An Antichain in P(n) has size at most $\binom{n}{\lfloor n/2 \rfloor}$.

Proof

We first show that there is a Partition pf P(n) into $\binom{n}{\lfloor n/2 \rfloor}$. chains.

To do that we show that there are Complete matchings between subsequent layers of the Power set (which is isomorphic to the Discrete cube). To show these exist, we show that the condition for Hall's theorem is satisfied. This is done by counting the number of edges between them, which depend on the sizes of the sets in Hall's theorem..

First proof of Sperner’s Lemma. This is now easy. A chain and an antichain meet in at most one element. We have partitioned P(n) into $\binom{n}{\lfloor n/2 \rfloor}$chains, so no antichain can have more than $\binom{n}{\lfloor n/2 \rfloor}$ elements.

See full proof here

Sperner’s Lemma tells us the maximal size of an antichain, but what can we say about uniqueness? And what happens if we start using sets of different sizes? The LYM inequality gives a much more refined picture.

See also Dilworth's lemma