Problem

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Tags: floor function, combinatorics proposed, combinatorics



suppose that $\mathcal F\subseteq p(X)$ and $|X|=n$. we know that for every $A_i,A_j\in \mathcal F$ that $A_i\supseteq A_j$ we have $3\le |A_i|-|A_j|$. prove that: $|\mathcal F|\le \lfloor\frac{2^n}{3}+\frac{1}{2}\dbinom{n}{\lfloor\frac{n}{2}\rfloor}\rfloor$ (20 points)