arXiv:1709.02027v1 [] 6 Sep 2017

LARGE SETS IN BOOLEAN AND NON-BOOLEAN GROUPS
AND TOPOLOGY
OL’GA V. SIPACHEVA

Various notions of large sets in groups and semigroups naturally arise in dynamics and combinatorial number theory. Most familiar are those of syndetic, thick
(or replete), and piecewise syndetic sets. Apparently, the term “syndetic” was
introduced by Gottschalk and Hedlund in their 1955 book [1] in the context of
topological groups, although syndetic sets of integers have been studied long before
(they appear, e.g., in Khintchine’s 1934 ergodic theorem). During the past decades,
large sets in Z and in abstract semigroups have been extensively studied. It has
turned out that, e.g., piecewise syndetic sets in N have many attractive properties:
they are partition regular (i.e., given any partition of N into finitely many subsets,
at least one of the subsets is piecewise syndetic), contain arbitrarily long arithmetic
progressions, and are characterized in terms of ultrafilters on N (namely, a set is
piecewise syndetic it and only if it belongs to an ultrafilter contained in the minimal
two-sided ideal of βN). Large sets of other kinds are no less interesting, and they
have numerous applications to dynamics, Ramsey theory, the ultrafilter semigroup
on N, the Bohr compactification, and so on.
Quite recently Reznichenko and the author have found yet another application of
large sets. Namely, we introduced special large sets in groups, which we called fat,
and applied them to construct a discrete set with precisely one limit point in any
countable nondiscrete topological group in which the identity element has nonrapid
filter of neighborhoods. Using this technique and special features of Boolean groups,
we proved, in particular, the nonexistence of a countable nondiscrete extremally
disconnected group in ZFC (see [2]).
In this paper, we study right and left thick, syndetic, piecewise syndetic, and fat
sets in groups (although they can be defined for arbitrary semigroups). Our main
concern is the interplay between such sets in Boolean groups. We also consider
natural topologies closely related to fat sets, which leads to interesting relations
between fat sets and ultrafilters.
1. Basic Definitions and Notation
We use the standard notation Z for the group of integers, N for the set (or
semigroup, depending on the context) of positive integers, and ω for the set of
nonnegative integers or the first infinite cardinal; we identify cardinals with the
corresponding initial ordinals. Given a set X, by |X| we denote its cardinality, by
[X]k for k ∈ N, the kth symmetric power of X (i.e., the set of all k-element subsets
of X), and by [X]<ω , the set of all finite subsets of X.
Definition 1 (see [3]). Let G be a group. A set A ⊂ G is said to be
This work was financially supported by the Russian Foundation for Basic Research (Project
No. 15-01-05369).
1

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OL’GA V. SIPACHEVA

(a) right thick, or simply thick if, for every finite F ⊂ S, there exists a g ∈ G
(or, equivalently, g ∈ A [3, Lemma 2.2]) such that F g ⊂ A;
(b) right syndetic, or simply syndetic, if there exists a finite F ⊂ G such that
G = F A;
(c) right piecewise syndetic, or simply piecewise syndetic, if there exists a finite
F ⊂ G such that F A is thick.
Left thick, left syndetic, and left piecewise syndetic sets are defined by analogy;
in what follows, we consider only right versions and omit the word “right.”
Definition 2. Given a subset A of a group G, we shall refer to the least cardinality
of a set F ⊂ G for which G = F A as the syndeticity index, or simply index (by
analogy with subgroups) of A in G. Thus, a set is syndetic if and only if it is of
finite index. We also define the thickness index of A as the least cardinality of
F ⊂ G for which F A is thick.
A set A ⊂ Z is syndetic if and only if the gaps between neighboring elements of A
are bounded, and B ⊂ Z is thick if and only if it contains arbitrarily long intervals
of consecutive integers. The intersection of any such sets A and B is piecewise
syndetic; clearly, such a set is not necessarily syndetic or thick (although it may as
well be both syndetic and thick). The simplest general example of a syndetic set
in a group is a coset of a finite-index subgroup.
In what follows, when dealing with general groups, we use multiplicative notation, and when dealing with Abelian ones, we use additive notation.
Given a set A in a group G, by hAi we denote the subgroup of G generated by A.
As mentioned, we are particularly interested in Boolean groups, i.e., groups in
which all elements are self-inverse. All such groups are Abelian. Moreover, any
Boolean group G can be treated as a countable-dimensional vector space over the
two-element field Z2 ; therefore, for some set X (basis), G can be represented as the
free Boolean group B(X) on X, i.e., as [X]<ω with zero ∅, which we denote by 0,
and the operation of symmetric difference, or Boolean sum, which we denote by △:
A △ B = (A ∪ B) \ A ∩ B. The elements of B(X) (i.e., finite subsets of X) are called
words. The length of a word equals its cardinality. The basis X is embedded in
B(X) as the set of words of length 1. Given n ∈ ω, we use the standard notation
Bn (X) for the set of words
S of length at most n; thus, B0 (X) = {0}, B1 (X) =
X ∪ {0}, and B(X) = n∈ω Bn (X). For the set of words of length precisely n,
where n ∈ N, we use the notation B=n (X); we have B=n (X) = Bn (X) \ Bn−1 (X).
Any free filter F on an infinite set X determines a topological space XF =
X ∪{∗} with one nonisolated point ∗; the neighborhoods of this point are A∪{∗} for
A ∈ F . The topology of the free Boolean topological group B(XF ) = [X ∪ {∗}]<ω
on this space, that is, the strongest group topology that induces the topology of
XF on X ∪ {∗}, is described in detail in [4]. One of the possible descriptions is as
follows. For each n ∈ N, we fix an arbitrary sequence of neighborhoods Vn of ∗,
that is, of An ∪ {∗}, where An ∈ F , and set

U (Vn ) = {x △ y : x, y ∈ Vn }

for n ∈ N

LARGE SETS IN BOOLEAN AND NON-BOOLEAN GROUPS AND TOPOLOGY

3

and
[

U (Vn )n∈N =
(U (V1 ) △ U (V2 ) △ . . . △ U (Vn ))
n∈N

=

[

{x1 △ y1 △ . . . △ xn △ yn : xi , yi ∈ Ai for i ≤ n}.

n∈N

In particular, the subgroup generated by (A ∪ {∗}) △(A ∪ {∗}) is a neighborhood
of zero for any A ∈ F . Clearly, for n ∈ ω, a set Y ⊂ B=2n (XF ) is a trace on
B=2n (XF ) of a neighborhood of zero in
 B(XF ) if and only if it contains a set
of the form (A ∪ {∗}) △ . . . △(A ∪ {∗}) ∩ B=2n (XF ) = [A ∪ {∗}]2n , and a set1
{z
}
|
2n times

Y ⊂ B=2n (X) ⊂ B=2n (XF ) is a trace on B=2n (X) of a neighborhood
of zero in

B(XF ) if and only if it contains a set of the form A △ . . . △ A ∩ B=2n (X) = [A]2n .
| {z }
2n times

