Skip to main content
8 events
when toggle format what by license comment
Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Apr 13, 2017 at 12:19 history edited CommunityBot
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
May 5, 2014 at 17:45 comment added Michał Kukieła @Seb Destercke: Of course you are right. I corrected the proof, hopefully without introducing new mistakes. However I feel the question deserves a more elegant answer.
May 5, 2014 at 17:43 history edited Michał Kukieła CC BY-SA 3.0
corrected proof
Apr 26, 2014 at 8:24 comment added Seb Destercke Thank you for the nice proof, yet if I am not wrong for the two disjoint 2-element chains we have $$3\cdot 3 \cdot 2^{n-4}=(2^3+1)\cdot 2^{n-4}=2^{n-1}+2^{n-4} > 2^{n-1}+1$$
Apr 25, 2014 at 16:12 comment added Emil Jeřábek @VinceVatter: $2^{2-1}+2\ne3$. Anyway, there are $2^{n-1}$ antichains that are subsets of the nonminimal elements (including the empty antichain), and one antichain consisting of the minimal element.
Apr 25, 2014 at 12:31 vote accept Seb Destercke
Apr 25, 2014 at 10:20 history answered Michał Kukieła CC BY-SA 3.0