There may be no Hausdorff ultrafilters
classification
🧮 math.LO
keywords
hausdorffthereultrafiltersconsistentfunctionsmappingsomeultrafilter
read the original abstract
An ultrafilter U is Hausdorff if for any two functions f,g mapping N to N, f(U)=g(U) iff f(n)=g(n) for n in some X in U. We will show that it is consistent that there are no Hausdorff ultrafilters.
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