppose B x+z + B y+z (x + y) + G + Z and let u ((x + y) + G + Z) (B x+z + B y+z ).

18 340 Prerna Bihani, Renlin Jin Let b B x+z and c B y+z be such that b + c u ( mod ). Let n Z be such that b + c + n = u. Let B b = B b N (b + Z) and B + c+n = B c+n+n (c + n + Z). Then d b N (B b ) > 1 2 and d c+n+n(b + c+n ) > 1 2. Hence we have B c+n + (b + c + n B b ). This implies u = b + c + n B b + B+ c+n B x+z + B y+z, which contradicts the choice of u. (Claim 4.1.3) Claim Let x S be such that 0 x H 1 and let G [0, 1] be such that B x+z + B x+z = (G + N) (2x + Z). Then there is [c, d] [0, H] with c H 0 and d H 1 such that (G + N) [2c, 2d] (B + B) [2c, 2d]. Proof of Claim 4.1.4: Let c and d be defined by c = min{y [0, x] : (G + N) [2y, 2x] (B + B) [2y, 2x]} d = max{y [x, H] : (G + N) [2x, 2y] (B + B) [2x, 2y]}. The number c and d are well defined by Claim It suffices to show that c H 0 and d H 1. We show d H 1 first. The statement c H 0 then follows from symmetry. Suppose d H 1. Then either 2d + 1 or 2d + 2 belons to the set (G + N) (B + B). Suppose 2d + 1 (G + N) (B + B). Let u = min{h d, d}. Then u H 0 and [d u + 1, d + u] [0, H]. Let S = S [d u + 1, d + u]. Then 2d + 1 S [d u + 1, d + u]. Since S has Loeb measure one, then S (2d + 1 S ), which implies that there are y, z S such that 2d + 1 = y + z. Let G [0, 1] be such that B y+z + B z+z = (G + N) (y + z + Z). Then we have G = G by Claim Note that d x+y 1 N (B + B) < 2α and (G + N) (y + z 1 N) B + B by the maximality of d. Hence G G, which implies G = G because they have the same cardinality. So we have 2d + 1 (G + N) (2d Z) = (G + N) (x + y + Z) = B y+z + B z+z (B + B).

19 Kneser s theorem for upper Banach density 341 This contradicts the assumption that 2d + 1 (B + B). If 2d + 1 B + B and 2d + 2 (G + N) (B + B), then one needs only to replace [d u + 1, d + u] by [d u, d + u] in the aruments above. (Claim 4.1.4) By Claim we can now fix G [0, 1] and [c, d] such that the conclusions of Claim are true. Let x S be such that x + Z [c, d]. Note that ( A + A) (2x + Z) = (G + N) (2x + Z). Let F [0, 1] be minimal such that B[c, d] F + N. Note that F + F = G in Z/Z. Since = x,x, then by Lemma 2.2 and minimality of x,x it is impossible to have 2 F 1 > G because otherwise G would have a non-trivial stabilizer S in Z/Z, which contradicts the minimality of x,x. On the other hand, since A x+z F + N, then by Lemma 2.3 we have 2 F 1 = G. Claim B + B G + N. Proof of Claim 4.1.5: Suppose the claim is not true and let a (B + B) (G + N). Then there are u, v B such that u + v = a. Without loss of enerality we can assume u F + N. By Lemma 2.3 we have u + F + N G + N. Choose z S such that u + z + Z [2c, 2d]. Without loss of enerality we can assume that z B z+z and u + z G + N because otherwise z can be replaced by z + n for some n Z. Note that z + n is also in S. Let B z + = B (z + N). Then (u + B z + ) (G + N) =. Since then (u + B + z ) ((G + N) (u + z + Z)) (B + B) (u + z + Z), d u+z+n (B + B) m + d z+n(b + z ) 2α by Lemma 2.6, which contradicts (c) of Theorem 4.1. (Claim 4.1.5) Clearly, the conclusion (3) of the theorem follows from Claim and Claim (Theorem 4.1) 5. Comments and questions In Theorem 1.1 we can replace A + A by A + B as lon as BD(A) = BD(B) = α, BD(A + B) < 2α, and the sequence of intervals {[a n.b n ] : n N} satisfies lim (b n a n ) = and n lim n A(a n, b n ) b n a n + 1 = lim n B(a n, b n ) b n a n + 1 = α. However, the chane will make the theorem more tedious and less natural. Without the last condition above, the problem becomes complicated. The followin is a eneral question about the sum of multiple sets.

