The Ubiquity of Sidon Sets That Are Not $I_0$

0.16 MB·

by  Kathryn E. Hare

6
1
0 The Ubiquity of Sidon sets that are not I
2 0

n
Kathryn E. Hare and L. Thomas Ramsey
a
J

6 Abstract. Weprovethateveryinfinite,discreteabeliangroupadmitsapair
2 of I0 sets whose union is not I0. In particular, this implies that every such
groupcontains aSidonsetthatisnotI0.
]
A
C
h. 1. Introduction
t
a A subset E of a discrete abelian group Γ with compact dual group G is said
m
to be a Sidon set if every bounded E-sequence can be interpolated by the Fourier
[ transform of a measure on G. If the measure can be chosen to be discrete, E is

called an I set. Finite sets in any group Γ and Hadamard sequences of integers
1 0
v are examples of these interpolation sets in the group Γ=Z.
1 Clearly, I0 sets are Sidon. However, the converse is not true since the class of
4 Sidon sets is known to be closed under finite unions, but the class of I sets is not.
0
2 Indeed,in[12],M´elagaveanexampleofapairofHadamardsetsinZwhoseunion
4
is not I .
0 0
. In this note we prove that every infinite, discrete abelian group admits a pair
2
of I sets whose union is not I . Consequently, every such group admits a Sidon
0 0 0
setthatisnotI . Ourmethodisconstructiveandestablishesevenmore: Weprove
6 0
1 that given any infinite subset F Γ there are I0 sets E F and E′ F +F F,
⊆ ⊆ ⊆ −
: whose union is not I (but is, of course, Sidon). In fact, we show that the sets
v 0
E, E′ have stronger interpolation properties than just I . These depend upon the
i 0
X algebraic properties of the initial set F.
r
a 2. Preliminaries
Let G be a compact abelian group and Γ its discrete abelian dual group. One
example is G= T, the circle group of complex numbers of modulus one, with dual
group Γ= Z.
Definition 1. (i) A subset E Γ is said to be Sidon if for every bounded
⊆
function φ : E C there is a measure µ on G with µ(γ) = φ(γ) for all γ E. If
→ ∈
the interpolating measure µ can always be chosen to be discrete, then the set E is
said to be I . b
0
2000 Mathematics Subject Classification. Primary43A46.
Key words and phrases. Sidonset,I0 set,Kroneckerset.
ThisresearchissupportedinpartbyNSERC#44597.
Thispaperisinfinalformandnoversionofitwillbesubmittedforpublicationelsewhere.
1
2 KATHRYNE.HAREANDL.THOMASRAMSEY
(ii) A subset E Γ is said to be ε-Kronecker if for every φ : E T there
⊆ →
exists x G such that
∈
(2.1) φ(γ) γ(x) <ε for all γ E
| − | ∈
and is called weak ε-Kronecker if the strict inequality is replaced by .
≤
HadamardsetsE = n N withHadamardratioq =infn /n >1areI
j j+1 j 0
{ }⊆
sets [11], thus both 3j and 3j +j are I subsets of Z. However, M´ela,
j≥1 j≥1 0
{ } { }
in [12], proved that their union is not I . This shows that not all Sidon sets in Z
0
are I since it is a deep result, first obtained by Drury [1], with a later proof given
0
by Pisier [13], that any finite union of Sidon sets (in any group Γ) is Sidon. It is
an open problem whether every Sidon set is a finite union of I sets.
0
Hadarmard sets with ratio q >2 are also known to be weak ε-Kronecker with
ε= 1 eiπ(q−1) . In fact, every ε-Kronecker set is I if ε<√2 [5] and is Sidon if
0
−
ε<2(cid:12) [8]. (cid:12)
(cid:12) (cid:12)
Itisknownthateveryinfinite subsetofΓcontainsinfinite I setsandifΓdoes
0
not contain any elements of order 2, then every infinite subset contains an infinite
weak 1-Kronecker set; c.f., [4], [6], [7] and the references cited there. Of course, if
Γ consists of only elements of order two, then it does not contain any ε-Kronecker
set with ε<√2.
