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Oct 26, 2021 at 9:13 vote accept Dominic van der Zypen
Oct 25, 2021 at 23:47 comment added Wojowu @BjørnKjos-Hanssen That's a good point!
Oct 25, 2021 at 23:37 comment added Bjørn Kjos-Hanssen @Wojowu actually you need condition 1, since otherwise $E$ could consist of all the singletons.
Oct 25, 2021 at 23:34 answer added Bjørn Kjos-Hanssen timeline score: 2
Oct 25, 2021 at 22:55 comment added Wojowu It also seems to me that the condition 2 alone, for any one pair $a,b$ incomparable under $\subsetneq$, itself is inconsistent. For instance if they are disjoint, we can just map them by some bijection to some pair of disjoint subsets of the same sizes and with equal sums.
Oct 25, 2021 at 22:52 comment added Wojowu Isn't your notion of "non-nested" simply an antichain?
Oct 25, 2021 at 22:39 history asked Dominic van der Zypen CC BY-SA 4.0