Sperner's Theorem

We know that if more than a half of subsets of an n-set A have been selected, there are bound to be at least two of which one contains another. This is proven with the help of the Pigeonhole Principle. Below we prove by far a stronger result - the Sperner's Theorem. It appears that a half is a few too many; there is no need to take this many subsets.

To remind, C(n, m) is a binomial coefficient

that appears in the Binomial Theorem which, for an integer exponent, can be written as

For an even n, there are an odd number of binomial coefficients with the middle coefficient C(n,n') corresponding to n' = n/2. For n odd, the number of coefficients is even and there are two equal coefficients, one corresponding to n' = (n-1)/2 and the other to n' = (n+1)/2. For even n, [(n+1)/2] = n/2, where [] means "the integer part". At this time, I shall assume the unimodal behavior of the sequence of binomial coefficients: C(n,k) is not decreasing as k changes from k = 0 to k = n' and is not increasing afterwards. Therefore, for any n, C(n,n') with n' = [(n+1)/2] is the largest binomial coefficient:

For all 0kn, C(n,k)C(n,n').

C(n,k) is the number of ways to select k (unordered) elements out of n. It follows that the number of sets in any antichain in which all sets have the same number of elements is bounded above by C(n,n').

A maximal chain in the n-set A is a sequence X₀,X₁,...,X_n of subsets of A such that for all i card(X_i), cardinality or the number of elements in the set X_i, is i; and X_iX_i+1, for i=0,...,n-1. Obviously, there is a 1-1 correspondence between permutations of the set A and its maximal chains: any permutation {x₁ x₂ ...x_n} gives rise to a maximal chain

{x1}{x1,x2}...{x1,x2,...,xn} = A

and vice versa. Let B = {B₁,B₂,...,B_t} be an antichain in A, with card(B_i) = b_i, i=1,...,t. By definition, any chain in A - maximal or not - can contain at most one of the sets B_i. Now, there are b_i! ways of forming a chain from to B_i, and there are (n -b_i )! ways of extending this chain to a maximal one. Thus the number of maximal chains that contain the set B_i is b_i!(n -b_i )!. We therefore have the inequality

Suppose that among the cardinalities {b_i} there are p₁ 1s, p₂ 2s, ..., p_n n's. So that p_k's are nonnegative integers that sum up to t. The previous inequality can be rewritten as

This is known as the Lubell-Yamamoto-Meshalkin (LYM) inequality. From here it follows at once that

or tC(n,n').

References

1. V.K.Balakrishnan, Theory and Problems of Combinatorics, Schaum's Outline Series, McGraw-Hill, 1995