The intersection of a neighborhood of zero with B=k (XF ) may be empty for all
odd k.
In what follows, we deal with rapid, κ-arrow, and Ramsey filters and ultrafilters.
Definition 3 ([5]). A filter F on ω is said to be rapid if every function ω → ω is
majorized by the increasing enumeration of some element of F .
Clearly, any filter containing a rapid filter is rapid as well; thus, the existence
of rapid filters is equivalent to that of rapid ultrafilters. Rapid ultrafilters are
also known as semi-Q-point, or weak Q-point, ultrafilters. Both the existence and
nonexistence of rapid ultrafilters is consistent with ZFC (see, e.g., [6] and [7]).
The notions of κ-arrow and Ramsey filters are closely related to Ramsey theory, more specifically, to the notion of homogeneity with respect to a coloring, or
partition. Given a set X and positive integers m and n, by an m-coloring of [X]n
we mean any map c : X → Y of X to a set Y of cardinality m. Any such coloring determines a partition of X into m disjoint pieces, each of which is assigned
a color y ∈ Y . A set A ⊂ X is said to be homogeneous with respect to c, or
c-homogeneous, if c is constant on [A]n . The celebrated Ramsey theorem (finite
version) asserts that, given any positive integers k, l, and m, there exists a positive
integer N such that, for any k-coloring c : [X]l → Y , where |X| ≥ N and |Y | = k,
there exists a c-homogeneous set A ⊂ X of size m.
We consider κ-arrow and Ramsey filters on any, not necessarily countable, infinite
sets. For convenience, we require these filters to be uniform, i.e., nondegenerate in
the sense that all of their elements have the same cardinality (equal to that of the
underlying set).
Definition 4. Let κ be an infinite cardinal, and let F be a uniform filter on a set
X of cardinality κ.
(i) We say that F is a Ramsey filter if, for any 2-coloring c : [X]2 → {0, 1}, there
exists a c-homogeneous set A ∈ U .
(ii) Given an arbitrary cardinal λ ≤ κ, we say that F is a λ-arrow filter if, for
any 2-coloring c : [X]2 → {0, 1}, there exists either a set A ∈ F such that
c([A]2 ) = {0} or a set S ⊂ X with |S| ≥ λ such that c([S]2 ) = {1}.

1Recall that X ⊂ X
F = X ∪ {∗}, and, therefore, B(X) (without topology) is naturally
embedded in B(XF ) as a subgroup.

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OL’GA V. SIPACHEVA

Any filter F on X which is Ramsey or λ-arrow for λ ≥ 3 is an ultrafilter. Indeed,
let S ⊂ X and consider the coloring c : [X]2 → {0, 1} defined by
(
0 if x, y ∈ S or x, y ∈ X \ S,
c({x, y}) =
1 otherwise.
Clearly, any c-homogeneous set containing more than two points is contained entirely in S or in X \ S; therefore, either S or X \ S belongs to F , so that F is an
ultrafilter.
According to Theorem 9.6 in [8], if U is a Ramsey ultrafilter on X, then, for
any n < ω and any 2-coloring c : [X]n → {0, 1}, there exists a c-homogeneous set
A∈U.
It is easy to see that if F is λ-arrow, then, for any A ∈ F and any c : [A]2 →
{0, 1}, there exists either a set B ∈ F such that B ⊂ A and c([B]2 ) = {0} or a set
S ⊂ A with |S| ≥ λ such that c([S]2 ) = {1}.
In [9], where k-arrow ultrafilters for finite k were introduced, it was shown that
the existence of a 3-arrow (ultra)filter on ω implies that of a P -point ultrafilter;
therefore, the nonexistence of κ-arrow ultrafilters for any κ ≥ 3 is consistent with
ZFC (see [10]).
On the other hand, the continuum hypothesis implies the existence of k-arrow
ultrafilters on ω for any k ≤ ω. To formulate a more delicate assumption under
which k-arrow ultrafilters exist, we need more definitions. Given a uniform filter
F on ω, a set B ⊂ ω is called a pseudointersection of F if the complement A \ B
is finite for all A ∈ F . The pseudointersection number p is the smallest size of a
uniform filter on ω which has no infinite pseudointersection. It is easy to show that
ω1 ≤ p ≤ 2ω , so that, under the continuum hypothesis, p = 2ω . It is also consistent
with ZFC that, for any regular cardinals κ and λ such that ω1 ≤ κ ≤ λ, 2ω = λ
and p = κ (see [11, Theorem 5.1]). It was proved in [9] that, under the assumption
p = 2ω (which is referred to as P(c) in [9]), there exist κ-arrow ultrafilters on ω for
all κ ≤ ω. Moreover, for each k ∈ N, there exists a k-arrow ultrafilter on ω which
is not (k + 1)-arrow, and there exists an ultrafilter which is k-arrow for each k ∈ N
but is not Ramsey and hence not ω-arrow [9, Theorems 2.1 and 4.10].
In addition to the free group topology of Boolean groups on spaces generated by
filters, we consider the Bohr topology on arbitrary abstract and topological groups.
This is the weakest group topology with respect to which all homomorphisms to
compact topological groups are continuous, or the strongest totally bounded group
topology; the Bohr topology on an abstract group (without topology) is defined as
the Bohr topology on this group endowed with the discrete topology.
Finally, we need the definition of a minimal dynamical system.
Definition 5. Let G be a monoid with identity element e. A pair (X, (Tg )g∈G ),
where X is a topological space and (Tg )g∈G is a family of continuous maps X → X
such that Te is the identity map and Tgh = Tg ◦ Th for any g, h ∈ G, is called a
topological dynamical system. Such a system is said to be minimal if no proper
closed subset of X is Tg -invariant for all g ∈ G.
We sometimes identify sequences with their ranges.
All groups considered in this paper are assumed to be infinite, and all filters are
assumed to have empty intersection, i.e., to contain the Fréchet filter of all cofinite
subsets (and hence be free).

LARGE SETS IN BOOLEAN AND NON-BOOLEAN GROUPS AND TOPOLOGY

5

2. Properties of Large Sets
We begin with well-known general properties of large sets defined above. Let G
be a group.
1. A set A ⊂ G is thick if and only if the family {gA : g ∈ G} of all translates
of A has the finite intersection property.
Indeed,
this property means that, for every finite subset F of G, there exists an
T
h ∈ g∈F g −1 A, and this, in turn, means that gh ∈ A for each g ∈ F , i.e., F h ⊂ A.
2 [3, Theorem 2.4]. A set A is syndetic if and only if A intersects every thick
set nontrivially, or, equivalently, if its complement G \ A is not thick.
3. A set A is thick if and only if A intersects every syndetic set nontrivially, or,
equivalently, if its complement G \ A is not syndetic.
4 [3, Theorem 2.4]. A set A is piecewise syndetic if and only if there exists a
syndetic set B and a thick set C such that A = B ∩ C.
5 [12, Theorem 4.48]. A set A is thick if and only if
A

βG

= {p ∈ βG : A ∈ p}

(the closure of A in the Stone–Čech compactification βG of G with the discrete
topology) contains a left ideal of the semigroup βG.
6 [12, Theorem 4.48]. A set A is syndetic if and only if every left ideal of βG
βG
intersects A .
7. The families of thick, syndetic, and piecewise syndetic sets are closed with
respect to taking supersets.
8. Thickness, syndeticity, and piecewise syndeticity are translation invariant.
9 [3, Theorem 2.5]. Piecewise syndeticity is partition regular, i.e., whenever a
piecewise syndetic set is partitioned into finitely many subsets, one of these subsets
is piecewise syndetic.
10 [3, Theorem 2.4]. For any thick set A ⊂ G, there exists an infinite sequence
B = (bn )n∈N in G such that
FP(B) = {xn1 xn2 . . . xnk : k, n1 , n2 , . . . , nk ∈ N, n1 < n2 < · · · < nk }
is contained in A.
11. Any IP∗ -set in G, i.e., a set intersecting any infinite set of the form FP(B),
is syndetic. This immediately follows from properties 2 and 10.
3. Fat Sets
As mentioned at the beginning of this section, in [2], Reznichenko and the author
introduced a new2 class of large sets, which we called fat; they have played the key
role in our construction of nonclosed discrete subsets in topological groups.
Definition 6. We say that a subset A of a group G is fat in G if there exists
a positive integer m such that any m-element set F in G contains a two-element
subset D for which D−1 D ⊂ A. The least number m with this property is called
the fatness of A.
We shall refer to fat sets of fatness m as m-fat sets.
In a similar manner, κ-fat sets for any cardinal κ can defined.
2Later, we have found out that similar subsets of Z had already been used in [13]: the ∆∗ -sets
n
considered there and n-fat subsets of Z are very much alike.

