# Maximum cardinality of a set of subsets

Let $$N$$ be a system of subsets of the set $$X = \{1,2,3,\cdots ,n \}$$ such that there are no three elements $$A,B,C \in N$$ such that $$A \subset B \subset C$$. Prove that $$|N| \leq 2 \cdot {{n}\choose{ \lfloor n/2 \rfloor}}.$$

I have thought about the fact that a semi-independent system of subsets can have at most two elements in common with any chain from $$(\mathcal{P}(X), \subseteq),$$ but I do not know how to continue using Sperner's theorem.

Let $$\mathcal{M}$$ be the family of all maximal chains. (We call a chain $$\{\varnothing\subsetneq A_1\subsetneq A_2\subsetneq\cdots \subsetneq A_n=X\}\subset 2^X$$ of length $$n+1$$ as a maximal chain.) Define $$Q=\sum_{M\in \mathcal{M}}\sum_{A\in N} 1_{A\in M}$$ where $$1_{A\in M}=1$$ if $$A\in M$$ and is $$0$$ otherwise. Since for every maximal chain $$M\in\mathcal{M}$$, there are at most $$2$$ distinct $$A\in N$$ such that $$A\in M$$, we have $$Q\le\sum_{M\in \mathcal{M}}2=2|\mathcal{M}|=2\cdot n!.$$On the other hand, we have $$Q=\sum_{A\in N}\sum_{M\in \mathcal{M}}1_{A\in M}=\sum_{A\in N}|A|!(n-|A|)!$$ since the number of maximal chains containing $$A$$ is $$|A|!(n-|A|)!$$. Note that $$j!(n-j)!$$ is minimized when $$j=\lfloor \frac{n}{2}\rfloor$$. Thus, we have $$Q= \sum_{A\in N}|A|!(n-|A|)!\ge \lfloor \frac{n}{2}\rfloor!\left(n-\lfloor \frac{n}{2}\rfloor\right)!|N|$$ Gathering them together, we get $$\lfloor \frac{n}{2}\rfloor!\left(n-\lfloor \frac{n}{2}\rfloor\right)!|N|\le 2\cdot n!$$or $$|N|\le 2\cdot \frac{n!}{\lfloor \frac{n}{2}\rfloor!\left(n-\lfloor \frac{n}{2}\rfloor\right)!}=2\cdot \binom{n}{\lfloor \frac{n}{2}\rfloor},$$ as desired.
If $$X$$ does not contain a $$3$$-elament chain $$A\subset B\subset C$$, then $$X$$ is the union of two antichains. Namely, the set of all minimal elements of $$X$$ is an antichain, and the set of all non-minimal elements of $$X$$ is another antichain. By Sperner's theorem, an antichain of subsets of $$\{1,2,3,\dots,n\}$$ has at most $$\binom n{\lfloor n/2\rfloor}$$ elements, so your set $$X$$ has at most twice that many elements.