Solution to the Inverse Problem for Upper Asymptotic Density Renling Jin∗
Abstract Inverse problems study the structure of a set A when the “size” of A + A is small. In the article, the structure of an infinite set A of natural numbers with positive upper asymptotic density is characterized when A is not a subset of an infinite arithmetic progression of difference greater than one and A + A has the least possible upper asymptotic density. For example, if the upper asymptotic density α of A is strictly between 0 and 12 , the upper asymptotic density of A + A is equal to 32 α, and A is not a subset of an infinite arithmetic progression of difference greater than one, then A is either a large subset of the union of two infinite arithmetic progressions with the same common difference k = α2 or for every increasing sequence hn of positive integers such that the relative density of A in [0, hn] approaches α, the set A ∩ [0, hn] can be partitioned into two parts A ∩ [0, cn] and A ∩ [bn, hn ] such that cn /hn approaches 0, i.e. the size of A ∩ [0, cn] is asymptotically small comparing with the size of [0, hn], and (hn − bn )/hn approaches α, i.e. the size of A ∩ [bn , hn ] is asymptotically almost same as the size of the interval [bn , hn ]. The results here answer a question of the author in [8].
1
Introduction
Let N be the set of all natural numbers, including 0. Let A and B always denote the sets of natural numbers, and let a, b, c, h, i, j, k, m, n, x, y, z always denote natural numbers. For integers m, n and set A, we write [m, n] exclusively for the interval of integers {k ∈ N : m 6 k 6 n} and write A[m, n] for the set A ∩ [m, n] and A(m, n) for the number of elements in A[m, n]. Sometimes, A(1, n) is written as A(n). For sets A and B, we write A ± B for the set {a ± b : a ∈ A and b ∈ B} and write 2A for A + A. For a set A and a number b, we write A ± b for A ± {b} and write b ± A for {b} ± A. In this paper we often use (2A)(m, n) for | {x ∈ 2A : m 6 x 6 n} | and 2A(m, n) for 2 times A(m, n). We write a.p. as an abbreviation for “arithmetic progression”, which can be either finite or infinite. An a.p. of a single element has difference d for every Mathematics Subject Classification 2000 Primary 11B05, 11B13, 11U10, 03H15 Keywords: upper asymptotic density, inverse problem, nonstandard analysis ∗ The author was supported in part by the NSF grant DMS–#0070407.
1
non-negative integer d. We call a set J = I0 ∪I1 a bi-arithmetic progression, or b.p. for abbreviation, if I0 and I1 are two a.p. of the same difference d and the sets 2I0, 2I1 , and I0 + I1 are pairwise disjoint. The number d is called the difference of J . When we say that A is a subset of a b.p. I0 ∪ I1 we often implicitly assume that A ∩ I0 6= ∅ and A ∩ I1 6= ∅ unless we specifies otherwise. For a set A, the upper asymptotic density of A is defined by A(n) ¯ d(A) = lim sup . n n→∞ The main results of this paper characterize the structural properties of A when A is not a subset of an a.p. of difference > 1 and B is not a subset of an a.p. of ¯ ¯ . d(2A) = inf d(2B) : ¯ ¯ difference > 1 and d(B) > d(A) For the motivation of the main results, we would like to quote a few sentences from the preface of the book [11]: “The classical problems in additive number theory are direct problems, in which we start with a set A of integers and proceed to describe the h–fold sumset hA, that is, the set of all sums of h elements of A. In an inverse problems, we begin with the sumset hA and try to deduce information about the underlying set A. In the last few years, there has been remarkable progress in the study of inverse problems for finite sets in additive number theory. There are important inverse theorems due to Freiman, Kneser, Pl¨ unnecke, Vosper, and others. In particular, Ruzsa recently discovered a new method to prove a generalization of Freiman’s theorem.” Although the results in this paper are not directly related to the Freiman’s Theorem mentioned above, they share the same pattern, which says that if 2A is small, then A must have some structure. In fact, the idea of inverse problem occurs also in some of the theorems involving densities. The theorems about Shnirel’man’s pairs and Mann’s pairs in [5] deduce information about the Shnirel’man density of A and Shnirel’man density of B when the Shnirel’man density of A + B is small. Kneser’s Theorem (cf. [1, 4] deduces information about the arithmetic structure of A and B when the lower asymptotic density of A + B is small. In [8], the inverse problems for upper asymptotic density are considered. In [8] the structural properties of A are characterized when the upper asymptotic density of 2A + {0, 1} is small. However, adding {0, 1} to 2A seems to be a non–traditional requirement. The results should be more interesting if the requirement can be dropped. However, if A itself is an a.p. , then the upper asymptotic 2
density of A + A is exactly the same as the upper asymptotic density of A. Hence it is natural to assume that “A is not a subset of an a.p. I of difference > 1” because otherwise we can replace N by I and discuss the properties of A in I instead. ¯ Why do we need to add {0, 1} to 2A in the first place in [8]? Let α = d(A) and a0 = min A. By Lemma 1.5 below, one can prove the following: If α 6 12 and gcd(A − a0 ) = 1, i.e. the greatest common divisor of all numbers in A − a0 is one, ¯ ¯ then d(2A) > 3 α. If α > 1 , then d(2A) > α+1 . Note that the two inequalities above 2
2
2
are optimal. The equalities hold for the following two examples. Example 1.1 For every real number 0 6 α 6 1, let A=
∞ [
n
n
[d(1 − α)22 e, 22 ].
n=1
¯ ¯ Then d(A) = α, d(2A) =
1+α 2
¯ if α > 12 , and d(2A) = 32 α if α 6 12 .
Example 1.2 Let k, m, n ∈ N be such that k > 4 and 2m, 2n, m + n are pairwise distinct modulo k. Let A = {m + ik : i ∈ N} ∪ {n + ik : i ∈ N} . ¯ Then d(A) =
2 k
=α6
1 2
¯ and d(2A) =
3 k
= 32 α. It is easy to choose k, m, n such that
gcd(A − a0) = 1. At the time when [8] was written, we believed that if A is a set with positive upper asymptotic density α < 12 such that gcd(A − min A) = 1 and the upper asymptotic density of 2A reaches its smallest possible value, then A should be a set similar to the ¯ one in Example 1.1 or to the one in Example 1.2. If we require that d(2A+{0, 1}) = 3 α 2
¯ when d(A) = α 6 12 , then A cannot be the set similar to the one in Example 1.2. Hence we need only to show that A is a set similar to the one in Example 1.1 as done in [8]. This greatly simplifies the proof. Besides, adding {0, 1} to the set 2A makes it possible to apply Besicovitch’s Theorem [4, page 6] to the proof of [8, Lemma 2.1]. Without adding {0, 1}, we need not only to consider that A can be a set similar to the one in Example 1.2, but also to find a new way of proofs by-passing Besicovitch’s Theorem. These difficulties were overcome recently. We prove a key lemma in §2, which is inspired by Kneser’s Theorem mentioned above. The key lemma allows us by–passing Besicovitch’s Theorem in the proofs. 3
Next, we state the main theorem of this paper. Without loss of generality we always assume 0 ∈ A. ¯ Theorem 1.3 Let A be a set of natural numbers, 0 ∈ A, and d(A) = α > 0. 1 1+α ¯ Part I: Assume α > 2 . Then d(2A) = 2 implies that for every increasing A(0,hn ) sequence hhn : n ∈ Ni with limn→∞ hn +1 = α, we have lim
n→∞
Part II: Assume α
4 and c ∈ [1, k − 1] such that α =
2 k
and
A ⊆ {ik : i ∈ N} ∪ {c + ik : i ∈ N} or (b) for every increasing sequence hhn : n ∈ Ni with limn→∞
A(0,hn ) hn +1
= α, there exist
two sequences 0 6 cn 6 bn 6 hn such that A(bn , hn ) = 1, n→∞ hn − bn + 1 lim
cn = 0, n→∞ hn lim
and [cn + 1, bn − 1] ∩ A = ∅ for every n ∈ N. ¯ = 32 α implies that either Part III: Assume α = 12 and gcd(A) = 1. Then d(2A) (a) there exists c ∈ {1, 3} such that A ⊆ {4i : i ∈ N} ∪ {c + 4i : i ∈ N} or (b) for every increasing sequence hhn : n ∈ Ni with limn→∞
A(0,hn ) hn +1
= α, we have
(2A)(0, hn ) = α. n→∞ hn + 1 lim
Remark 1.4 (1) The proof of Part I of Theorem 1.3 is easy and will be omitted. (See [8, (2) of the remarks in Section 4] for an explanation.) (2) Part I and (b) of Part III cannot be improved so that the set A has the structure similar to the structure described in (b) of Part II. See [8, (1) of the remarks in Section 4] for the argument. 4
¯ (3) The value of d(2A) in each of the conditions of all three parts is the least ¯ possible value d(2A) can have. (4) In Part II (a) and Part III (a), A is a large subset of the union of the two a.p. in terms of upper asymptotic density because both have the same upper asymptotic density. (5) Recently G. Bordes [3] generalizes Part II of Theorem 1.3 for sets A with small ¯ upper asymptotic density. He characterizes the structure of A when d(A) 6 α0 for 5¯ ¯ some small positive number α0 and d(2A) < d(A). 3
In the next section §2, we will prove several lemmas including a key lemma necessary for the proof of Theorem 1.3. In §3, the proof of Theorem 1.3 is split into two nonstandard theorems, one for Part II and the other for Part III of Theorem 1.3. At the end of §3, a corollary is given. In both sections, the techniques from nonstandard analysis are used. Before getting into nonstandard analysis, we would like to list a few existing theorems as lemmas, which are needed in §2 and §3. Two of Freiman’s theorems (cf.[11, Theorem 1.15 and Theorem 1.16, page 28] or cf.[2, Proposition 1.1]) will be frequently cited. Lemma 1.5 (G. Freiman) Let A = {a0 , a1, . . . , ak−1 } be such that 0 = a0 < a1 < , then (2A)(0, 2n) > 3k − 3. If k > n+3 , · · · < ak−1 = n and gcd(A) = 1. If k 6 n+3 2 2 then (2A)(0, 2n) > k + n. Lemma 1.6 (G. Freiman) Let A ⊆ N be such that |A| = k > 2. If |2A| = 2k − 1 + b 6 3k − 4, then A is a subset of an a.p. of the length k + b 6 2k − 3. The next lemma is due to Lev and Smeliansky (see [9] and [11, p.118]). Lemma 1.7 (V. Lev & P. Y. Smeliansky) Let A and B be two finite sets of nonnegative integers. Suppose 0 ∈ A ∩ B, both A and B contain more than one element, gcd(A) = 1, m = max A, and n = max B 6 m. If m = n, then |A + B| > min {m + |B|, |A| + 2|B| − 3}. If m > n, then |A+B| > min {m + |B|, |A| + 2|B| − 2}. Note that Lemma 1.7 can be easily modified to the following form: Let A and B both be finite subsets of {dn : n ∈ N}. Suppose 0 ∈ A ∩ B, A and B contain more than one element, gcd(A) = d, m = max A, and n = max B 6 m. If 5
m = n, then |A + B| > min m + |B|, |A| + 2|B| − 3 . If m > n, then |A + B| > m d min d + |B|, |A| + 2|B| − 2 . The next lemma is due to van der Corput (see [4, p.18]). Lemma 1.8 (van der Corput) Let A, B be subsets of non-negative integers with 0 ∈ A ∩ B and let γ be a real number between 0 and 1. If for each n > 0, 1 + A(1, m) + B(1, m) inf : m = 1, 2, . . . , n > γ, m+1 then inf
2
1 + (A + B)(1, m) : m = 1, 2, . . . , n m+1
> γ.
