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Aug 9 at 20:28 vote accept Dominic van der Zypen
Aug 9 at 14:42 comment added Andreas Blass This argument essentially shows that sep$(\kappa)$ is the smallest $\lambda$ such that $2^\lambda\ge\kappa$. (This characterization of sep$(\kappa)$ would have been my comment on the question, but I put it here because you've provided all the ingredients of the proof.)
Aug 9 at 9:47 history edited Calliope Ryan-Smith CC BY-SA 4.0
Stronger, more precise result.
Aug 9 at 9:01 history answered Calliope Ryan-Smith CC BY-SA 4.0