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AND LIOUVILLE FIELDS

K. SENTHIL KUMAR, R. THANGADURAI AND M. WALDSCHMIDT

Abstract. Following earlier work by ´E. Maillet 100 years ago, we introduce the definition of a Liouville set, which extends the definition of a Liouville number. We also define a Liouville field, which is a field generated by a Liouville set. Any Liouville number belongs to a Liouville setS having the power of continuum and such thatQSis a Liouville field.

Update: May 25, 2013

1. Introduction

For any integerq and any real numberx∈R, we denote by kqxk= min

m∈Z|qx−m|

the distance ofqxto the nearest integer. Following ´E. Maillet [3, 4], an irrational real numberξis said to be aLiouville numberif, for each integern≥1, there exists an integerqn≥2 such that the sequence un(ξ)

n≥1 of real numbers defined by un(ξ) =−logkqnξk

logqn

satisfies lim

n→∞un(ξ) = ∞. If pn is the integer such that kqnξk =|ξqn−pn|, then the definition ofun(ξ) can be written

|qnξ−pn|= 1 qunn(ξ)

·

An equivalent definition is to saying that a Liouville number is a real number ξ such that, for each integern≥1, there exists a rational numberpn/qn withqn ≥2 such that

0<

ξ−pn

qn

≤ 1 qnn·

We denote byL the set of Liouville numbers. Following [2], any Liouville number is transcendental.

We introduce the notions of aLiouville setand of aLiouville field. They extend what was done by ´E. Maillet in Chap. III of [3].

Definition. ALiouville setis a subsetSofLfor which there exists an increasing sequence (qn)n≥1 of positive integers having the following property: for anyξ∈S, there exists a sequence bn

n≥1 of positive rational integers and there exist two positive constantsκ1 andκ2 such that, for any sufficiently largen,

(1) 1≤bn≤qκn1 andkbnξk ≤ 1 qnκ2n

·

1

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It would not make a difference if we were requesting these inequalities to hold for anyn≥1: it suffices to change the constantsκ1 andκ2.

Definition. ALiouville field is a field of the formQ(S) whereSis a Liouville set.

From the definitions, it follows that, for a real numberξ, the following conditions are equivalent:

(i)ξis a Liouville number.

(ii)ξbelongs to some Liouville set.

(iii) The set{ξ}is a Liouville set.

(iv) The fieldQ(ξ) is a Liouville field.

If we agree that the empty set is a Liouville set and that Qis a Liouville field, then any subset of a Liouville set is a Liouville set, and also (see Theorem 1) any subfield of a Liouville field is a Liouville field.

Definition. Letq= (qn)n≥1be an increasing sequence of positive integers and let u= (un)n≥1be a sequence of positive real numbers such thatun→ ∞as n→ ∞.

Denote bySq,u the set ofξ∈Lsuch that there exist two positive constantsκ1and κ2and there exists a sequence bn

n≥1 of positive rational integers with 1≤bn≤qnκ1 andkbnξk ≤ 1

qκn2un

·

Denote byn the sequence u= (un)n≥1 := (1,2,3, . . .) withun =n(n≥1). For any increasing sequence q= (qn)n≥1 of positive integers, we denote by Sq the set Sq,n.

Hence, by definition, a Liouville set is a subset of someSq. In section 2 we prove the following lemma:

Lemma 1. For any increasing sequenceq of positive integers and any sequenceu of positive real numbers which tends to infinity, the setSq,u is a Liouville set.

Notice that if (mn)n≥1 is an increasing sequence of positive integers, then for the subsequenceq0 = (qmn)n≥1 of the sequenceq, we haveSq0,u⊃Sq,u.

Example. Letu= (un)n≥1 be a sequence of positive real numbers which tends to infinity. Definef :N→R>0byf(1) = 1 and

f(n) =u1u2· · ·un−1 (n≥2),

so that f(n+ 1)/f(n) = un for n ≥ 1. Define the sequence q = (qn)n≥1 by qn=b2f(n)c. Then, for any real numbert >1, the number

ξt=X

n≥1

1 btf(n)c

belongs toSq,u. The set{ξt | t >1}has the power of continuum, since ξt1 < ξt2 fort1> t2>1.

