English

Exactly $n$-resolvable Topological Expansions

General Topology 2023-11-21 v2

Abstract

For κ\kappa a cardinal, a space X=(X,\sT)X=(X,\sT) is κ\kappa-{\it resolvable} if XX admits κ\kappa-many pairwise disjoint \sT\sT-dense subsets; (X,\sT)(X,\sT) is {\it exactly} κ\kappa-{\it resolvable} if it is κ\kappa-resolvable but not κ+\kappa^+-resolvable. The present paper complements and supplements the authors' earlier work, which showed for suitably restricted spaces (X,\sT)(X,\sT) and cardinals κλω\kappa\geq\lambda\geq\omega that (X,\sT)(X,\sT), if κ\kappa-resolvable, admits an expansion \sU\sT\sU\supseteq\sT, with (X,\sU)(X,\sU) Tychonoff if (X,\sT)(X,\sT) is Tychonoff, such that (X,\sU)(X,\sU) is μ\mu-resolvable for all μ<λ\mu<\lambda but is not λ\lambda-resolvable (cf. Theorem~3.3 of \cite{comfhu10}). Here the "finite case" is addressed. The authors show in ZFC for 1<n<ω1<n<\omega: (a) every nn-resolvable space (X,\sT)(X,\sT) admits an exactly nn-resolvable expansion \sU\sT\sU\supseteq\sT; (b) in some cases, even with (X,\sT)(X,\sT) Tychonoff, no choice of \sU\sU is available such that (X,\sU)(X,\sU) is quasi-regular; (c) if nn-resolvable, (X,\sT)(X,\sT) admits an exactly nn-resolvable quasi-regular expansion \sU\sU if and only if either (X,\sT)(X,\sT) is itself exactly nn-resolvable and quasi-regular or (X,\sT)(X,\sT) has a subspace which is either nn-resolvable and nowhere dense or is (2n)(2n)-resolvable. In particular, every ω\omega-resolvable quasi-regular space admits an exactly nn-resolvable quasi-regular expansion. Further, for many familiar topological properties \PP\PP, one may choose \sU\sU so that (X,\sU)\PP(X,\sU)\in\PP if (X,\sT)\PP(X,\sT)\in\PP.

Keywords

Cite

@article{arxiv.1008.5371,
  title  = {Exactly $n$-resolvable Topological Expansions},
  author = {W. W. Comfort and Wanjun Hu},
  journal= {arXiv preprint arXiv:1008.5371},
  year   = {2023}
}
R2 v1 2026-06-21T16:07:37.046Z