Lemmas in Nonstandard Analysis
As mentioned in §1, the techniques from nonstandard analysis are needed in §2 and §3. One of the advantages of nonstandard methods is that an asymptotic argument such as upper asymptotic density in the standard world can be translated into one ∗ finite argument in a nonstandard world so that instead of dealing with a sequence of intervals in an upper asymptotic density argument, we can deal with only one interval of ∗finite length in the nonstandard world. For basic knowledge of nonstandard analysis, the reader is recommended to consult [10], [6], or [7]. We work within a fixed ℵ1 –saturated nonstandard universe ∗V throughout this article. For each standard set A, we write ∗A for the nonstandard version of A in ∗V . For example, ∗N is the set of all natural numbers in ∗V , and if A is the set of all even numbers in N, then ∗A is the set of all even numbers in ∗N. If we do not specify that A, B are sets of standard natural numbers, A, B are always assumed to be internal sets of (standard and nonstandard) natural numbers. Now a, b, c, h, i, j, k, m, n, x, y, z can take values in ∗N. The integers in ∗N r N are called hyperfinite integers. The letters H, K and N are exclusively used for hyperfinite integers. The Greek letters α, β, γ, δ, and are reserved exclusively for standard real numbers. For the convenience of handling nonstandard arguments, we would like to introduce some notation of comparisons. For real numbers r, s in ∗V , by r ≈ s we mean that r − s is an infinitesimal, by r s (r s) we mean that r < s (r > s) and r 6≈ s, and by r / s (r ' s) we mean r < s (r > s) or r ≈ s. Given a hyperfinite integer H and two real numbers r, s, by r ∼H s we mean that 6
s−r H
≈ 0, by r ≺H s (r H s) we
mean that r < s (r > s) and r 6∼H s, and by r H s (r H s) we mean that r ≺H s (r H s) or r ∼H s. It is often said that a is insignificant with respect to H if a ∼H 0. In the most of the cases the subscript H is clearly given so that it will be dropped as a subscript for convenience. Note that the comparison relations , , ≈, etc. can be interpreted in terms of ≺, , ∼, etc. or vice versa when H is clearly given. For example,
a H
b H
/
iff a b. We use, for example, more often than the use of /
because fractions can be avoided. When using ∼, ≺, , etc. insignificant quantities can often be neglected. For example, instead of using A(0, H) ∼ α(H +1), we can use its equivalent form A(0, H) ∼ αH. For another example, when a 6 c 6 b, we often write A(a, c) ∼ A(a, b) + A(b, c) instead of writing A(a, c) = A(a, b) + A(b + 1, c). For a real number r ∈ ∗R bounded by a standard real number, let st(r), the standard part of r, be the unique standard real number α such that r ≈ α. Next lemma is stated for showing how upper asymptotic density can be translated into a nonstandard version. ¯ Lemma 2.1 Let A ⊆ N and let 0 6 α 6 1. Then d(A) > α if and only if there is a hyperfinite integer H such that ∗A(0, H) αH. Proof: “⇒” Let h1 < h2 < · · · be an increasing sequence in N such that 1 . n
∗
∗A(0,h
n)
A(0,hn ) hn +1
>
1 } n
α− Then the internal set X = {n ∈ N : hn +1 > α − contains all standard natural numbers. Hence X must contain a hyperfinite integer N . This implies that ∗A(0,h
N)
hN +1
>α−
1 N
≈ α. ∗
“⇐” Let > 0 and k ∈ N. We have A(0,H) > α − . This means that the H+1 ∗A(0,x) ∗ sentence “there exists x ∈ N with x > k such that x+1 > α − ” is true in ∗V . By the transfer principle the sentence is also true in V . Hence there exists x ∈ N with A(0,x) x+1
> α − . Since > 0 can be arbitrarily small and k ∈ N can be ¯ arbitrarily large, we have d(A) > α. 2(Lemma 2.1)
x > k such that
Lemma 2.2 Let A ⊆ [0, H] and 0 6 α 6 1. If A(0, x) α(x + 1) and A(0, H) α(H + 1), then there exists a y x such that A(0, y) ∼ α(y + 1) and either y ∼ H or for every y ≺ z 6 H, A(0, z) α(z + 1). Proof: Let
z β = sup st H +1
: z ∈ [0, H] and A(0, z) α(z + 1) . 7
By the completeness of the standard real line, β is well defined. Let y ∈ [0, H] be such that
y H+1
≈ β. Clearly y x by the definition of β.
It is easy to see that if y ≺ H, then A(0, z) α(z + 1) for every y ≺ z 6 H by the supremality of β. Suppose A(0, y) α(y + 1). For every positive n ∈ N there is a z < y with z>y−
H n
such that A(0, z) < (α + n1 )(z + 1). Let Xn be the internal set containing
all z ∈ [y − Hn , y] such that A(0, z) < (α + n1 )(z + 1). Then X1 ⊇ X2 ⊇ · · · are all non-empty. By ℵ1 –saturation1 there is a z ∼ y such that A(0, z) α(z + 1). Hence A(0, y) ∼ A(0, z) α(z + 1) ∼ α(y + 1), a contradiction. By a similar argument we can also show that A(0, y) ≺ α(y + 1) is impossible.
2(Lemma 2.2)
Remark 2.3 In Lemma 2.2 one can get various versions by reversing the direction of the inequalities or replacing the set A by H − A. Lemma 2.4 Let A ⊆ [0, H]. Suppose 0, H ∈ A. If 0 6 x1 ≺ x2 6 H satisfy the following (1) (2A)(2x1 , 2x2) 3A(x1, x2), (2) if 0 ≺ x1, then A[0, x] is not a subset of an a.p. of difference d > 2 and A(0, x) 12 (x + 1) for some x ∼ x1, (3) if x2 ≺ H, then A[x, H] is not a subset of an a.p. of difference d > 2 and A(x, H) 12 (H − x + 1) for some x ∼ x2, then |2A| 3|A|. Proof: By Lemma 1.5, we have (2A)(0, 2x1 ) 3A(0, x1) and (2A)(2x2 , 2H) 3A(x2, H). Hence (2A)(0, 2H) ∼ (2A)(0, 2x1 ) + (2A)(2x1 , 2x2) + (2A)(2x2 , 2H) 3A(0, x1 ) + 3A(x1, x2) + 3A(x2 , H) ∼ 3A(0, H). 2(Lemma 2.4) 1
One form of ℵ1 –saturation is that the intersection of any countable sequence of non-empty internal sets X1 ⊇ X2 ⊇ · · · is non-empty.
8
The next lemma appears but is not clearly stated in [8]. The full proof can be found there. Since the proof in [8] is not well organized, we would like to include it here just for the reader’s convenience. Lemma 2.5 Let A ⊆ [0, H] for a hyperfinite integer H and 0 ≺ a ≺ H. If 0 ∈ A, A(0, x) 12 x for every 0 ≺ x ≺ a, A(0, a) ∼ 12 a, and (2A)(a, c) ∼ 0 for some c a, then there is b ∼ 12 a such that A[0, a] ⊆ [0, b]. Proof: Let b = max A[0, a]. Choose two standard natural numbers p < q such that gcd(p, q) = 1 and 2a a+ + 2 < c. p a i]. Then For each i = 0, 1, . . . , 2p, let xi = [ 2p
A[xp−i, xp−i+1 ] + A[xp+i, xp+i+1 ] ⊆ (2A)[a − 2, c]. Hence for each of i = 0, 1, . . . , p − 1, A(xp−i , xp−i+1 ) 0 implies A(xp+i , xp+i+1 ) = ∅ and A(xp+i , xp+i+1) 0 implies A(xp−i, xp−i+1 ) = ∅. Note that when i = 0, one has A(xp, xp+1 ) ∼ 0. Since A(0, a) ∼ 12 a, then there are exactly half of the i’s in {1, 2, . . . , p − 1} ∪ {p + 1, p + 2, . . . , 2p − 1} such that A(xi, xi+1 ) ∼ xi+1 − xi and for the rest of the i’s, A[xi, xi+1 ] = ∅. Since A(xp, xp+1) ∼ 0, then A(x0, x1) ∼ x1 − x0. a By the same procedure, one can define yj = [ 2q j] for j = 0, 1, . . . , 2q so that there are exactly half of the j’s in {1, 2, . . . , q − 1} ∪ {q + 1, q + 2, . . . , 2q − 1} such that
A(yj , yj+1 ) ∼ yj+1 −yj and for the rest of the j’s, A[yj , yj+1 ] = ∅. Since A(yq , yq+1 ) ∼ 0, then A(y0, y1) ∼ y1 − y0. Since p and q are relatively prime, for i ∈ {1, 2, . . . , p − 1} and j ∈ {1, 2, . . . , q − 1}, one has xi 6∼ yj . Hence there is no i ∈ {1, 2, . . . , p − 1} such that A(xi−1 , xi ) 6∼ A(xi, xi+1 ) because otherwise, one can take a j ∈ {1, 2, . . . , q − 1} such that xi ∈ [yj , yj+1 ], which would make A(yj , yj+1 ) 6∼ 0 and A(yj , yj+1 ) 6∼ yj+1 − yj at the same time. Since A(x0, x1) ∼ x1 −x0, then A(0, xp ) ∼ xp and A[xp+1, a] = ∅. So xp 6 b < xp+1 . By the fact that p and q can be chosen arbitrarily large in N, we have b ∼ 12 a. This ends the proof of the lemma.
2(Lemma 2.5)
The next lemma is trivial and will be frequently referred as the pigeonhole principle. 9
Lemma 2.6 Let A ⊆ ∗ N and x, y, t ∈ ∗ N. If A(x, x + t) + A(y − t, y) > t + 1, then x + y ∈ (2A). The next three lemmas are needed in the proof of the key lemma. Lemma 2.7 Let A ⊆ [0, H] be a hyperfinite set such that A[0, H − 1] is a subset of a b.p. I0 ∪ I1 of difference d > 3, A[0, H − 1] is not a subset of an a.p. of difference > 1, and A is not a subset of a b.p. of difference d. If |A ∩ I0| < 14 |A| or |A ∩ I1| < 14 |A|, then
|2A| |A|
' 3 + 14 .
Proof: Let Ai = Ii ∩ A[0, H − 1] and ai = min Ai for i = 0, 1. Suppose H ≡ a0 (mod d). Then 2a0 , 2a1 , a0 +a1 are not pairwise distinct modulo d because otherwise A is a subset of a b.p. (I0 ∪ {H}) ∪ I1. Clearly a0 6≡ a1 (mod d) because otherwise A[0, H − 1] is a subset of an a.p. of difference d > 1. Hence 2a0 6≡ a0 + a1 (mod d). By the same reason 2a1 6≡ a0 + a1 (mod d). So we have 2a0 ≡ 2a1 (mod d). If d is an odd number, then a0 ≡ a1 (mod d). If d is an even number, then a0 ≡ a1 (mod d2 ). But each of them contradicts that A[0, H − 1] is not a subset of an a.p. of difference > 1. So we can assume that H 6≡ a0 (mod d). By the same reason we can also assume H 6≡ a1 (mod d). Suppose |A0| < 14 |A|. Since (H + A1 ) ∩ (2A1 ∪ (A0 + A1)) = ∅, then |2A| > |2A1 | + |A0 + A1| + |H + A1| > 3|A1| + 3|A0 | + |A1| − 2|A0 | − 2 1 > 3|A| − 5 + |A| − 1, 4 which implies the lemma. The proof of the case when |A1|
3, A[0, H − 1] is not a subset of an a.p. of difference > 1, and A is not a subset of a b.p. of difference d. If |A ∩ Ii | > 14 |A| for i = 0, 1, then
|2A| |A|
' 3 + 14 .
Proof: We use the same notation as in Lemma 2.7. Again we can assume that H is distinct from a0, a1 modulo d. If H +a0 ≡ 2a1 (mod d) and H +a1 ≡ 2a0 (mod d), then 3(a1 − a0) ≡ 0 (mod d). If d is a multiple of 3, then a1 − a0 ≡ 0 (mod d3 ), which implies 10
that A[0, H −1] is a subset of an a.p. of difference
d 3
> 1. If d is not a multiple of 3, then
a1 − a0 ≡ 0 (mod d), which implies that A[0, H − 1] is a subset of an a.p. of difference of d > 1. So we can assume that either H + a0 6≡ 2a1 (mod d) or H + a1 6≡ 2a0 (mod d). Suppose H + a0 6≡ 2a1 (mod d). Then (H + A0) ∩ ((2A0) ∪ (2A1 ) ∪ (A0 + A1)) = ∅. Hence |2A| > |2A0 | + |2A1| + |A0 + A1| + |H + A0| 1 > 4|A0 | + 3|A1 | − 3 > 3|A| − 6 + |A|. 4 The proof of the case H + a1 6≡ 2a0 (mod d) is symmetric.
2(Lemma 2.8)
Lemma 2.9 Let A ⊆ [0, H] be a hyperfinite subset of a b.p. I0 ∪I1 of difference d > 1. Let ai = min(A ∩ Ii). Suppose gcd((A0 − a0) ∪ (A1 − a1)) = d. If gcd(Ai − ai) = di > d ' 3 + 14 . and |Ai | > 14 |A| for either i = 0 or i = 1, then |2A| |A| Proof: Assume gcd(A0 − a0) = d0 > d and |A0| > 14 |A|. The lemma is true because |A0 +A1 | > 2|A0 |+|A1|−2. Same for the case gcd(A1 −a1 ) = d1 > d and |A1| > 14 |A|. 2(Lemma 2.9) Next we introduce more notation and definitions, which are needed in the key lemma. An infinite initial segment U of ∗N is called a cut if U + U ⊆ U . Clearly U = N and U = ∗N are cuts. A cut U 6= ∗N is an external set. For example, N is external. For a hyperfinite integer H, the set UH =
\
[0, [H/n]]
n∈N
is an external cut. From now on, the only cut we need is UH for a given H. Hence when H is clearly given, the letter U is always used for UH . We often write x > U for x ∈ ∗N r U and write x < U for x ∈ U . When we say that “for all sufficiently large x < U ... ” we mean the statement “there is a y < U such that for every x > y in U ...” and when we say “there is a sufficiently large x < U ...” we mean the statement “for every y < U , there is an x < U , y 6 x, such that ...”. Note that if x < U and y > U , then
x y
≈ 0.