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OL’GA V. SIPACHEVA

Definition 7. Given a cardinal κ, we say that a subset A of a group G is κ-fat
in G if any set S ⊂ G with |S| = κ contains a two-element subset D for which
D−1 D ⊂ A.
The notions of an ω-fat and a k-fat set are very similar to but different from
those of ∆∗ - and ∆∗k -sets. ∆∗ -Sets were introduced and studied in [3] for arbitrary
semigroups, and ∆∗k -sets with k ∈ N were defined in [13] for the case of Z.
Definition 8. Given a finite of countable cardinal κ and a sequence (gn )n∈κ in a
group G, we set

−1
gn }
∆ (gn )n∈κ = {x ∈ G : there exist m < n < κ such that x = gm
and


−1
}.
∆D (gn )n∈κ = {x ∈ G : there exist m < n < κ such that x = gn gm


∗
(g
)
A subset of a group G is
called
a
right
(left
)
∆
-set
if
it
intersects
∆
n
n∈κ
I
κ

(respectively, ∆D (gn )n∈κ ) for any one-to-one sequence (gn )n∈κ in G. ∆∗ω -sets are
referred to as ∆∗ -sets.
Remark. For any one-to-one sequence S = (gn )n∈κ in a Boolean group with zero
0, we have ∆I (S) = ∆D (S) = (S △ S) \ {0}. Hence any κ-fat set in such a group is
a right and left ∆∗κ -set. Moreover, the only difference between ∆∗κ - and κ-fat sets
in a Boolean group is in that the latter must contain 0.
The most obvious feature distinguishing fatness among other notions of largeness
is symmetry (fatness has no natural right and left versions). In return, translation
invariance is sacrificed. Thus, in studying fat sets, it makes sense to consider also
their translates.
Clearly, a 2-fat set in a group must coincide with this group. The simplest
nontrivial example of a fat set is a subgroup of finite index n; its fatness equals
n + 1 (any (n + 1)-element subset has two elements x and y in the same coset, and
both x−1 y and y −1 x belong to the subgroup).
It seems natural to refine the definition of fat sets by requiring A ∩ F −1 F to be
of prescribed size rather than merely nontrivial. However, this (and even a formally
stronger) requirement does not introduce anything new.
Proposition 1 ([2, Proposition 1.1]). For any fat set A in a group G and any
positive integer n, there exists a positive integer m such that any m-element set F
in G contains an n-element subset F ′ for which F ′−1 F ′ ⊂ A.
Indeed, considering the coloring c : [G]2 → {0, 1} defined by c({x, y}) = 1 ⇐⇒
x y, y −1 x ∈ A and applying the finite Ramsey theorem, we find a c-homogeneous
set F of size n (provided that m is large enough). If n is no smaller than the fatness
of A (which we can assume without loss of generality), then c([F ]2 ) = {1}.
There is yet another important distinguishing feature of fat sets, namely, the
finite intersection property. Neither thick, syndetic, nor piecewise syndetic sets
have this property.
(Indeed, the disjoint
sets of even and odd numbers are syndetic
S
S
in Z, and i≥0 [22i , 22i+1 ) ∩ Z and i≥1 [22i−1 , 2i ) are thick.) The following theorem
is valid.
−1

Theorem 2 ([2]). Let G be a group.
(i) If A ⊂ G is fat, then so is A−1 .

LARGE SETS IN BOOLEAN AND NON-BOOLEAN GROUPS AND TOPOLOGY

7

(ii) If A ⊂ B ⊂ G and A is fat, then so is B.
(iii) If A ⊂ G and B ⊂ G are fat, then so is A ∩ B.
Assertions (i) and (ii) are obvious, and (iii) follows from Proposition 1.
Proposition 3. If G is a group, S ⊂ G, and S ∩ (SS ∪ S −1 S −1 ) = ∅, then G \ S
is 3-fat.
Proof. Take any three different elements a, b, c ∈ G. We must show that the identity
element e belongs to G\S (which is true by assumption) and either (a−1 b)±1 ∈ G\S,
(b−1 c)±1 ∈ G \ S, or (c−1 a)±1 ∈ G \ S. Assume that, on the contrary, (a−1 b)ε ∈ S
(i.e., a−1 b ∈ S ε ), b−1 c ∈ S δ , and c−1 a ∈ S γ for some ǫ, δ, γ ∈ {−1, 1}. At least two
of the three numbers ε, δ, and γ are equal. Suppose for definiteness that ε = δ.
Then we have c−1 a = c−1 bb−1 a ∈ S −ε S −ε , which contradicts the assumption
S ∩ (S 2 ∪ S −2 ) = ∅.

We see that the family of fat sets in a group resembles, in some respects, a base
of neighborhoods of the identity element for a group topology. However, as we shall
see in the next section, it does not generate a group topology even in a Boolean
group: any Boolean group has a 3-fat subset A containing no set of the form B △ B
for fat B. On the other hand, very many groups admit of group topologies in which
all neighborhoods of the identity element are fat; for example, such are topologies
generated by normal subgroups of finite index. A more precise statement is given
in the next section. Before turning to related questions, we consider how fat sets
fit into the company of other large sets.
We begin with a comparison of fat and syndetic sets.
Proposition 4 (see [2, Proposition 1.7]). Let G be any group with identity element e. Any fat set A in G is syndetic, and its syndeticity index is less than its
fatness.
Proof. Let n denote the fatness of A. Take a finite set F ⊂ G with |F | = n − 1 such
that x−1 y ∈
/ A or y −1 x ∈
/ A for any different x, y ∈ F . Pick any g ∈ G \ F . Since
|F ∪ {g}| = n, it follows that x−1 g ∈ A and g −1 x ∈ A for some x ∈ F , whence
g ∈ xA, i.e., G \ F ⊂ F A. By definition, the identity element of G belongs to A,
and we finally obtain G = F A.

Examples of nonfat syndetic sets are easy to construct: any coset of a finite-index
subgroup in a group is syndetic, while only one of them (the subgroup itself) is fat.
However, the existence of syndetic sets with nonfat translates is not so obvious. An
example of such a set in Z can be extracted from [13].
Example 1. There exists a syndetic set in Z such that none of its translates is fat.
This is, e.g., the set constructed in [13, Theorem 4.3]. Namely, let C = {0, 1}Z, and
let τ : C → C be the shift, i.e., the map defined by τ (f )(n) = f (n + 1) for f ∈ C.
It was proved in [13, Theorem 4.3] that if M ⊂ C is a minimal closed τ -invariant
subset3 and the dynamical system (M, (τ n )n∈Z ) satisfies a certain condition4, then
the support of any f ∈ M is syndetic but not piecewise Bohr; the latter means
that it cannot be represented as the intersection of a thick set and a set having
nonempty interior in the Bohr topology on Z. Clearly, any translate of supp f has
3Then the support of each f ∈ M is syndetic in Z (see, e.g., [14]).
4Namely, is weakly mixing; see, e.g., [14]

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OL’GA V. SIPACHEVA

these properties as well. On the other hand, according to Theorem II in [13], any
∆∗n -set in Z (i.e., any set intersecting the set of differences {kj − ki : i < j ≤ n} for
each n-tuple (k1 , . . . , kn ) of different integers) is piecewise Bohr. Since every n-fat
set is a ∆∗n -set, it follows that the translates of supp f cannot be fat.
Bearing in mind our particular interest in Boolean groups, we also give a similar
example for a Boolean group.
Example 2. We construct a syndetic set in the Boolean group B(Z) with nonfat
translates. Let S be a syndetic set in Z all ofSwhose translates are not ∆∗n -sets
for all n (see Example 1). By definition, Z = k≤r (sk + S) for some r ∈ N and
different s1 , . . . , sr ∈ Z. We set
Sk′ = {x1 △ . . . △ xn : n ∈ N, xi ∈ Z for i ≤ n, xi 6= xj for i 6= j,
X
{x1 , . . . , xn } ∩ {s1 , . . . , sr } = {sk },
xi ∈ 2sk + S},

k ≤ r,

i≤n

and

S′ =

[

Sk′ .

k≤r

We have

sk △ Sk′ = {x1 △ . . . △ xn : n ∈ N, xi ∈ Z for i ≤ n, xi 6= xj for i 6= j,
X
{x1 , . . . , xn } ∩ {s1 , . . . , sr } = ∅,
xi ∈ sk + S},

k ≤ r.

i≤n

Since
[

S

k≤r (sk

+ S) = Z, it follows that

(sk △ S ′ )

k≤r

⊂ {x1 △ . . . △ xn : n ∈ N, xi ∈ Z for i ≤ n, {x1 , . . . , xn } ∩ {s1 , . . . , sr } = ∅}.

