Say $A_1,A_2,...,A_k\subseteq \{1,2,...,11\}$ are such a distinct sets that for any three of them at least two are not comparable by inclusion. What is a maximum of $k$?
I'v got idea for this problem from Sperner theorem. It is clear that family of all $5$ and $6$ element sets satisfy this condition, so the result would be $${11\choose 5} +{11\choose 6} = 924$$ But now i'm not sure how to prove that if $k\geq 925$ then such family does not exsist. Perhaps Dilworth theorem would help? If $k=925$ possible then we should have $3$ sets from different antichains, so we must cover this family of sets with at least three antichains and thus there must be a chain of size at least $3$ but that is something we don't want.
Is this correct?
Hmm, I'm almost sure now this is not correct. Else we could prove Sperner theorem like this, but I have never seen something like this.
By Sperner's theorem, the maximum size of an antichain in $\mathcal P(\{1,\dots,11\})$ is $\binom{11}5=\binom{11}6=462$. It follows by Dilworth's theorem that $P(\{1,\dots,11\})$ is the union of $462$ chains. Since your set $\mathcal A$ contains at most $2$ elements from each chain, it follows that $|\mathcal A|\le462\cdot2=924$.