The setsSq,u have the following property (compare with Theorem I3 in [3]):

Theorem 1. For any increasing sequence q of positive integers and any sequence uof positive real numbers which tends to infinity, the setQ∪Sq,u is a field.

We denote this field byKq,u, and byKq for the sequenceu=n. From Theorem 1, it follows that a field is a Liouville field if and only if it is a subfield of someKq. Another consequence is that, ifSis a Liouville set, thenQ(S)\Qis a Liouville set.

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It is easily checked that if

lim inf

n→∞

un

u0n >0, thenKq,u is a subfield ofKq,u0. In particular if

lim inf

n→∞

un n >0, thenKq,u is a subfield ofKq, while if

lim sup

n→∞

un

n <+∞

thenKq is a subfield ofKq,u.

If R ∈ Q(X1, . . . , X`) is a rational fraction and if ξ1, . . . , ξ` are elements of a Liouville setSsuch thatη=R(ξ1, . . . , ξ`) is defined, then Theorem 1 implies that η is either a rational number or a Liouville number, and in the second caseS∪ {η}

is a Liouville set. For instance, if, in addition,R is not constant andξ1, . . . , ξ` are algebraically independent over Q, then η is a Liouville number and S∪ {η} is a Liouville set. For`= 1, this yields:

Corollary 1. LetR∈Q(X)be a rational fraction and letξbe a Liouville number.

ThenR(ξ) is a Liouville number and{ξ, R(ξ)} is a Liouville set.

We now show thatSq,u is either empty or else uncountable and we characterize such sets.

Theorem 2. Letq be an increasing sequence of positive integers andu= (un)n≥1 be an increasing sequence of positive real numbers such that un+1≥un+ 1. Then the Liouville set Sq,u is non empty if and only if

lim sup

n→∞

logqn+1 unlogqn

>0.

Moreover, if the setSq,u is non empty, then it has the power of continuum.

Lettbe an irrational real number which is not a Liouville number. By a result due to P. Erd˝os [1], we can writet=ξ+ηwith two Liouville numbersξandη. Let qbe an increasing sequence of positive integers andube an increasing sequence of real numbers such thatξ∈Sq,u. Since any irrational number in the fieldKq,u is in Sq,u, it follows that the Liouville numberη =t−ξdoes not belong toSq,u.

One defines a reflexive and symmetric relationRbetween two Liouville numbers byξRη if{ξ, η}is a Liouville set. The equivalence relation which is induced by R is trivial, as shown by the next result, which is a consequence of Theorem 2.

Corollary 2. Let ξandη be Liouville numbers. Then there exists a subset ϑofL having the power of continuum such that, for each such %∈ϑ, both sets{ξ, %}and {η, %} are Liouville sets.

In [3], ´E Maillet introduces the definition of Liouville numbers corresponding to a given Liouville number. However this definition depends on the choice of a given sequence q giving the rational approximations. This is why we start with a sequenceqinstead of starting with a given Liouville number.

The intersection of two nonempty Liouville sets maybe empty. More generally, we show that there are uncountably many Liouville sets Sq with pairwise empty intersections.

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Proposition 1. For0< τ <1, defineq(τ) as the sequence(q(τ)n )n≥1 with qn(τ)= 2n!bnτc (n≥1).

Then the sets Sq(τ), 0 < τ < 1, are nonempty (hence uncountable) and pairwise disjoint.

To prove that a real number is not a Liouville number is most often difficult.

But to prove that a given real number does not belong to some Liouville set S is easier. If q0 is a subsequence of a sequence q, one may expect that Sq0 may often contain strictlySq. Here is an example.

Proposition 2. Define the sequences q,q0 and q00 by

qn= 2n!, qn0 =q2n= 2(2n)! and qn00=q2n+1= 2(2n+1)! (n≥1), so thatqis the increasing sequence deduced from the union of q0 andq00. Letλn be a sequence of positive integers such that

n→∞lim λn=∞ and lim

n→∞

λn n = 0.

Then the number

ξ:=X

n≥1

1 2(2n−1)!λn belongs toSq0 but not to Sq. Moreover

Sq =Sq0∩Sq00.

