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The existence of free ultrafilters on ω does not imply the extension of filters on ω to ultrafilters
Author(s) -
Hall Eric J.,
Keremedis Kyriakos,
Tachtsis Eleftherios
Publication year - 2013
Publication title -
mathematical logic quarterly
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.473
H-Index - 28
eISSN - 1521-3870
pISSN - 0942-5616
DOI - 10.1002/malq.201100092
Subject(s) - ultrafilter , mathematics , extension (predicate logic) , nowhere dense set , filter (signal processing) , countable set , combinatorics , set (abstract data type) , discrete mathematics , computer science , computer vision , programming language
Let X be an infinite set and let BPI ( X ) and UF ( X ) denote the propositions “ every filter on X can be extended to an ultrafilter ” and “ X has a free ultrafilter ”, respectively. We denote by S ( X ) the Stone space of the Boolean algebra of all subsets of X . We show: For every well‐ordered cardinal number ℵ, UF (ℵ) iff UF (2 ℵ ).UF ( ω ) iff “ 2 ωis a continuous image ofS ( ω ) ” iff “ S ( ω )has a free open ultrafilter ” iff “ every countably infinite subset ofS ( ω )has a limit point ”.BPI ( ω ) implies “ every open filter on2 ωextends to an open ultrafilter ” implies “ 2 ω has an open ultrafilter ” implies UF ( ω ) .It is relatively consistent with ZF that UF (ω) holds, whereas BPI (ω) fails. In particular, none of the statements given in (2) implies BPI (ω).