The main result of this paper is to show that every infinite, discrete abelian
group admits a pair of I sets that are weak ε-Kroneckerfor suitable ε, but whose
0
union is not I . The number ε can always be chosen to be at most √2 and often
0
canbetakentobearbitrarilysmall,suchaswhenΓistorsionfree. Asfarasweare
aware,thisisthefirstproofthateveryinfinite,discreteabeliangroupadmitsSidon
sets that are not I . Knowing the existence of such sets is useful, for instance, in
0
studying the space ofweaklyalmostperiodic functions onΓ; c.f., [3, p. 17],as well
as [2] for further background.
Like M´ela’s original proof, our argument is constructive and relies upon the
well known Hartman/Ryll-Nardzewski characterization of I sets in terms of the
0
Bohr topology on Γ, the Bohr compactification of Γ.
Proposition 1. [9] A subset E Γ is I if and only if whenever E and E
0 1 2
⊆
are disjoint subsets of E, then E and E have disjoint closures in Γ.
1 2
With this, one can quickly prove the following generalization of M´ela’s argu-
ment.
Lemma 1. Suppose that 0 is a cluster point of χ ∞ Γ in the Bohr
{ n}n=1 ⊆
compactification of Γ. If the sets E = γ Γ and E′ = γ +χ Γ are
{ n} ⊆ { n n} ⊆
disjoint, then E E′ is not an I set.
0
∪
Proof. Assume the subnet χ (indexed by α) converges to 0 in Γ. Since
{ nα}
Γ is compact, there is a subnet (not relabelled) such that γ convergesto some
{ nα}
γ Γ. Since χ alsoconvergesto0alongthissubnet,itfollowsthat γ +χ
∈ { nα} { nα nα}
converges to γ. But that means the disjoint sets E and E′ both have γ in their
closures. By Proposition 1, their union is not I . (cid:3)
0
In the case of Γ = Z, one can take χ = N. For the more general situation,
∞ { n}
we will take χ = H , where the sets H are constructed in the following
{ n} m m
mS=1
lemma.
UNION PROPERTY 3
Lemma 2. If F is any infinite subset of Γ, then there is a countable subset
H (F F)(cid:31) 0 which has 0 as a cluster point in the Bohr topology. Indeed, we
⊆ − {∞}
can take H = H where for each positive integer m, H is a finite subset of
m m
mS=1
(F F)(cid:31) 0 having the property that for all x ,...,x G there is some γ H
1 m m
− { } ∈ ∈
with
1
sup γ(x ) 1 < .
j
| − | m
1≤j≤m
Proof. To begin, we claim that there is some φ Γ with the property that if
∈
V isanopenneighbourhoodofφ,thenV F isinfinite. Ifnot,thenforeveryφ Γ
∩ ∈
there is an open neighbourhood of φ, say U , such that U F is finite. As the
φ φ
∩
open sets U cover Γ and Γ is compact, we can choose a finite subcover, U J .
φ { φj}j=1
But that contradicts the assumption that F is infinite and hence provesthe claim.
In particular, if x ,...,x G, then
1 m
∈
1
V = γ Γ: sup γ(x ) φ(x ) <
j j
(cid:26) ∈ | − | 2m(cid:27)
j=1,...,m
is a neighbourhood of φ and hence must contain infinitely many elements from F.
Foreachsuchcollectionx ,...,x ,choosef =f inF suchthat f (x ) φ(x ) <
1 m 1 2 i j j
6 | − |
1/(2m) for all j = 1,...,m. Then f (x ) f (x ) < 1/m for i = 1,2 and all
1 j 2 j
| − |
j =1,...,m. Since f f F F Γ, and is therefore a continuous characteron
1 2
− ∈ − ⊆
G, there is a neighbourhood of (x ,...,x ) Gm, denoted W(f ,f ), such that if
1 m 1 2
∈
(y ,...,y ) W(f ,f ), then
1 m 1 2
∈
1
sup f (y ) f (y ) < .
1 j 2 j
| − | m
j=1,...,m
As Gm is compact,there are finitely many suchneighbourhoods, W(f(k),f(k)), for
1 2
k =1,...,N , so that Gm = NmW(f(k),f(k)). The set
m 1 2
kS=1
H = f(k) f(k) :k =1,...,N
m { 2 − 1 m}
meets the requirements of the lemma.