Suppose U ⊆ [0, H] is a cut. Given a function f : [0, H] 7→ ∗R (not necessarily internal) bounded by a standard real number, the lower U –density of f is defined by 11
the following: dU (f ) = sup {inf {st(f (n)) : n ∈ U r [0, m]} : m ∈ U } . Given a set A ⊆ [0, H], let fA (x) = of A is defined by
A(0,x) x+1
for every x ∈ [0, H]. The lower U –density
dU (A) = dU (fA ). For each x ∈ ∗N, we define the lower (x + U )–density of A by the following: dx+U (A) = dU ((A − x) ∩ ∗ N). Remark 2.10 (1) Suppose we replace U by N in the definition of dU . Then for every A ⊆ N, d(A) = dN (∗A), where d(A) = lim inf n→∞
A(n) n
is the standard definition of
the lower asymptotic density of A. (2) It is easy to check that for every a ∈ U , dU (A + a) = dU (A) and dU (A r [0, a]) = dU (A). (3) Let H be hyperfinite and A ⊆ [0, H]. If dU (A) > γ, then there are x ∈ U and y ∈ [0, H] r U such that for every x 6 z 6 y, A(0,z) > γ. This is true because the z+1 following reason: Let > 0 be such that dU (A) > γ + . Then we can find an x ∈ U such that for every z > x in U , A(0,z) > γ. Now the internal set X = {y > x : ∀z ∈ z+1 > γ } contains all elements in U r [0, x − 1]. Hence X must contain an [x, y] A(0,z) z+1 element y > U , which gives the desired result. (4) If dU (A) = α, then there is an x ∈ U such that for every x < y < U , one has A(0,y) ' α. This y+1 1 α− n } for each n
is true because the following reason: Let Xn = {y ∈ U :
A(0,y) y+1
>
∈ N. Then there exists an an ∈ U such that U r[0, an −1] ⊆ Xn . By T∞ ℵ1 –saturation we have that n=1 [an , Hn ] 6= ∅. Let x be in the intersection. It is easy to see that x ∈ U and x is greater than every an for n ∈ N. Hence for every y > x in U , we have
A(0,y) y+1
' α. Note that the reason above also shows that the cofinality of U
is uncountable, i.e. every countable increasing sequence in U is upper bounded in U . Another important notation needed is called e–transform (cf.[11, p.42]). It is also called τ –transformation (cf.[4, p.58]). Let A, B ⊆ ∗N and a ∈ A. An ea–transform of 12
(A, B) is the pair (A0, B 0) = ea(A, B) such that A0 = A ∪ (B + a) and B 0 = B ∩ (A − a). Since in the key lemma we mainly concern the part of a set in U , we always assume a ∈ A ∩ U when an ea–transform is applied to the pair (A, B). Remark 2.11 The following are important properties of the ea –transform. Let (A0, B 0) = ea (A, B): (1) A0 ⊇ A and B 0 ⊆ B. (2) A0 + B 0 ⊆ A + B. Hence dU (A0 + B 0) 6 dU (A + B). (3) If x ∈ U and
a x
≈ 0, then for every y with x < y < U we have A0(0, y) + B 0 (0, y) A(0, y) + B(0, y) ≈ . y+1 y+1
Hence dU (fA0 + fB 0 ) = dU (fA + fB ) when a ∈ U . (4) If 0 ∈ A ∩ B, then 0 ∈ A0 ∩ B 0 because a ∈ A. We often write E(A, B) for a finite sequence E of e–transforms ean ◦ ean−1 ◦· · · ◦ ea1 successively applied to (A, B) such that each step of the application is well defined. We say that ea occurs in E if ea is in the sequence. If (A0, B 0) = E(A, B), then (1), (2), and (4) of Remark 2.11 are still true and (3) of Remark 2.11 is true for all sufficiently large x ∈ U . Let U be a cut. A b.p. J = I0 ∪ I1 is called U –unbounded if both I0 ∩ U and I1 ∩ U are upper unbounded in U . Note that a U –unbounded b.p. has its difference at least 3. We are now ready to prove the key lemma. The key lemma is inspired by Kneser’s Theorem (cf [4]). In the proof of key lemma and the proofs in the next section, we number the claims, subclaims, cases, subcases, subsubcases, etc., so that the reader can see how they are nested. Lemma 2.12 Let H be hyperfinite, U = UH , and A ⊆ [0, H] be such that 0 < dU (A) = α < 23 . If A ∩ U is neither a subset of an a.p. of difference greater than 1 nor a subset of a U –unbounded b.p. , then there is a γ > 0 such that for every N > U , there is a K ∈ A, U < K < N , such that (2A)(0, 2K) 3A(0, K) > + γ. 2K + 1 2(K + 1) 13
Proof: We prove the lemma by proving a sequence of claims and cases. Without loss of generality, we assume 0 ∈ A. In the proof we only deal with the numbers in U . Hence the comparison relations ≺, , , with respect to H are no longer useful because for every x ∈ U we have x ∼ 0. Claim 2.12.1: If dU (2A) > 32 α, then the lemma is true. Proof of Claim 2.12.1: There are x < U and y > U in A such that for every x < z < 2y, (2A)(0,z) > 32 α + 2, where = (dU (2A) − 32 α)/3. Let γ = 12 . For every z+1 < α + . This implies N > U choose a K > U such that K < N , K < y, and A(0,K) K+1 that
(2A)(0, 2K) 3 3 1 3A(0, K) > α + 2 = (α + ) + > + γ. 2K + 1 2 2 2 2(K + 1)
Note that K can be chosen from A because otherwise one can replace K by the smallest integer in A which is greater than K. This ends the proof.
2(Claim
2.12.1) If
1 2
< α < 23 , then there is an x ∈ U such that U r [0, x − 1] ⊆ (2A) by the
pigeonhole principle. Hence dU (2A) = 1 > 32 α. By Claim 2.12.1 we can assume that 0 < α 6 12 . Claim 2.12.2: If for every x < U , there is a y ∈ A with x < y < U such that (2A)(0, 2y) 3A(0, y) , 2y + 1 2(y + 1) then the lemma is true. Proof of Claim 2.12.2: By the fact that the cofinality of U is uncountable, one can find γ > 0 such that the set Y ∩ U is unbounded in U , where 3A(0, y) (2A)(0, 2y) > +γ Y = y∈A: 2y + 1 2(y + 1) is clearly internal. For every N > U let K be the largest element in Y ∩ [0, N − 1]. Then K ∈ Y and U < K < N . This shows the lemma is true.
2(Claim 2.12.2)
Claim 2.12.3: For every finite sequence E of e–transforms with (A0, B 0) = E(A, A), there is an x ∈ U such that for all y ∈ B 0 , x < y < U , we have A(0, y) B 0(0, y) ' . y+1 2(y + 1)
14
Proof of Claim 2.12.3: Suppose not. Then there is a finite sequence E of e– transforms such that (A0 , B 0) = E(A, A) and for every x ∈ U , there is a y > x in B 0 ∩ U such that
B 0 (0,y) y+1
A(0,y) . 2(y+1)
Then
(A0 + B 0)(0, 2y) (2A)(0, 2y) ' 2y + 1 2y + 1 0 0 A0(0, y) A (0, y) + |A [0, y] + y| ≈2 ' 2y + 1 2y + 1 0 0 0 B (0, y) A (0, y) + B (0, y) − ≈2 2y + 1 y+1 2A(0, y) B 0(0, y) − ≈2 2y + 1 y+1 A(0, y) 3A(0, y) A(0, y) − ≈ . 2 y+1 2(y + 1) 2(y + 1) Hence by Claim 2.12.2, the lemma is true.
2(Claim 2.12.3)
Next we need to define two functions f and g on subsets of ∗N. For every X ⊆ ∗N let
f (X) =
min{|x − y| − 1 : x, y ∈ X ∩ U and x 6= y},
∞,
if there are x, y ∈ X ∩ U such that x 6= y and |x − y| is finite; otherwise.
g(X) =
gcd(X ∩ U ), if gcd(X ∩ U ) is finite;
∞,
otherwise.
Clearly g(C) 6 f (C) + 1 for every C. Note that if ea (C, D) = (C 0, D0 ), then f (D0 ) > f (D), f (C 0) 6 f (C), g(D0 ) > g(D), and g(C 0) 6 g(C). Let there is a finite sequence E of e–transforms 0 S = f (B ) : . such that E(A, A) = (A0, B 0) If S contains ∞ or S is unbounded in N, then there is a sequence E of e–transforms such that (A0, B 0) = E(A, A) and f (B 0) > α3 . By Remark 2.11 there is an a ∈ U such that for every x > a in U ,
15
(2A)(0, x) x+1 (A0 + B 0)(0, x) A0 (0, x) ' ' x+1 x+1 A0(0, x) + B 0(0, x) B 0(0, x) ' − x+1 x+1 0 A(0, x) B (0, x) − ≈2 x+1 x+1 3 1 α ' 2α − = α + α. 3 2 6 Hence dU (2A) > 32 α + 16 α, which, together with Claim 2.12.1, implies the lemma. By the argument above we can now assume that the set S is bounded in N. Let (A0 , B 0) = E(A, B) for some 0 t = min f (A ) : . finite sequence E of e–transforms Since S is upper bounded in N, we can fix a finite sequence E0 of e–transforms with (A0, B 0) = E0 (A, A) such that (1) f (A0 ) = t; (2) f (B 0 ) = l and f (B 00) = l for each E with E(A0 , B 0) = (A00, B 00); (3) g(B 0 ) = m and g(B 00) = m for each E with E(A0, B 0) = (A00, B 00); (4) let F = {k ∈ [0, m − 1] : ∃a ∈ A0 ∩ U (a ≡ k (mod m))} such that for each E with E(A0, B 0) = (A00, B 00) and F 00 = {k ∈ [0, m − 1] : ∃a ∈ A00 ∩ U (a ≡ k (mod m))} we have F = F 00. Let a ¯ = max{a ∈ U : ea occurs in E0 }. The rest of the proof is divided into four cases in terms of the size of F . From now on let’s assume that the key lemma is not true. We will derive a contradiction in each of the following cases. Let |F | = n and let F = {0 = k0 < k1 < · · · < kn−1 }. For each i = 0, 1, . . . , n − 1 let Ii = {ki + xm : x ∈ [0, H]}, Ai = A ∩ Ii, A0i = A0 ∩ Ii, and ai = min Ai . Let T = {ai : i = 0, 1, . . . , n − 1}. Without loss of generality we assume T + B 0 ⊆ A0 because otherwise we can first perform n successive e–transforms on (A0, B 0) to get (A00, B 00) so that T + B 00 ⊆ A00 and then replace (A0, B 0) by (A00, B 00). Case 2.12.1:
|F | > 4.
For all sufficiently large x ∈ B 0 ∩ U , (A0 + B 0)(0, 2x) (2A)(0, 2x) ' 2x + 1 2x + 1 16
|T + B 0[0, x]| + |T + x + B 0[0, x]| 2x + 1 0 8B (0, x) 4B 0(0, x) ' ' 2x + 1 x+1 3A(0, x) A(0, x) . '2 x+1 2(x + 1) '
The second last inequality is a consequence of Claim 2.12.3. By Claim 2.12.2, the lemma is true, which contradicts the assumption. 2(Case 2.12.1) Case 2.12.2: |F | = 3. By Claim 2.12.1 and gcd(A ∩ U ) = 1 together with Lemma 1.5, we have dU (2A) = 3 α. 2
Since |F | = 3, for every sufficiently large x ∈ B 0 ∩ U with gcd(B 0[0, x]) = m, we
have that (2A)(0, 2x) (A0 + B 0)(0, 2x) ' 2x + 1 2x + 1 |T + B 0[0, x]| + |T + x + B 0[0, x]| ' 2x + 1 0 3A(0, x) 6B (0, x) ' . ' 2(x + 1) 2(x + 1) 3A(0,x) , then the lemma If there exist sufficiently large x ∈ U such that (2A)(0,2x) 2x+1 2(x+1) follows from Claim 2.12.2. So we can assume that for all sufficiently large x ∈ U , 2B 0 (0,x) x+1
≈
A(0,x) . x+1
Let F0 = {k ∈ [0, m − 1] : ∃a ∈ A ∩ U (a ≡ k (mod m))} . Since 0 ∈ A, then 0 ∈ F0 . If F0 = {0}, then A is a subset of an a.p. of difference m > 3. If |F0| = 2, then A ∩ U is either a subset of an a.p. of difference m > 1 or a subset of 2 a U –unbounded b.p. (this is because if A∩U is a subset of a b.p. of difference m, which is not U –unbounded, then there is an e–transform ea with (A00, B 00) = ea (A0, B 0) such that B 00 is U –bounded, which contradicts Claim 2.12.3). Hence we can assume that |F0| = 3 and so F0 = F . Let F = {0 = k0 < k1 < k2 }. Since T + B 0 ⊆ A0, then for every sufficiently large x ∈ B 0 ∩ U , A0 (0, x) B 0 (0, x) / i x+1 x+1 for i = 0, 1, 2. Fix such an x ∈ U with
a ¯ x
≈ 0. By Claim 2.12.3,
A0 (0, x) B 0 (0, x) ≈ i x+1 x+1 17
for i = 0, 1, 2. Since B 0[0, x] ⊆ A0 [0, x] ⊆ A00[0, x], then A1 (0,x)+A2 (0,x) x+1
≈
A0 (0,x) x+1
≈
A(0,x) . 2(x+1)
This implies
A0 (0,x) . x+1
Let C, D ⊆ [0, H], a ∈ C ∩ U , and D0 ⊆ D. Let (C 0, D0 ) = ea (C, D). We call that a number c ∈ C 0 r C is obtained from D0 via ea if there is a d ∈ D0 such that c = a + d. Let E be a finite sequence of e–transforms such that (C 0, D0 ) = E(C, D). Let D0 ⊆ D and c ∈ C 0 r C. We call that c is obtained from D0 via E if c is obtained from D0 via an e–transform ea, which occurs in E. For i = 0, 1, 2 let zi = max Ai[0, x] and zi0 = max A0i [0, x]. Since T + x ⊆ A0, then we have that (2A)(0, 2x) (A0 + B 0)(0, 2x) ' 2x + 1 2x + 1 0 0 |B [0, x] + B [0, x]| + |{a1, a2 , x + a1 , x + a2 } + B 0[0, x]| ' 2x + 1 3A(0, x) 6B 0 (0, x) ' . ' 2x + 1 2(x + 1) If the lemma is not true, then by Claim 2.12.2 we have that 2B 0 (0, x) |B 0[0, x] + B 0[0, x]| ≈ . 2x + 1 2x + 1 0
(0,x) ≈ m1 by Lemma 1.6. Hence This implies that Bx+1 0 z consequence we have xi ≈ 1 for i = 1, 2.