When q is the increasing sequence deduced form the union of q0 and q00, we always have Sq ⊂ Sq0 ∩Sq00; Proposition 1 gives an example where Sq0 6= ∅ and Sq00 6=∅, whileSq is the empty set. In the example from Proposition 2, the set Sq coincides withSq0∩Sq00. This is not always the case.

Proposition 3. There exists two increasing sequencesq0andq00of positive integers with unionqsuch that Sq is a strict nonempty subset of Sq0∩Sq00.

Also, we prove that given any increasing sequenceq, there exists a subsequence q0 ofq such thatSq is a strict subset ofSq0. More generally, we prove

Proposition 4. Let u= (un)n≥1 be a sequence of positive real numbers such that for every n≥1, we have√

un+1 ≤un+ 1≤un+1. Then any increasing sequence qof positive integers has a subsequenceq0 for which Sq0,u strictly containsSq,u. In particular, for any increasing sequence q of positive integers has a subsequence q0 for whichSq0 is strictly contains Sq.

Proposition 5. The setsSq,u are notGδ subsets ofR. If they are non empty, then they are dense inR.

The proof of lemma 1 is given in section 2, the proof of Theorem 1 in section 3, the proof of Theorem 2 in section 4, the proof of Corollary 2 in section 5. The proofs of Propositions 1, 2, 3 and 4 are given in section 6 and the proof of Proposition 5 is given in section 7.

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2. Proof of lemma 1

Proof of Lemma 1. Givenqandu, define inductively a sequence of positive integers (mn)n≥1 as follows. Let m1 be the least integerm ≥1 such that um >1. Once m1, . . . , mn−1 are known, define mn as the least integer m > mn−1 for which um> n. Consider the subsequence q0 ofqdefined byqn0 =qmn. ThenSq,u ⊂Sq0,

henceSq,u is a Liouville set.

Remark 1. In the definition of a Liouville set, if assumption (1) is satisfied for some κ1, then it is also satisfied with κ1 replaced by any κ01 > κ1. Hence there is no loss of generality to assume κ1>1. Then, in this definition, one could add to (1) the conditionqn≤bn. Indeed, if, for somen, we havebn < qn, then we set

b0n= qn

bn

bn,

so that

qn≤b0n≤qn+bn≤2qn. Denote byan the nearest integer tobnξand set

a0n= qn

bn

an.

Then, forκ02< κ2 and, for sufficiently largen, we have b0nξ−a0n

= qn

bn

bnξ−an

≤ qn

qnκ2n

≤ 1 (qn)κ02n· Hence condition (1) can be replaced by

qn ≤bn≤qκn1 andkbnξk ≤ 1 qκn2n

.

Also, one deduces from Theorem 2, that the sequence bn

n≥1 is increasing for sufficiently largen. Note also that same way we can assume that

qn≤bn ≤qnκ1 andkbnξk ≤ 1 qnκ2un

.

3. Proof of Theorem1 We first prove the following:

Lemma 2. Let q be an increasing sequence of positive integers and u= (un)n≥1 be an increasing sequence of real numbers. Letξ∈Sq,u. Then 1/ξ∈Sq,u.

As a consequence, ifSis a Liouville set, then, for anyξ∈S, the setS∪ {1/ξ}is a Liouville set.

Proof of Lemma 2. Letq= (qn)n≥1 be an increasing sequence of positive integers such that, for sufficiently largen,

k bnξk ≤q−un n,

wherebn≤qκn1. Writekbnξk=|bnξ−an|withan∈Z. Sinceξ6∈Q, the sequence (|an|)n≥1 tends to infinity; in particular, for sufficiently large n, we have an 6= 0.

Writing

1 ξ −bn

an =−bn

ξan

ξ−an

bn

,

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one easily checks that, for sufficiently largen,

k |an−1k ≤ |an|−un/2 and 1≤ |an|< b2n≤qn1.

Proof of Theorem 1. Let us check that for ξ andξ0 in Q∪Sq,u, we have ξ−ξ0 ∈ Q∪Sq,u andξξ0∈Q∪Sq,u. Clearly, it suffices to check

(1) Forξin Sq,u andξ0 in Q, we haveξ−ξ0∈Sq,u andξξ0∈Sq,u. (2) Forξin Sq,u andξ0 in Sq,u withξ−ξ0 6∈Q, we haveξ−ξ0∈Sq,u. (3) Forξin Sq,u andξ0 in Sq,u withξξ06∈Q, we haveξξ0 ∈Sq,u.

