Thin Subsets of Groups
For a group G and a natural number m; a subset A of G is called m-thin if, for each finite subset F of G; there exists a finite subset K of G such that |F g ∩ A| ≤ m for all g ∈ G \ K: We show that each m-thin subset of an Abelian group G of cardinality ℵ n ; n = 0, 1,… can be split into ≤ m n+1...
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| author | Protasov, I. V. Slobodianiuk, S. V. Протасов, І. В. Слободянюк, С. В. |
| author_facet | Protasov, I. V. Slobodianiuk, S. V. Протасов, І. В. Слободянюк, С. В. |
| author_sort | Protasov, I. V. |
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| datestamp_date | 2020-03-18T19:17:02Z |
| description | For a group G and a natural number m; a subset A of G is called m-thin if, for each finite subset F of G; there exists a finite subset K of G such that |F g ∩ A| ≤ m for all g ∈ G \ K: We show that each m-thin subset of an Abelian group G of cardinality ℵ n ; n = 0, 1,… can be split into ≤ m n+1 1-thin subsets. On the other hand, we construct a group G of cardinality ℵ ω and select a 2-thin subset of G which cannot be split into finitely many 1-thin subsets. |
| first_indexed | 2026-03-24T02:24:43Z |
| format | Article |
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UDC 512.5
I. V. Protasov, S. Slobodyanyuk (Kyiv Nat. Taras Shevchenko Univ.)
THIN SUBSETS OF GROUPS
ТОНКI ПIДМНОЖИНИ ГРУП
For a group G and a natural number m, a subset A of G is called m-thin if, for each finite subset F of G, there exists a
finite subset K of G such that |Fg ∩ A| 6 m for all g ∈ G \K. We show that each m-thin subset of an Abelian group
G of cardinality ℵn, n = 0, 1, . . . can be split into 6 mn+1 1-thin subsets. On the other hand, we construct a group G of
cardinality ℵω and select a 2-thin subset of G which cannot be split into finitely many 1-thin subsets.
Нехай G — група, m — натуральне число. Пiдмножина A ⊆ G називається m-тонкою, якщо для кожної скiнченної
пiдмножини F групи G знайдеться така скiнченна пiдмножина K, що |Fg ∩A| 6 m для всiх g ∈ G \K. Доведено,
що m-тонку пiдмножину абелевої групи G потужностi ℵn, n = 0, 1, . . . , можна розбити на 6 mn+1 1-тонких
пiдмножин. Побудовано групу G потужностi ℵω i 2-тонку пiдмножину G, яку не можна розбити на скiнченне число
1-тонких пiдмножин.
Let G be a group, κ and µ be cardinals, |G| > κ > ℵ0 and µ 6 κ, [G]<κ = {X ⊂ G : |X| < κ}.
We say that a subset A of G is (κ, µ)-thin if, for every F ∈ [G]<κ, there exists K ∈ [G]<κ such
that
|Fg ∩A| 6 µ
for each g ∈ G \K.
If κ is regular, then A is (κ, 1)-thin if (see Lemma 1) and only if, for each g ∈ G, g 6= e, e is the
identity of G, we have ∣∣{a ∈ A : ga ∈ A}
∣∣ < κ.
An (ℵ0, 1)-thin subset is called thin. For thin subsets, its modifications and applications see [1 – 7].
For m ∈ N, the (ℵ0,m)-thin subsets appeared in [3] under name m-thin in attempt to characterize
the ideal in the Boolean algebra of subsets of G generated by the family of thin subsets of G. If a
subset A of G is a union of m thin subsets, then A is m-thin. On the other hand, if G is countable
and A is m-thin then A can be partitioned into 6 m thin subsets. Thus, the ideal generated by thin
subsets of a countable group G coincides with the family of all m-thin, m ∈ N subsets of G. Does
this characterization remain true for all infinite groups? In other words, can every m-thin subset of an
uncountable group G be partitioned in m (finitely many) thin subsets? In this paper we give answer
to these questions.
The paper consists of 5 sections. In Section 1 we see that the thin subsets can be defined in the
much more general context of balleans, the counterparts of the uniform topological spaces. From this
point of view, a thin subset is a counterpart of a uniformly discrete subset of a uniform space. As
a corollary of some ballean statement (Theorem 1), we get that, for each infinite regular cardinal κ
and each m ∈ N, every (κ,m)-thin subset of a group G of cardinality κ can be partitioned into 6 m
(κ, 1)-thin subsets.
