The existence of free ultrafilters on ω does not imply the extension of filters on ω to ultrafilters

Mathematical Logic Quarterly 59 (4-5):258-267 (2013)
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Let X be an infinite set and let and denote the propositions “every filter on X can be extended to an ultrafilter” and “X has a free ultrafilter”, respectively. We denote by the Stone space of the Boolean algebra of all subsets of X. We show: For every well‐ordered cardinal number ℵ, (ℵ) iff (2ℵ). iff “ is a continuous image of ” iff “ has a free open ultrafilter ” iff “every countably infinite subset of has a limit point”. implies “every open filter on extends to an open ultrafilter” implies “has an open ultrafilter” implies It is relatively consistent with that (ω) holds, whereas (ω) fails. In particular, none of the statements given in (2) implies (ω).



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References found in this work

The Axiom of Choice.Thomas J. Jech - 1973 - Amsterdam, Netherlands: North-Holland.
On generic extensions without the axiom of choice.G. P. Monro - 1983 - Journal of Symbolic Logic 48 (1):39-52.

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