Ramsey-product subsets of a group
Igor Protasov, Ksenia Protasova

TL;DR
This paper introduces the concept of Ramsey-product subsets in infinite groups, showing they form a filter and relate to a specific subsemigroup of ultrafilters, advancing the understanding of algebraic and combinatorial properties of groups.
Contribution
It defines Ramsey-product subsets in groups, proves they form a filter, and links them to a subsemigroup of ultrafilters, providing new insights into group combinatorics.
Findings
The family of all Ramsey-product subsets forms a filter.
Ramsey-product subsets define a subsemigroup of ultrafilters.
The structure of these subsets relates to algebraic properties of the group.
Abstract
We say that a subset of an infinite group is a Ramsey-product subset if, for any infinite subsets , of , there exist and such that and . We show that the family of all Ramsey-product subsets of is a filter and defines the subsemigroup of the semigroup of all free ultrafilters on .
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Taxonomy
TopicsAdvanced Topology and Set Theory · Limits and Structures in Graph Theory · Mathematical and Theoretical Analysis
2010 MSC: 22A15, 03E05
Ramsey-product subsets of a group
Igor Protasov and Ksenia Protasova
Abstract.
We say that a subset of an infinite group is a Ramsey-product subset if, for any infinite subsets , of , there exist and such that and . We show that the family of all Ramsey-product subsets of is a filter and defines the subsemigroup of the semigroup of all free ultrafilters on .
Key words and phrases:
Stone-ech compactification, product of ultrafilters, strongly prime ultrafilter, sparse set, Ramsey product set.
All groups under consideration are supposed to be infinite; a countable set means a countably infinite set.
We say that a subset of a group is
- •
a *Ramsey-product subset * if, for any infinite subsets of , there exist and such that and ;
- •
a *Ramsey-square subset * if, for every infinite subset of there exist distinct elements such that , ;
- •
a *Ramsey-quotient subset * if, for every infinite subset of , there exist distinct elements such that and .
We show that above defined subsets of arise naturally in context of the Stone-ech compactification of the group endowed with the discrete topology. We identify with the set of all ultrafilters on . Then the family , where , forms a base for the topology of . Given a filter on , we denote , so defines the closed subset of , and every non-empty closed subset of can be defined in this way .
We use the standard extension [3, Section 4.1] of the multiplication on to the semigroup multiplication on . Given two ultrafilters , we choose and, for each , pick . Then and the family of these subsets forms a base of the product . We note that the set of all free ultrafilters of is a closed subsemigroup of , and the closure of is a subsemigroup of . An ultrafilter is called strongly prime.
By [5, Propositions 3, 4], the family of all Ramsey-quotient subsets of a group is a filter and is the smallest closed subset of containing all ultrafilters of the form , , where . Analogously, the closure of is defined by the filter of all Ramsey-square subsets of . In both cases, we used the classical Ramsey theorem [2, p. 16].
In this note, the key technical part plays the following version of Ramsey theorem (Theorem 6 from [2, p.98]).
Let be a finite coloring . Then there exists an infinite set and colors , , (not necessarily distinct) such that if , if and if .
Lemma 1. If is a Ramsey-product subsets of a group then, for any countable subsets of , there exist disjoint countable subsets and such that , .
Proof. Passing to subsets of and , we may suppose that . We enumerate , and define a -coloring of by the rule: if , , , and otherwise. By Theorem 6 from [2, p.98], there exists a countable subset of such that is monochrome on all , such that , , . We partition into two infinite subsets and denote , . Since is a Ramsey-product set, there are , such that , . By the choice of , we have , for all , .
Lemma 2. For a free ultrafilter on a group , if and only if, for every subset , there exist two countable subsets and of such that
Proof. We assume that , take an arbitrary and choose such that . We choose and , such that . We take an arbitrary countable subset of and choose an injective sequence in such that , . Then .
On the other hand, let and , are chosen so that . We take an arbitrary free ultrafilters such that , . Then , so and .
We recall [1] that a subset of a group is sparse if, for every infinite subset of , there exists a non-empty finite subset such that is finite. By [1, Theorem 9], an ultrafilter is strongly prime if and only if there exists a sparse subset of such that . For sparse subsets see also [4], [6].
Teorem. For every group , the following statements hold:
* the family of all Ramsey-product subsets of is a filter;*
* ;*
* if and only if is sparse.*
Proof. We take two Ramsey-product subsets and of and show that Let be infinite subsets of . By Lemma 1, there exist disjoint countable subsets and such that , for all , . Since is a Ramsey-product subsets, there exist , such that , . Then , , so .
To show that , we suppose the contrary, choose and take such that . By Lemma 2, there are countable sets and such that . We use Lemma 1 to choose disjoint countable subsets , such that , . To get a contradiction, we take an arbitrary and pick such that .
Now we take an arbitrary and prove that . Given any , by Lemma 1, there exist two disjoint countable subsets of such that, for any countable subsets and , either or . We enumerate , and define a -coloring of by the rule: if and only if and or and . Applying Theorem 6 from [2, p. 98], we get an infinite subset of such that the restriction of to is monochrome, and the restriction of to is monochrome. By the choice of and , either or . In the first case, we choose two injective sequences in and in such that . In the second case, we choose two injective sequences in and in such that . Applying Lemma 2, we conclude that .
We apply and Theorem 9 from [1].
We recall that a subset of a group is *large (extralarge) * if there is a finite subset of such that ( is large for every large subset of ). By Theorem 4.1 from [4] , the complement of sparse subset is extralarge. Applying , we see that each Ramsey-product subset is extralarge.
By the statement , every Ramsey-product subset of is a member of every idempotent of the semigroup , in particular, every idempotent from the minimal ideal of , so is combinatorially very rich (see [3, Section 14]).
If is an amenable group and is a sparse subset of , by Theorem 5.1 from [4], for every left invariant Banach measure on . Applying (iii), we see that for every Ramsey-product subset of and every left invariant Banach measure on .
We conclude the note with the following formal generalization. We fix two infinite subset of a group and say that a subset of is a *Ramsey -product subset * if, for any infinite subsets and , there exist and such that , . Then the family of all Ramsey -product subsets of is a filter and .
The reference list from the paper itself. Each links out to its DOI / PubMed record.
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- 2[2] R. Graham, B. Rotschild, J. Spencer, Ramsey Theory , Willey, New York, 1980.
- 3[3] Hindman N., Strauss D. Algebra in the Stone- C ˇ ˇ 𝐶 \check{C} ech compactification , de Gruyter, Berlin, 1998.
- 4[4] Ie. Lutsenko, I. Protasov, Sparse, thin and other subsets of groups , Algebra Computation 19 (2009), 491-510.
- 5[5] I. Protasov, K. Protasova, On recurrence in G 𝐺 G -spaces, Algebra Discrete Math. (to appear), preprint (arxiv: 1703, 00696).
- 6[6] I. Protasov, S. Slobodianiuk, Ultracompanions of subsets of a group, Comment. Math. Univ. Carolin. 55.2 (2014), 257-265.
