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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Dec 13, 2009 at 23:17 comment added Harrison Brown Ah, wait, I think I see why that doesn't work. Never mind, then.
Dec 13, 2009 at 23:14 comment added Harrison Brown "The only proof I know uses linear algebra": I haven't thought this through all the way, but I would think that $B_n/G$ would inherit a weighting of its elements from the LYM inequality, from which the fact that $B_n/G$ is Sperner would follow from some easy counting argument. Does this not work? Or is it what you were referring to?
Dec 13, 2009 at 23:14 history edited Qiaochu Yuan CC BY-SA 2.5
added 182 characters in body
Dec 13, 2009 at 23:07 history answered Qiaochu Yuan CC BY-SA 2.5