Note on a problem of Ruzsa by

LXIX.2 (1995)

Note on a problem of Ruzsa

by

Norbert Hegyv´ ari (Budapest)

1. Introduction. Let B = {1 ≤ b ₁ < b ₂ < . . .} be an infinite sequence of integers. For any integer n we define the counting function of B up to n to be the number of elements of B not exceeding n; we denote it by B(n).

The lower asymptotic density dB and the upper asymptotic density dB are defined by

dB = liminf

n→∞ B(n)/n, dB = limsup

n→∞ B(n)/n.

If dB = dB, we say that B has asymptotic density dB, given by the common value.

In [3] I. Z. Ruzsa proved that if A = {1 ≤ a ₁ < a ₂ < . . .} is an infinite sequence of integers and if a n+1 ≤ 2a n for all but at most finitely many values of n, then P (A) has an asymptotic density, where P (A) is the set of all sums of the form P

ε _i a _i , ε _i = 0 or 1. Ruzsa conjectured that for every pair of numbers 0 ≤ α ≤ β ≤ 1 there exists A = {1 ≤ a 1 < a 2 < . . .}

for which d(P (A)) = α and d(P (A)) = β. He also mentioned that an easy argument shows the case β = 1.

In this paper we prove Ruzsa’s conjecture:

Theorem. Let 0 ≤ α ≤ β ≤ 1. Then there exists an A = {a ₁ < a ₂ < . . .}

such that

(1) d(P (A)) = α and d(P (A)) = β.

The finite version of this question may be the following: for which t is it possible to find a sequence a ₁ < . . . < a _n so that there are exactly t distinct integers of the form P _n

i=1 ε _i a _i , ε _i = 0 or 1. It was raised in [1] and solved in [2].

Acknowledgements. I would like to thank the referee for his helpful comments and suggestions.

Supported by CNRS Laboratoire de Math´ematiques Discr`etes, Marseille.

[113]

2. The construction. If α = β 6= 0 then it is easy to see that d(P (A))

= α for A = {[2 ⁿ /α] | n ∈ N}; if α = β = 0 then clearly d(P (A)) = 0 for A = {2 ²

| n ∈ N}. So assume that 0 ≤ α < β ≤ 1.

We use the following notation: A = {a ₁ < a ₂ < . . .}, A _n = {a ₁ <

. . . < a _n }; s _n = P _n

i=1 a _i ; % _n = |P (A _n )|/s _n ; p _n (x) = |P (A _n ) ∩ [1, x]|; τ _n = p n−1 (a n )/a n .

Let A = S _∞

i=0 B i , where the blocks B i will be determined by an iterative process.

First let us define the block B 0 . Let k 0 = max{[18β/α] + 8, [18β] + 2, 2/(β − α)} and let B ₀ = {a ₁ < . . . < a _k

}, where

a i =

 2 ⁱ if 0 ≤ i ≤ k 0 − 1,

min{x | (2 ^k

⁺¹ − 2)/(x + 2 ^k

) ≤ β} for i = k ₀ . Thus P (A k

) = [1, 2 ^k

− 1] ∪ [x + 1, x + 2 ^k

− 1] and so

% _k

= (2 ^k

⁺¹ − 1)/(x + 2 ^k

) ≤ β and an easy calculation shows that % k

≥ β − 1/k 0 .

Assume now that the blocks B 1 , . . . , B j−1 have been defined such that for each 1 ≤ m ≤ j − 1,

B _m = {a _N

1

< . . . < a _N

2

< . . . < a _N

3

}

where s _N

3

< a _N

1

with N ₁ ^(m) = N ₃ ^(m−1) + 1 and a _k

< a _N

with N ₁ (1) = k ₀ + 1. Furthermore, if k := k ₀ + m, then for every m, 1 ≤ m ≤ j, the following properties are true:

(2) α ≤ τ _N

1

≤ α + 1/k,

(3) % _N

2

> β/3,

(4) β − 1/k ≤ % _N

3

≤ β.

Our task is to define blocks B j , B j+1 , . . . so that the properties (2)–(4) remain valid for k = k ₀ + m, m ≥ j as well. We verify these parallel with the construction.

In the last section we prove that for x > s _k

,

(5) α ≤ |P (A) ∩ [1, x]|/x ≤ β.

Now we note that (2), (4) and (5) imply

d(P (A)) = α and d(P (A)) = β.

