TITLE: Non-containing subsets of two sizes
QUESTION [6 upvotes]: Let $T=\{1,2,\dots,2n+1\}$. What is the largest $k$ such that we can choose $k$ subsets of size $n$ and $k$ subsets of size $n+1$ of $T$ so that no chosen subset contains another?
$k=\binom{2n}{n-1}$ is possible: Choose all sets of size $n$ containing $1$, and all sets of size $n+1$ not containing $1$. Does it follow from Sperner's theorem that this is the best possible?

REPLY [2 votes]: Here's another approach:  Given families $\mathcal{A}$ and $\mathcal{B}$ of $n$-sets and $(n+1)$-sets (respectively) that obey the condition, replace $\mathcal{B}$ with the family $\mathcal{B}'$ of complements of sets in $\mathcal{B}$.
Now the condition that no set in $\mathcal{A}$ is contained in one in $\mathcal{B}$ becomes that every set in $\mathcal{A}$ intersects non-trivially with every set in $\mathcal{B}'$.  Such a system is called cross-intersecting.
Now you can apply combinatorial shifting to $\mathcal{A}$ and $\mathcal{B}'$, and show that it preserves the intersection property.  This reduces showing the proof of your inequality to shifted families, where it is likely to be much easier.  (I haven't worked out all the details here.)