∞
TheobservationthatH = H clustersat0followsdirectlyfromthefact
m=1 m
that any neighbourhood of 0 iSn the Bohr topology contains a subset of the form
γ : sup γ(x ) 1 < 1/m for some positive integer m and x ,...,x
G{. j=1,...,m| j − | } 1 m (cid:3)∈
Beforeturning tothe detailsofthe proofofourmainresult,we listsomeother
elementaryfactsaboutKroneckerandI setswhichcanbe foundin[7]andwillbe
0
used in the proof of our main result.
Lemma 3. (i) Suppose π : Γ Λ is a homomorphism that is an injection on
→
E Γ. If π(E) is weak ε-Kronecker (or I ) as a subset of the group π(Γ), then E
0
⊆
is weak ε-Kronecker (resp., I ) as a subset of Γ.
0
(ii) Suppose Λ is a subgroup of Γ and that E is weak ε-Kronecker (or I ) as a
0
subset of the group Λ. Then E is also weak ε-Kronecker (resp., I ) as a subset of
0
Γ.
4 KATHRYNE.HAREANDL.THOMASRAMSEY
3. The Main Result
3.1. Statement and outline of the proof. Here is the statement of our
main result.
Theorem 1. If F is any infinite subset of a discrete abelian group Γ, there
are countable, disjoint sets E F and E′ F +F F such that both E and E′
⊆ ⊆ −
are I , but E E′ is not I . Furthermore, the sets E,E′ can be chosen to be weak
0 0
∪
ε-Kronecker for suitable ε √2.
≤
Remark1. Thechoiceofεwillbeclearfromtheproofanddependsonalgebraic
properties of F. As will be seen in the proof, in many situations ε can be chosen to
be arbitrarily small.
Since the union of any two Sidon sets is again Sidon, we immediately obtain
the following corollary.
Corollary 1. Every infinite, discrete abelian group admits a Sidon set that
is not I .
0
The remainder of the paper will be devoted to proving the theorem. Its proof
depends upon the general structure theory for abelian groups.
Theorem 2. (see [7, p. 165]) Given any discrete abelian group Γ, there is an
index set such that Γ is isomorphic to a subgroup of
I
Ω= Ω ,
α
αL∈I
where for each α either Ω = Q or there is a prime number p such that Ω =
α α α
C(p∞), the group of all pn-th roots of unity.
α α
Throughout the remainder of the paper π will denote the projection from Ω
α
onto the factor group Ω . Our proof of Theorem 1 will be constructive and will
α
depend on the following two cases:
Case 1: There is some index α such that π (F) is infinite. This case will
α
∈ I
be handled by Lemma 4 when Ω =Q and Lemma 5 when Ω =C(p∞). We will
α α α
seethat wecanevenarrangeforthe setsE andE′ to be ε-Kroneckerfor anygiven
ε>0. Making the choice with ε<√2 ensures E, E′ are both I .
0
Case 2: Otherwise, π (F) is finite for all indices α and then there must
α
∈ I
be an infinite subset such that for each α there is some λ F with
J ⊆I ∈ J ∈
π (λ) = 0. The existence of such an infinite subset of indices allows us to either
α
6
construct sets E, E′ that are weak ε-Kronecker for some ε < √2 (and hence I )
0
or to construct sets E, E′ that are both independent (and hence I ) and weak
0
√2-Kronecker. The choice of construction depends on the orders of the non-zero
characters π (λ). This argument can be found in Lemma 6.
α
In both cases, the two sets we construct will be disjoint and have the form
E = γ , E′ = γ +χ where χ clusters at 0. Thus the fact that E E′ is
{ n} { n n} { n} ∪
not I will follow immediately from Lemma 1.
0
We now turn to handling these two cases.
3.2. Proof of the Theorem in Case 1.
Lemma 4. Suppose there exists an index α such that π (F) is infinite and
α
∈I
Ω =Q. Givenanyε>0,thereareinfinitedisjoint setsE F andE′ F+F F
α
⊂ ⊂ −
such that E and E′ are weak ε-Kronecker and I , but E E′ is not I .
