Theorem 1   Let $P$ be a poset of height $h$ . Then $P$ can be partitioned into $h$ antichains and furthermore at least $h$ antichains are required.

Proof. Induction on $h$ . If $h=1$ , then no elements of $P$ are comparable, so $P$ is an antichain. Now suppose that $P$ has height $h\geq 2$ and that the theorem is true for $h-1$ . Let $A_1$ be the set of maximal elements of $P$ . Then $A_1$ is an antichain in $P$ and $P-A_1$ has height $h-1$ since we have removed precisely one element from every chain. Hence, $P-A_1$ can be paritioned into $h-1$ antichains $A_2,A_3,\dots A_h$ . Now we have the partition $A_1\cup A_2\cup\dots\cup A_h$ of $P$ into $h$ antichains as desired.

The necessity of $h$ antichains is trivial by the pigeonhole principle; since $P$ has height $h$ , it has a chain of length $h$ , and each element of this chain must be placed in a different antichain of our partition.