2008
DOI: 10.1007/s11856-008-1017-y
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Resolvability and monotone normality

Abstract: A space X is said to be κ-resolvable (resp., almost κ-resolvable) if it contains κ dense sets that are pairwise disjoint (resp., almost disjoint over the ideal of nowhere dense subsets). X is maximally resolvable if and only if it is ∆(X)-resolvable, where ∆(X) = min{|G| : G = ∅ open}.We show that every crowded monotonically normal (in short: MN) space is ω-resolvable and almost µ-resolvable, where µ = min{2 ω , ω 2 }. On the other hand, if κ is a measurable cardinal then there is a MN space X with ∆(X) = κ su… Show more

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Cited by 9 publications
(8 citation statements)
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“…Let F = G ∩ V 1 and consider the space X = X(F ). As it was observed in [6], spaces obtained as X(H) from some filter H are monotonically normal and ωresolvable.…”
mentioning
confidence: 63%
See 1 more Smart Citation
“…Let F = G ∩ V 1 and consider the space X = X(F ). As it was observed in [6], spaces obtained as X(H) from some filter H are monotonically normal and ωresolvable.…”
mentioning
confidence: 63%
“…The space X = X(U) is monotonically normal by [6,Theorem 3.1]. An ultrafilter U is λ-descendingly complete if {U ζ : ζ < λ} = ∅ for each decreasing sequence {U ζ : ζ < λ} ⊂ U.…”
mentioning
confidence: 99%
“…Construction. We use the following, somewhat different topology, inspired by filtration spaces [6] and resolutions [13].…”
Section: Non-reconstructible Spaces and Propertiesmentioning
confidence: 99%
“…Indeed, this space is countable, and hence it is trivially a σ-space. Moreover, it is monotonically normal by Theorem 3.2 of [39]. If F is a Ramsey ultrafilter (which exists, for example, if one assumes CH), then Seq(F ) is even a topological group (see [63]).…”
Section: D-separabilitymentioning
confidence: 99%