0 0
∪
UNION PROPERTY 5
Proof. Let H = ∞ H = χ ∞ (F F)(cid:31) 0 be a set that clusters
m=1 m { n}n=1 ⊆ − { }
at 0, constructed in LeSmma 2. Fix 0<ε<√2 and assume we can find a sequence
of characters λ π (F) such that:
n α
∈
(a) V = λ and V′ = λ +π (χ ) are weak ε-Kroneckersets in Ω ; and
{ n} { n α n } α
(b) For n 6= n′ we have λn 6= λn′, λn 6= λn′ +πa(χn′), and λn +πα(χn) 6=
λn′ +πa(χn′).
Then,foreachλ choosesomeγ F suchthatπ (γ )=λ . SetE = γ
n n ∈ α n n { n}⊆
F and E′ = γ +χ F +F F. By construction, π is one-to-one from E
{ n n} ⊆ − α
to V and one-to-one from E′ to V′. Condition (b) and the fact that H consists of
nonzerocharactersimplies thatE andE′ consistofdistincttermsandaredisjoint.
By Lemma 3(i), E and E′ inherit the weak ε-Kronecker property from V and
V′ respectively. Since ε<√2, both E and E′ are I . Furthermore, because 0 is a
0
cluster point of the set H, Lemma 1 implies that E E′ is not I .
0
∪
Thus the proof of the lemma will be complete if we can construct a sequence
of characters satisfying the two conditions (a) and (b). This will be an induction
argument which depends on whether π (F ) is a subset of a group isomorphic to
α
Z or it is not.
First, suppose that there is some integer bound B > 0 such that for all λ
∈
π (F), there are are integers b>0 and a such that π (λ)=a/b and b B. Then
α α
≤
π (F) is a subset of the (additive) subgroup 1 Z of Q. Because H F F, we
α B! ⊆ −
also have π (H) contained in this subgroup.
α
Given ε > 0, choose an integer q > 2 such that π/(q 1) < ε. As π (F) is
α
−
infinite, we may inductively choose λ π (F), sufficiently large in modulus, so
n α
∈
that both V and V′ are Hadamard sequences in Q with Hadamard ratio q and
≥
condition(b) is satisfied. By [7, Prop. 2.2.6], V and V′ are both weak ε-Kronecker
subsets of 1 Z and by Lemma 3(ii) they are also both weak ε-Kronecker sets in
B!
Ω . Thus condition (a) is satisfied.
α
Otherwise, for every positive integer B, there is some s/t π (F) with t>B
α
∈
and gcd(s,t) = 1. (We will say s/t is in reduced form.) In this case we need to
carefully account for the denominators of rational numbers. Note that any x Q
∈
has a unique reduced form, s/t, and we will write D(x) for the denominator t.
Given ε> 0, choose an integer q > 2 such that π/q < ε. Let B =D(π (χ ))
0 α 1
and choose λ π (F) such that D(λ ) > qB !. Assuming λ ,...,λ have been
1 α 1 0 1 n
∈
inductively constructed for n 1, let
≥
B =2max D(λ ), D(π (χ )): 1 i n, 1 j n+1 .
n { i α j ≤ ≤ ≤ ≤ }
Now choose λ π (F) so that D(λ ) > qB !. This choice ensures that λ
n+1 α n+1 n i
∈
and λ +π (χ ) for 1 i n, as well as π (χ ), all belong to 1 Z, while
i α i ≤ ≤ α n+1 Bn!
λ and λ +π (χ ) are outside 1 Z. It follows that condition (b) will be
n+1 n+1 α n+1 Bn!
satisfied for V and V′.