The idea of the proof is as follows. When ξ ∈ Sq,u is approximated by an/bn

and whenξ0 =r/s∈Q, thenξ−ξ0 is approximated by (san−rbn)/bn andξξ0 by ran/sbn. When ξ∈ Sq,u is approximated by an/bn and ξ0 ∈Sq,u by a0n/b0n, then ξ−ξ0 is approximated by (anb0n−a0nbn)/bnb0n and ξξ0 byana0n/bnb0n. The proofs which follow amount to writing down carefully these simple observations.

Letξ00=ξ−ξ0 andξ=ξξ0. Then the sequence (a00n) and (b00n) are corresponding toξ00; Similarly (an) and (bn) corresponds toξ.

Here is the proof of (1). Letξ∈Sq,u andξ0 =r/s∈Q, withrandsinZ,s >0.

There are two constants κ1 and κ2 and there are sequences of rational integers an

n≥1 and bn

n≥1 such that

1≤bn ≤qnκ1 and 0<

bnξ−an

≤ 1

qnκ2un

· Let ˜κ1> κ1and ˜κ2< κ2. Then,

b00n =bn=sbn. a00n =san−rbn, an =ran.

Then one easily checks that, for sufficiently largen, we have 0<

b00nξ00−a00n =s

bnξ−an ≤ 1

qκn002un

,

0<

bnξ−an =|r|

bnξ−an ≤ 1

qκn2un

·

Here is the proof of (2) and (3). Let ξ and ξ0 be inSq,u. There are constants κ01, κ02 κ001 and κ002 and there are sequences of rational integers an

n≥1, bn

n≥1, a0n

n≥1 and b0n

n≥1 such that

1≤bn ≤qnκ01 and 0<

bnξ−an ≤ 1

qnκ02un

,

1≤b0n≤qκ

00

n1 and 0<

b0nξ0−a0n ≤ 1

qnκ002un

· Define ˜κ101001 and let ˜κ2>0 satisfy ˜κ2<min{κ02, κ002}. Set

b00n=bn=bnb0n, a00n=anb0n−bna0n, an=ana0n.

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Then for sufficiently largen, we have

b00nξ00−a00n=b0n bnξ−an

−bn b0nξ0−a0n and

bnξ−an=bnξ b0nξ0−a0n

+a0n bnξ−an , hence

b00nξ00−a00n ≤ 1

qκn˜2un

and

bnξ−an ≤ 1

qn˜κ2un

· Also we have

1≤b00n≤qκn˜1 and 1≤bn≤qnκ˜1.

The assumptionξ−ξ06∈Q(respξξ0 6∈Q) impliesb00nξ006=a00n (respectively,bnξ6=

an). Henceξ−ξ0 andξξ0 are inSq,u. This completes the proof of (2) and (3).

It follows from (1), (2) and (3) thatQ∪Sq,u is a ring.

Finally, ifξ∈Q∪Sq,u is not 0, then 1/ξ∈Q∪Sq,u, by Lemma 2. This completes

the proof of Theorem 1.

Remark 2. Since the field Kq,u does not contain irrational algebraic numbers, 2 is not a square in Kq,u. For ξ ∈ Sq,u, it follows that η = 2ξ2 is an element in Sq,u which is not the square of an element inSq,u. According to [1], we can write

√2 =ξ1ξ2with two Liouville numbersξ1, ξ2; then the set{ξ1, ξ2}is not a Liouville set.