In Section 2 we show (Theorem 3) that, for every infinite regular cardinal κ,m ∈ N and n ∈ ω,
each (κ,m)-thin subset of an Abelian group G of cardinality κ+n can be partitioned into 6 mn+1
(κ, 1)-thin subsets. Here, κ+0 = κ, κ+(n+1) = (κn)+. In particular, everym-thin subset of an Abelian
c© I. V. PROTASOV, S. SLOBODYANYUK, 2013
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 9 1245
1246 I. V. PROTASOV, S. SLOBODYANYUK
group G of cardinality ℵn can be partitioned into 6 mn+1 thin subsets. Clearly, in this case the ideal
generated by thin subsets also coincides with the family of all m-thin subsets, m ∈ N. In Theorem 4
we describe (κ, µ)-thin groups that can be partitioned into µ (κ, 1)-thin subsets.
In Section 3 one can find two auxiliary combinatorial theorems (of independent interest!) on
coloring of the square G×G of a group G which will be used in the next section.
Answering a question from [3], G. Bergman constructed a group G of cardinality ℵ2 and a 2-thin
subset A of G which cannot be partitioned into two thin subsets. With kind permission of the author,
we reprint in Section 4 his letter with this remarkable construction (Example 1). Then we modify the
Bergman’s construction to show (Example 2) that for each natural number m > 2 there exist a group
Gn of cardinality ℵn, n =
m(m+ 1)
2
− 1, and a 2-thin subset A of G which cannot be partitioned
into m-thin subsets. And finally (Example 3), we point out a group G of cardinality ℵω and a 2-thin
subset of G which cannot be finitely partitioned into thin subsets.
We conclude the paper with some observations on interplay between thin subsets and ultrafilters
in Section 5.
1. Ballean context. A ball structure is a triple B = (X,P,B), where X, P are non empty sets
and, for any x ∈ X and α ∈ P, B(x, α) is a subset of X which is called a ball of radius α around
x. It is supposed that x ∈ B(x, α) for all x ∈ X and α ∈ P. The set X is called the support of B, P
is called the set of radii. Given any x ∈ X,A ⊆ X,α ∈ P we put
B∗(x, α) =
{
y ∈ X : x ∈ B(y, α)
}
, B(A,α) =
⋃
a∈A
B(a, α).
Following [8], we say that a ball structure B = (X,P,B) is a ballean if
for any α, β ∈ P, there exist α′, β′ such that, for every x ∈ X,
B(x, α) ⊆ B∗(x, α′), B∗(x, β) ⊆ B(x, β′);
for any α, β ∈ P, there exists γ ∈ P such that, for every x ∈ X,
B
(
B(x, α), β
)
⊆ B(x, γ).
We note that a ballean can also be defined in terms of entourages of diagonal in X ×X. In this
case it is called a coarse structure [9].
A ballean B is called connected if, for any x, y ∈ X, there exists α ∈ P such that y ∈ B(x, α).
All balleans under consideration are supposed to be connected. Replacing each ball B(x, α) to
B(x, α) ∩ B∗(x, α), we may suppose that B(x, α) = B∗(x, α) for all x ∈ X,α ∈ P. A subset
Y ⊆ X is called bounded if there exist x ∈ X and α ∈ P such that Y ⊆ B(x, α).
We use a preordering 6 on the set P defined by the rule: α 6 β if and only if B(x, α) ⊆ B(x, β)
for every x ∈ X. A subset P ′ ⊆ P is called cofinal if, for every α ∈ P, there exists α′ ∈ P ′ such
that α 6 α′. A ballean B is called ordinal if there exists a cofinal subset P ′ ⊆ P well ordered by 6 .
Let B = (X,P,B) be a ballean, µ be a cardinal. We say that a subset A ⊆ X is µ-thin if, for
every α ∈ P, there exists a bounded subset Y ⊆ X such that
∣∣B(x, α)∩A
∣∣ 6 µ for every x ∈ G\Y.
A 1-thin subset is called thin.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 9
THIN SUBSETS OF GROUPS 1247
Lemma 1. Let B = (X,P,B) be a ballean, µ be a cardinal. A subset A ⊆ X is µ-thin if and
only if the set {
a ∈ A : |B(a, α) ∩A| > µ
}
is bounded.
Proof. The “if” part is evident. To verify the “only if ”, we take an arbitrary α ∈ P and choose
β ∈ P such that B
(
B(x, α), α
)
⊆ B(x, β) for each x ∈ X. By the assumption, the set Y =
{
a ∈
∈ A :
∣∣B(a, α) ∩ A
∣∣ > µ
}
is bounded. We put Z = B(Y, α) and take an arbitrary x ∈ X \ Z. If∣∣B(x, α)∩A
∣∣ > µ and a ∈ B(x, α)∩A then
∣∣B(a, β)∩A
∣∣ > µ because B(x, α) ⊆ B(a, β). Hence,
a ∈ Y and x ∈ Z. This contradiction shows that
∣∣B(x, α) ∩A
∣∣ 6 µ and A is µ-thin.