We now show that either
z1 x
≈ 1 or
z2 x
A0i (0,x) x+1
≈
1 m
for i = 0, 1, 2. As a
≈ 1. Suppose not and let z = max{z1, z2}. A0 (z+1,x)
A0 (z+1+¯ a ,x)
Then for i = 1, 2, Ai [z + 1, x] = ∅. Since 1 x−z ≈ m1 , then 1 x−z ≈ m1 . Since 0 (z+1,x) 0 (0,x) ≈ 0 because Bx+1 ≈ m1 , then there must be some elements also A0 (z+1,x)−B x−z in A0[z + 1 + a ¯, x] that are obtained from A1 [z + 1, x] ∪ A2[z + 1, x] via E0. This contradicts that A1[z + 1, x] ∪ A2[z + 1, x] = ∅ Claim 2.12.2.1 F0 ∪ {m} is an a.p. of difference k1 . Proof of Claim 2.12.2.1 Suppose not. Then either 2k1 6∈ {k2 , m} or k1 + k2 6= m. We want to show that
(2A)(0,2x) 2x+1
3A(0,x) , 2(x+1)
which contradicts the assumption that the
lemma is not true by Lemma 2.12.2. Case 2.12.2.1.1: 2k1 6∈ {k2 , m}. Hence (2A1) ∩ (A0 + A) = ∅. We want to show that this case is impossible. Subcase 2.12.2.1.1.1:
A1 (0,x) x+1
0 and 18
z1 x
1.
Hence
z2 x
≈ 1. This implies that A01[z1, x] obtains numbers from A2 via E0 . So
there is a c ∈ {0, k1 , k2 } such that k1 ≡ c + k2 (mod m). Clearly there is only one possible value for c, i.e. c = k2 . So we have 2k2 = m + k1 . If k1 + k2 6= m, then 2A1 , A1 + A2, and A0 + A are pairwise disjoint. Hence 4A0(0, x) + 4A1 (0, x) + 2A2 (0, x) 3A(0, x) (2A)(0, 2x) ' . 2x + 1 2x + 1 2(x + 1) So we can assume k1 + k2 = m. But that, together with 2k2 = m + k1 , implies that F ∪ {m} is an a.p. of difference k1 . 2(Subcase 2.12.2.1.1.1) (0,x) Subcase 2.12.2.1.1.2: A1x+1 0 and zx1 ≈ 1. We now have |A0[0, x] + {a1, z1}| 2A0 (0, x) (A0 + A1)(0, 2x) ' ≈ . 2x + 1 2x + 1 2x + 1 Hence 5A0(0, x) + A2 (0, x) + 2A1(0, x) 3A(0, x) (2A)(0, 2x) ' . 2x + 1 2x + 1 2(x + 1) 2(Subcase 2.12.2.1.1.2) (0,x) Subcase 2.12.2.1.1.3: A1x+1 ≈ 0. 0 Clearly A1[0, x] obtains numbers from A2. Hence 2k2 = m + k1 . This implies
2k2 6= m. If k1 + k2 6= m, then (A1 + A2), 2A2 , 2A0, and A0 + A2 are pairwise disjoint. Hence
(2A)(0, 2x) 3A0 (0, x) + 4A2(0, x) 3A(0, x) ' . 2x + 1 2x + 1 2(x + 1) If k1 + k2 = m, then F ∪ {m} is an a.p. of difference k1 . 2(Case 2.12.2.1.1) Case 2.12.2.1.2: 2k1 ∈ {k2 , m}.
If 2k1 = m, then k1 + k2 > m. If 2k1 = k2 , then k1 + k2 = m implies that F ∪ {m} is an a.p. of difference k1 . Hence we can assume that k1 + k2 6= m. Subcase 2.12.2.1.2.1:
A2 (0,x) x+1
0 and
z1 x
≈ 1.
We have (A0 + A1)(0, 2x) |A0[0, x] + {a1, z1}| 2A0 (0, x) ' ' . 2x + 1 2x + 1 2x + 1 Hence (A0 + A)(0, 2x) + (A1 + A2 )(0, 2x) (2A)(0, 2x) ' 2x + 1 2x + 1 3A(0, x) 5A0 (0, x) + 2A2 (0, x) + A1(0, x) . ' 2x + 1 2(x + 1) 19
2(Subcase 2.12.2.1.2.1) A2 (0,x) x+1
Subcase 2.12.2.1.2.2: z2 x
0 and
z1 x
1.
A01 [z1, x]
Then ≈ 1. Clearly, obtains numbers from A2 via E0 . Hence we have 2k2 = m + k1 . If 2k1 = k2 , then F ∪ {m} is an a.p. of difference k1 . Hence we can assume that 2k1 = m. If
A1 (0,x) x+1
0, then (A0 + A)(0, 2x) + (A1 + A2 )(0, 2x) (2A)(0, 2x) ' 2x + 1 2x + 1 5A0 (0, x) + 2A1 (0, x) + A2(0, x) 3A(0, x) ' . 2x + 1 2(x + 1)
If
A1 (0,x) x+1
≈ 0, then
A2 (0,x) x+1
≈
1 . m
Hence
(2A0 )(0, 2x) + (A0 + A2)(0, 2x) (2A)(0, 2x) ' 2x + 1 2x + 1 (A1 + A2 )(0, 2x) + (2A2 )(0, 2x) + 2x + 1 3A(0, x) 4A0 (0, x) + 3A2(0, x) . ' 2x + 1 2(x + 1) 2(Subcase 2.12.2.1.2.2) Subcase 2.12.2.1.2.3: A1 (0,x) x+1
1 m
A2 (0,x) x+1 A02[0, x]
≈ 0.
Then ≈ and obtains numbers from A1 via E0 . This implies k2 = 2k1 . Now we have that 2A0 , A0 + A1, 2A1 , and A1 + A2 are pairwise disjoint. Hence
(2A)(0, 2x) 7A0 (0, x) 3A(0, x) ' . 2x + 1 2x + 1 2(x + 1)
This ends the proof of the claim.
2(Claim 2.12.2.1)
If k1 > 1, then A is a subset of an a.p. with the difference k1 > 1 by Claim 2.12.2.1. So we can assume m = 3 and F = {0, 1, 2}. Let z < U be sufficiently large with
A(0,z) z+1
≈ α and let x ∈ B 0 be such that x > z
and [z, x − 1] ∩ B 0 = ∅. By Claim 2.12.3, (2A)(0, 2x) (A0 + B 0 )(0, 2x) ' 2x + 1 2x + 1 0 0 |B [0, x] + B [0, x]| + |a1 + B 0[0, x] + B 0[0, x]| ' 2x + 1 20
|a2 + B 0[0, x] + B 0[0, x]| 2x + 1 0 6B (0, x) 3A(0, x) ' ' . 2x + 1 2(x + 1) +
Hence we have
3A(0, x) (2A)(0, 2x) 2x + 1 2(x + 1)
unless
|B 0[0, x] + B 0[0, x]| 2B 0 (0, x) ≈ . 2x + 1 2x + 1 But the latter implies, by Lemma 1.6, that B 0[0, x] is a subset of an a.p. of length ≈ 0. Hence A(0,x) ≈ A(0,z) ≈ α. Note that B 0(0, x) + b with xb ≈ 0. This implies x−z x x+1 z+1
gcd(B 0) = m = 3. So we have
B 0 (0,x) x+1
≈ 13 . This implies
(2A)(0, 2x) 6B 0 (0, x) 3 3A(0, x) ' ≈1 α≈ . 2x + 1 2x + 1 2 2(x + 1) By Claim 2.12.2, the lemma is true, which contradicts the assumption. 2.12.2)
2(Case
Case 2.12.3: |F | = 2. Since A ∩ U is neither a subset of an a.p. with the difference greater than 1 nor a subset of a U –unbounded b.p. , then we have m = 2 and F = {0, 1}. Let a be the least odd number in A. We can again assume B 0 ⊆ A0 and a + B 0 ⊆ A0. Let z ∈ U be sufficiently large with gcd(B 0[0, z]) = 2 such that
A(0,z) z+1
≈ α and
A(0,y) y+1
' α for every y > z in U . Let x ∈ B 0 ∩U be such that x > z and [z, x−1]∩B 0 = ≈ β. Since B 0 ⊆ A0, a + B 0 ⊆ A0 and B 0 ∩ (a + B 0) = ∅, then ∅. Let A(0,x) x+1 A0 (0,x) x+1 A0 (0,x) x+1
0
B 0 (0,x) ≈ . This implies x+1 1 ≈ β + , β + > 2(β − ), and, by Claim 2.12.3, β − > 2 β. Hence 2 6 β 6 3. Note that β 6 12 because x−z ≈ 0 implies β = α 6 12 and x−z 0 implies that x x A0 (z,x) A(z,x) A(z,x) A(0,x) 1 ≈ 2 x−z , which implies x−z / 2 , hence again β ≈ x+1 / max{α, 12 } = 12 . x−z Let A00 be the set of all even numbers in A0 and A01 be the set of all odd numbers A0 (0,x) A0 (0,x) in A0 . Let 0x+1 ≈ β0 > 0 and 1x+1 ≈ β1 > 0. Then β0 + β1 = β + . By Lemma 1.7 in which let B 0[0, x] be A and A0i[0, x] be B, we have (0,x) ' 2 Bx+1 . Hence there is an > 0 such that β −
(A0 + B 0)(0, 2x) (2A)(0, 2x) ' 2x + 1 2x + 1 21
(A00 + B 0)(0, 2x) + (A01 + B 0)(0, 2x) 2x + 1 0 0 |A0[0, x] + B [0, x]| + |A01[0, x] + B 0[0, x]| ' 2x + 1 2β0 + 1 2β1 + 1 β − β− ' min , β0 + + min , β1 + . 4 2 4 2 ≈
Since
2β0 + 1 2β1 + 1 β++1 3β + 3 + = > β, 4 4 2 2 2 2β0 + 1 β− + β1 + 4 2 2β0 + 1 + 4β1 + 2β − 2 4β + 1 + 2β1 3β + β1 3β = = > , 4 4 2 2 β − 2β1 + 1 + 2 4 4β0 + 2β − 2 + 2β1 + 1 4β + 2β0 + 1 3β = = , 4 4 2 3 β − β− β0 + + β1 + = β + + β − = 2β β, 2 2 2
β0 +
then
(2A)(0, 2x) 3 3A(0, x) β≈ , 2x + 1 2 2(x + 1) which, by Claim 2.12.2, contradicts the assumption that the key lemma is not true. 2(Case 2.12.3) Case 2.12.4: |F | = 1. This is the last case and is the most difficult one. We will prove several claims in this case. Since A is not a subset of a a.p. with the difference greater than 1, then we have m = g(B 0 ) = gcd(B 0) = 1. Claim 2.12.4.1: For every sufficiently large x < U with gcd(B 0[0, x]) = 1 such 0 (0,x) ≈ α, we have Bx+1 ≈ α. that A(0,x) x+1 Proof of Claim 2.12.4.1: Suppose not. For every y < U , there is an x > y in U 0
0
(0,x) (0,x) with A(0,x) ≈ α and Bx+1 α. Then we have Ax+1 α. Let z > x be such that x+1 0 (0,z) A(0,z) 0 ≈ α − . [x, z − 1] ∩ B = ∅. Let z+1 ≈ β and let > 0 be such that Bz+1
Note that
x z
12 .
≈ 1. Then by Lemma 1.7 (2A)(0, 2x) (2A)(0, 2z) ≈ 2x + 1 2z + 1 0 |A [0, z] + B 0[0, z]| ' 2z + 1 α− α++1 , ' min α + + 2 2 3A(0, x) 3α + . > 2 2(x + 1)
This contradicts the assumption that Lemma 2.12 is false. Suppose
1 2
6
Note that
Since
B 0 (0,x) x+1
x z
1. A0(x, z) A0(x, z) + B 0(x, z) A(x, z) ≈ ≈2 . z+1 z+1 z+1
α, then
A0 (0,x) x+1
α≈
A(0,x) . x+1
Hence
(A0 + B 0 )(0, 2z) (2A)(0, 2z) ' 2z + 1 2z + 1 |A0[0, z] + B 0 [0, z]| ' 2z + 1 min {2A0(0, z) + B 0 (0, z), A0(0, z) + z} ' 2z + 1 0 2A (0, x) + 2A0(x, z) + B 0(0, x) A0(0, x) + A0(x, z) + x + (z − x) , ≈ min 2z + 1 2z + 1 0 0 2A(0, x) + 4A(x, z) + A (0, x) A (0, x) + 2A(x, z) + 2A(0, x) + A(x, z) , ' min 2z + 1 2z + 1 3A(0, z) 3A(0, z) 3A(0, z) , = . min 2(z + 1) 2(z + 1) 2(z + 1) By Claim 2.12.2, we have a contradiction.
