Minimal chain covering of all subsets of {$1,2,3,4,....,2n,2n+1$}

Question

Let $n\in\mathbb{N^*}$

Let $M=${$1,2,3,....,2n,2n+1$}

Let a chain be a family of $2n+2$ subsets of $M$ such that if $F=${$A_0,A_1,A_2,...A_{2n+1}$} is a chain, then $|A_i|=i$ for every $i\in${$0,1,2,....,2n+1$} and $A_i\subset A_{i+1}$ for every $i\in${$0,1,2,....,2n$}

Prove that there is a set of ${2n+1}\choose n$ chains such that every subset of $M$ is in at least one chain.

My attempt:

My best idea is to construct the chains iductively but it is either impossible or very hard because we would need the induction to go $2$ steps at a time. I know the theorem that says that the minimal chain covering is as big as the biggest antichain but I can't prove that this is the length of the biggest antichain.

Answer 1

This is Sperner's theorem. A maximal antichain on $\{1,2,\ldots,k\}$ has size $$\binom{k}{\lfloor k/2 \rfloor}.$$