Obviously, the set on the right-hand side of this inclusion is syndetic; therefore, so
is S ′ .
Let us show that no translate of S ′ is fat. Suppose that, on the contrary, k, n ∈ N,
z1 , . . . , zk ∈ Z, w = z1 △ . . . △ zk , and w △ S ′ is n-fat. Take any different k1 , . . . , kn ∈
Z larger than the absolute values of all elements of w (which is a finite subset of Z)
and of all si , i ≤ r. We set
F = {k1 , k1 △(−k1 ) △ k2 , k1 △(−k1 ) △ k2 △(−k2 ) △ k3 ,
. . . , k1 △(−k1 ) △ k2 △(−k2 ) △ . . . △ kn−1 △(−kn−1 ) △ kn }.
Suppose that there exist different x, y ∈ F for which x △ y ∈ w △ S ′ , i.e., there exist
i, j ≤ n for which i < j and
k1 △(−k1 ) △ . . . △ ki−1 △(−ki−1 ) △ ki △ k1 △(−k1 ) △ . . . △ kj−1 △(−kj−1 ) △ kj
= ki △ ki △(−ki ) △ ki+1 △(−ki+1 ) △ . . . △ kj−1 △(−kj−1 ) △ kj = w △ s ∈ w △ S ′ ,
where s is an element of S ′ and hence belongs to Sl′ for some l ≤ r, which means,
in particular, that s contains precisely one of the letters s1 , . . . , sr , namely, sl .
There are no such letters among ±ki , . . . , ±kj−1 , kj . Therefore, one of the letters
zm (say z1 ) is sl . The other letters of w do not equal ±ki , . . . , ±kj−1 , kj either and,
therefore, are canceled with letters of s ∈ S ′ in the word w + s. By the definition of

LARGE SETS IN BOOLEAN AND NON-BOOLEAN GROUPS AND TOPOLOGY

9

the set S ′ containing s, one letter of the word w (namely, z1 = sl ) belongs to the set
{s1 , . . . , sr } and the other letters do not. Since the sum (in Z) of the integer-letters
of s belongs to 2sl + S (by the definition of Sl′ ) and sl = z1 , it follows that the sum
of letters of w + s belongs to S + z1 − z2 − · · · − zk and the letter z1 is determined
uniquely for the given word w. To obtain a contradiction, it remains to recall that
the translates of S (in particular, S + z1 − z2 − · · · − zk ) are not ∆∗n -sets in Z and
choose k1 , . . . , kn so that {kj − ki : i < j ≤ n} ∩ (S + z1 − z2 − · · · − zk ) = ∅.
Example 3. There exist fat sets which are not thick and thick sets which are not
fat. Indeed, as mentioned, any proper finite-index group is fat, but it cannot be
thick by the first property in the list of properties of large sets given above.
An example of a nonfat thick set is, e.g., any thick nonsyndetic set. In an infinite
Boolean group G, such a set can be constructed as follows. Take any
S basis X in G
(so that G = B(X)), fix any nonsyndetic thick set T in N (say T = n ([an , bn ]∩N),
where the an and bn are numbers such that the bn − an and the an+1 − bn increase
without bound), and consider the set
A = {x1 △ . . . △ xn ∈ B(X) : n ∈ T, xi ∈ X for i ≤ n, xi 6= xj for i 6= j}
of all words in B(X) whose lengths belong to T . The thickness of this set is obvious
(by the same property 1), because the translate of A by any
S word g ∈ B(X) of any
length l surely contains all words whose lengths belong to n ([an +l, bn −l]∩N) ⊂ T
and, therefore, intersects A. However,
A is not fat, because it misses all words
S
whose lengths belong to the set n ((bn , an+l ) ∩ N). The last set contains at least
one even positive integer 2k. It remains to choose different points x1 , x2 , . . . in X,
set B = {xkn+1 △ xkn+2 . . . △ xkn+k : n ∈ ω}, and note that all nonempty words in
B △ B have length 2k. Therefore, A is disjoint from B △ B (much more from F △ F
for any finite F ⊂ B). Note that the translates of A are not fat either, because
both thickness and (non)syndeticity are translation invariant.
Proposition 5. Let G be any group with identity element e.
(i) If a set A in G is 3-fat, then (G \ A)−1 (G \ A) ⊂ A.
(ii) If a set A in G is 3-fat, then either AA−1 = G or A is a subgroup of index 2.
Proof. (i) Suppose that A is a 3-fat subset of a group G with identity element e.
Take any different x, y ∈
/ A (if there exist no such elements, then there is nothing to
prove). By definition, the set {x, y, e} contains a two-element subset D for which
D−1 D ⊂ A. Clearly, D 6= {x, e} and D 6= {y, e}. Therefore, x−1 y ∈ A and
y −1 x ∈ A (and e ∈ A, too), whence (G \ A)−1 (G \ A) ⊂ A.
(ii) If AA−1 6= G, then there exists a g ∈ G for which gA ∩ A = ∅. If A is, in
addition, 3-fat, then (ii) implies (gA)−1 gA = A−1 A ⊂ A, which means that A is a
subgroup of G. According to (i), A is syndetic of index at most 2; in fact, its index
is precisely 2, because A does not coincide with G.

4. Quotient sets
−1

In [3] sets of the form AA or A−1 A were naturally called quotient sets. We
shall refer to the former as right quotient sets and to the latter as left quotient
sets. Thus, a set in a group G is m-fat if it intersects nontrivially the left quotient
set of any m-element subset of G. Quotient sets play a very important role in
combinatorics, and their interplay with large sets is quite amazing.

10

OL’GA V. SIPACHEVA

First, the passage to right quotient sets annihilates the difference between syndetic and piecewise syndetic sets.
Theorem 6 (see [3, Theorem 3.9]). For each piecewise syndetic subset A of a
group G, there exists a syndetic subset B of G such that BB −1 ⊂ AA−1 and the
syndeticity index of B does not exceed the thickness index of A.
Briefly, the construction of B given in [3] is as follows: we takeTa finite set T such
that T A is thick and, for each finite F ⊂ G, let ΦF = {ϕ ∈ T G : x∈F x−1 ϕ(x)A 6=
∅}. Then we pick ϕ∗ in the intersection of all ΦF (which exists since the product
space T G is compact) and let B = {ϕ∗ (x)−1 x : x ∈ G}. Since ϕ∗ (G) ⊂ T , it follows
that T B = G, which means that B is syndetic and its index does not exceed |T | = t.
Moreover, for any finite F ⊂ B, there exists a g ∈ G such that F g ⊂ A, and this
implies BB −1 ⊂ AA−1 .
In Theorem 6, right quotient sets cannot be replaced by left ones: there are
examples of piecewise syndetic sets A such that A−1 A does not contain B −1 B for
any syndetic B. One of such examples is provided by the following theorem.
Theorem 7. (i) If a subset A of a group G is syndetic of index s, then A−1 A is
fat, and its fatness does not exceed s + 1.
(ii) If a subset A of an Abelian group G is piecewise syndetic of thickness index
t, then A − A is fat, and its fatness does not exceed t + 1.
(iii) There exists a group G and a thick (in particular, piecewise syndetic) set
A ⊂ G such that A−1 A is not fat and, therefore, does not contain B −1 B for
any syndetic set.
(iv) If a subset A of a group G is thick, then AA−1 = G.
Proof. (i) Suppose that F A = G, where F = {g1 , . . . , gs }. Any (s + 1)-element
subset of G has at least two points x and y in the same “coset” gi A. We have
x = gi a′ and y = gi a′′ , where a′ , a′′ ∈ A. Thus, x−1 y, y −1 x ∈ A−1 A.
Assertion (ii) follows immediately from (i) and Theorem 6.
Let us prove (iii). Consider the free group G on two generators a and b and let
A be the set of all words in G whose last letter is a. Then A is thick (given any
finite F ⊂ G, we have F an ⊂ A for sufficiently large n). Clearly, all nonidentity
words in A−1 A contain a or a−1 . Therefore, if F ⊂ G consists of words of the form
bn , then the intersection F −1 F ∩ A−1 A is trivial, so that A−1 A is not fat.
Finally, to prove (iv), take any g ∈ G. We have A ∩ gA 6= ∅ (by property 1 in
our list of properties of large sets). This means that g ∈ AA−1 .