LetNbe a positive integer such thatNcannot be written as a sum of two squares of an integer. Let us show that, for%∈Sq,u, the Liouville numberN %2∈Sq,uis not the sum of two squares of elements inSq,u. Dividing by%2, we are reduced to show that the equationN =ξ2+ (ξ0)2 has no solution (ξ, ξ0) inSq,u×Sq,u. Otherwise, we would have, for suitable positive constantsκ1 andκ2,

ξ−an

bn

≤ 1

qnκ2un+1

, 1≤bn ≤qnκ1,

ξ0−a0n b0n

≤ 1

qκn2un+1

, 1≤b0n ≤qnκ1, hence

ξ2−a2n b2n

≤ 2|ξ|+ 1 qκn2un+1

,

0)2−(a0n)2 (b0n)2

≤ 2|ξ0|+ 1 qκn2un+1

and

ξ2+ (ξ0)2− anb0n2

+ a0nbn2 bnb0n2

≤ 2(|ξ|+|ξ0|+ 1) qnκ2un+1

· Usingξ2+ (ξ0)2=N, we deduce

N bnb0n2

− anb0n2

− a0nbn2 <1.

The left hand side is an integer, hence it is 0:

N bnb0n2

= anb0n2

+ a0nbn2 .

This is impossible, since the equation x2+y2 =N z2 has no solution in positive rational integers.

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Therefore, if we writeN =ξ2+ (ξ0)2 with two Liouville numbersξ, ξ0, which is possible by the above mentioned result from P. Erd˝os [1], then the set{ξ, ξ0}is not a Liouville set.

4. Proof of Theorem2

We first prove the following lemma which will be required for the proof of part (ii) of Theorem 2.

Lemma 3. Let ξ be a real number,n, q andq0 be positive integers. Assume that there exist rational integerspandp0 such that p/q6=p0/q0 and

|qξ−p| ≤ 1

qun, |q0ξ−p0| ≤ 1 (q0)un+1· Then we have

either q0≥qun or q≥(q0)un. Proof of Lemma 3. From the assumptions we deduce

1

qq0 ≤ |pq0−p0q|

qq0

ξ−p q

+

ξ−p0 q0

≤ 1

qun+1 + 1 (q0)un+2, hence

qun(q0)un+1≤(q0)un+2+qun+1. Ifq < q0, we deduce

qun≤q0+ q

q0 un+1

< q0+ 1.

Assume nowq≥q0. Since the conclusion of Lemma 3 is trivial ifun = 1 and also ifq0= 1, we assumeun>1 andq0≥2. From

qun(q0)un+1≤(q0)un+2+qun+1≤(q0)2qun+qun+1 we deduce

(q0)un+1−(q0)2≤q.

From (q0)un−1>(q0)un−2 we deduce (q0)un−1≥(q0)un−2+ 1, which we write as (q0)un+1−(q0)2≥(q0)un.

Finally

(q0)un≤(q0)un+1−(q0)2≤q.

Proof of Theorem 2. Suppose lim sup

n→∞

logqn+1

unlogqn

= 0. Then, we get,

n→∞lim

logqn+1

unlogqn = 0.

Suppose Sq,u 6= ∅. Let ξ ∈ Sq,u. From Remark 1, it follows that there exists a sequence bn

n≥1of positive integers and there exist two positive constantsκ1and κ2such that, for any sufficiently largen,

qn≤bn ≤qnκ1 andkbnξk ≤q−κn 2un·

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Letn0be an integer≥κ1such that these inequalities are valid forn≥n0and such that, for n≥n0, qn+1κ1 < qnun (by the assumption). Since the sequence (qn)n≥1 is increasing, we haveqnκ1< qun+1n forn≥n0. From the choice ofn0we deduce

bn+1≤qκn+11 < qnun≤bunn and

bn≤qκn1 < qn+1un ≤bun+1n

for any n ≥ n0. Denote by an (resp. an+1) the nearest integer to ξbn (resp. to ξbn+1). Lemma 3 with q replaced by bn and q0 by bn+1 implies that for each n≥n0,

an

bn

=an+1

bn+1

·

This contradicts the assumption thatξis irrational. This proves that Sq,u=∅.

Conversely, assume

lim sup

n→∞

logqn+1 unlogqn

>0.

Then there exists ϑ > 0 and there exists a sequence (N`)`≥1 of positive integers such that

qN` > qϑ(uN N`−1)

`−1

for all`≥1. Define a sequence (c`)`≥1 of positive integers by 2c` ≤qN` <2c`+1.

Lete= (e`)`≥1 be a sequence of elements in{−1,1}. Define ξe=X

`≥1

e`

2c`·

It remains to check thatξe∈Sq,u and that distincteproduce distinctξe.

