Recall Sperners's Theorem:
Sperners's Theorem: Let $M$ be a finite set and note that its power set $\mathcal{P}(M)$ is a partial ordered set by inclusion. The maximal size of an antichain in $\mathcal{P}(M)$ is $\binom{n}{\lfloor n/2 \rfloor}$.
Prover Sperner's Theorem in the following way:
a) Find an antichain of size $\binom{n}{\lfloor n/2 \rfloor}$.
b) If $\mathcal{P}(M)$ is union of $r$ chains, then an antichain in $\mathcal{P}(M)$ is has size at most $r$.
c) Let $G := (\mathcal{P}(M),E)$ be the graph such that there is an edge between $A,B \subseteq M$ if and only if $A \subset B$ and $\lvert A \setminus B \rvert = 1$. Let furthermore be $\mathcal{X}_k$ the subsets of cardinality $k$ of $M$ and let $G_k := G[\mathcal{X}_k \cup \mathcal{X}_{k+1}]$ for $0 \le k \le n-1$. Show that $G_k$ has a matching that saturates $\mathcal{X}_k$ if $k < n/2$ and that $G_k$ has a matching that saturates $\mathcal{X}_{k+1}$ if $k+1 > n/2$.
d) Connect the matchings to write $\mathcal{P}(M)$ as a union of (pairwise disjoint) chains, argue that there are at most $\binom{n}{\lfloor n/2 \rfloor}$ chains.
I realise that $\mathcal{P}(M)$ being the union of at most $\binom{n}{\lfloor n/2 \rfloor}$ chains by point d) implies Sperner's Theorem by point b).
Now to the subpoints:
a) The set $\big\{ A \subseteq M \mid \lvert A \rvert = \binom{n}{\lfloor n/2 \rfloor} \big\}$ is such an antichain.
b) Any set of of $r+1$ subsets has to contain two elements of the same chain by the pigeon hole principle.
c) We note that every $A \in \mathcal{X}_k$ has $n-k$ neighbours and that $\lvert \mathcal{X}_k \rvert = \binom{n}{k}$ and $\lvert \mathcal{X}_{k+1} \rvert = \binom{n}{k+1}$. Furthermore we note that for fixed $n$ the binomial coefficient $\binom{n}{k}$ reaches the maximum at $n/2$ for even $n$ and at $\lfloor n/2 \rfloor, \lceil n/2 \rceil$ for odd $n$. So the condition $k < n/2$ implies $\lvert \mathcal{X}_k \rvert < \lvert \mathcal{X}_{k+1} \rvert$ and the condition $k+1 > n/2$ implies $\lvert \mathcal{X}_k \rvert > \lvert \mathcal{X}_{k+1} \rvert$. However, I do not see how to argue that these matchings exist from there, maybe somehow via Hall's theorem?
d) Here I can see that there are at most $\binom{n}{\lfloor n/2 \rfloor}$ chains, as $\binom{n}{\lfloor n/2 \rfloor}$ is the maximum value of the binomial coefficients $\binom{n}{k}$ for fixed $n$ (as detailed in c) ). But I do not understand why these chains should be disjoint as there are only $n$ subsets of $M$ with cardinality $1$ to serve as a start for a chain.
Could you please help me?