We see that the right quotient sets AA−1 of thick sets A are utmostly fat, while
the left quotient sets A−1 A may be rather slim. In the Abelian case, the difference
sets of all thick sets coincide with the whole group.
It is natural to ask whether condition (i) in Theorem 7 characterizes fat sets in
groups. In other words, given any fat set A in a group, does there exist a syndetic
(or, equivalently, piecewise syndetic) set B such that B −1 B ⊂ A (or BB −1 ⊂ A)?
The answer is no, even for thick 3-fat sets in Boolean groups. The idea of the
following example was suggested by arguments in paper [13] and in John Griesmer’s
note [15], where the group Z was considered.
Example 4. Let G be a countable Boolean group with zero 0. Any such group
can be treated as the free Boolean group on Z. We set
A = G \ {m △ n = {m, n} : m, n ∈ Z, m < n, n − m = k 3 for some k ∈ N}.

LARGE SETS IN BOOLEAN AND NON-BOOLEAN GROUPS AND TOPOLOGY

11

Clearly, A is thick (if F ⊂ G is finite and a word g ∈ G is sufficiently long, then
all words in the set F △ g have more than two letters and, therefore, belong to
A). Let us prove that A is 3-fat. Take any different a, b, c ∈ G. We must show
that a △ b ∈ A, b △ c ∈ A, or a △ c ∈ A. We can assume that c = 0; otherwise,
we translate a, b, and c by c, which does not affect the Boolean sums. Thus, it
suffices to show that, given any different nonzero x, y 6∈ G, we have x △ y ∈ A. The
condition x, y 6∈ G means that x = {k, l}, where k < l and l − k = r3 for some
r ∈ Z, and y = {m, n}, where m < n and n − m = s3 for some s ∈ Z. Suppose
for definiteness that n > l or n = l and m > k. If x △ y 6∈ A, then either k = m
and l − n = t3 for some t ∈ N, l = m and n − k = t3 for some t ∈ N, or l = n and
m − k = t3 for some t ∈ N. In the first case, we have l − k = l − n + n − m, i.e.,
r3 = t3 + s3 ; in the second, we have n − k = n − m + l − k, i.e., t3 = s3 + r3 ; and in
the third, we have l − k = n − m + m − k, i.e., r3 = s3 + t3 . In any case, we obtain
a contradiction with Fermat’s theorem.
It remains to prove that there exists no syndetic (and hence no piecewise syndetic) B ⊂ G for which B △ B ⊂ A. Consider any syndetic set B. Let F =
{f1 , . . . , fk } ⊂ G be a finite set for which F B = G, and let m be the maximum
absolute value of all letters of words in F (recall that all letters are integers). To
each n ∈ Z with |n| > m we assign a word fi ∈ F for which n ∈ fi △ B; if there are
several such words, then we choose any of them. Thereby, we divide the set of all
integers with absolute value larger than m into k pieces I1 , . . . , Ik . To accomplish
our goal, it suffices to show that there is a piece Ii containing two integers r and s
such that r − s = z 3 for some z ∈ Z. Indeed, in this case, we have r ∈ fi △ B and
s ∈ fi △ B, so that r △ s ∈ B △ B. On the other hand, r △ s 6∈ A.
From now on, we treat the pieces I1 , . . . , Ik as subsets of Z. We have Z =
{−m, −m + 1, . . . , 0, 1, . . . , m} ∪ I1 ∪ · · · ∪ Ik . Since piecewise syndeticity is partition
regular (see property 9 of large sets), one of the sets Ii , say Il , is piecewise syndetic.
Therefore, by Theorem 5, Il − Il ⊃ S − S for some syndetic set S ⊂ Z.
Let d∗ (S) denote the upper Banach density of S, i.e.,
|S ∩ {n, n + 1, · · · , n + d}|
.
n→∞ d→∞
d
The syndeticity of S in Z implies the existence of an N ∈ N such that every interval
of integers longer than N intersects S. Clearly, we have d∗ (S) ≥ 1/N . Proposition 3.19 in [14] asserts that if X is a set in Z of positive upper Banach density
and p(t) is a polynomial taking on integer values at the integers and including 0 in
its range on the integers, then there exist x, y ∈ X, x 6= y, and z ∈ Z such that
x − y = p(z) (as mentioned in [14], this was proved independently by Sárközy).
Thus, there exist different x, y ∈ S and a z ∈ Z for which x − y = z 3 . Since
S − S ⊂ Il − Il , it follows that z 3 = r − s for some r, s ∈ Il , as desired.
d∗ (S) = lim lim sup

5. Large Sets and Topology
In the context of topological groups quotient sets arise again, because for each
neighborhood U of the identity element, there must exist a neighborhood V such
that V −1 V ⊂ U and V V −1 ⊂ U . Thus, if we know that a group topology consists
of piecewise syndetic sets, then, in view of Theorem 6, we can assert that all open
sets are syndetic, and so on. Example 4 shows that if G is any countable Boolean
topological group and all 3-fat sets are open in G, then some nonempty open sets
in this group are not piecewise syndetic. Thus, all syndetic or piecewise syndetic

12

OL’GA V. SIPACHEVA

subsets of a group G do not generally form a group topology. Even their quotient
(difference in the Abelian case) sets are insufficient; however, it is known that double
difference sets of syndetic (and hence piecewise syndetic) sets in Abelian groups are
neighborhoods of zero in the Bohr topology.5 These and many other interesting
results concerning a relationship between Bohr open and large subsets of abstract
and topological groups can be found in [16, 17]. As to group topologies in which
all open sets are large, the situation is very simple.
Theorem 8. For any topological group G with identity element e, the following
conditions are equivalent:
(i) all neighborhoods of e in G are piecewise syndetic;
(ii) all open sets in G are piecewise syndetic;
(iii) all neighborhoods of e in G are syndetic;
(iv) all open sets in G are syndetic;
(v) all neighborhoods of e in G are fat;
(vi) G is totally bounded.
Proof. The equivalences (i) ⇔ (ii) and (iii) ⇔ (iv) follow from the obvious translation invariance of piecewise syndeticity and syndeticity. Theorem 6 implies (i) ⇔
(iii), Theorem 7 (i) implies (iii) ⇒ (v), and Proposition 4 implies (v) ⇒ (iii). The
implication (iii) ⇒ (i) is trivial. Finally, (vi) ⇔ (iii) by the definition of total
boundedness.

Thus, the Bohr topology on a (discrete) group is the strongest group topology
in which all open sets are syndetic (or, equivalently, piecewise syndetic, or fat).
For completeness, we also mention the following corollary of Theorem 7 and
Theorem 3.12 in [3], which relates fat sets to topological dynamics.
Corollary 9. If G is an Abelian group with zero 0, X is a compact Hausdorff space,
and (X, (Tg )g∈G ) is a minimal dynamical system, then the set {g ∈ G : U ∩Tg−1 U 6=
∅} is fat for every nonempty open subset U of X.
6. Fat and Discrete Sets in Topological Groups
As mentioned above, fat sets were introduced in [2] to construct discrete sets in
topological groups. Namely, given a countable topological group G whose identity
element e has nonrapid filter F of neighborhoods, we can construct a discrete set
with precisely one limit point in this group as follows. The nonrapidness of F
means that, given any sequence (mn )n∈N of positive integers, there exist finite sets
Fn ⊂ G, n ∈ N, such that each neighborhood of e intersects some Fn in at least mn
points (see [7, Theorem 3 (3)]).TThus, if we have a decreasing sequence of closed
mn -fat sets An in G such that An = {e}, then the set
[
D=
{a−1 b : a 6= b, a, b ∈ Fn , a−1 b ∈ An }
n∈N

is discrete (because e ∈
/ D and each g ∈ G\{e} has a neighborhood of the form G\An
which contains only finitely many elements of D), and e is the only limit point of D
(because, given any neighborhood U of e, we can take a neighborhood V such that

5It follows, in particular, that, given any piecewise syndetic set A in an Abelian group, there
exists an infinite sequence of fat sets A1 , A2 , . . . such that A1 − A1 ⊂ A + A − A − A and
An+1 − An+1 ⊂ An for all n (because all Bohr open sets are syndetic).