Letκ1= 1 and let κ2 be in the interval 0< κ2< ϑ. For sufficiently largen, let

`be the integer such thatN`−1≤n < N`. Set bn= 2c`−1, an =

`−1

X

h=1

eh2c`−1−ch, rn= an

bn· We have

1 2c` <

ξe−rn

=

ξe−X

h≥`

eh

2ch

≤ 2 2c`· Sinceκ2< ϑ,nis sufficiently large and n≤N`−1, we have

4qκn2un≤4qNκ2uN`−1

`−1 ≤qN`, hence

2 2c` < 4

qN` < 1 qnκ2un

for sufficiently largen. This provesξe∈Sq,u and henceSq,u is not empty.

Finally, ifeande0are two elements of{−1,+1}Nfor whicheh=e0hfor 1≤h < ` and, say,e`=−1,e0`= 1, then

ξe<

`−1

X

h=1

eh

2ch < ξe0,

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henceξe6=ξe0. This completes the proof of Theorem 2.

5. Proof of Corollary2

The proof of Corollary 2 as a consequence of Theorem 2 relies on the following elementary lemma.

Lemma 4. Let (an)n≥1 and(bn)n≥1 be two increasing sequences of positive inte- gers. Then there exists an increasing sequence of positive integers(qn)n≥1satisfying the following properties:

(i)The sequence(q2n)n≥1 is a subsequence of the sequence(an)n≥1. (ii)The sequence (q2n+1)n≥0 is a subsequence of the sequence(bn)n≥1. (iii)Forn≥1,qn+1≥qnn.

Proof of Lemma 4. We construct the sequence (qn)n≥1 inductively, starting with q1 =b1 and with q2 the least integerai satisfyingai ≥b1. Once qn is known for somen≥2, we take forqn+1 the least integer satisfying the following properties:

•qn+1∈ {a1, a2, . . .}ifnis odd,qn+1∈ {b1, b2, . . .} ifnis even.

•qn+1≥qnn.

Proof of Corollary2. Letξandη be Liouville numbers. There exist two sequences of positive integers (an)n≥1 and (bn)n≥1, which we may suppose to be increasing, such that

kanξk ≤a−nn and kbnηk ≤b−nn

for sufficiently largen. Letq= (qn)n≥1 be an increasing sequence of positive inte- gers satisfying the conclusion of Lemma 4. According to Theorem 2, the Liouville setSq is not empty. Let%∈Sq. Denote byq0 the subsequence (q2, q4, . . . , q2n, . . .) ofqand byq00the subsequence (q1, q3, . . . , q2n+1, . . .). We have%∈Sq =Sq0∩Sq00. Since the sequence (an)n≥1 is increasing, we haveq2n ≥an, henceξ ∈Sq0. Also, since the sequence (bn)n≥1 is increasing, we haveq2n+1 ≥bn, henceη ∈Sq00. Fi- nally,ξand%belong to the Liouville setSq0, whileη and%belong to the Liouville

setSq00.

6. Proofs of Propositions1,2,3 and 4

Proof of Proposition1. The fact that for 0 < τ < 1 the set Sq(τ) is not empty follows from Theorem 2, since

n→∞lim

logq(τ)n+1 nlogqn(τ)

= 1.

In fact, if (en)n≥1 is a bounded sequence of integers with infinitely many nonzero terms, then

X

n≥1

en

qn(τ)

∈Sq(τ).

Let 0< τ1< τ2<1. For n≥1, define

q2n=qn1)= 2n!bnτ1c and q2n+1=qn2)= 2n!bnτ2c. One easily checks that (qm)m≥1 is an increasing sequence with

logq2n+1

nlogq2n

→0 and logq2n+2

nlogq2n+1

→0.

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From Theorem 2 one deducesSq1 )∩Sq2 )=∅.

Proof of Proposition2. For sufficiently largen, define

an=

n

X

m=1

2(2n)!−(2m−1)!λm.

Then 1

q(2n+1)λ2n n+1

< ξ− an

q2n = X

m≥n+1

1

2(2m−1)!λm ≤ 2 q2n(2n+1)λn+1

· The right inequality with the lower boundλn+1≥1 proves thatξ∈Sq0.