2(Claim 2.12.4.1)
Recall that
there is a finite sequence of e–transforms 00 . t = f (A ) = min f (A ) : E with (A00, B 00) = E(A, A) 0
Claim 2.12.4.2: If t = 0, then the lemma is true. Hence we can assume t > 0. Proof of Claim 2.12.4.2: Suppose t = 0. Then A0 ∩ U contains two consecutive numbers. We first show by induction that for every k ∈ N, there is a finite sequence 23
E 0 of e–transforms such that (A00, B 00) = E 0(A0, B 0 ) and A00 ∩ U contains k consecutive numbers. It is clearly true for k = 2. Suppose there is a finite sequence E1 of e–transforms with (A1, B1 ) = E1 (A0, B 0) and A1 ∩ U contains k − 1 consecutive numbers a, a + 1, . . . , a + k − 2. Now we need the next subclaim to show that B1 ∩ U contains two consecutive numbers. Without loss of generality we can assume [a, a+k−2]+B1 ⊆ A1. Subclaim 2.12.4.2.1: For every finite set T ⊆ A1 ∩ U and every number x < U , there is a y ∈ B1 ∩ U , y > x, such that T + y ⊆ B1 . Proof of Subclaim 2.12.4.2.1: Suppose the subclaim is not true. Without loss of generality we assume T + B1 ⊆ A1. Let |T | = s. For each b ∈ B1 , let h(b) = min((T + b) r B1) when the set (T + b) r B1 is not empty and let h(b) = H otherwise. Since the subclaim is not true, then for every b ∈ B1 ∩ U , h(b) ∈ (A1 r B1 ) ∩ U is well defined. So for every i ∈ (A1 r B1 ) ∩ U , |h−1 (i)| 6 s. Since B1 ⊆ A1, we have B1 (0, x) 1 B1 (0, x) A1 (0, x) ' (1 + ) x+1 x+1 s x+1 for all sufficiently large x < U with
A(0,x) x+1
≈ α, which contradicts Claim 2.12.4.1. Note ¯ 0, B 0 ) that Subclaim 2.12.4.2.1 is also true if A1 is replaced by A00 where (A00, B 00) = E(A for some sequence E¯ of e–transforms. 2(Subclaim 2.12.4.2.1) Since A1 ∩ U contains k − 1 consecutive numbers [a, ak − 2], then by the subclaim above, we can assume that B1 ∩ U contains k − 1 consecutive numbers b + [a, a + k − 2] for some b ∈ B1 . Hence [a, a + k − 2] + b + [a, a + k − 2] ⊆ [a, a + k − 2] + B1 ⊆ A1. Hence A1 contains 2k − 3 > k consecutive numbers. Note that using the same idea we can prove that if A0 ∩ U contains two numbers of difference d, then for every k ∈ N there is a finite sequence E of e–transforms with (A00, B 00) = E(A0 , B 0) such that A00 ∩ U contains an a.p. of difference d and length k. Next we conduct the second half of the proof of Claim 2.12.4.2. Subclaim 2.12.4.2.2: If t = 0, then dU (2A) > 2α. Proof of Subclaim 2.12.4.2.2: The idea of the proof can be found in [4, p.61]. We include the proof here because nonstandard setting is involved. Suppose A0 ∩ U contains k consecutive numbers. Without loss of generality, we assume [0, k − 1] ⊆ A0 and [0, k − 1] + B 0 ⊆ A0. It suffices now to show dU (A0 + B 0) > 24
k γ k+1
for every γ < 2α. Since dU (fA0 + fB 0 ) = dU (2fA ) = 2dU (A) = 2α > γ, then there exists an x0 ∈ U such that for every x ∈ U with x > x0, A0(0, x) + B 0 (0, x) > γ. x+1 Let x0 be the least such number. Then A0(x0, x0 + x) + B 0(x0 , x0 + x) >γ x+1 for every x ∈ U . It is easy to see that x0 ∈ (A0 ∪ B 0) ∩ U ⊆ A0 . Let x1 = min {z ∈ B 0 : z > x0} and let ¯ = (B 0 − x1). ¯ = (A0 − x0) and B A ¯ and It is easy to check that 0 ∈ A¯ ∩ B ¯ = d (A0 + B 0). dU (A¯ + B) U ¯ x) + B(1, ¯ x) > By Lemma 1.8 we need only to show that for every x ∈ U , 1 + A(1, k γ(x k+1
+ 1).
Suppose x >
k+1 γ
− 1.
Then γ(x + 1) > k + 1 and hence (k + 1)γ(x + 1) − (k + 1) > kγ(x + 1). This implies γ(x + 1) − 1 >
k γ(x + 1). k+1
Since ¯ x) = A0(x0, x0 + x) − 1 A(1, and ¯ x) = B 0(x1 + 1, x1 + x) > B 0 (x0, x1 + x) − 1 > B 0(x0, x0 + x) − 1, B(1, then ¯ x) + B(1, ¯ x) 1 + A(1, > A0 (x0, x0 + x) + B 0(x0 , x0 + x) − 1 k γ(x + 1). > γ(x + 1) − 1 > k+1 25
Suppose x < x1 − x0. Since B 0(x0 , x0 + x) = 0, then ¯ x) + B(1, ¯ x) 1 + A(1, > A0(x0 , x0 + x) = A0(x0 , x0 + x) + B 0(x0 , x0 + x) k γ(x + 1). > γ(x + 1) > k+1 Suppose x1 − x0 6 x 6 x1 − x0 + k − 1. Since x1 + [0, k − 1] = [x1, x1 + k − 1] ⊆ [0, k − 1] + B 0 ⊆ A0, then ¯ x) + B(1, ¯ x) 1 + A(1, > A0(x0 , x1 − 1) + x − (x1 − x0 ) + B 0(x0, x1 − 1) + 1 > γ(x1 − x0) + x − (x1 − x0 ) + 1 > γ(x1 − x0) + γ(x − (x1 − x0) + 1) k γ(x + 1). = γ(x + 1) > k+1 Suppose x1 − x0 + k 6 x
A0(x1, x1 + k − 1) k γ(x + 1). =k> k+1 Now we have that for all x < U ¯ x) + B(1, ¯ x) 1 + A(1, k > γ. x+1 k+1 ¯ > k γ. Since k ∈ N and γ < 2α are arbitrary, By Lemma 1.8, we have dU (A¯ + B) k+1 ¯ > 2α. Hence by Claim 2.12.1 the lemma is true. This we have d (2A) = d (A¯ + B) U
U
finishes the proof of Claim 2.12.4.2. 2(Claim 2.12.4.2) We can now assume t > 0 and fix c such that both c and c + t + 1 lie in A0. Claim 2.12.4.3: If T ⊆ (2A) ∩ U is a finite set, then there is an a ∈ A ∩ U and a sequence E of e–transforms such that (A00, B 00) = E(A0, B 0) and a + T ⊆ A00. Proof of 2.12.4.3: Let T = {a1 + b1, a2 + b2, . . . , ak + bk }, where ai, bi ∈ A for i = 1, 2, . . . , k. Without loss of generality we can assume ai + B 0 ⊆ A0 for i = 1, 2, . . . , k. 26
By Subclaim 2.12.4.2.1 we can assume that there is an a ∈ B 0 ∩ U ⊆ A such that a + {b1 , b2, . . . , bk } ⊆ B 0. Since ai + B 0 ⊆ A0, then a + T ⊆ {a1 + a + b1, a2 + a + b2, . . . , ak + a + bk } ⊆
k [
(ai + B 0) ⊆ A0.
i=1
2(Subclaim 2.12.4.3) Claim 2.12.4.4: For all a1 < a2 in A0 ∩ U such that a2 − a1 is in N, i.e. a1 and a2 are finite units apart, we have (t + 1) | (a2 − a1), i.e. a2 − a1 is a multiple of t + 1. Proof of Claim 2.12.4.4: Suppose the claim is not true. Without loss of generality we can assume that {a1, a2}+B 0 ⊆ A0. By the same argument as in the proof of Claim 2.12.4.2, for k ∈ N and (t+1)k > a2 −a1, we can assume that A0 ∩U contains an a.p. of length k + 1 with difference t + 1. Let T = {a, a + (t + 1), . . . , a + k(t + 1)} ⊆ A0 ∩ U . By Subclaim 2.12.4.2.1, we can assume T + b ⊆ B 0 for some b ∈ B 0. We now have (a1 + b+ T ) ∪ (a2 + b+ T ) ⊆ A0. Note that a1 + b+ a < a2 + b+ a < a1 + b+ a + k(t +1), i.e. the element a2 + b + a is between the first and last elements of the a.p. a1 + b + T . Since (t + 1) - (a2 − a1), then a2 + b + a 6∈ a1 + b + T . This gives two elements in A0 ∩ U with the difference < t + 1, which contradicts the minimality of t.
2(Claim
2.12.4.4) From Claim 2.12.4.4 we can see that for all a ∈ A0, (A0 − a) ∩ N is a subset of {k(t + 1) : k ∈ N}. Claim 2.12.4.5: For each x < U , the set (A ∩ U ) r [0, x] is not a subset of an a.p. of difference k > 1. Proof of Claim 2.12.4.5: Since A ∩ U is not a subset of an a.p. of difference > 1, then there is a ∈ A ∩ U such that A ∩ (a + A) ∩ U is bounded in U . Hence if (A00, B 00) = ea (A, A), then B 00 ∩ U is bounded in U , which contradicts Claim 2.12.3. 2(Claim 2.12.4.5) Claim 2.12.4.6: For each x < U , the set (A ∩ U ) r [0, x] is not a subset of a U –unbounded b.p. . Proof of Claim 2.12.4.6: Suppose the claim is not true. Let (A∩U )r[0, x] ⊆ I1 ∪ I2, where I1 ∩U = {a + uk : u ∈ U } and I2 ∩U = {b + uk : u ∈ U } with a, b ∈ [0, k−1] and {2a, 2b, a + b} are, modulo k, pairwise distinct. By Claim 2.12.4.5 we can assume A ∩ I1 and A ∩ I2 are unbounded in U . Since A ∩ U is not a subset of a U –unbounded b.p. , there is a c 6 x in A such that c 6≡ a, b (mod k). For i = 1, 2 let Ai = A ∩ Ii and 27
li = min Ai . Without loss of generality we assume {l1, l2, c} + B 0 ⊆ A0. Since B 0 ⊆ A0, then B 0 r [0, x] ⊆ I1 ∪ I2. Let Bi0 = B 0 ∩ Ii. Given sufficiently large y ∈ U such that A(0,y) y+1
≈ α, suppose we have B10 (0, y) 6 B20 (0, y). Then we can find an s ∈ {l1 , l2, c}
such that (s + B20 ) ∩ U ∩ (I1 ∪ I2) = ∅. Hence B 0(0, y) + B20 (0, y) 3B 0 (0, y) A0(0, y) ' ' . y+1 y+1 2(y + 1) This implies
B 0 (0,y) y+1
α, which contradicts Claim 2.12.4.1
2(Claim 2.12.4.6)
In order to finish the proof of Case 2.12.4 we construct two internal increasing sequences of integers {xj : j ∈ J } and {yj : j ∈ J } in A for some interval J = [0, K] such that the following are true: (1) xj < yj < xj+1 < yj+1 , (2) A(yj , xj+1 ) = 2, A(x ,y )
(2A)(2x ,2y )
j α (3) either yj −xj j +1 6 , where = 400 , or A(xj ,yj j ) j > 3 + (4) {xj , yj : j ∈ J } ∩ U is upper unbounded in U .
1 100
for all xj , yj ∈ U ,
Claim 2.12.4.7: The existence of the sequences described above leads to a contradiction to the assumption that the key lemma is not true. Proof of Claim 2.12.4.7: Let y ∈ U be sufficiently large such that y = yj0 . By Claim 2.12.2, we need only to show that (2A)(0,2y) 3A(0,y) . 2y+1 2(y+1) P0 Pj0 for the sum over all j’s such that In the sum j=0 below we write P00 and write for the sum over the rest of j’s. Let β =
A(0,y) . y+1
Then β > α. Also
β≈ implies
P00
A(xj ,yj ) y+1
P0
P00 P00 A(xj , yj ) + A(xj , yj ) A(xj , yj ) 6 + y+1 y+1
' β − . Hence we have
Pj0 (2A)(0, 2y) j=0 (2A)(2xj , 2yj ) ' 2y + 1 2y + 1 P00 1 (3 + 100 ) A(xj , yj ) 1 β− ' ' (3 + ) 2y + 1 100 2 α 301 399 3 3A(0, y) 1 β − 400 ) > · β β≈ . ' (3 + 100 2 100 800 2 2(y + 1) 2(Claim 2.12.4.7) 28
A(xj ,yj ) yj −xj +1
6
Now we construct the sequences xj and yj . Let x0 = 0. If we have found yj in A, then xj+1 > yj is just the least element in A, which is greater than yj . Suppose we have found xj . We need to find yj . x > xj and A[xj , x] is not a subset z = min x ∈ A : . of a b.p. with the difference > 1
Let
when z is well defined. Otherwise the process stops. By Claim 2.12.4.5 and Claim 2.12.4.6, if xj ∈ U , then z ∈ U . Also by Claim 2.12.4.4 and the assumption that t > 0, we have that if xj ∈ U , then z − xj must be a hyperfinite integer. If there exists a z 0 ∈ A[xj + 100, z] such that
A(xj ,z 0 ) z0 −xj +1
< , then let yj be the least such z 0 .