LARGE SETS IN BOOLEAN AND NON-BOOLEAN GROUPS AND TOPOLOGY

13

V −1 V ⊂ U ; we have |V ∩Fn | ≥ mn for some n, and hence (V ∩Fn )−1 (V ∩Fn )∩An 6=
∅, so that U ∩ D 6= ∅). It remains to find a family of closed fat sets with trivial
intersection and make it decreasing.
The former task is easy to accomplish in any topological group: by Proposition 3,
in any topological group G, the complements to open neighborhoods gU of all g ∈ G
satisfying the condition gU ∩ (U 2 ∪ (U −1 )2 ) = ∅ form a family of closed 3-fat sets
with trivial intersection. In countable groups, this family can be made decreasing by
using Theorem 2, according to which the family of fat sets has the finite intersection
property. Unfortunately, no similar argument applies in the uncountable case,
because countable intersections of fat sets may be very small. Thus, in Zω
2 , the
intersection of the 3-fat sets Hn = {f ∈ Zω
:
f
(n)
=
0}
(each
of
which
is a
2
subgroup of index 2 open in the product topology) is trivial.
7. Large Sets in Boolean Groups
In the case of Boolean groups, many assertions concerning large sets can be
refined. For example, properties 10 and 11 of large sets are stated as follows.
Proposition 10. (i) For any thick set T in a Boolean group G with zero 0, there
exists an infinite subgroup H of G for which T ∪ {0} ⊃ H.
(ii) Any set which intersects nontrivially all infinite subgroups in a Boolean group
G is syndetic.
Note that this is not so in non-Boolean groups: the set {n! : n ∈ N} intersects
any infinite subgroup in Z, but it is not syndetic, because the gaps between neighboring elements are not bounded. The complement of this set contains no infinite
subgroups, and it is thick by property 2 of large sets.
Another specific feature of thick sets in Boolean groups is given by the following
proposition.
Proposition 11. For any thick set T in a countable Boolean group G with zero 0,
there exists a set A ⊂ G such that T ∪ {0} = A △ A (and A △ A △ A △ A = G by
Theorem 7 (iv)).
Proposition 11 is an immediate corollary of Lemma 4.3 in [3], which says that
any thick set in a countable Abelian group equals ∆I (gn )∞
n=1 for some sequence
(gn )∞
n=1 .
In view of Example 4, we cannot assert that the set A in this proposition is large
(in whatever sense), even for the largest (3-fat) nontrivial thick sets T .
The following statement can be considered as a partial analogue of Propositions 10 and 11 for ∆∗ - (in particular, fat) sets in Boolean groups.
Theorem 12. For any ∆∗ -set A in a Boolean group G with zero 0, there exists a
B ⊂ G with |B| = |A| such that B △ B ⊂ A ∪ {0}.
Proof. First, note that |A| = |G|. Any Boolean group is algebraically free; therefore,
we can assume that G = B(X) for a set X with |X| = |A|. Let

A2 = A ∩ B=2 (X) = {x, y} = x △ y ∈ A : x, y ∈ X

be the intersection of A with the set of words of length 2. We have |A2 | = |X|,
because A must intersect nontrivially each countable set of the form Y △ Y for

14

OL’GA V. SIPACHEVA

Y ⊂ X. Consider the coloring c : [X]2 → {0, 1} defined by
(
0 if {x, y} ∈ A2 ,
c({x, y}) =
1 otherwise.
According to the well-known Erdös–Dushnik–Miller theorem κ → (κ, ℵ0 )2 (see,
e.g., [18]), there exists either an infinite set Y ⊂ X for which [Y ]2 ∩ A2 = ∅ or a
set Y ⊂ X of cardinality |X| for which [Y ]2 ⊂ A2 . The former case cannot occur,
because [Y ]2 = Y △ Y in B(X), [Y ]2 ⊂ B=2 (X), and A is a ∆∗ -set. Thus, the
latter case occurs, and we set B = Y .

We have already distinguished between fat sets and translates of syndetic sets in
Boolean groups (see Example 2). For completeness, we give the following example.
Example 5. The countable Boolean group B(Z) contains an IP∗ -set (see property 11 of large sets) which is not a ∆∗ -set. An example of such a set is constructed
from the corresponding example in Z (see [14, p. 177] in precisely the same way as
Example 2.
8. Large Sets in Free Boolean Topological Groups
As shown in Section 5, given any Boolean group G, the filter of fat sets in G
cannot be the filter of neighborhoods of zero for a group topology, because not all fat
and even 3-fat sets are neighborhoods of zero in the Bohr topology. Moreover, if we
fix any basis X in G, so that G = B(X), then not all traces of 3-fat sets on the set
B=2 (X) of two-letter words contain those of Bohr neighborhoods of zero. However,
there are natural group topologies on B(X) such that the topologies which they
induce on B2 (X) contain those generated by n-fat sets.
Theorem 13. Let k ∈ N , and let F be a filter on an infinite set X. Then the
following assertions hold.
(i) For k 6= 4, the trace of any k-fat subset of B(X) on B2 (X) ⊂ B2 (XF ) contains
that of a neighborhood of zero in the free group topology of B(XF ) if and only
if F is a k-arrow filter.
(ii) If the trace of any 4-fat set on B2 (X) contains that of a neighborhood of zero
in the free group topology of B(XF ), then F is a 4-arrow filter, and if F is
a 4-arrow filter, then the trace of any 3-fat set on B2 (XF ) contains that of a
neighborhood of zero in the free group topology of B(XF ).
(iii) The trace of any ω-fat set on B2 (X) contains that of a neighborhood of zero
in the free group topology of B(XF ) if and only if F is an ω-arrow ultrafilter.
The proof of this theorem uses the following lemma.
Lemma 14. (i) If k 6= 4, w1 , . . . , wk ∈ B(X), and wi △ wj ∈ B=2 (X) for any
i < j ≤ k, then there exist x1 , . . . , xk ∈ X such that wi △ wj = xi △ xj for any
i < j ≤ k.
(ii) If k = 4, w1 , w2 , w3 , w4 ∈ B(X), and wi △ wj ∈ B=2 (X) for any i < j ≤ 4,
then there exist either
(a) x1 , x2 , x3 , x4 ∈ X such that wi △ wj = xi △ xj for any i < j ≤ 4 or