Letκ1 andκ2 be positive numbers,na sufficiently large integer,san integer in the intervalq2n+1 ≤s≤qκ2n+11 and ran integer. Sinceλn+1 < κ2nfor sufficiently largen, we have

q2n(2n+1)λn+1 < q2nκ2n(2n+1)=q2n+1κ2n ≤sκ2n. Therefore, ifr/s=an/q2n, then

ξ−r

s =

ξ− an q2n

> 1 q2n(2n+1)λn+1

> 1 sκ2n· On the other hand, forr/s6=an/q2n, we have

ξ−r

s ≥

an

q2n −r s

ξ− an

q2n

≥ 1

q2ns− 2 q(2n+1)λ2n n+1

·

Sinceλn→ ∞, for sufficiently largenwe have

4q2ns≤4q2nq2n+1κ1 = 4q1+κ2n 1(2n+1)≤q2n(2n+1)λn+1 hence

2 q(2n+1)λ2n n+1

≤ 1 2q2ns· Further

2q2n< q2n+1< q2n+1κ2n−1≤sκ2n−1. Therefore

ξ−r

s ≥ 1

2q2ns > 1 sκ2n,

which shows thatξ6∈Sq00.

Proof of Proposition3. Let (λs)s≥0 be a strictly increasing sequence of positive rational integers withλ0= 1. Define two sequences (n0k)k≥1and (n00h)h≥1of positive integers as follows. The sequence (n0k)k≥1 is the increasing sequence of the positive integers n for which there exists s ≥ 0 with λ2s ≤ n < λ2s+1, while (n00h)h≥1 is the increasing sequence of the positive integersnfor which there existss≥0 with λ2s+1≤n < λ2s+2.

Fors≥0 andλ2s≤n < λ2s+1, set

k=n−λ2s2s−1−λ2s−2+· · ·+λ1. Thenn=n0k.

Fors≥0 andλ2s+1≤n < λ2s+2, set

h=n−λ2s+12s−λ2s−1+· · · −λ1+ 1.

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Thenn=n00h.

For instance, whenλs=s+ 1, the sequence (n0k)k≥1 is the sequence (1,3,5. . .) of odd positive integers, while (n00h)h≥1 is the sequence (2,4,6. . .) of even positive integers. Another example isλs=s!, which occurs in the paper [1] by Erd˝os.

In general, forn=λ2s, we writen=n0k(s) where

k(s) =λ2s−1−λ2s−2+· · ·+λ1< λ2s−1. Notice thatλ2s−1 =n00h withh=λ2s−k(s).

Next, define two increasing sequences (dn)n≥1and q= (qn)n≥1 of positive inte- gers by induction, withd1= 2,

dn+1 =

(kdn ifn=n0k, hdn ifn=n00h

for n ≥ 1 andqn = 2dn. Finally, let q0 = (qk0)k≥1 and q00 = (qh00)h≥1 be the two subsequences ofq defined by

q0k=qn0

k, k≥1, q00h=qn00

h, h≥1.

Henceq is the union of theses two subsequences. Now we check that the number ξ=X

n≥1

1 qn

belongs toSq0TSq00. Note that by Theorem 2 thatSq6=∅ asSq0 6=∅andSq00 6=∅.

Define

an=

n

X

m=1

2dn−dm. Then

1 qn+1

< ξ−an

qn

= X

m≥n+1

1 qm

< 2 qn+1

· Ifn=n0k, then

ξ−an0k

qk0

< 2 (q0k)k while ifn=n00h, then

ξ−an00

h

qh00

< 2 (qh00)h· This provesξ∈Sq0∩Sq00.

Now, we chooseλs= 22s fors≥2 and we prove thatξ does not belong toSq. Notice thatλ2s−1=√

λ2s. Letn=λ2s=n0k(s). We havek(s)<√ λ2sand

ξ−an

qn

> 1

qn+1 = 1 qk(s)n

> 1 q

n n

·

Let κ1 and κ2 be two positive real numbers and assume s is sufficiently large.

