The question is simply: let $|S| = n,$ how big can a subset $I$ of $2^S$ be such that for any $A, B \in I,$ $A\not\subset B$ (so, this is an independent set in the graph whose vertices are subsets, and edges correspond to inclusion in some direction). It is clear that $|I| \geq \binom{n} {\lfloor \frac{n}{2} \rfloor},$ but is that sharp? 


**Update** As Gerhard points out, the answer is YES. Now a followup: what about a family which has no $k$-element chain $A\subset B \subset \dotsc \subset C?$



Independent sets of subsets
To address the updated question, a family of subsets of $[n]$ is called *$k$-Sperner* if it does not contain a chain of length $k+1$. By taking all sets whose size lies in the middle $k$ values of $[n]$, there exist $k$-Sperner families who size is the sum of the $k$ middle bịnomial coefficients. Erdős proved that this bound is tight in this [paper](https://projecteuclid.org/euclid.bams/1183507531) (see Theorem 5). The extremal example is also essentially unique (for parity reasons there may be two intervals of middle $k$ values). 


