Exactly -resolvable Topological Expansions
arXiv:1008.5371
Abstract
For a cardinal, a space $X=(X,\sT)$ is -{\it resolvable} if admits -many pairwise disjoint $\sT$-dense subsets; $(X,\sT)$ is {\it exactly} -{\it resolvable} if it is -resolvable but not -resolvable. The present paper complements and supplements the authors' earlier work, which showed for suitably restricted spaces $(X,\sT)$ and cardinals that $(X,\sT)$, if -resolvable, admits an expansion $\sU\supseteq\sT$, with $(X,\sU)$ Tychonoff if $(X,\sT)$ is Tychonoff, such that $(X,\sU)$ is -resolvable for all but is not -resolvable (cf. Theorem~3.3 of \cite{comfhu10}). Here the "finite case" is addressed. The authors show in ZFC for : (a) every -resolvable space $(X,\sT)$ admits an exactly -resolvable expansion $\sU\supseteq\sT$; (b) in some cases, even with $(X,\sT)$ Tychonoff, no choice of $\sU$ is available such that $(X,\sU)$ is quasi-regular; (c) if -resolvable, $(X,\sT)$ admits an exactly -resolvable quasi-regular expansion $\sU$ if and only if either $(X,\sT)$ is itself exactly -resolvable and quasi-regular or $(X,\sT)$ has a subspace which is either -resolvable and nowhere dense or is -resolvable. In particular, every -resolvable quasi-regular space admits an exactly -resolvable quasi-regular expansion. Further, for many familiar topological properties $\PP$, one may choose $\sU$ so that $(X,\sU)\in\PP$ if $(X,\sT)\in\PP$.