Further, letu/v∈Qwithv≤qnκ1. Ifu/v=an/qn, then

ξ−u

v =

ξ−an

qn

> 1 q

n n

> 1 qnκ2n

·

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On the other hand, ifu/v6=an/qn, then

ξ−u

v ≥

u v −an

qn

ξ−an

qn with

u v −an

qn

≥ 1 vqn

≥ 1 qnκ1+1

> 2 q

n n

and

ξ−an

qn

< 1 q

n n

· Hence

ξ−u

v > 1

q

n n

> 1 qκn2n

·

This proves Proposition 3.

Proof of Proposition4. Let u = (un)n≥1 be a sequence of positive real numbers such that√

un+1≤un+ 1≤un+1. We prove more precisely that for any sequence q such that qn+1 > qnun for alln ≥1, the sequence q0 = (q2m+1)m≥1 hasSq0,u 6=

Sq,u. This implies the proposition, since any increasing sequence has a subsequence satisfyingqn+1> qnun.

Assumingqn+1> qnun for alln≥1, we define dn=

(qn for evenn, qb

unc

n−1 for oddn.

We check that the number

ξ=X

n≥1

1 dn

satisfiesξ∈Sq0,uandξ6∈Sq,u. Setbn=d1d2· · ·dn and

an=

n

X

m=1

bn

dm

=

n

X

m=1

Y

1≤i≤n,i6=m

di,

so that

ξ−an bn

= X

m≥n+1

1 dm

·

It is easy to check from the definition of dn and qn that we have, for sufficiently largen,

bn ≤q1· · ·qn≤qun−1n−1qn≤q2n and

1 dn+1

≤ξ−an

bn

≤ 2 dn+1

· For oddn, sincedn+1=qn+1≥qunn, we deduce

ξ−an

bn

≤ 2 qunn

,

henceξ∈Sq0,u.

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For evenn, we plainly have

ξ−an

bn

> 1

dn+1 = 1 qb

un+1c n

·

Letκ1 andκ2 be two positive real numbers, and let nbe sufficiently large. Lets be a positive integer withs≤qnκ1 and letrbe an integer. Ifr/s=an/bn, then

ξ−r

s =

ξ−an bn

> 1 qκn2un

· Assume nowr/s6=an/bn. From

ξ−an

bn

≤ 2

qb

un+1c n

≤ 1 2qκn1+2

,

we deduce 1 qnκ1+2

≤ 1 sbn

≤ r s −an

bn

≤ ξ−r

s +

ξ−an

bn

≤ ξ−r

s + 1

2qκn1+2

,

hence

ξ−r

s ≥ 1

2qnκ1+2

> 1 qκn2un

·

This completes the proof thatξ6∈Sq,u.

7. Proof of Proposition 5

Proof of Proposition5. IfSq,u is non empty, let γ∈Sq,u. By Theorem 1,γ+Qis contained inSq,u, henceSq,u is dense inR.

Lett be an irrational real number which is not Liouville. Hence t 6∈Kq,u, and therefore, by Theorem 1,Sq,u∩(t+Sq,u) =∅. This implies thatSq,u is not a Gδ

dense subset ofR.

References

[1] P. Erd˝os,Representation of real numbers as sums and products of Liouville numbers, Michigan Math. J,9(1) (1962) 59–60.

[2] J. Liouville,Sur des classes tr`es ´etendues de quantit´es dont la valeur n’est ni alg´ebrique, ni eme r´eductible des irrationnelles alg´ebriques, J. Math. Pures et Appl.18(1844) 883–885, and 910–911.

[3] ´E. Maillet,Introduction la th´eorie des nombres transcendants et des propri´et´es arithm´etiques des fonctions, Paris, 1906.

[4] ´E. Maillet,Sur quelques propri´et´es des nombres transcendants de Liouville, Bull. S.M.F50 (1922), 74–99.

(K. Senthil Kumar and R. Thangadurai)Harish-Chandra Research Institute, Chhatnag Road, Jhunsi, Allahbad, 211019, India

(M. Waldschmidt)Institut de Math´ematiques de Jussieu, Th´eorie des Nombres Case courrier 247, Universit´e Pierre et Marie Curie (Paris 6), PARIS Cedex 05 FRANCE

E-mail address, K. Senthil Kumar: senthil@hri.res.in E-mail address, R. Thangadurai:thanga@hri.res.in E-mail address, M. Waldschmidt: miw@math.jussieu.fr

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