Lemma 1 is proved.
Theorem 1. Let B = (X,P,B) be a ballean, µ be a cardinal, A ⊆ X. Then the following
statements hold;
(i) if A is a union of µ thin subsets and a union of µ bounded subsets of X is bounded, then A
is µ-thin;
(ii) if B is ordinal and A is µ-thin, µ ∈ N, then A can be partitioned into 6 µ thin subsets.
Proof. (i) Let A =
⋃
λ6µAλ and each Aλ is thin, α ∈ P. For each λ 6 µ, we pick a bounded
subset Yλ such that
∣∣B(x, α)∩A
∣∣ 6 1 for each x ∈ X\Yλ. We put Y =
⋃
λ6µ Yλ. By the assumption,
Y is bounded. Clearly,
∣∣B(x, α) ∩A
∣∣ 6 µ for each x ∈ X \ Y so A is µ-thin.
(ii) Apply Lemma 1 and [3] (Theorem 1.2).
Theorem 1 is proved.
Theorem 2. Let G be a group, κ be an infinite regular cardinal, |G| = κ. Then the following
statements hold:
(i) each (κ,m)-thin subset A of G, m ∈ N is a union of 6 m (κ, 1)-thin subsets;
(ii) the ideal generated by the family of (κ, 1)-thin subsets coincides with the family of all (κ,m)-
thin subsets, m ∈ N.
Proof. Clearly, (ii) follows from (i). To prove (i), we consider a ballean B(G, κ) =
(
G, [G]<κ, B
)
,
where B(x, F ) = Fx ∪ {x} for all x ∈ G,F ∈ [G]<κ. We enumerate G = {gα : α < κ} and put
Fα = {gβ : β < α}. Since κ is regular, {Fα : α < κ} is cofinal in [G]<κ so B(G, κ) is ordinal. To
apply Theorem 1 (i), it suffices to note that A is (κ,m)-thin if and only if A is m-thin in the ballean
B(G, κ).
Theorem 2 is proved.
In view of [8] (Chapter 1), the balleans can be considered as asymptotic counterparts of the
uniform spaces. For uniform spaces see [10] (Chapter 8). Now we describe the uniform counterparts
of thin subsets.
Let U be a uniformity on a set X. For an entourage U ∈ U and x ∈ X, we put U(x) =
{
y ∈
∈ X : (x, y) ∈ U
}
. Let A be a subset of X, µ be a cardinal. We say that A is (U , µ)-discrete if there
exists U ∈ U such that |U(x)∩A| 6 µ for each x ∈ X, a (U , 1)-discrete subset is called U-discrete.
We show that each (U , µ)-discrete subset A of X can be partitioned into 6 µ U-discrete subsets.
We fix an entourage U ∈ U such that
∣∣U(x) ∩ A
∣∣ 6 µ for each x ∈ X and choose a symmetric
entourage V ∈ U such that V 2 ⊆ U. Then we consider a graph Γ with the set of vertices A and the
set of edges E defined by the rule: (x, y) ∈ E if and only if there exists z ∈ X such that x, y ∈ V (z).
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 9
1248 I. V. PROTASOV, S. SLOBODYANYUK
Since |U(x)∩A| 6 µ for each x ∈ X and V 2 ⊆ U, each unit ball in Γ is of cardinality 6 µ. Hence,
the chromatic number χ(Γ) does not exceed µ. We take a partition P of A such that |P| = χ(Γ) and
each P ∈ P has no incident vertices. If x ∈ P then V (x)∩P = {x}. It follows that P is U-discrete.
2. Partitions.
Lemma 2. Let G be a group, κ be an infinite cardinal, κ 6 |G|, m ∈ N. Let A be a (κ,m)-thin
subset of G, S ⊆ G, |S| > κ. Then there exists a subgroup H of G such that S ⊆ H, |H| = |S| and∣∣Hx ∩A∣∣ 6 m for each x ∈ G \H. In particular, A is (κ′,m)-thin for each κ′ > κ, κ′ 6 |G|.
Proof. We may suppose that S is a subgroup. LetH0 = S, [H0]
m+1 =
{
X ⊂ H0 : |X| = m+1
}
,
|S| = κ′. For each X ∈ [H0]
m+1, we choose K0(X) ∈ [G]<κ such that |Xg ∩ A| 6 m for
every g ∈ G \ K0(X). We put K0 = ∪
{
K0(X) : X ∈ [H0]
m+1
}
and note that |K0| 6 κ′ and
|H0x ∩A| 6 m for each x ∈ G \K0.
