dual of Dilworth’s theorem
Theorem 1.
Let be a poset of height . Then can be partitioned into antichains and furthermore at least antichains are required.
Proof.
Induction on . If , then no elements of are comparable
, so is an antichain. Now suppose that has height and that the theorem is true for . Let be the set of maximal elements
of . Then is an antichain in and has height since we have removed precisely one element from every chain. Hence, can be paritioned into antichains . Now we have the partition
of into antichains as desired.
The necessity of antichains is trivial by the pigeonhole principle; since has height , it has a chain of length , and each element of this chain must be placed in a different antichain of our partition.∎