Indeed, by (2) and (4) we have

k→∞ lim |P (A _N

)|/a _N

= α, lim

k→∞ |P (A _N

)|/s _N

= β,

and by (5) we get (1).

3. Proof of the Theorem. Now we prepare the block B k .

We use the abbreviations N _i = N _i ^(k) for i = 1, 2, 3 and let N ₀ = N ₃ ^(k−1) . In the first step we make the sequence less dense. Let

(6) a _N

= max{y | y > s _N

, |P (A _N

)|/y ≥ α}.

Since 1/2 ^k

< 1/k 0 < β − α, y exists.

Since N 1 > k 0 + j, this definition implies

0 ≤ τ _N

− α = |P (A _N

)|/a _N

− α < |P (A _N

)|/a _N

− |P (A _N

)|/(a _N

+ 1)

= |P (A N

)|/{a N

(a N

+ 1) < α/a N

< 1/N 1 < 1/k, 1 ≤ k ₀ < k, showing (2).

In the next step we do two things: we “stabilize” the density of our sequence and then we make it more dense up to β/3.

Let M = a _N

. Let

(7) a N

+i = (i + 1)a N

for i = 1, . . . , M and if t := [a _N

/s _N

] ≥ 2 then let (8) a _N

_{+M +i} = (M + i + 1)a _N

+ s _N

for i = 1, . . . , t − 1. The elements defined in (7) stabilize the density and the ones defined in (8) will make the density close to β/3, which we now show.

Let N 2 = N 1 + M + t − 1. Then % N

≥ β/3. Indeed, if t < 2 then by (4), (7) and since k > k ₀ > 3/β + 1, we have

% N

> % N

−1 /2 = % N

/2 > (β − 1/(k − 1))/2 > β/3.

Let now t ≥ 2 and let M + t ≤ j ≤ ^{M +1} ₂

. Clearly P (A t ) = P (A t−1 ) ∪ {a _t + P (A _t−1 )} for every t ∈ N. Since a _N

> s _N

and by (7) we see that w ∈ P (A _N

) ∩ [ja _N

, (j + 1)a _N

] if and only if there exist v ∈ P (A _N

) and z, 1 ≤ z ≤ ^{M +1} ₂

, so that w = za N

+ v. So we have (9) |P (A _N

) ∩ [ja _N

, (j + 1)a _N

)| = t|P (A _N

)|

and by (8),

(10) s N

≤

 M + t + 2 2

 a N

. Furthermore, if ^{M +1} ₂

≤ j ≤ ^{M +t+1} ₂

then it is easy to check that

(11) P (A _N

) ∩ [ja _N

, (j + 1)a _N

] = ja _N

+ {us _N

+ P (A _N

) | 0 ≤ u ≤ t}.

Hence

2 ≤ t ≤ a N

/s N

= {|P (A N

)|/s N

}{a N

/|P (A N

)|} = % N

/τ N

< β/α so we get

(12) β > 2α.

Since M > a N

> N 1 > k 0 by (9), (10) and (12) we get (13) % _N

= |P (A N

)|/s N

≥

 M + 1 2



− (M + t)



t|P (A N

)|

 M + t + 2 2

 a N

≥

M +1 2

− (M + t)

M +t+2 2

(a _N

/s _N

− 1)|P (A _N

)|/a _N

≥ ((1 − 2t/M ) ² − 2/M )(|P (A N

)|/s N

− |P (A N

)|/a N

)

≥ {(1 − 2β/(αk ₀ )) ² − 2/k ₀ }{β − 1/k ₀ − α}

≥ (1 − 4β/(αk ₀ ) − 2/k ₀ )(β/2 − 1/k ₀ ) ≥ β/3.

For the last inequality we use k ₀ > 16β/α+8 and thus 1−4β/(αk ₀ )−2/k ₀ >

3/4; furthermore, k 0 > 18/β and thus β/2 − 1/k 0 > 4β/9. This proves (3).

In the next step we achieve that the sequence will be more dense, satis- fying (4). Let v = s N

. Let

(14) a _N

_+i = 2 ⁱ s _N

for i = 1, . . . , v. This definition implies that a _N

_+i > s _N

_+i−1 and so

(15) % _N

_+v = % _N

.