We consider a subgroup H1 of G generated by H0 ∪ K0. Clearly, |H1| = κ′. For each X ∈
∈ [H1]
m+1, we take K1(X) ∈ [G]<κ such that
∣∣Xg ∩ A∣∣ 6 m for every g ∈ G \K1(X). We put
K1 = ∪
{
K1(X) : X ∈ [H1]
m+1
}
and note that
∣∣H1x ∩A
∣∣ 6 m for each x ∈ G \K1.
After ω steps we get an increasing sequence of {Hn : n ∈ ω} of subgroups of G and a sequence
{Kn : n ∈ ω} of subsets of G such that |Hn| = κ′, |Kn| 6 κ′. Since ∪n∈ωKn ⊆ ∪n∈ωHn, for the
subgroup H = ∪n∈ωHn we get a desired statement.
To show that A is (κ′,m)-thin, we take S ∈ [G]<κ
′
. If |S| < κ then there exists K ∈ [G]<κ such
that
∣∣Sx ∩ A∣∣ 6 m for each x ∈ G \K because A is (κ,m)-thin. If |S| > κ, we apply the previous
statement.
Lemma 2 is proved.
For a cardinal κ and n ∈ ω, we use the following notations from [11]: κ+0 = κ, κ+(n+1) =
= (κ+n)+. In particular, ℵ+n0 = ℵn for each n ∈ ω.
Theorem 3. Let κ be an infinite regular cardinal, m ∈ N, n ∈ ω, G be an Abelian group of
cardinality κ+n. Each (κ,m)-thin subset A of G can be partitioned into 6 mn+1 (κ, 1)-thin subsets.
Proof. We use an induction by n. For n = 0, apply Theorem 2. Let A be a (κ,m)-thin subset of
G and |G| = κ+(n+1). By Lemma 2, A is
(
κ+(n+1),m
)
-thin. Applying Theorem 2, we can partition
A in 6 m
(
κ+(n+1), 1
)
-thin subsets. We suppose that A itself is (κ+(n+1), 1)-thin and show that A
can be partitioned in 6 mn+1 (κ, 1)-thin subsets.
Since A is
(
κ+(n+1), 1
)
-thin, we use Lemma 2 to write G as a union of increasing chain of
subgroups
{
Hα : α < κ+(n+1)
}
such that H0 = {e}, |Hα| = κ+n and
∣∣Hαx ∩ A
∣∣ 6 1 for all
x ∈ G \ Hα, α > 0, Hβ = ∪α<βHα for each limit ordinal β < κ+(n+1). Clearly, G \ {e} =
= ∪α<κ+(n+1)Hα+1 \Hα.
For each α < κ+(n+1), α > 0, we put Aα = A ∩Hα. Since Aα is (κ,m)-thin and |Hα| = κ+n,
by the inductive assumption, each Aα can be partitioned in kα 6 mn+1 (κ, 1)-thin subsets of Hα.
Admitting empty sets of the partition, we suppose that kα = mn+1 for each α < κ+(n+1) and write
Aα = Aα(1) ∪ . . . ∪Aα(mn+1),
where each Aα(i) is (κ, 1)-thin.
For all α < κ+(n+1) and i ∈
{
1, . . . ,mn+1
}
, we put
Bα(i) = Aα+1(i) \Aα(i), Bi =
⋃
α<κ+(n+1)
Bα(i).
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 9
THIN SUBSETS OF GROUPS 1249
Since A \ {e} = ∪
{
Bi : i ∈
{
1, . . . ,mn+1
}}
, it suffices to verify that each subset Bi is (κ, 1)-
thin. It turns out, since κ is regular, in view of Lemma 1, it suffices to show that, for each g ∈ G,
g 6= e, ∣∣{x ∈ Bi : gx ∈ Bi}∣∣ < κ.
We take the minimal α < κ+(n+1) such that g ∈ Hα+1 \Hα. If x ∈ Bi \Hα+1, then gx /∈ Bi by
the choice of Hα+1. Since Aα+1(i) is (κ, 1)-thin,
∣∣{x ∈ Aα+1(i)\Aα(i) : gx ∈ Aα+1(i)\Aα(i)}
∣∣ <
< κ. If x, y ∈ Aα+1(i) \ Aα(i) and gx, gy ∈ Aα(i), then x−1y ∈ Aα(i). Since G is Abelian,
(x−1yx) = y and, by the choice of Hα, x = y. If x, y ∈ Aα(i) and gx, gy ∈ Aα+1(i) \ Aα(i),
replacing g to g−1, we get the previous case.
Theorem 3 is proved.