LARGE SETS IN BOOLEAN AND NON-BOOLEAN GROUPS AND TOPOLOGY

15

(b) x1 , x2 , x3 ∈ X such that
w1 △ w4 = w2 △ w3 = x2 △ x3 ,
w2 △ w4 = w1 △ w3 = x1 △ x3 ,
w3 △ w4 = w1 △ w3 = x1 △ x3 .
(iii) If w1 , w2 , · · · ∈ B(X) and wi △ wj ∈ B2 (X) for any i < j, then there exist
x1 , x2 , · · · ∈ X such that wi △ wj = xi △ xj for any i < j.
Proof. We prove the lemma by induction on k. There is nothing to prove for k = 1,
and for k = 2, assertion (i) obviously holds.
Suppose that k = 3. For some y1 , y2 , y3 , y4 ∈ X, we have w1 △ w2 = y1 △ y2 and
w2 △ w3 = y3 △ y4 . Since w1 △ w3 = w1 △ w2 △ w2 △ w3 ∈ B=2 (X), it follows that
either y1 = y3 , y1 = y4 , y2 = y3 , or y2 = y4 . If y1 = y3 , then w1 △ w3 = y2 △ y4 and
w2 △ w3 = y1 △ y4 , so that we can set x1 = y2 , x2 = y1 , and x3 = y4 . If y2 = y3 ,
then w1 △ w3 = y1 △ y4 and w2 △ w3 = y2 △ y4 , and we set x1 = y1 , x2 = y1 , and
x3 = y4 . The remaining cases are treated similarly.
Suppose that k = 4 and let x1 , x2 , x3 ∈ X be such that wi △ wj = xi △ xj for
i = 1, 2, 3. There exist y, z ∈ X for which w1 △ w4 = y △ z. We have w2 △ w4 =
w1 △ w2 △ w1 △ w4 = x1 △ x2 △ y △ z ∈ B2 (X). Therefore, either x1 = y, x2 = y,
x1 = z, or x2 = z.
If x1 = y or x1 = z, then the condition in (ii) (a) holds for x4 = z in the former
case and x4 = y in the latter.
Suppose that x1 6= y and x1 6= z. Then x2 = y or x2 = z. Let x2 = y. Then
w1 △ w4 = x2 △ z, and we have w3 △ w4 = w1 △ w3 △ w1 △ w4 = x1 △ x3 △ x2 △ z ∈
B2 (X), whence x3 = z (because x1 , x2 6= z), so that w1 △ w4 = x2 △ x3 = w2 △ w3 ,
w2 △ w4 = w1 △ w2 △ w1 △ w4 = x1 △ x3 = w1 △ w3 , and w3 △ w4 = x1 △ x3 =
w1 △ w3 , i.e., assertion (ii) (b) holds. The case x2 = z is similar. Note for what
follows that, in both cases x2 = y and x2 = z, we have w4 = w1 △ w2 △ w3 .
Let k > 4. Consider the words w1 , w2 , w3 , and w4 . Let x1 , x2 , x3 ∈ X be such
that wi △ wj = xi △ xj for i = 1, 2, 3. As previously, there exist y, z ∈ X for which
w1 △ w4 = y △ z and either x1 = y, x2 = y, x1 = z, or x2 = z.
Suppose that x1 6= y and x1 6= z; then w4 = w1 △ w2 △ w3 . In this case, we
consider w5 instead of w4 . Again, there exist y ′ , z ′ ∈ X for which w1 △ w5 = y △ z
and either x1 = y ′ , x2 = y ′ , x1 = z ′ , or x2 = z ′ . Since w5 6= w4 , it follows that
w5 6= w1 △ w2 △ w3 , and we have x1 = y ′ or x1 = z ′ . In the former case, we set
x5 = z ′ and in the latter, x5 = y ′ . Consider again w4 ; recall that w1 △ w4 = y △ z.
We have wi △ w4 = w1 △ wi △ w1 △ w4 = x1 △ xi △ y △ z ∈ B=2 (X) for i ∈ {2, 3, 5}.
Since x2 6= x5 and x3 6= x5 , it follows that x1 = y, which contradicts the assumption.
Thus, x1 = y or x1 = z. As above, we set x4 = z in the former case and x4 = y
in the latter; then the condition in (ii) (a) holds.
Suppose that we have already found the required x1 , . . . , xk−1 ∈ X for w1 , . . . ,
wk−1 . There exist y, z ∈ X for which w1 △ wk = y △ z. We have wi △ wk =
w1 △ wi △ w1 △ wk = x1 △ xi △ y △ z ∈ B=2 (X) for i ≤ k − 1. If x1 6= y and x1 6= z,
then we have xi ∈ {y, z} for 2 ≤ i ≤ k−1, which is impossible, because k > 4. Thus,
either x1 = y or x1 = z. In the former case, we set xk = z and in the latter, xk = y.
Then w1 △ wk = x1 △ wk and, for any i ≤ k − 1, wi △ wk = w1 △ wi △ w1 △ wk =
x1 △ xi △ x1 △ xk = xi △ xk .
The infinite case is proved by the same inductive argument.


16

OL’GA V. SIPACHEVA

Proof of Theorem 13. (i) Suppose that F is a k-arrow filter on X. Let C be a
k-fat set in B(XF ). Consider the 2-coloring of [X]2 defined by
(
0 if {x, y} = x △ y ∈ C,
c({x, y}) =
1 otherwise.
Since F is k-arrow, there exists either an A ∈ F for which c([A]2 ) = {0} and hence
[A]2 ⊂ C ∩ B2 (XF ) or a k-element set F ⊂ X for which c([F ]2 ) = {1} and hence
[F 2 ] ∩ C = [F 2 ] ∩ C ∩ B=2 (XF ) = ∅. The latter case cannot occur, because C is
k-fat. Therefore, C ∩B2 (XF ) contains the trace [A]2 ∪{0} = ((A∪{∗}) △(A∪{∗}))
of the subgroup hA ∪ {∗}i, which is an open neighborhood of zero in B(XF ).
Now suppose that k 6= 4 and the trace of each k-fat set on B2 (X) contains the
trace on B2 (X) of a neighborhood of zero in B(XF ), i.e., a set of the form A △ A
for some A ∈ F . Let us show that F is k-arrow. Given any c : [X]2 → {0, 1}, we
set

C = x △ y : c({x, y}) = 1
and C ′ = B(XF ) \ C.
′
If C is not k-fat, then there exist w1 , . . . , wk ∈ B(X) such that wi △ wj ∈ C for
i < j ≤ k. By Lemma 14 (i) we can find x1 , . . . , xk ∈ X such that xi △ xj ∈ C
(and hence xi 6= ∗) for i < j ≤ k. This means that, for F = {x1 , . . . , xk }, we have
c([F ]2 ) = {1}. If C ′ is k-fat, then, by assumption, there exists an A ∈ F for which
A △ A \ {0} ⊂ C ′ ∩ B2 (X) = C, which means that c([A]2 ) = {0}.
The same argument proves (ii); the only difference is that assertion (ii) of
Lemma 14 is used instead of (i).
The proof of (iii) is similar.

Let Rr (s) denote the least number n such that, for any r-coloring c : [X]2 → Y ,
where |X| ≥ n and |Y | = r, there exists an s-element c-homogeneous set. By the
finite Ramsey theorem, such a number exists for any positive integers r and s.
Theorem 15. There exists a positive integer N (namely, N = R36 (R6 (3)) + 1)
such that, for any uniform ultrafilter U on a set X of infinite cardinality κ, the
following conditions are equivalent:
(i) the trace of any N -fat subset of B(X) on B4 (X) ⊂ B4 (XU ) contains that of
a neighborhood of zero in the free group topology of B(XU );
(ii) all κ-fat sets in B(X) are neighborhoods of zero in the topology induced from
the free topological group B(XU );
(iii) U is a Ramsey ultrafilter.
Proof. Without loss of generality, we assume that X = κ.
(i) ⇒ (iii) Suppose that N is as large as we need and the trace of each N -fat set
on B4 (κU ) contains the trace on B4 (κ) of a neighborhood of zero in B(XU ), which,
in turn, contains a set of the form (A △ A △ A △ A) ∩ B=4 (κ) for some A ∈ U . Let
us show that U is a Ramsey ultrafilter. Consider any 2-coloring c : [κ]2 → {0, 1}.
We set

C = α1 △ α2 △ α3 △ α4 : αi ∈ κ for i ≤ 4, α1 < α2 < α3 < α4 ,
c({α1 , α2 }) 6= c({α3 , α4 }), c({α1 , α3 }) 6= c({α2 , α4 }),

c({α1 , α4 }) 6= c({α2 , α3 })
and
C ′ = B(X) \ C.