Write for short N = N ₂ + v; W = s _N and Y = [W min{1/2, β/% _N − 1}]

and L = s ² _N

. Let now

(16) K W (z) = |P (A N ) ∩ (P (A N ) + W − Y + z)|

for 0 ≤ z ≤ L.

Lemma. There exists a z ^∗ ∈ [0, L] such that K W (z ^∗ ) ≤ Y (% ² _N + 3/s N

).

P r o o f. Let K _W = P _L

z=0 K _W (z)/L. By (14) we have (17) |P (A _N ) ∩ [t, t + L]| < % _N L + s _N

= (% + 1/s _N

)L for t = 0, . . . , s N − L and

(18) |P (A _N ) ∩ [W − Y, W ]| < Y (% _N + 1/s _N

).

Write

H = P (A N ) ∩ [W − Y, W ], L z = W − Y + z + P (A N ).

Then by (17) and (18) we have K _W ≤

X L z=0

X

u∈H∩L

1 ≤ (% _N + 1/s _N

) ² · L · Y /L < (% ² _N + 3/s _N

)Y.

This implies that

K _W (z ^∗ ) := min

0≤z≤L K _W (z) ≤ K _W < (% ² _N + 3/s _N

)Y, which proves the lemma.

Let

(19) a N +1 = W − Y + z ^∗ .

Now we deduce a lower estimate for % _{N +1} . By the Lemma we get

% _{N +1} = 2W % N − K W (z ^∗ )

2W − Y + z ^∗ ≥ 2W % N − (% ² _N + 3/s N

)Y 2W − Y + z ^∗ (20)

≥ % _N + Y % _N (1 − % _N )/(2W − Y + z ^∗ )

− z ^∗ /(2W − Y ) − 3Y /(s N

(2W − Y )).

Let

ω N = z ^∗ /(2W − Y ) − 3Y /(s N

(2W − Y )).

Clearly

N →∞ lim ω _N = 0.

First case: Y = [W min{1/2, β/% _N − 1}] = [W (β/% _N − 1)]. Then by (20)

(21) % _{N +1}

≥ % N + W (β/% N −1)% N (1−% N )/{2W − W (β/% N −1) + z ^∗ } − (ω N + 1/W )

= % _N + (β − % _N )% _N (1 − % _N )/{(3% _N − β) + % _N z ^∗ /W } − (ω _N + 1/W ).

Since β/3 < % N < β ≤ 1 the relation (β − % N )% N (1 − % N )/{(3% N − β) +

% _N z ^∗ /W } > 0 holds. This implies that if W (and so N ) is large enough we have

(22) % _{N +1} > % _N .

Repeating the previous process we define by (14) and (19) the sequence a N +2 , a N +3 , . . . More precisely, let N ⁽¹⁾ = N + 1 and define a _N

by (14) and a _N

by (19) and if N ⁽¹⁾ , N ⁽²⁾ , . . . , N ^(2r) have been defined then let N ^(2r+1) = N ^(2r) + 1 and define a _N

by (14) and a _N

by (19). Then (22) yields that β/% _{N +1} − 1 < β/% _N so at each step of the iterative process described above we always fall in the first case. Since % _N

≤ β and also by (22) we conclude that lim i→∞ % _N

= λ exists and clearly λ ≤ β. Thus by (21) we get

λ ≥ λ + (β − λ)λ(1 − λ)/(3λ − β),

which implies λ = β. Hence there is an i ∈ N such that β − 1/k ≤ % _N

≤ β.

So choosing N ₃ = N ⁽ⁱ⁾ we get (4).

Second case: Y = [W/2]. Then by (20) we have

(23) % _{N +1} ≥ % _N + (W/2)% _N (1 − % _N )/(2W − W/2) + ω ⁰ _N ,

where lim ω ⁰ _N = 0. This implies that % _{N +1} ≥ % _N if W (and so N ) is large enough. Repeating the previous processes which are defined by (14) and (19) we see that lim _i→∞ % _N

= µ exists. By (23) we conclude that µ ≥ 1 thus there is an i ∈ N for which min{1/2, β/% _N

− 1} = β/% _N

− 1 so we can use case 1.

4. Proof of property (5). We divide the interval [s _k

, ∞) into the union

[s k

, ∞) = [

k≥k

[s _N

, s _N

3

).

We now prove by induction on k that if s _N

3

≤ x < s _N

for some k then (5) is true.