Otherwise let yj = z. We now check that yj is what we want. We need only to check when yj ∈ U . If A(xj ,yj ) yj −xj +1
A(xj ,yj ) yj −xj +1
< , then we are done. Suppose
> . Let A[xj , yj − 1] be a subset
of a b.p. I1 ∪ I2 of difference k. Here we allow I2 ∩ A[xj , yj − 1] = ∅. For that case A[xj , yj − 1] is a subset of an a.p. of difference > 1. Let Ai = A[xj , yj − 1] ∩ Ii , li = min Ai = min Ii, and ui = max Ai = max Ii for i = 1, 2 when they are well defined. If k > 3, then by the minimality of z we have A2 6= ∅. By Lemma 2.7 and Lemma 2.8 we have
(2A)(2xj ,2yj ) A(xj ,yj )
>3+
1 . 100
Subcase 2.12.4.1 k = 3. Again A2 6= ∅ by the minimality of z. Without loss of generality, we assume l1 ≡ 0 (mod 3), l2 ≡ 1 (mod 3), yj ≡ 2 (mod 3), and l1 < l2 . Let A0 = A[xj , yj ]. If |A2 | < 14 |A0|, then (2A)(2xj , 2yj ) |2A1| + |A1 + A2| + |yj + A1| ' 0 |A | |A0 | |A1| − 2|A2| 1 4|A1| + |A2| ≈3+ '3+ . ' 0 0 |A | |A | 4 So we can assume |A2| > 14 |A0|. By the same reason we can assume |A1| > 14 |A0|. We now want to show that (2A)(2xj , 2yj ) 1 '3+ . 0 |A | 100 Suppose
(2A)(2xj ,2yj ) |A0 |
3+
1 . 100
Then there exist non-negative integers b1, b2, and
b3 such that |2A1 | + |2A2| + |A1 + A2| 29
= (2|A1| − 1 + b1) + (2|A2 | − 1 + b2) + (|A1| + |A2| − 1 + b3) 1 |A0| − 3. 6 |2A0| < 3(|A1| + |A2|) + 100 |A0 | 6 |A25i | for i = 1, 2. By Lemma 1.6, Ai 100 i most 26 |Ai|, which means ui −l + 1 6 26 |Ai|. 25 3 25
This implies bi 6
is a subset of an a.p. with
the length at
Note that
1 1 25 u1 − l1 + 1). |A2| > |A0| > |A1 | > ( 4 3 78 3 If l1 < l2 < u2 < u1 , then u1 − l1 is hyperfinite and |2A1 | + |1 + 2A2 | > 2|A1 | − 1 + 2|A2| − 1 50 u1 − l1 u1 − l1 + )−2 > ( 26 3 9 200 u1 − l1 50 4(u1 − l1 ) )−2 > ( + 1) − 5 > ( 26 9 78 3 2u1 − 2l1 > + 1. 3 Since (2A1) ∪ (1 + 2A2) ⊆ [2l1, 2u1 ] ∩ {3u : u ∈ U }, then (2A1) ∩ (1 + (2A2)) 6= ∅. This implies that (2A1) ∪ (2A2 ) contains two consecutive numbers. Hence 2A contains two consecutive numbers, which contradicts t > 0 by Claim 2.12.4.3. If l1 < u1 < l2 < u2 , then (yj + A2) ∩ (2A1 ) = ∅. Hence |2A1 | + |2A2| + |A1 + A2 | + |yj + A2| 1 (2A)(2xj , 2yj ) > '3+ . 0 0 |A | |A | 4 Finally we assume l1 < l2 < u1 < u2. If u1 − l2 >
1 (u2 20
− l1), then
|2A1| + |1 + 2A2| > 2|A1| − 1 + 2|A2 | − 1 50 u1 − l1 u2 − l2 + + 2) − 2 > ( 26 3 3 50 u1 − l2 u2 − l1 > ( + + 2) − 2 26 3 3 50 21(u2 − l1 ) + 2) − 2 > ( 26 60 2u2 − 2l1 > + 2. 3 The last inequality uses the fact that u2 − l1 is hyperfinite. Hence again (2A1 ) ∩ (1 + (2A2)) 6= ∅. This implies 2A contains two consecutive numbers, which is again a contradiction. 30
Suppose u1 − l2
u1 − (u2 − l2 + u1 − l1 ) 20 26 1 > u1 − (3 · |A0|) 20 25 39 · 4|A2|) > u1 − ( 250 78 u2 − l2 > u1 − ( + 1) 125 3 1 > u1 − (u2 − l2). 3
l2 > u1 −
This shows 23 (u2 −l2) < u2 −u1, which implies 2u1 < l2 +u2. Hence (yj +A2)∩(2A1) = ∅. So again we have (2A)(2xj , 2yj ) 1 '3+ . A(xj , yj ) 4 2(Subcase 2.12.4.1) Subcase 2.12.4.2 k = 2. Suppose A2 6= ∅. Without loss of generality, we assume l1 < l2 and A1 is a set of even numbers. Since 2I1 and 2I2 are disjoint, then u1 < l2. Clearly yj must be an even number. If |A2 | >
1 A(xj , yj ) 100
− 1, then
|2A1 | + |2A2 | + |A1 + A2| + |yj + A2| 1 (2A)(2xj , 2yj ) ' '3+ . A(xj , yj ) A(xj , yj ) 100 1 Hence we can assume |A2| < 100 A(xj , yj ) − 1, which implies |A1| > (2A)(2x ,2y ) 1 1 . If |2A1 | > 2|A1 | + 100 A(xj , yj ), then A(xj ,yj j ) j ' 3 + 100
99 A(xj , yj ). 100
1 If |2A1| 6 2|A1 | − 1 + 100 A(xj , yj ), then by Lemma 1.6, A1 is a subset of an a.p. of 1 length at most |A1| + 100 A(xj , yj ) < 100 |A1 |. 99 If gcd(A1 − l1) > 2, then |A1 + A2 | > 2|A1 |. Hence
can assume gcd(A1 − l1 ) = 2. This implies
u1 −l1 2
the distance between yj and u1 is large. n Let v = min x ∈ [l1, u1 ] : x is even and
+1 6
A1 (x,u1 ) u1 −x+2
99 u1 − l1 ( + 1) 6 |A1| 100 2 31
6
(2A)(2xj ,2yj ) 98 ' 3 + 100 . So we A(xj ,yj ) 100 |A1|. We want to show that 99
1 4
o
. Then
= A1(l1 , v − 2) + A1(v, u1) v − l1 u1 − v + 2 + 6 2 4 u1 − l1 u1 − v + 2 = +1− . 2 4 Hence u1 − v + 2 6
1 u1 −l1 ( 2 25
+ 1) and v >
49u1 +l1 +98 . 50
By the pigeonhole principle and
by the definition of v, all even numbers in [l1 + u1 , v + u1 − 2] are in 2A1. If l2 < u1 + v − l1, then 2A contains two consecutive numbers l1 + l2 and l1 + l2 − 1, which contradicts t > 0 by Claim 2.12.4.3. So we can assume that l2 > u1 + v − l1. Hence yj > u1 + v − l1. This shows yj − u1 > v − l1. So A1 (2u1 − yj , u1 ) > A1(u1 − v + l1, u1 ) 1 u1 − l1 > A1(l1 + [ ( + 1) − 2], u1) 25 2 1 u1 − l1 + 1) > |A1| − ( 50 2 99 u1 − l1 1 u1 − l1 > ( + 1) − ( + 1) 100 2 50 2 97 u1 − l1 = ( + 1) 100 2 97 97 × 99 > |A1| > A(xj , yj ) 100 100 × 100 95 > A(xj , yj ). 100 Hence (2A)(2xj , 2yj ) A(xj , yj ) |2A1 | + |A1 + A2| + |yj + A1[2u1 − yj , u1]| ' A(xj , yj ) 95 A(xj , yj ) − 2|A2| 93 '3+ . ' 3 + 100 A(xj , yj ) 100 Suppose A2 = ∅ and again assume A1 is a set of even numbers. Then yj is an odd number and l1 + yj < 2u1 because otherwise A[xj , yj ] is a subset of a b.p. of difference 2, which contradicts the minimality of z. If |2A1 | > 2|A1| +
1 |A1|, 100
1 3|A1 | + 100 |A1| (2A)(2xj , 2yj ) 1 ' '3+ . A(xj , yj ) A(xj , yj ) 100
32
then
So we can assume |2A1 | < 2|A1| +
1 |A1|. 100
By Lemma 1.6 we have
1 101 u1 − l1 + 1 6 |A1| + |A1| = |A1 |. 2 100 100 Let
1 A1 (l1, x) v = max x ∈ [l1, u1] : x is an even number and 6 x − l1 + 2 4
and
1 A1(x, u1) v = min x ∈ [l1, u1] : x is an even number and 6 u1 − x + 2 4 0
Then v − l1 6
4 u1 −l1 ( 2 101
+ 1) and u1 − v 0 6
4 u1 −l1 ( 2 101
.
+ 1). Hence v 0 − v is hyperfinite.
By the pigeonhole principle we have {x ∈ [l1 + v + 2, u1 + v 0 − 2] : x is even } ⊆ 2A1 . Hence all even numbers in [2l1 +v+2, 2u1 +v 0 −2] are elements of (3A1 ) = A1 +A1 +A1. Since all odd numbers in yj + [l1 + v + 2, u1 + v 0 − 2] are elements of (3A) and since 2l1 + v + 2 < yj + u1 + v 0 − 2 and yj + l1 + v + 2 < 2u1 + v 0 − 2, then there are x1, x2, x3 , z1, z2 ∈ A1 such that x1 + x2 + x3 = z1 + z2 + yj + 1. Next we want to show that there is sequence E of e–transforms with (A00, B 00) = E(A, A) such that A00 ∩ U contains two consecutive numbers, which contradicts the assumption t > 0. Let R = {x1 + x2, x3 , z1 + z2, yj }. Then R ⊆ 2A. By Claim 2.12.4.3 there is c1 ∈ A ∩ U and a sequence E1 of e–transforms such that (A1 , B1) = E1 (A0, B 0) and c1 + R ⊆ A1 . By Subclaim 2.12.4.2.1 there is c2 ∈ B1 ∩ U such that c2 + c1 + R ⊆ B1 . Hence (c2 + c1 + R) + (c2 + c1 + R) ⊆ 2B1 ⊆ A1 + B1 ⊆ 2A. Again by Claim 2.12.4.3 there is c3 ∈ A ∩ U and a sequence E2 of e–transforms with (A00, B 00) = E2 (A1, B1 ) such that c3 + (c2 + c1 + R) + (c2 + c1 + R) ⊆ A00. This shows that A00 contains two consecutive numbers c3 + 2c2 + 2c1 + x1 + x2 + x3 and c3 + 2c2 + 2c1 + z1 + z2 + yj . This ends the proof of Case 2.12.4 for |F | = 1. Hence the lemma is proven.
3
2(Lemma 2.12)
Proofs of the Main Theorem
In order to use nonstandard techniques, we first translate Theorem 1.3 into nonstandard forms. We translate the part II of Theorem 1.3 into Theorem 3.1 and the part 33
III of Theorem 1.3 into Theorem 3.2. Then we prove Theorem 3.1 and Theorem 3.2. 1 2
Theorem 3.1 Let A ⊆ [0, H], 0, H ∈ A, and 0 < α
α by Lemma hK 3 ¯ 2.1. (4) is true because otherwise d(2A) > α by Lemma 2.1. A(hn ) hn
2
Now for every hyperfinite integer K either (a) or (b) of Theorem 3.1 is true with H = hK and A replaced by B = ∗A ∩ [0, H]. If there is a hyperfinite integer K such that (a) of Theorem 3.1 is true for H = hK and B = ∗A ∩ [0, hK ], then there are a, d ∈ N such that α = 2d and H H ∪ a + dn : 0 6 n 6 . A ⊆ B ⊆ dn : 0 6 n 6 d d So clearly (a) of Part II of Theorem 1.3 is true for A. Otherwise for every hyperfinite integer K, (b) of Theorem 3.1 is true for H = hK and B = ∗A ∩ [0, H]. Given a standard positive integer k and let Xk be the set of all non-negative integers n such that there are 0 6 c 6 b 6 hn satisfying
c hn
< 1k ,
34
B(b,hn ) hn −b
> 1− k1 , and [c+1, b−1]∩B =
∅, where B = ∗A ∩ [0, hn ]. Then Xk is internal and contains every hyperfinite integer. Hence there is a standard positive integer mk such that Xk contains all positive integers greater than or equal to mk . Without loss of generality we can assume that m1 < m2 < · · ·. For each k = 1, 2, . . . and for each n ∈ [mk , mk+1 − 1] choose cn = c n) > 1 − k1 , and [c + 1, b − 1] ∩ A = ∅. and bn = b so that 0 6 c 6 b 6 hn , hcn > k1 , A(b,h hn −b Choose cn = bn = 0 for every n ∈ [0, m1 − 1]. It is easy to see that limn→∞ [cn + 1, bn − 1] ∩ A = ∅, and
n ,hn ) limn→∞ hA(b n −bn +1
= 1.
