A probabilistic proof for the lym-inequality

Here we give a short, inductive argument yielding (1). First note that (1) is evident if n = 1, and also if X E S (in the latter case necessarily 9 -= 1.x) holds). Now assume (1) is true for n 1, !F is an antichain and X 4 9. Let x be a random variable which takes the values 1,. . . , n; each with probability l/n. Let us define s(x) = {FE ZF: x9! fl. For any function g(x), we denote by E(g(x)) its expectation. As 9(x) is an antichain on X-(x}, for 9(.x) (1) hoids with n 1 instead of n. We infer (~(FE 9:(x):r denotes the probability that FE @ belongs to the random family s(x), thus it equals (n-IF/)/n):