Given an infinite group G and infinite cardinal κ, κ 6 |G|, we denote by µ(G, κ) the minimal
cardinal µ such that G can be partitioned in µ κ-thin subsets. For a cardinal γ, cf γ is a cofinality of
γ, γ+ is the cardinal successor of γ. By [5],
µ(G, κ) =
γ, if |G| is non-limit cardinal and |G| = γ+;
|G|, if |G| is a limit cardinal and either
κ < |G| or |G| is regular,
cf |G|, if |G| is singular, κ = |G| and cf |G|
is a limit cardinal.
If |G| is singular, κ = |G| and cf |G| is a non-limit cardinal, cf |G| = γ+, then µ(G, κ) ∈
∈ {γ, γ+}.
Now let γ be a cardinal, γ 6 κ. Then G is (κ, γ)-thin if and only if κ = γ+. Applying above
formulae for µ(G, κ), we get the following statement.
Theorem 4. Let G be a group, γ be an infinite cardinal, κ = γ+, |G| > γ. Then G can be
partitioned in γ (κ, 1)-thin subsets if and only if |G| = κ.
3. Colorings. For a group G and g ∈ G, we say that {G} × {g} is a horizontal line in G ×G,
{g} ×G is a vertical line in G×G,
{
(x, gx) : x ∈ G
}
is a diagonal in G×G.
The statement (i) in the following theorem was proved by G. Bergman, (ii) by the first author.
Theorem 5. For a group G with the identity e, the following statements hold:
(i) if |G| > ℵ2 and χ : G × G → {1, 2, 3}, then there is g ∈ G, g 6= e such that either some
horizontal line G × {g} has infinitely many points of color 1, or some vertical line {g} × G has
infinitely many points of color 2, or some diagonal
{
(x, gx) : x ∈ G
}
has infinitely many points of
color 3;
(ii) if |G| 6 ℵ1, then there is a coloring χ : G×G→ {1, 2, 3} such that each horizontal line has
only finite number of points of color 1, each vertical line has only finite number of points of color 2,
each diagonal has only finite number of points of color 3.
Proof. (i) We suppose the contrary and fix a corresponding coloring χ : G×G→ {1, 2, 3}. Let
G0, G1 be subgroups of G such that G0 ⊂ G1, |G0| = ℵ0, |G1| = ℵ1. Since the set G × G1 has at
most ℵ1 points of color 1, there is g ∈ G, g 6= e such that gG1×G1 has no points of color 1. The set
gG0 ×G1 has at most ℵ0 points of color 2, so there is h ∈ G1 , h 6= g, h 6= e such that gG0 × hG0
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 9
1250 I. V. PROTASOV, S. SLOBODYANYUK
has no points of color 2. Hence, the set
{
(gx, hx) : x ∈ G0
}
consists of color 3 and is contained in
the diagonal
{
(y, g−1hy) : y ∈ G
}
.
(ii) We proceed in three steps.
Step 1. Let X be a countable set, {An : n ∈ ω}, {Bn : n ∈ ω} be partitions of X such that
An∩Bm is finite for all n,m ∈ ω. Then there is a coloring χ : X → {1, 2} such that each subset An
has only finite number of elements of color 1 and each subset Bn has only finite number of elements
of color 2.
We color A0 in 2, B0 \ A0 in 1, A1 \ B0 in 2, B1 \ (A0 ∪ A1) in 1, A2 \ (B0 ∪ B1) in 2,
B2 \ (A0 ∪A1 ∪A2) in 1, and so on.
Step 2. Let H be a countable group, K be a subgroup of H. Applying Step 1, we define a
coloring χ :
(
(H ×H) \ (K ×K)
)
→ {1, 2, 3} such that
χ
(
(H \K) × (H \K)
)
= {1, 2} and each horizontal line in this set has only finite number of
points of color 1, and each vertical line has only finite number of points of color 2;
χ
(
K × (H \K)
)
= {1, 3} and each horizontal line in this set has only finite number of points
of color 1, and each diagonal has only finite number of points of color 3;
χ
(
(H \K) ×K
)
= {2, 3} and each vertical line in this set has only finite number of points of
color 2, and each diagonal has only finite number of points of color 3.
Step 3. To prove (ii), we may suppose that |G| = ℵ1 so write G as a union of an increasing chain
{Gα : α < ω1}, G0 = {e}, e is the identity of G, such that Gβ =
⋃
α<β Gα for each limit ordinal
β < ω1. We put χ0(e) = 1 and, for each α < ω1 use a coloring χα
(
(Gα+1 ×Gα+1) \ (Gα ×Gα)
)
defined at Step 2. We put χ =
⋃
α<ω1
χα and verify that χ : G×G→ {1, 2, 3} is thin.