cn hn
= 0,
2(Part II of Theorem 1.3)
Theorem 3.2 Let A ⊆ [0, H] and 0, H ∈ A be such that (1) gcd(A[0, x]) = 1 for every hyperfinite x, (2) A(0, H) ∼ 12 H, (3) for every hyperfinite x 6 H, A(0, x) 12 x, (4) for every hyperfinite x 6 H, (2A)(0, 2x) 32 x. Then (a) either there is an a ∈ {0, 3} such that H H A ⊆ 4n : 0 6 n 6 ∪ a + 4n : 0 6 n 6 4 4 (b) or (2A)(0, H) ∼ A(0, H). Proof of Part III of Theorem 1.3 from Theorem 3.2: Let A ⊆ N be such 1 3 ¯ ¯ that 0 ∈ A and gcd(A) = 1 such that d(A) = 2 and d(2A) = 4 . Take an increasing sequence hn ∈ N such that limn→∞
A(hn ) hn
=
1 . 2
Again we can assume hn ∈ A for
every n ∈ N. Let’s fix an arbitrary hyperfinite integer K. Let H = hK and let B = ∗A ∩ [0, H]. By the same idea as in the proof above, we have 0, H ∈ B and can check that (1)–(4) of Theorem 3.2 are true for B in the place of A. Now it is easy to see that (a) of Theorem 3.2 for B implies (a) of Part III of Theorem 1.3 for A and (b) of Theorem 3.2 for B implies (b) of part III of Theorem 1.3 for A. 2(Part III of Theorem 1.3) Now we are ready to prove Theorem 3.1 and Theorem 3.2. Proof of Theorem 3.1: Note that (4) of Theorem 3.1 implies |2A| ∼ 3|A|. Suppose A is a subset of a b.p. I0 ∪ I1 of difference d > 1. Let Ai = A ∩ Ii , li = min Ai, and ui = max Ai for i = 0, 1. Without loss of generality we assume 0 = l0. By (1) of Theorem 3.1 we have l1 ∈ N. Since |2A| ∼ |2A0| + |2A1| + |A0 + A1| 2|A0 | + 2|A1| + |A0| + |A1| ∼ 3|A|, 35
then |2Ai| ∼ 2|Ai |. Hence by Lemma 1.6 we have |Ai | ∼ Ii (li, ui ). Suppose d = 2. Then u0 < l1. Hence 1 1 |A| ∼ |A1| ∼ (u1 − l1 ) ∼ H, 2 2 which contradicts α < 12 . Suppose d = 3. If u0 ∼ 0, then by the argument above we have α = 13 . Now 2 (2A)(0, H) ∼ |0 + A1 | + |l1 + A1| ∼ 2|A1 | ∼ H 3 contradicts (4) of Theorem 3.1 for x = [ H2 ]. By the same idea we can show that u1 ∼ 0 is also impossible. Let u = min {u0 , u1}. Then A(0, u) ∼ A0(0, u) + A1(0, u) ∼ 23 u, which contradicts (3) of Theorem 3.1. Suppose d > 4. If u0 ∼ 0, then α 6 1d . This implies 3 2 H ∼ (2A)(0, H) αH. d 2 Hence
1 d
>α>
4 3d
> 1d , which is absurd. If 0 ≺ u0 ≺ H, then u1 = H and 1 (u0 + H) ∼ |A0| + |A1| = |A| ∼ αH d
implies α < 2d . Now we have A(0, u0) ∼ d2 u0 αu0, which contradicts (3) of Theorem 3.1. By symmetry, u1 ≺ H is impossible. Hence u0 ∼ u1 ∼ H, which implies α = 2d . Hence (a) of Theorem 3.1 is true. From the arguments above we can assume that A is not a subset of a b.p. of difference > 1. For notational convenience we deal with the set B = H − A in the rest of the proof. Case 3.1.1: dU (B) < α. Then there is an x 0 such that B(0, x) ≺ αx. This implies A(0, H − x) α(H − x), which contradicts (3) of Theorem 3.1. 2(Case 3.1.1) Case 3.1.2: dU (B) > 12 . Let 0 ≺ b ≺ H be such that B(0, b) ∼ 12 b and for every 0 ≺ x ≺ b, B(0, x) 12 x. Then (2B)(0, b) ∼ b. Subcase 3.1.2.1 B(b, H) ∼ 0. 36
Since (2B)(0, 2H) (2B)(0, b) + (2B)(b, H) + (2B)(H, H + b) ∼ 3|B| + (2B)(b, H), then (2B)(b, H) ∼ 0. By Lemma 2.5 we have max B[0, b] = ¯b ∼
1 b. 2
If c =
min B[b, H] ∼ H, then (b) of Theorem 3.1 is true. If c ≺ H, then |2B| b + |c + B[0, H − c]| + |H + B[0, b]| ∼ 3|B| + B(0, H − c) 3|B|, which contradicts (4) of Theorem 3.1. 2(Subcase 3.1.2.1) Subcase 3.1.2.2 B(b, H) 0. Let c ∈ B[b, H] be such that B(b, c) ∼ 0 and for every c ≺ x 6 H, B(c, x) 0. By (1) and (3) of Theorem 3.1 and Lemma 1.5 we have (2B)(2c, 2H) 3B(c, H). Choose an x ∈ B with c ≺ x ≺ c + b. Then |2B| (2B)(0, c) + |c + B[0, x − c]| +|x + B[0, 2c − x]| + (2B)(2c, 2H) 2B(0, c) + B(0, x − c) + B(0, 2c − x) + 3B(c, H) ∼ 3|B| + B(0, x − c) − B(2c − x, c). Note that B(0, x − c) 12 (x − c) and 1 B(2c − x, c) ∼ B(2c − x, b) B(b + c − x, b) ≺ (x − c). 2 Then we have |2B| 3|B|, a contradiction. 2(Case 3.1.2) Case 3.1.3: α 6 dU (B) 6 12 . We divide the proof into three subcases according the arithmetic structure of B ∩ U. Subcase 3.1.3.1: B ∩ U is neither a subset of an a.p. of difference > 1 nor a subset of a U –unbounded b.p. . By Lemma 2.12 there is a y ∈ B with 0 ≺ y ≺ H such that (2B)(0, 2y) 3B(0, y). By (1) and (3) of Theorem 3.1 we have B(y, H) α(H − y) and gcd(B[y, H] − y) = 1. By Lemma 2.4 we have |2B| 3|B|. 2(Subcase 3.1.3.1) Subcase 3.1.3.2 B ∩ U is a subset of an a.p. of difference d > 1.
37
Let a = min {x ∈ B : gcd(B[0, x]) = 1} and c = max B[0, a − 1]. Since 0 ∈ B, then a > U . If c ∈ U , then B[c + 1, a − 1] = ∅, which implies B(a,H) H−a+1
B(0,a) a+1
≈ 0. Hence
α, which contradicts (3) of Theorem 3.1. So we have c > U or equivalently
c 0. Let d = gcd(B[0, c]). Note that d > 1 is a standard integer because otherwise B(0,c) B(c,H) / 1d ≈ 0, which implies H−c+1 α. The proof is easy if 1d < 32 α because c+1 (2B)(0, c) 1d c ≺ 32 αc implies (2B)(c, 2H) 32 α(2H − c), which contradicts (4) of Theorem 3.1. So we assume than the other cases.
1 d
> 32 α. The proof of this case is much more tedious
Firstly, we can assume that a ≺ H by the following reason: Suppose a ∼ H. Then 3|B| ∼ |2B| |B[0, c] + B[0, c]| + |a + B[0, c]| ∼ 3|B| implies B(0, c) ∼
1 c d
by Lemma 1.6. If c ∼ H, then |B| ∼
1 H d
3 αH, 2
which
contradicts (3) of Theorem 3.1. If c ≺ H, then by (1) of Theorem 3.1 there exists a b ∼ H such that b 6≡ a (mod d). Hence |2B| (2B)(0, H) + | {a, b} + B| ∼ B(0, c) + |c + B[0, H − c]| + 2|B| ∼ 3|B| + B(0, H − c) 3|B|. Secondly, we can assume that B(a, H) 0 by the following reason: Suppose B(a, H) ∼ 0. If |B[0, c] + B[0, c]| 2B(0, c), then |2B| |B[0, c] + B[0, c]| + |a + B[0, c]| 3|B|. So we can assume |B[0, c] + B[0, c]| ∼ 2B(0, c). By Lemma 1.6 we have B(0, c) ∼ 1d c. This implies B(a + c − H, c) 0. Hence we have |2B| |B[0, c] + B[0, c]| +|a + B[0, c]| + |H + B[a + c − H, c]| 3B(0, c) + B(a + c − H, c) ∼ 3|B| + B(a + c − H, c) 3|B|. Now we are ready to prove the subcase. The proof of Subcase 3.1.3.2 is divided into three subsubcases. Subsubcase 3.1.3.2.1 d = 2. 38
Since we have |B[a, H] + B[a, H]| 3B(a, H), then 3|B| ∼ |2B| (2B)(0, 2c) + |a + B[0, c]| + (2B)(2a, 2H) 3|B| implies that B(0, c) ∼ 12 c by Lemma 1.6. Without loss of generality we can assume c, c − 2, c − 4 ∈ B. Suppose c ≺ a. Choose an x a in B such that either B(a, x) ∼ 0 or x − a < a − c. Then we have |2B| 2B(0, c) + |a + B[0, c]| +|x + B[a + c − x, c]| + |B[a, H] + B[a, H]| 3B(0, c) + 3B(a, H) + B(a + c − x, c) 3|B|. Hence we can assume c ∼ a. Recall that we have B(0, c) 0, B(a, H) 0, gcd(B[0, c]) = 2, gcd(B[a, H] − a) = 1, and
B(a,H) H−a+1
/ α.
Subsubsubcase 3.1.3.2.1.1 da+U (B) = 0. Choose an x ∈ B with x a such that B(a, x) < 18 (x − a + 1). Then |B[x, H] + B[x, H]| 3B(x, H) by (3) of Theorem 3.1 and Lemma 1.5. Hence |2B| |B[0, c] + B[0, c]| + |a + B[0, c]| +|x + B[a + c − x, c]| + |B[x, H] + B[x, H]| 1 3B(0, c) + (x − a + 1) + 3B(x, H) 2 1 3|B| + (x − a + 1) − 3B(a, x) 3|B|. 2 2(Subsubsubcase 3.1.3.2.1.1) Subsubsubcase 3.1.3.2.1.2 da+U (B) > 12 . By the same proof as in Case 3.1.2 we have that either (2B)(2a, 2H) 3B(a, H) or there are ¯b < c0 in B[a, H] such that B[a, H] ⊆ [a, ¯b] ∪ [c0, H], ¯b − a ∼ B(a, H), and c0 ∼ H. The first possibility above implies |2B| 3|B| by Lemma 2.4 and the second also implies |2B| 3|B| because 2¯b ≺ a + H and |2B| (2B)(0, 2a) + (2B)(2a, 2¯b) +(2B)(2¯b, a + H) + |H + B[a, H]| |B[0, c] + B[0, c]| + |a + B[0, c]| +2B(a, ¯b) + |H + B[2¯b − H, a]| + B(a, H) 3B(0, c) + 3B(a, H) + B(2¯b − H, a) = 3|B| + B(2¯b − H, a) 3|B|. 39
The last inequality holds because c ∼ a 2¯b−H and B(0, c) ∼ 12 c.