We argue next that V′ is ε-Kronecker in Ω . To this end, let φ : V′ T, say
α
→
φ(λ +π (χ )) = t T. We need to prove there is some character g Q, the
n α n n ∈ ∈
dual of Q, such that
b
(3.1) g(λ +π (χ )) t <ε for all n.
| n α n − n|
Asexplainedin[10,25.5],elementsofQcanbeidentifiedwithsequences ω T,
n
{ }⊂
subject to the constraints that ωn+1 =ω , with the understanding that g(1/n!)=
n+1 b n
6 KATHRYNE.HAREANDL.THOMASRAMSEY
ω . Clearly it will be sufficient to satisfy the consistency condition
n
(3.2) ωBn+1!/Bn! =ω ,
Bn+1 Bn
provided that for j / B , say B <j <B , one specifies
n n n+1
∈{ }
ω =ωBn+1!/j!.
j Bn+1
To start the specification of g, set ω = 1. This ensures that if k Z, then
B0 ∈
g(k)=ωkB0!, hence g(Z)=1. (This will be helpful in the next lemma as it allows
B0
us to interpret g as a characteron Q/Z.) Since π (χ )=s/B ! for some integer s,
α 1 0
it follows that
g(π (χ ))=ωs =1.
α 1 B0
By Equation (3.2) one may choose ω to be any J-th root of unity, where J =
B1
B !/B !, in other words, one can choose any integer K [0,J 1] and specify
1 0
∈ −
ω =e2πiK/J.
B1
Because λ 1 Z, the reduced form of λ is s/t with t dividing B !. Thus, with
1 ∈ B1! 1 1
y/z the reduced form of B !/t, we have
0
g(λ )= e2πiK/J sB1!/t =e2πiKsB0!/t =e2πiKsy/z.
1
(cid:16) (cid:17)
Sincet=D(λ )>qB !, the reducedformofB !/tis y/z withz >q. Bothsandy
1 0 0
are relatively prime to z, hence the exponential e2πisy/z is a primitive z-th root of
unity. By definition, B 2t and t > qB !, thus J =B !/B ! B > t z. This
1 0 1 0 1
≥ ≥ ≥
means we can choose any of the z-th roots of unity as the value for g(λ ). If we
1
make a choice that is closest to t , then the angular difference between g(λ ) and
1 1
t is at most π/z and thus
1
π π
g(λ +π (χ )) 1 = g(λ ) t < < <ε.
| 1 α 1 − | | 1 − 1| z q
We proceed to define ω inductively. Assume that for 1 j n we have
{ Bn} ≤ ≤
specifiedω sothat g(λ +π (χ )) t <ε. BythedefinitionofB weknowthat
Bj | j α j − j| n
π (χ )isin 1 Z. Thereforeg hasalreadybeenspecifiedatπ (χ ). However,
α n+1 Bn! α n+1
λ / 1 Z,butisinside 1 Z,sotheselectionofω willdetermineg(λ ).
n+1 ∈ Bn! Bn+1! Bn+1 n+1
By Equation (3.2), ω can be chosen to be any J-th root of ω , where
Bn+1 Bn
J = B !/B !. If we write eiθ for ω , then we are free to choose any integer
n+1 n Bn
K [0,J 1] and define
∈ −
ω =eiθ/J e2πiK/J.
Bn+1 ·
Let s/t be the reduced form of λ and y/z the reduced form for B !/t. Since t
n+1 n
divides B !, we have
n+1
g(λ )=ωsBn+1!/t =eiθsBn!/t e2πiKsBn!/t =eiθsBn!/t e2πiKsy/z.
n+1 Bn+1 · ·
As before we see that any of the z-th roots of unity can be used to help define
g(λ ) and we make the choice (of K) so that the corresponding z-th root of
n+1
unity differs in angle by at most π/q from
−1
eiθsBn!/t g(π (χ ))−1t .
α n+1 n+1
(cid:16) (cid:17)
UNION PROPERTY 7
(We remind the reader that the first two factors above are known as they have
already been determined by ω .) Therefore
Bn
−1
g(λ +π (χ )) t = e2πiKsy/z eiθsBn!/t g(π χ )−1t
(cid:12) n+1 α n+1 − n+1(cid:12) (cid:12)(cid:12) −(cid:16) (cid:17) α n+1 n+1(cid:12)(cid:12)
(cid:12) (cid:12) (cid:12)π π (cid:12)
<(cid:12) < <ǫ. (cid:12)
z q
Since the choice of g satisfies (3.1), it follows that V′ is ε-Kronecker.