Clearly, (G× {e}) ∩ χ−1(1) ⊆ G1 × {e}, ({e} ×G) ∩ χ−1(2) ⊆ {e} ×G1. If g ∈ Gα+1 \Gα
then (
G× {g}
)
∩ χ−1(1) ⊆ Gα+1 × {g},
(
{g} ×G
)
∩ χ−1(2) ⊆ {g} ×Gα+1.
Thus, each horizontal line in G × G has only finite number of points of color 1, and each vertical
line in G×G has only finite number of points of color 2.
At last, if (x, y) ∈ (Gα+1 \Gα)×Gα or (x, y) ∈ Gα × (Gα+1 \Gα) then x−1y ∈ Gα+1 \Gα.
It follows that each diagonal has only finite number of points of color 3.
Theorem 5 is proved.
In [12, 13] R. Davies proved the following theorem (see also [11], Theorem 1.7).
For every n ∈ N, the following statements are equivalent:
1) 2ℵ0 6 ℵn;
2) there is a sequence L0, . . . Ln+1 of lines in the plane R2 and a coloring χ : R2 → {0, . . . , n+
+ 1} such that, for each i ∈ {0, . . . , n + 1}, every line in R2 parallel to Li intersects χ−1(i) in
finitely many points.
We note that the group R of real numbers is the direct sum of 2ℵ0 copies of the group Q of
rational numbers and use a part of this theorem in the following form.
Theorem 6. Let n ∈ N, H = ⊕ℵnQ, a0, . . . , an, b0, . . . , bn be rational numbers such that, for
each i, either ai 6= 0 or bi 6= 0. Then, for every coloring χ : H × H → {0, . . . , n}, there exist
i ∈ {0, . . . , n}, h ∈ H, h 6= 0 and infinitely many pairs a, b ∈ H such that χ(a, b) = i and
aia+ bib = h.
Proof. For i ∈ {0, . . . , n}, let Li =
{
(x, y) ∈ H ×H : aix+ biy = 0
}
. Apply Davies’ theorem
to the lines L0, . . . , Ln.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 9
THIN SUBSETS OF GROUPS 1251
4. Examples. With minor changes, the first example is a reprint of the original G. Bergman’s
letter to the first author (15 May 2011).
Example 1. Let H,K be groups of cardinality ℵ2, G = H ×K. We construct a 2-thin subset A
of G which cannot be partitioned into two thin subsets.
One can find a thin subset X ⊂ K of cardinality ℵ2. For instance, do a recursion over ℵ2,
selecting at each step an element not in the subgroup generated by those that precede. Since X has
cardinality ℵ2, we can index it by pairs of elements of H : X = {x{a,b} : a, b ∈ H}. After choosing
such an indexing, we let
A =
{
x{a,b}, ax{a,b}, bx{a,b} : a, b ∈ H
}
, A ⊆ H ×K = G.
We claim first that A is 2-thin. For this it suffices to show that for every 3-element subset F
of G, only finitely many right translates of F lie in A. In proving this, we may, by an initial right
translation assume that e ∈ F, e is the identity of G.
Assume that F lay in HK but not in H. Then every one of its right translates Fg : g ∈ G
has elements lying in more than one left coset of H; hence if such a right translate is contained in
A, its elements do not all have the same second coordinate in X. Since X is thin in K, we have
{g ∈ G : Fg ⊂ A} is finite.
We are left with the case F ⊂ H. In this case, it is not hard to see that F has exactly 6 right
translates contained in A; namely, these are obtained by taking the 6 arrangement of the elements of
F as an ordered 3-tuple, applying to each the right translate by a member of H that puts it in the
form (e, a, b) and then right translating this by x{a,b} to get a 3-tuple of members of A.
Finally, let us show that A cannot be partitioned into two thin subsets, A1 and A2. Suppose we
had such a partition. Then let us color H ×H as follows. Color an element (a, b) ∈ H ×H
with color 1 if x{a,b} and ax{a,b} lie in the same one of A1 and A2, and bx{a,b} lies in the other
one;
with color 2 if x{a,b} and bx{a,b} lie in the same one of A1 and A2, and ax{a,b} lies in the other
one;
with color 1 if ax{a,b} and bx{a,b} lie in the same one of A1 and A2, and x{a,b} lies in the other
one;
with any of these three colors if x{a,b}, ax{a,b}, and bx{a,b} all lie in the same set A1 or A2.
Now if some vertical line {a} ×H in H ×H had infinitely many points of color 1, then there
would be infinitely many b such that x{a,b} and ax{a,b} lay in the same one of A1 and A2. Hence
one of the latter sets, say Ai, contains infinitely many 2-element sets {x{a,b}, ax{a,b}}. This gives
infinitely many right translates of the pair {1, a} in Ai, contradicting the assumption of thinness.