2(Subsubsubcase
3.1.3.2.1.2) Subsubsubcase 3.1.3.2.1.3 0 < da+U (B) 6 12 . Since c ∼ a, then c − 4, c − 2, c, a ∈ (c − 4 + U ) and dc−4+U (B) = da+U (B). Hence gcd((B[c − 4, H] − c + 4) ∩ U ) = 1 and (B[c − 4, H] − c + 4) ∩ U is not a subset of a U –unbounded b.p. of difference d > 1. By Lemma 2.12 there exists a y a in B such that (2B)(2(c − 4), 2y) 3B(c − 4, y). Note that B(y, H) ≺ 12 (H − y) and gcd(B[y, H]−y) = 1 when H −y 6∈ N. Hence by Lemma 2.4 |B[c−4, H]+B[c−4, H]| 3B(c − 4, H), which implies |2B| 3|B|. This ends the proof of Subsubcase 3.1.3.2.1. 2(Subsubcase 3.1.3.2.1) Subsubcase 3.1.3.2.2 d = 3. By the same reasons as in Subsubcase 3.1.3.2.1 we can assume that B(0, c) ∼ 13 c and c ∼ a. Since B is not a subset of a b.p. of difference > 1, we can define b = min {x ∈ B : x 6∈ {0, a} (mod 3)} . Let B0 = B∩{3n : n ∈ ∗ N}, Ba = B∩{a + 3n : n ∈ ∗ N}, and Bb = B∩{b + 3n : n ∈ ∗ N}. Let l0, la, lb be the least element of B0 , Ba , Bb , respectively. Let u0, ua , ub be the largest element of B0 , Ba , Bb , respectively. Subsubsubcase 3.1.3.2.2.1 b ∼ H. We have |B| ∼ |B0| + |Ba |. We can also assume |Ba | 0 because otherwise |2B| |2B0 | + |a + B0 | + |b + B0 | = 4|B0 | = 4|B|. Since B0 ∪ Ba is a subset of a b.p. , then B0 (l0, u0) ∼ 1 (ua 3
1 (u0 3
− l0) and B(la, ua ) ∼
− la). This implies ua ≺ H or u0 ≺ H because otherwise B(a, H) ∼ 23 (H − a),
which contradicts (3) of Theorem 3.1. Suppose ua ≺ H and ua 6 u0 . Then |2B| |2B0 | + |2Ba | +|B0 + Ba | + |b + B0[ua + u0 − b, u0]| 3|B| + B0 (ua + u0 − b, u0) 3|B|. By the same reason, if u0 ≺ H and u0 6 ua, then |2B| 3|B|. Note that if both u0 ≺ H and ua ≺ H are true, then either u0 6 ua or ua 6 u0 . 3.1.3.2.2.1) 40
2(Subsubsubcase
Subsubsubcase 3.1.3.2.2.2 b ≺ H. Note that we have gcd(B[b, H] − b) = 1 and B(b, H) α(H − b). Let us redefine u0 = max(B0 [0, b − 1]) and ua = max(Ba [0, b − 1]). Then by Lemma 1.6 we have B0(l0 , u0) ∼ 13 (u0 − l0) and Ba(la , ua) ∼ 13 (ua − la ). By the same reason as in Subsubsubcase 3.1.3.2.2.1 we can assume u0 , ua ∼ b and Ba(la , ua) 0. If db+U (B) = 0, then there is an x ∈ B, x b such that either x − b < u0 and B(b, x)
1 (x 10
− b), or B(b, x) ∼ 0. Hence |2B| |B0 [l0, u0 ] + B0 [l0, u0]| +|Ba[la, ua ] + Ba [la, ua ]| + |B0 [l0, u0 ] + Ba [la, ua ]| +|B[x, H] + B[x, H]| + |x + B0 [2u0 − x, u0]| 3B(0, b) + 3B(x, H) + B0 (2u0 − x, u0) 1 3B(0, b) + 3B(x, H) + (x − u0 ) 3 1 3 3|B| + (x − u0 ) − (x − b) 3|B|. 3 10
If db+U (B) > 12 , then by the same reason as in the proof of Case 3.1.2 we have that either (2B)(2b, 2H) 3B(b, H), which implies |2B| 3|B|, or B[b, H] ⊆ [b, ¯b]∪[c0, H], where ¯b − b ∼ B(b, H) and c0 ∼ H. The latter again implies, by the same argument as in the proof of Subsubsubcase 3.1.3.2.1.2, |2B| 3|B|. Now we can assume 0 < db+U (B) 6 12 . Let u = min {u0, ua }. Then u ∼ b. Since B0(0, u0 ) ∼ 13 (u0 − l0 ) and Ba (la, ua) ∼ 13 (ua − la), we can assume u − 3 ∈ B. Hence (B − (u − 3)) ∩ U is neither a subset of an a.p. of difference > 1 nor a subset of a U –unbounded b.p. of difference > 1. By Lemma 2.12 there exists a y ∈ B with y b such that (2B)(2(u − 3), 2y) 3B(u − 3, y). Finally by Lemma 2.4, |2B| 3|B|. 2(Subsubcase 3.1.3.2.2) Subsubcase 3.1.3.2.3 d > 4. Since B is not a subset of a b.p. of difference > 1, the number b = min {x ∈ B : x 6∈ {0, a} (mod d)} is well defined. If b ≡ 2a (mod d), then a + b 6≡ 0 (mod d) because otherwise B[0, b] is a subset of an a.p. with difference d3 > 1. Hence we have either b 6≡ 2a (mod d) or a + b 6≡ 0 (mod d). This implies (b + B0 [0, u0]) ∩ (B[0, b − 1] + B[0, b − 1]) = ∅ 41
or (b + Ba [la, ua]) ∩ (B[0, b − 1] + B[0, b − 1]) = ∅. Again it is easy to check that ua ∼ b, u0 ∼ b, Ba(la , ua) 0, B0 (0, u0 ) 0, B0(0, u0 ) ∼ 1 u d 0
and Ba (la, ua ) ∼ 1d (ua − la). Hence |2B| |B0 [0, u0] + B0 [0, u0]| + |Ba [la, ua] + Ba [la, ua]| +|B0 [0, u0] + Ba [la, ua ]| + |B[b, H] + B[b, H]| + min {|b + B0[0, u0 ]|, |b + Ba[la, ua ]|} 3B(0, b) + 3B(b, H) + min {B0 (0, u0 ), Ba(la , ua)} 3|B|.
This ends the proof of Subcase 3.1.3.2.
2(Subcase 3.1.3.2)
Subcase 3.1.3.3 B ∩ U is a subset of a U –unbounded b.p. of difference d. Let b = min {x ∈ [0, H] : B[0, x] is not a subset of a b.p. of difference d.} and let B[0, b − 1] be a subset of a b.p. I1 ∪ I2 of difference d. Let Bi = B[0, b − 1] ∩ Ii , li = min Bi , and ui = max Bi . Note that B(0, b − 1) αb by (3) of Theorem 3.1. By Subcase 3.1.3.2 we can assume gcd(B[0, b − 1]) = 1. If d > 3, then the subcase follows from Lemma 2.7, Lemma 2.8, and Lemma 2.4. Suppose d = 3. Since |2B| ∼ 3|B|, then |B[0, b − 1] + B[0, b − 1]| ∼ 3B(0, b − 1) implies Bi (li, , ui ) ∼ 13 (ui − li) by Lemma 1.6. Since B ∩ U is already a subset of a U –unbounded b.p. of difference 3, then l1, l2 ∈ U and u1, u2 6∈ U . Hence dU (B) =
2 , 3
which contradicts the assumption
1 2
dU (B) 6 for Case 3.1.3. This completes the proof of Case 3.1.3 as well as the proof of Theorem 3.1. 2(Theorem 3.1) Proof of Theorem 3.2: Suppose A is a subset of a b.p. I0 ∪ I1 of difference d > 1. Let Ai = A ∩ Ii , li = min Ai , and ui = max Ai for i = 0, 1. Without loss of generality we assume 0 = l0. By (1) of Theorem 3.2 we have l1 ∈ N. Since |2A| ∼ |2A0| + |2A1| + |A0 + A1| 2|A0 | + 2|A1| + |A0| + |A1| ∼ 3|A|, then by Lemma 1.6 we have |Ai | ∼ Ii (li, ui ). Suppose d = 2. Then u0 < l1. Hence (2A)(0, H) ∼ |0 + A1| + |l1 + A1| ∼ H 34 H, which contradicts (4) of Theorem 3.2. Suppose d = 3. If u0 ∼ 0 or u1 ∼ 0, then |A| 13 H, which contradicts (2) of Theorem 3.2. So we have ui 0 for i = 0, 1. Since |2A| ∼ 3|A|, then Ai(li , ui) ∼ 1 (ui 3
− li) for i = 0, 1. This implies dU (A) = 23 , a contradiction. 42
Suppose d > 4. Then |A| 2d H ≺ 12 H, which contradicts (3) of Theorem 3.2. Suppose d = 4. If u0 ≺ H, then u1 = H and 1 1 |A| ∼ |A0| + |A1| (u0 + H) ≺ H, 4 2 which contradict (3) of Theorem 3.2. By symmetry, u1 ≺ H is also impossible. Hence u0 ∼ u1 ∼ H, which implies (a) of Theorem 3.2. From the arguments above we can assume that A is not a subset of a b.p. of difference > 1. For notational convenience we again deal with the set B = H − A. Case 3.2.1 dU (B) < 12 . Same as in Case 3.1.1. 2(Case 3.2.1) Case 3.2.2: dU (B) > 12 . Suppose for every 0 ≺ x ≺ H we have B(0, x) 12 x, then (2B)(0, H) ∼ H ∼ 2B(0, H). Hence by (4) of Theorem 3.2 we have B(0, H) (2B)(H, 2H) |H + B[0, H]| ∼ B(0, H), which implies (2B)(H, 2H) ∼ B(0, H). Hence (b) of Theorem 3.2 is true. Otherwise there is 0 ≺ b ≺ H such that B(0, b) ∼ 12 b and B(0, x) 12 x for every 0 ≺ x ≺ b. Now the proof is the same as the proof of Subcase 3.1.2.2.
2(Case
3.2.2) Case 3.2.3: dU (B) = 12 . If B ∩ U is neither a subset of an a.p. of difference > 1 nor a subset of a U – unbounded b.p. , then by the same proof as in Subcase 3.1.3.1 we have |2B| 3|B|. Suppose B ∩ U is a subset of an a.p. of difference d > 1. Then d = 2. Since B is not a subset of an a.p. of difference > 1, then B contains an odd number. Let b be the least odd number in B. Then b 0. If B(0, b) ≺ 12 b, then B(b, H) 12 (H − b), which contradicts (3) of Theorem 3.2. Hence B(0, b) ∼ 12 b. Since (2B)(0, 2H) ∼ 32 H and (2B)(0, b − 1) ∼ 12 b, then 1 3 (2B)(b, 2H) ∼ H − b 2 2 1 3 3 = (2H − b) + b (2H − b), 4 4 4 which contradicts (4) of Theorem 3.2. Suppose B ∩ U is a subset of a U –unbounded b.p. I0 ∪ I1 of difference d. Then d > 2. Let c = min {x ∈ B : B[0, x] is not a subset of a b.p. of difference d.}. By the argument above we can assume gcd(B[0, c − 1]) = 1. 43
If d > 4, then by Lemma 2.7 and Lemma 2.8 we have (2B)(0, 2c) 3B(0, c). By Lemma 2.4 we have |2B| 3|B|. Suppose d = 3. Without loss of generality we assume B[0, c − 1] = B0 ∪ B1 , where Bi = {x ∈ B[0, c − 1] : x ≡ i (mod 3)} for i = 0, 1 and c ≡ 2 (mod 3). Let li = min Bi and ui = max Bi . Since we can assume that B ∩ U is not a subset of an a.p. of difference > 1 by the argument above we have l1 ∼ 0. Since (2B)(0, 2c − 1) ∼ |2B0 | + |2B1 |+|B0 +B1 | 3B(0, c−1), then |2B| ∼ 3|B| implies (2B)(0, 2c−1) ∼ 3B(0, c−1), which implies |Bi | ∼ 13 (ui − li ). If u0 ∼ 0 or u1 ∼ 0, then dU (B) 6 13 . Otherwise dU (B) > 23 . But each of them contradicts dU (B) = 12 .
2(Theorem 3.2)
Next we would like to present a corollary of Theorem 1.3. Let 2 Q= :k>4 . k ¯ = Corollary 3.3 Suppose A is not a subset of an a.p. of difference > 1 and d(A) α < 1. If α 6∈ Q and
B is not a subset of an a.p. ¯ ¯ : d(2A) = min d(2B) ¯ of difference > 1 and d(B) >α
,
then d(A) = 0. Proof: The corollary is trivially true if α = 0. So we can assume α > 0. Since α 6∈ Q, then A cannot have the structure characterized in the conclusion of Part II (a) or Part III (a) of Theorem 1.3. If α < 12 , then A has the structure characterized in the conclusion of Part II (b) of Theorem 1.3. It is then easy to see that limn→∞ limn→∞ cbnn = 0. Hence d(A) = 0. Suppose α >
1 2
A(bn ) bn
6
and d(A) = β > 0. Without loss of generality we assume 0 ∈ A.
n) Let hn ∈ A be such that limn→∞ hn = ∞ and limn→∞ A(h = α. Let K be an hn ∗ arbitrary hyperfinite integer and let H = hK . Let B = A[0, H]. Since α 6∈ Q, then
(2B)(0, H) ∼ B(0, H). This implies that for each a with 0 ≺ a 6 H, (2B)(0, a) ∼ B(0, a). By the transfer principle we have that for every hyperfinite x 6 H, Let
B(0,x) x+1
' β.
B(0, x) γ = inf st :0≺x6H . x+1
Then γ > β. If γ > 12 , then by the pigeonhole principle we have that [0, H] r U ⊆ 2B. This implies (2B)(0, H) ∼ H αH ∼ B(0, H) because α < 1. Hence we have a contradiction because (2B)(0, H) ∼ B(0, H). So we can assume γ 6 12 . 44
Choose an a with 0 ≺ a 6 H such that
B(0,a) a+1
/ 43 γ and let b = max x ∈ B : x 6 12 a .
If b ≺ a, then hai hai (2B)(0, ) B(0, b) + |b + B[0, − b]| 2 h i 2 h i a a ) + B(0, − b) B(0, 2i 2 ha ). B(0, 2 This contradicts (2B)(0, a2 ) ∼ B(0, a2 ). So we can assume b ∼ a2 . By Lemma 1.5, we have |B[0, b] + B[0, b]| min {3B(0, b), b + B(0, b)}. Then B(0, a) 4 γ' 3 a+1 (2B)(0, a) |B[0, b] + B[0, b]| ≈ ' a +1 2b + 1 3B(0, b) b + B(0, b) ' min , 2b + 1 2b + 1 3 3 1+γ γ, ' γ, ' min 2 2 2 which is absurd. This ends the proof.
2(Corollary 3.3)
If α ∈ Q, then the corollary is no longer true. The following is an example for showing that. Example 3.4 Let α ∈ Q and α = k2 . Note that α 6 12 . For each 0 6 β 6 α, we need ¯ ¯ = α, d(A) = β, and d(2A) to construct A such that d(A) = 32 α. Let B = {kn : n ∈ N} ∪ {1 + kn : n ∈ N} . ¯ Then d(B) = α. If β = 0, then let A=Br
[
2n
[22 , 22
2n+1
],
n∈N
and if 0 < β 6 α, then let [ β n n A=Br [ 22 , 22 ]. α n∈N ¯ ¯ It is easy to check that 0 ∈ A, gcd(A) = 1, d(A) = α, d(2A) = 32 α, and d(A) = β. 45
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