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Dec 12, 2015 at 6:39 history edited Daniel Soltész
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Jul 24, 2014 at 16:03 history edited Daniel Soltész
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Jun 3, 2014 at 22:28 answer added t3suji timeline score: 6
Jun 3, 2014 at 22:08 answer added Kellen Myers timeline score: 4
Jun 3, 2014 at 20:33 comment added Daniel Soltész @eins6180 A little better but still simple upper bound would be that all the $2 \binom{|\mathcal{A}|}{2}$ differences are different so there can not be more than $2^n$ of them which means that $|\mathcal{A}|$ can not be asymptotically bigger than $2^{n/2}$.
Jun 3, 2014 at 20:26 comment added eins6180 Here is a simple and probably much too large upper bound: Any maximal antichain in the lattice of all subsets of $[n]$ together with two other subsets doesn't work. So the family $\mathcal{A}$ cannot be bigger than ${n \choose [n/2]} + 1$.
Jun 3, 2014 at 20:20 comment added Daniel Soltész @Seva Nice construction, you can also add [1,m] and [m+1,n] to obtain a system of $\lfloor n/2 \rfloor +2$ elements.
Jun 3, 2014 at 19:52 comment added Seva Constructing $m:=\lfloor n/2\rfloor$ sets is easy: assuming that the ground set is $[1,n]$, take $A_i:=\{i\}\cup[m+1,n]\setminus\{m+i\}\ (1\le i\le m)$.
Jun 3, 2014 at 19:37 history edited Daniel Soltész CC BY-SA 3.0
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Jun 3, 2014 at 18:04 history asked Daniel Soltész CC BY-SA 3.0