LARGE SETS IN BOOLEAN AND NON-BOOLEAN GROUPS AND TOPOLOGY

17

If C ′ is not N -fat, then there exist w1 , . . . , wN ∈ B(κ) such that wi △ wj ∈ C for
i < j ≤ N . We can assume that wN = 0 (otherwise, we translate all wi by wN ).
Then wi ∈ C ⊂ B4 (κ), i < N . Let wi = αi1 △ αi2 △ αi3 △ αi4 for i < N and consider
the 36-coloring of all pairs {wi , wj }, i < j < N , defined as follows. Since wi △ wj is
a four-letter word, it follows that wi △ wj = β1 △ β2 △ β3 △ β4 , where βi ∈ κ. Two
letters among β1 , β2 , β3 , β4 (say β1 and β2 ) occur in the word wi and the remaining
two (β3 and β4 ) occur in wj . We assume that β1 < β2 and β3 < β4 . Let us
denote the numbers of the letters β1 and β2 in wi (recall that the letters in wi are
numbered in increasing order) by i′ and i′′ , respectively, and the numbers of the
letters β3 and β4 in wj by j ′ and j ′′ . To the pair {wi , wj } we assign the quadruple
(i′ , i′′ , j ′ , j ′′ ). The number of all possible quadruples is 36, so that this assignment
is a 36-coloring. We choose N ≥ R36 (N ′ ) + 1 for N ′ as large as we need. Then
there exist two pairs i′0 , i′′0 and j0′ , j0′′ and N ′ words win , where n ≤ N ′ and is < it
for s < t, such that i′ = i′0 , i′′ = i′′0 , j ′ = j0′ , and j ′′ = j0′′ for any pair {wi , wj } with
i, j ∈ {i1 , . . . , iN ′ } and i < j. Clearly, if N ′ ≥ 3, then we also have j0′ = i′0 and
j0′′ = i′′0 . In the same manner, we can fix the position of the letters coming from
wi and wj in the sum wi △ wj : to each pair {wis , wit }, s, t ∈ {1, . . . , N ′ }, s < t,
we assign the numbers of the i′0 th and i′′0 th letters of wis in the word wis △ wit
(recall that the letters of all words are numbered in increasing order); the positions
of the letters of wit in wis △ wit are then determined automatically. There are
six possible arrangements: 1, 2, 1, 3, 1, 4, 2, 3, 2, 4, and 3, 4. Thus, we have a
6-coloring of the symmetric square of the N ′ -element set {wi1 , . . . , wiN ′ }, and if
N ′ ≥ R6 (3) (which we assume), then there exists a 3-element set {wk , wl , wm }
homogeneous with respect to this coloring, i.e., such that all pairs of words from
this set are assigned the same color. For definiteness, suppose that this is the color
1, 2; suppose also that i′0 = 1, i′′0 = 2, k < l < m, and wt = αt1 △ αt2 △ αt3 △ αt4 for
t = k, l, m. Then wk , wl , wm ∈ C, wk △ wl = αk1 △ αk2 △ αl1 △ αl2 ∈ C, wl △ wm =
m
m
k
k
m
αl1 △ αl2 △ αm
1 △ α2 ∈ C, and wk △ wm = α1 △ α2 △ α1 △ α2 ∈ C. By the definition
k
k
l
l
l
l
m
k
k
of C we have c(α1 △ α2 ) 6= c(α1 △ α2 ), c(α1 △ α2 ) 6= c(α1 △ αm
2 ), and c(α1 △ α2 ) 6=
m
m
c(α1 △ α2 ), which is impossible, because c takes only two values. The cases of
other colors and other numbers i′0 and i′′0 are treated in a similar way.
Thus, C ′ is N -fat and, therefore, contains (A △ A △ A △ A) ∩ B4 (κ) for some
A ∈ U . Take any α ∈ A and consider the sets A′ = {β > α : c({α, β}) = {0}} and
A′′ = {β > α : c({α, β}) = {1}}. One of these sets belongs to U , because U is
uniform. For definiteness, suppose that this is A′ . By Theorem 13 U is 3-arrow.
Therefore, there exists either an A′′ ⊂ A′ for which c([A′′ ]2 ) = {0} or β, γ, δ ∈ A′ ,
β < γ < δ, for which c([{β, γ, δ}]2 ) = {1}. In the former case, we are done. In
the latter case, we have α, β, γ, δ ∈ A, α < β < γ < δ, c({β, γ}) = c({γ, δ}) =
c({β, δ}) = 1, and c({α, β}) = c({α, γ}) = c({α, δ}) = 0 (by the definition of A′ ).
Therefore, α △ β △ γ △ δ ∈ C, which contradicts the definition of A.
(iii) ⇒ (ii) Suppose that U is a Ramsey ultrafilter on X and C is a κ-fat set in
B(X). Take any n ∈ N and consider the coloring c : [X]2n → {0, 1} defined by
(
0 if {x1 , . . . , x2n } = x1 △ . . . △ x2n ∈ C,
c({x1 , . . . , x2n }) =
1 otherwise.
Since U is Ramsey, there exists either a set An ∈ U for which [A]2n ⊂ C or
a set Y ⊂ X of cardinality κ for which [Y ]2n ∩ C = ∅. In the latter case, for
Z = [Y ]n ⊂ B(X), we have (Z △ Z) ∩ C ⊂ {0}, which contradicts C being κ-fat.

18

OL’GA V. SIPACHEVA

Hence the former case occurs, and C ∩ B2n (X) contains the trace [An ]2n ∩ B=2n (X)
of the open subgroup h(An ∪ {∗}) △(An ∪ {∗})i of B(XF ).
Thus, for each n ∈ N,Twe have found A1 , A2 , . . . , An ∈ F such that [Ai ]2i ∩
B=2i (X) ⊂ C. Let A = i≤n Ai . Then A ∈ U and [A]2i ∩ B=2i (X) ⊂ C for all
i ≤ n. Hence C ∩ B2n (X) contains the trace on B2n (X) of the open subgroup
h(A ∪ {∗}) △(A ∪ {∗})i of B(XU ) (recall that 0 ∈ C). This means that, for each n,
C ∩ B2n (X) is a neighborhood of zero in the topology induced from B(XU ).
If κ = ω, then B(XUS
) has the inductive limit topology with respect to the
decomposition B(XU ) = n∈ω Bn (XF ), because F is Ramsey (see [4]). Therefore,
in this case, C ∩ B(X) is a neighborhood of zero in the induced topology.
If κ > ω, then the ultrafilter U is countably complete [8, Lemma 9.5 and Theorem 9.6], i.e., any countable intersection of elements
Sof U belongs to U . Hence
T
A = n∈N An ∈ U , and h(A ∪ {∗}) △(A ∪ {∗})i ∩
n∈ω B2n (X) ⊂ C. Thus,
C ∩ B(X) is a neighborhood of zero in the induced topology in this case, too.
The implication (ii) ⇒ (i) is obvious.



Theorem 13 has the following purely algebraic corollary.
Corollary 16 (p = c). Any Boolean group contains ω-fat sets which are not fat
and ∆∗ -sets which are ∆∗k -sets for no k.
Proof. Theorem 4.10 of [9] asserts that if p = c, then there exists an ultrafilter
U on ω which is k-arrow for all k ∈ N but not Ramsey and, therefore, not ωarrow [9, Theorem 2.1]. By Theorem 13 the traces of all fat sets on B2 (ω) contain
those of neighborhoods of zero in B(ωU ), and there exist ω-fat sets whose traces
do not. This proves the required assertion for the countable Boolean group. The
case of a group B(X) of uncountable cardinality κ reduces to the countable case
by representing B(X) as B(κ) = B(ω) × B(κ); it suffices to note that a set of the
form C × B(κ), where C ⊂ B(ω), is λ-fat in B(ω) × B(κ) for λ ≤ ω if and only if
so is C in B(ω).

The author is unaware of where there exist ZFC examples of such sets in any
groups.
Acknowledgments
The author is very grateful to Evgenii Reznichenko and Anton Klyachko for
useful discussions.
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Department of General Topology and Geometry, Lomonosov Moscow State University, Leninskie Gory 1, Moscow 119991, Russia
E-mail address: o-sipa@yandex.ru