If some horizontal line H × {a} or diagonal
{
(h, ah) : h ∈ H
}
had infinitely many points of
color 2, respectively 3, we would get a contradiction in the same way. The case of horizontal line is
like that of vertical line, so let us check the diagonal case. Suppose that for some a infinitely many
of the pairs hx{h,ah} and ahx{h,ah} lay in the same of our thin sets. Then at least one of those sets
would contain infinitely many of these pairs; but this shows that the set would contain infinitely many
right translates of the pair {e, a}, contradicting thinness.
The above arguments show that our coloring of H ×H contradicts Theorem 5 (i); so A cannot,
as assumed, be decomposed into two thin sets.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 9
1252 I. V. PROTASOV, S. SLOBODYANYUK
Example 2. For each natural number m > 2, we construct an Abelian group G of cardinality
ℵn, n =
m(m+ 1)
2
− 1, and find a 2-thin subset A of G which cannot be partitioned into m-thin
subsets.
We put H = ⊕ℵnQ, take an arbitrary Abelian group K of cardinality ℵn and let G = K ⊕H.
Then we choose a thin subset X of K, |X| = ℵ2 and enumerate X by the pairs of elements of H :
X =
{
x{a,b} : a, b ∈ H
}
. After choosing such an indexing, we let
A =
{
x{a,b} + ka+ k2b : a, b ∈ H, k ∈ {0, . . . ,m}
}
.
To see that A is 2-thin, it suffices to show that, for any two distinct non-zero elements x, y ∈ G,
the set
A(x, y) =
{
a ∈ A : a+ x ∈ A, a+ y ∈ A
}
is finite. We write x = x1 + x2, y = y1 + y2, x1, x2 ∈ K, y1, y2 ∈ H. If either x1 = 0 or x2 = 0,
A(x, y) is finite because X is thin. Let x1 = x2 = 0. If x{a,b} + ia+ i2b ∈ A(x, y), then
x2 + ia+ i2b = ja+ j2b,
y2 + ia+ i2b = ka+ k2b
for some distinct i, k ∈ {0, . . . ,m}. In this system of relations, a, b are uniquely determined by i, j,
k. Since we have only finite number of possibilities to choose i, j, k, A(x, y) is finite.
Now assume that A is partitioned A = A1 ∪ . . . ∪ Am. To show that at least one cell of the
partition is not thin, we define a coloring χ : H ×H →
{
(k, l) : 0 6 k < l 6 m
}
by the following
rule: for a, b ∈ H, we choose k, l so that x{a,b} + ka+ k2b, x{a,b} + la+ l2b lie in the same cell of
the partition A1 ∪ . . . ∪Am and put χ(a, b) = (k, l). We note that(
x{a,b} + ka+ k2b
)
−
(
x{a,b} + la+ l2b) = (k − l)a+ (k2 − l2
)
b.
Since
m(m+ 1)
2
= n + 1, by Theorem 6, there exist h ∈ H,h 6= 0, k < l, and infinitely many
monochrome pairs a, b such that
(k − l)a+
(
k2 − l2
)
b = h.
By the definition of χ, there are a cell Ai of the partition and infinitely many pairs a, b such that
x{a,b} + ka+ k2b ∈ Ai, x{a,b} + ka+ k2b+ h inAi,
so Ai is not thin.
Example 3. We construct a group G of cardinality ℵω and point out a 2-thin subset A of G
which cannot be partitioned into m thin subsets for each m ∈ N.
For each m > 2, we take a group Gn, n =
m(m+ 1)
2
− 1 from Example 2, put N =
=
{m(m+ 1)
2
− 1: m > 2
}
, take a 2-thin subset An of Gn which cannot be partitioned into
m-thin subsets, and denote
G = ⊕n∈NGn, A = ∪n∈NAn.
We take any distinct x, y ∈ G \ {0} and, in notation of Example 2, show that A(x, y) is finite, so A
is 2-thin. If x, y ∈ Gn for some n, then A(x, y) is 2-thin because A2 is 2-thin. If x /∈ Gn for each n
then |A∩ (A+x)| 6 1 so A(x, y) is also finite. By the choice of {An : n ∈ N}, A cannot be finitely
partitioned into thin subsets.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 9
THIN SUBSETS OF GROUPS 1253
5. Ultrafilter context. Let G be a discrete group, βG be the Stone – Čech compactification of G.
We take the elements of βG to be ultrafilters on G identifying G with the set of principal ultrafilters,
so G∗ = βG \ G is the set of free ultrafilters. The topology of βG can be defined by the family{
A : A ⊆ G
}
as a base for open sets, A =
{
p ∈ βG : A ∈ p
}
.
