Given that the set S which has odd number of elements, is there any special way to prove that there are exactly two antichains length of $\binom{n}{\lfloor n/2 \rfloor}$?
For example, for S={1,2,3,4,5,6,7}, there are exactly two antichains of length 35, anti chain of 3-subsets and antichain of 4-subsets.
At first, I thought showing that no antichain of 1,2,5,6,7-subsets would not have length of 35 is enough for the proof.
But at the second thought, it seems to have a lot of problems in that proof.
Thank you.
Best Answer
In the spirit of Dilworth's theorem, but without using it, one can show the two obvious antichans you mentioned are of maximal size, by partitioning $\mathcal P(S)$ into chains that each meet both of your antichains (obviously in a unique point). Then there are as much chains as elements in each of your antichains, and no antichain can have more elements than that (lest it would have at least two elements in common with one of the chains by the pigeonhole principle, which would be absurd).
Here is an explicit partition, defined by two operations $e,f$ on $\mathcal P(S)$, that (under given conditions) will add respectively remove an element from a subset (so they always produce a comparable subset), and such that whenever one operation has acted the other can be applied to the result, and has the inverse effect. Given this it is easy to see that convertibility under these operation defines an equivalence relation whose equivalence classes are chains; moreover it will turn out that each equivalence class contains a subset with $\lfloor n/2\rfloor$ elements and one with $\lceil n/2\rceil$ elements.
The elements of $S$ are totally ordered, so given a subset $U$ of $S$ one can consider for each inital part $[1,k]$ of $S$, the sets $U_k=U\cap[1,k]$ and $U'_k=[1,k]\setminus U$ and in terms of them define the number $l_k(U)=|U_k|-|U'_k|$. If $l_k(U)<0$ for any $k$, then the operation $e$ is defined for $U$, and operates by adding to $U$ the smallest value $k$ for which $l_k(U)$ achieves its most negative value (it is clear that this $k$ was absent from$~U$). The operation $f$ is defined whenever there exists $k\geq0$ for which $l_k(U)<l_n(U)$, and removes $k$ where $k-1$ is the largest value (possibly $0$) for which $l_{k-1}(U)$ achieves its smallest value (it is clear that this $k$ must be present in$~U$).
Since $e$, if it adds $k$ to $U$, raises the value of $l_i$ by$~2$ for all $i\geq k$, and $f$, if it removes $k$ decreases the value of $l_i$ by$~2$ for all $i\geq k$, it can be easily seen that each operation reverses the effect of the other (notably that it removes/adds the same vaue that was just added/removed). Also $e$ applies to all subsets with less than $n/2$ elements (since then $l_n(U)<0$), and $f$ applies to all subsets with more than $n/2$ elements (since $l_0(U)<l_n(U)$), so we one can always move in the direction of the more balanced subsets.