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Dense Exact Factorizations of Countable Topological Groups
Abstract
We prove that every countably infinite group endowed with a topology of countable weight in which every nonempty open set is infinite admits an exact factorization into two dense infinite subsets. More generally, every finite partial exact factorization with nonempty factors extends to an exact factorization of the whole group in which both factors meet every member of a prescribed at most countable family of infinite subsets in infinitely many points. This settles the countable case of a question of Protasov and Slobodianiuk. The proof uses the finite graph Ramsey theorem, invariant finitely additive probability measures on amenable groups, and a structural lemma on quadratic coverings based on Neumann's coset covering lemma.
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BibTeX
@misc{kabenyuk2026dense,
title = {{Dense Exact Factorizations of Countable Topological Groups}},
author = {Mikhail Kabenyuk},
howpublished = {arXiv},
year = {2026},
month = nov,
doi = {10.48550/arxiv.2611.08171},
eprint = {2611.08171},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2611.08171},
}
RIS
TY - UNPB
TI - Dense Exact Factorizations of Countable Topological Groups
AU - Mikhail Kabenyuk
PY - 2026
DA - 2026/11/05
PB - arXiv
DO - 10.48550/arxiv.2611.08171
UR - https://arxiv.org/abs/2611.08171
AB - We prove that every countably infinite group endowed with a topology of countable weight in which every nonempty open set is infinite admits an exact factorization into two dense infinite subsets. More generally, every finite partial exact factorization with nonempty factors extends to an exact factorization of the whole group in which both factors meet every member of a prescribed at most countable family of infinite subsets in infinitely many points. This settles the countable case of a question of Protasov and Slobodianiuk. The proof uses the finite graph Ramsey theorem, invariant finitely additive probability measures on amenable groups, and a structural lemma on quadratic coverings based on Neumann's coset covering lemma.
ER -