TheproofthatV isε-Kroneckerissimilar,buteasier,asthefactorsg(π (χ ))
α n+1
are not present. This shows that condition (a) also holds and that completes the
proof of the Lemma. (cid:3)
Lemma 5. Suppose there exists an index α such that π (F) is infinite
α
∈ I
and Ω = C(p∞). Given any ε > 0, there are infinite disjoint sets E F and
α
⊂
E′ F +F F such that E, E′ are weak ε-Kronecker and I , but E E′ is not
0
⊂ − ∪
I .
0
Proof. LetH = χ (F F)(cid:31) 0 beacountablesetthatclustersat0,as
{ n}⊆ − { }
inthepreviouslemma. WeidentifyC(p∞)withasubgroupofQ/Z,sothatforevery
λ in the subgroupgeneratedby F there is some x Q such that π (λ)=x +Z.
λ α λ
∈
Because π (F) is infinite, the set of minimal denominators D(x ) : λ F
α λ
{ ∈ }
must be unbounded. The proof of the second part of Lemma 4 shows that given
0<ε<√2thereisasequenceV = x suchthatbothV andV′ = x +x
{ λn} { λn χn}
are ε-Kronecker sets in Q, with the interpolation being done by characters g Q
∈
such that g(Z)=1, and hence by characters on Q/Z. These can also be viewed as
b
characters on C(p∞) if the domain is suitably restricted. Of course, g(π (λ )) =
α n
g(x +Z) = g(x ) and g(π (λ +χ )) = g(x +x ). It follows that both
λn λn α n n λn χn
π (λ ) and π (λ +χ ) are ε-Kroneckersubsets of Ω .
{ α n } { α n n } α
The construction in the proof of the previous lemma also ensures that the
pullbacks, E = λ and E′ = λ +χ , are disjoint, have distinct terms and
{ n} { n n}
are ε-Kronecker in Γ. That their union is not I follows immediately from Lemma
0
1. (cid:3)
Remark 2. We note that similar arguments can be used to prove that under
the assumption that π (F) is infinite for some α there are infinite sets, E,E′, that
α
are I and have the property that for each ε>0 the sets E and E′ contain cofinite
0
subsets that are ε-Kronecker and whose union is not I . The latter statement is a
0
consequence of the fact that the union of an I set and a finite set is known to be
0
I (see [7, p.63]).
0
3.3. Proof of the Theorem in Case 2.
Lemma 6. Suppose π (F) is finite for all α , but that is infinite for
α q
∈ I I
some q 2, where
≥
= α : λ F s.t. π (λ) has order at least q .
q α
I { ∈I ∃ ∈ }
Let exp(iπ/q) 1 =ε . Thereareinfinitedisjoint setsE F andE′ F+F F
q
| − | ⊂ ⊂ −
such that E and E′ are both weak ε -Kronecker and I , but E E′ is not I .
q 0 0
∪
Proof. Again, let H = χ ∞ (F F)(cid:31) 0 be a countable set that
{ n}n=1 ⊆ − { }
clusters at 0.
8 KATHRYNE.HAREANDL.THOMASRAMSEY
First, assume is infinite for some q 3. Let β be chosen with the
Iq ≥ 1 ∈ I
property that π (χ ) = 0 and there is some λ F with π (λ ) having order
β1 1 1 ∈ β1 1
q. We can do this since there are only finitely many indices β with π (χ )= 0.
≥ β 1 6
Nowinductively chooseβ andλ F suchthat π (λ )has orderatleastq,
n ∈I n ∈ βn n
π (χ )=0 for all m n and π (λ )=0 for all m<n.
βn m ≤ βn m
SetΠequaltotheprojectionfromΩontoΛ= Ω . Then Π(λ ) and Π(λ +
β n n
n { } {
χn)} are sequences of distinct elements of ΛLsuch that Π(λm) 6= Π(λn +χn) if
m = n. Since χ = 0, the sets E = λ and E′ = λ + χ are disjoint.
