Let $A$ be a set of size $k$ and $B$ be a set of size $\ell$, for fixed $k$ and $\ell$, and such that $A\cap B=\emptyset$. What is the (or a) Sperner family $\mathcal{F}$ on $A\cup B$ for which $\mathcal{F}_B=\{C\cap B ~:~ C\in\mathcal{F}\}$ is maximized?

I actually just need an upper bound to $|\mathcal{F}_B|$ (possibly something better than $2^\ell$, which seems to be loose if $2^k<\ell$)

Any hint or reference where this kind of information or relevant material could be found would be much appreciated. Thanks.


1 Answer 1


The correct upper bound is the sum of the $2^k$ most central binomial coefficients: $$|\mathcal{F}_B|\leq\binom \ell {\lfloor (\ell - 2^k)/2+1\rfloor} + \dots + \binom \ell {\lfloor(\ell + 2^k)/2\rfloor},$$ or just $|\mathcal{F}_B|\leq 2^\ell$ if $2^k\geq \ell+1$. The sets $\{B\cap C\mid C\in\mathcal F\text{ and }A\cap C=A'\}$ are antichains. By the LYM inequality the union of $2^k$ antichains cannot be larger than the sum of the largest $2^k$ binomial coefficients. To achieve the bound, let $A=\{a_0,\dots,a_{k-1}\}$, and let $$\mathcal F=\left\{C\subseteq A\cup B \mid \lfloor(\ell + 2^k)/2\rfloor- \sum_{i\colon a_i\in C} 2^i=|C\cap B| \right\}.$$

  • $\begingroup$ Thank you, Colin. I believe it is the correct answer, as we got the same result in a much more convoluted way. $\endgroup$
    – Matteo
    Oct 24, 2012 at 2:12

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