First, note that if we choose a _N

at each step (a _N

is the initial element of the block B k ) then since a _N

1

> s _N

1

−1 we infer that the “density” of A will not be affected in the interval [s _N

3

, s _N

3

) if we select further elements a _N

3

+1 , a _N

+2 , . . .

For x = a _k

by the definition of a _k

we get α ≤ p(a _k

)/a _k

≤ β.

Now let k > k 0 and assume that s _N

3

≤ x ≤ s _N

3

. We use the abbrevia- tions N i = N _i ^(k) , i = 1, 2, 3, and N 0 = N ₃ ^(k−1) again.

1. Let s _N

≤ x < a _N

. Since p _N

(x)/x is a decreasing function of x in this interval we have, by (6),

α ≤ p(a _N

)/a _N

≤ p(x)/x ≤ p _N

(s _N

)/s _N

≤ β.

2. Let a N

≤ x ≤ s N

and let ja N

≤ x < (j + 1)a N

for some 1 ≤ j ≤

M +t+1 2

. Let x ⁰ = x−ja _N

. By the definition of a _N

₊₁ , . . . , a _N

we conclude by (9) and (11) that

p _N

(x)/x = {j|P (A _N

)| + ε|P (A _N

)| + p _N

(x ⁰ )}/x, where ε = 0 if 1 ≤ j ≤ ^{M +1} ₂

and ε = t − 1 if ^{M +1} ₂

≤ j ≤ ^{M +t+1} ₂
(i.e. if t = [a _N

/s _N

] ≥ 2). The inductive hypothesis

a _N

α ≤ |P (A _N

)| ≤ a _N

β, αs _N

≤ p(s _N

) ≤ βs _N

, and αx ⁰ ≤ p _N

(x ⁰ ) ≤ βx ⁰ yield

p _N

(x)/x ≤ {jts _N

β + βx ⁰ }/x ≤ β{j(a _N

/s _N

)s _N

+ x ⁰ }/x = β

and

p N

(x)/x ≥ {ja N

α + αx ⁰ }/x = α{ja N

+ x ⁰ }/x = α.

3. s N

< x ≤ s N where N = N 2 + v was defined in (14). By (14), (24) [a N

+i , a N

+i+1 ] ∩ P (A N ) = a N

+i + P (A N

).

Thus if a _N

_+i ≤ x < 2a _N

_+i and x ⁰ = x − a _N

_+i then by the inductive hypothesis again and by (24),

(25) p N (x)/x ≤ {βa N

+i + x ⁰ β}/x = β and

(26) p _N (x)/x ≥ {αa _N

_+i + x ⁰ α}/x = α.

4. Finally, let x ∈ [a _{N +1} , s _{N +1} ]. Since a _{N +1} < s _N it follows that if x ≤ s _N then p _N (x)/x ≥ α. This implies that

(27) p _{N +1} (x)/x ≥ α

for every x ∈ [a _{N +1} , s _{N +1} ].

Now we only have to prove that p _{N +1} (x)/x ≤ β. If x ≥ W then by Y ≤ W (β/% N − 1) we have

p N +1 (x)/x ≤ {% N x + Y % N }/x = % N + Y % N /x

≤ % _N + Y % _N /W ≤ % _N + (β − % _N ) = β.

If a _{N +1} ≤ x < W then

(28) p N +1 (x)/x = % N x + (x − a N +1 )% N /x < % N + Y % N /W = β.

Now, to define a N +2 , a N +3 , . . . by (14) and (19) we can apply the same ideas as in items 3 and 4; in this way we conclude that (25)–(28) hold for every x with s _N ≤ x ≤ s _N

, so that (5) holds and this completes the proof of the Theorem.

References

[1] P. E r d ˝o s and E. S z e m e r´ed i, On sums and products of integer, in: Studies in Pure Mathematics, To the Memory of P. Tur´an, Akad´emiai Kiad´o, 1983, 213–218.

[2] N. H e g y v ´a r i, On two problems in the combinatorial number theory (in Hungarian), to appear.

[3] I. Z. R u z s a, The density of the set of sums, Acta Arith. 58 (1991), 169–172.

DEPARTMENT OF MATHEMATICS, ELTE TFK L. E ¨OTV ¨OS UNIVERSITY

MARK ´O U. 29.

1055 BUDAPEST, HUNGARY

Received on 15.6.1993

and in revised form on 28.2.1994 (2416)