The multiplication on G can be naturally extended to βG (see [14], Chapter 4). By this extension,
the product pq of ultrafilters p and q can be defined as follows. Take an arbitrary P ∈ p and, for each
g ∈ P, pick Qg ∈ q. Then ∪g∈P gQg ∈ pq and each member of pq contains a subset of this form. In
particular, if g ∈ G and q ∈ βG then gq = {gQ : Q ∈ q}.
The proofs of all propositions in this section can be easily extracted from corresponding defini-
tions.
Proposition 1. For each m ∈ N, a subset A of a group G is m-thin if and only if |Gp∩A| 6 m
for each p ∈ G∗.
By [1], a subset A of a group G is sparse if for any infinite subset X ⊂ G there exists a finite
subset F such that ∩g∈F gA is finite. An ultrafilter p ∈ G∗ has a sparse member if and only if
p /∈ G∗G∗.
Proposition 2. A subset A of a group G is sparse if and only if the set Gp∩A is finite for each
p ∈ G∗.
Proposition 3. A subset A of a group G can be partitioned in finite number of thin subsets if
and only if each ultrafilter p ∈ A has a thin member.
Now take a group G of cardinality ℵω from Example 3 and corresponding 2-thin subset A. Since
A cannot be finitely partitioned into thin subsets, by Proposition 3, there is p ∈ A with no thin
members. Hence, p has a base consisting of 2-thin but not thin subsets.
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225.
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4. Protasov I. V. Selective survey on subset combinatorics of groups // Ukr. Math. Bull. – 2010. – 7. – P. 220 – 257.
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Received 08.06.12
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|
| id | umjimathkievua-article-2505 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:24:43Z |
| publishDate | 2013 |
| publisher | Institute of Mathematics, NAS of Ukraine |
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| resource_txt_mv | umjimathkievua/d4/cf24583e51c358346a071b37802d5fd4.pdf |
| spelling | umjimathkievua-article-25052020-03-18T19:17:02Z Thin Subsets of Groups Тонкі підмножини груп Protasov, I. V. Slobodianiuk, S. V. Протасов, І. В. Слободянюк, С. В. For a group G and a natural number m; a subset A of G is called m-thin if, for each finite subset F of G; there exists a finite subset K of G such that |F g ∩ A| ≤ m for all g ∈ G \ K: We show that each m-thin subset of an Abelian group G of cardinality ℵ n ; n = 0, 1,… can be split into ≤ m n+1 1-thin subsets. On the other hand, we construct a group G of cardinality ℵ ω and select a 2-thin subset of G which cannot be split into finitely many 1-thin subsets. Нехай $G$ — група, $m$ — натуральне число. Пщмножина$A \subseteq G$ називається $m$-тонкою, якщо для кожної скінченної підмножини $F$ групи $G$ знайдеться така скінченна пщмножина $K$, що $|F_g ∩ A| ≤ m$ для всіх $g ∈ G \ K$. Доведено, що $m$-тонку підмножину абелевої групи $G$ потужності $ℵ_n;\; n = 0, 1,…$, можна розбити на $≤ m^{n+1}$ 1-тонких підмножин. Побудовано групу $G$ потужності $ℵ_n$ i 2-тонку підмножину $G$, яку не можна розбити на скінченне число 1-тонких підмножин. Institute of Mathematics, NAS of Ukraine 2013-09-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2505 Ukrains’kyi Matematychnyi Zhurnal; Vol. 65 No. 9 (2013); 1245–1253 Український математичний журнал; Том 65 № 9 (2013); 1245–1253 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2505/1775 https://umj.imath.kiev.ua/index.php/umj/article/view/2505/1776 Copyright (c) 2013 Protasov I. V.; Slobodianiuk S. V. |
| spellingShingle | Protasov, I. V. Slobodianiuk, S. V. Протасов, І. В. Слободянюк, С. В. Thin Subsets of Groups |
| title | Thin Subsets of Groups |
| title_alt | Тонкі підмножини груп |
| title_full | Thin Subsets of Groups |
| title_fullStr | Thin Subsets of Groups |
| title_full_unstemmed | Thin Subsets of Groups |
| title_short | Thin Subsets of Groups |
| title_sort | thin subsets of groups |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2505 |
| work_keys_str_mv | AT protasoviv thinsubsetsofgroups AT slobodianiuksv thinsubsetsofgroups AT protasovív thinsubsetsofgroups AT slobodânûksv thinsubsetsofgroups AT protasoviv tonkípídmnožinigrup AT slobodianiuksv tonkípídmnožinigrup AT protasovív tonkípídmnožinigrup AT slobodânûksv tonkípídmnožinigrup |