20 342 Prerna Bihani, Renlin Jin Question 5.1. Let A i N for i = 1, 2,..., k and k k BD( A i ) < BD(A i ). i=1 Let {[a (i) n, b (i) n ] : n N} be such that lim n (b(i) n a (i) n ) = and i=1 lim n A i (a (i) n, b (i) n ) b (i) n a (i) n + 1 = BD(A i) for i = 1, 2,..., k. How well can we characterize the structure of k i=1 A i in k [ n=1 i=1 a (i) n, k i=1 b (i) n ]? The followin question for finite Abelian roups and the results of the paper seem to have some similar flavor. Question 5.2. Let G = Z/Z. Let F 0, F 1 G be such that F 0 = F 1 = k, F 0 + F 0 = 2k 1, and F 0 + F 1 = 2k 1. Can we conclude that there is an h G such that F 1 = h + F 0? Inverse problems for the addition of two sets A+A concern the structure of A when the size of A+A is relatively small. G. A. Freiman proved a series of important theorems on the inverse problems for a finite set A (cf.[2, 11]). For an infinite set A N, we use densities to measure the size of A. Lemma 2.3 can be viewed as a theorem for the inverse problems for infinite sets, which says rouhly that if d(a+a) < 2d(A) (i.e., the lower asymptotic density of A + A is small), then A F + N and d(a) > F 1 2 (i.e., A is a lare subset of the union of arithmetic sequences with a common difference ). In [8, 9] a theorem is proven for the inverse problem about upper asymptotic density. We can also state a corollary of Theorem 1.1 as a result for the inverse problem about the upper Banach density. Corollary 5.1. Let A be a set of non-neative inteers such that BD(A) = α and BD(A + A) < 2α. Then there are N and F [0, 1] such that A F + N and F 1 F 2 < α. There is a eneralization of Theorem 1.2 for A + A by Y. Bilu [1]. It is interestin to see whether we can also derive a theorem on the inverse problems about the upper Banach density parallel to Bilu s result. In [1] a condition lac(a) < Λ is imposed on A. Note that in the first part of Lemma 3.5 we actually have d x+n ( A) = d x N ( A) = d x+n ( A) = d x N ( A). Since d(a) = d(a) implies lac(a) = 1, then the condition lac(( A x) N) Λ is automatically satisfied when we apply some existin theorems about the lower asymptotic density to the set A x+n in Lemma 3.5.

21 Kneser s theorem for upper Banach density 343 Question 5.3. Suppose BD(A + A) cbd(a) for some c 2. should be the structure of A? What References [1] Y. Bilu, Addition of sets of inteers of positive density. The Journal of Number Theory 64 (1997), No. 2, [2] Y. Bilu, Structure of sets with small sumset. Asterisque 258 (1999), [3] H. Furstenber, Recurrence in Erodic Theory and Combinatorial Number Theory. Princeton University Press, [4] H. Halberstam, K. F. Roth, Sequences. Oxford University Press, [5] C. W. Henson, Foundations of nonstandard analysis A entle introduction to nonstandard extension in Nonstandard Analysis: Theory and Applications. Ed. by N. J. Cutland, C. W. Henson, and L. Arkeryd. Kluwer Academic Publishers, [6] R. Jin, Nonstandard methods for upper Banach density problems. The Journal of Number Theory, 91 (2001), [7] R. Jin, Standardizin nonstandard methods for upper Banach density problems in the DI- MACS series Unusual Applications of Number Theory, edited by M. Nathanson. Vol. 64 (2004), [8] R. Jin, Inverse problem for upper asymptotic density. The Transactions of American Mathematical Society 355 (2003), No. 1, [9] R. Jin, Solution to the Inverse problem for upper asymptotic density. Journal für die reine und anewandte Mathematik 595 (2006), [10] T. Lindstrom, An invitation to nonstandard analysis in Nonstandard Analysis and Its Application. Ed. by N. Cutland. Cambride University Press, [11] M. B. Nathanson, Additive Number Theory Inverse Problems and the Geometry of Sumsets. Spriner, [12] K. Petersen, Erodic Theory. Cambride University Press, Prerna Bihani Department of Mathematics University of Notre Dame Notre Dame, IN 46556, U.S.A. pbihani@nd.edu Renlin Jin Department of Mathematics Collee of Charleston Charleston, SC 29424, U.S.A. jinr@cofc.edu