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 thatQ∪Sis 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
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.
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.
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.
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
,
one easily checks that, for sufficiently largen,
k |an|ξ−1k ≤ |an|−un/2 and 1≤ |an|< b2n≤qn2κ1.
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 (a∗n) and (b∗n) 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 =b∗n=sbn. a00n =san−rbn, a∗n =ran.
Then one easily checks that, for sufficiently largen, we have 0<
b00nξ00−a00n =s
bnξ−an ≤ 1
qκn002un
,
0<
b∗nξ∗−a∗n =|r|
bnξ−an ≤ 1
qκn∗2un
·
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 ˜κ1=κ01+κ001 and let ˜κ2>0 satisfy ˜κ2<min{κ02, κ002}. Set
b00n=b∗n=bnb0n, a00n=anb0n−bna0n, a∗n=ana0n.
Then for sufficiently largen, we have
b00nξ00−a00n=b0n bnξ−an
−bn b0nξ0−a0n and
b∗nξ∗−a∗n=bnξ b0nξ0−a0n
+a0n bnξ−an , hence
b00nξ00−a00n ≤ 1
qκn˜2un
and
b∗nξ∗−a∗n ≤ 1
qn˜κ2un
· Also we have
1≤b00n≤qκn˜1 and 1≤b∗n≤qnκ˜1.
The assumptionξ−ξ06∈Q(respξξ0 6∈Q) impliesb00nξ006=a00n (respectively,b∗nξ∗6=
a∗n). 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
,
(ξ0)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.
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·
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,
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=qn(τ1)= 2n!bnτ1c and q2n+1=q(τn2)= 2n!bnτ2c. One easily checks that (qm)m≥1 is an increasing sequence with
logq2n+1
nlogq2n
→0 and logq2n+2
nlogq2n+1
→0.
From Theorem 2 one deducesSq(τ1 )∩Sq(τ2 )=∅.
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−λ2s+λ2s−1−λ2s−2+· · ·+λ1. Thenn=n0k.
Fors≥0 andλ2s+1≤n < λ2s+2, set
h=n−λ2s+1+λ2s−λ2s−1+· · · −λ1+ 1.
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
·
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.
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 mˆeme 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