Clarification about this proof about Dilworth's Theorem

Question

Theorem 6.1. Let P be a partially ordered finite set. The minimum number m of disjoint chains which together contain all elements of P is equal to the maximum number M of elements in an antichain of P. Proof: (i) It is trivial that m ≥ M. (ii) We use induction on |P|. If |P| = 0, there is nothing to prove. Let C be a maximal chain in P. If every antichain in P\C contains at most M − 1 elements, we are done. So assume that {a1, . . . , aM} is an antichain in P\C. Now define S− := {x ∈ P : ∃i[x ≤ ai]}, and define S+ analogously. Since C is a maximal chain, the largest element in C is not in S− and hence by the induction hypothesis, the theorem holds for S−. Hence S− is the union of M disjoint chains S − 1 , . . . , S − M, where ai ∈ S − i . Suppose x ∈ S − i and x > ai. Since there is aj with x ≤ aj , we would have ai < aj , a contradiction. This shows that ai is the maximal element of the chain S − i , i = 1, . . . , m. We do the same for S+. By combining the chains the theorem follows. Could someone explain the proof in a simpler manner and what is exactly meant by the term maximal chain in P ?

Answer 1

A maximal chain is one that cannot be extended any more. That means any other element you add to a maximal chain will never result in a longer chain. The above proof is by induction on the size of the poset $P$. The idea is to remove a maximal chain $C$ from $P$ to have a smaller poset. Then use the induction hypothesis.