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Feb 28, 2017 at 7:37 comment added Asaf Karagila Now I'm not sure whether or not the standard notation is that $\operatorname{FA}_\kappa$ means there is a filter meeting every $\kappa$ dense sets, or if I should just add the requirement that $\kappa>\aleph_1$.
Feb 27, 2017 at 18:56 comment added Asaf Karagila You're right. There's also some discrepancy in the example I gave in the question itself. I'll fix this later when I'm at a computer again. Thanks!
Feb 27, 2017 at 18:37 comment added Miha Habič The way you wrote it, PFA is compatible with CH (in particular it doesn't imply $\kappa=\aleph_2$). To make things nontrivial we should just always require that $\kappa$ is at least $\aleph_2$ when talking about $\mathrm{FA}_\kappa$ (this applies to your question too).
Feb 27, 2017 at 18:18 answer added Mohammad Golshani timeline score: 3
Feb 27, 2017 at 16:10 answer added Will Brian timeline score: 7
Feb 27, 2017 at 15:32 history asked Asaf Karagila CC BY-SA 3.0