6 n 6 { n} { n n}
MoreoverΠ(λ +χ )=π (λ )+ρ where π (ρ )=0 for all k n.
n n βn n n βk n ≥
We claim that Π(λ +χ ) is weak ε -Kronecker. To prove this, note that a
{ n n } q
character g on Λ is specified as g = g , with each g a character on Ω . Let
n n β
{ } n
M be the order of π (λ ). For any M -th root of unity, ω , there is a character
n β n n n
n
g such that g (π (λ ))=ω . Thus given t T, we may inductively specify
n n β n n n
n { }⊆
g so that
n
g(π (λ )) g(ρ )−1t exp(iπ/M ) 1 ε .
| βn n − n n|≤| n − |≤ q
Thus
g(Π(λ +χ )) t = g(π (λ ))g(ρ ) t ε ,
| n n − n| | βn n n − n|≤ q
which proves the claim.
The argument that Π(λ ) is weak ε -Kronecker is similar. By Lemma 3, E
n q
{ }
and E′ are also weak ε -Kronecker. As q 3, we have ε 1 and hence E and E′
q q
≥ ≤
are I . As before, their union is not I .
0 0
Otherwise, we can assume , but not , is infinite. Then there is a finite
2 3
I I
(possibly empty) set such that for all λ F and all α (cid:31) , either
J ⊆I ∈ ∈ I J
π (λ)=0orhasorder2. Repeattheconstructionasabove,butthis time withthe
α
additional requirement that β (cid:31) . As before, the sets E, E′ that arise from
n ∈ I J
the construction are disjoint and they are both weak √2-Kronecker. Moreover,
Π(E) and Π(E′) are both independent sets of elements of order 2 and one can
easily verify that such sets are I (c.f. [7, p. 66]). It follows from Lemma 3 that E
0
and E′ are both I sets, while their union is not. (cid:3)
0
Thesethree lemmascomplete the proofofthe maintheoremsincethe assump-
tion that F is infinite guarantees that either π (F) is infinite for some α, or there
α
are infinitely many indices α with π (F) not trivial and in that case is infinite
α q
I
for some q 2.
≥
References
[1] S.Drury,Surles ensembles de Sidon, C.R.Acad.Sci.Paris271A(1970), 162-163.
[2] M. Filali and J. Galindo, Approximable WAP and LUC-interpolation sets, Adv. in Math.
233(2013),87-114.
[3] M.FilaliandJ.Galindo,Interpolation setsandthe sizeof quotientsof functionspaces ona
locally compact group,preprint2016.
[4] J.GalindoandS.Hernandez, The concept of boundedness and the Bohr compactification of
a MAPabelian group,Fund.Math.15(1999),195-218.
[5] C.C.GrahamandK.E.Hare,ε-Kroneckerand I0 setsinabelian groups, I:arithmeticprop-
ertiesof ε-Kroneckersets,Math.Proc.Camb.Philo.Soc.140(2006),475-489.
[6] C.C.GrahamandK.E.Hare,Existenceoflargeε-KroneckersetsandFZI0(U)setsindiscrete
abeliangroups,ColloqMath. 127(2012),1-15.
[7] C.C. Graham and K.E. Hare, Interpolation and Sidon Sets for Compact Groups, Springer-
Verlag,NewYork,2013.
UNION PROPERTY 9
[8] K.E.HareandL.T.Ramsey,Therelationshipbetween ε-KroneckerandSidonsets,toappear
Can.Math.Bull.
[9] S.HartmanandC.Ryll-Nardzewski,Almost periodic extensionsof functions,Colloq.Math.
12(1964),23-39.
[10] E.HewittandK.A.Ross,Abstractharmonic analysis, volumeI,Springer-Verlag,NewYork,
1979.
[11] K.Kunen and W. Rudin, Lacunarity and the Bohr topology, Math. Proc. Camb. Phil.Soc.
126(1999),117-137.
[12] J.-F.M´ela,Surlesensemblesd’interpolationdeC.Ryll-NardzewskietdeS.Hartman,Studia
Math.(1968), 168-193.
[13] G.Pisier,Conditions d’entropie etcaracterisations arithmetiquedesensemblesde Sidon.In
Proc.Conf.onmoderntopicsinharmonicanalysis,Inst.deAltaMath,Torino,1982,911-941.
Dept. of Pure Mathematics,University of Waterloo, Waterloo, Ontario, Canada
N2L3G1
E-mail address: [email protected]
Dept. ofMathematics,University of Hawaii, Honolulu,HI., USA96822
E-mail address: [email protected]