Dilworth’s theorem and extremal set theory 張雁婷 國立交通大學應用數學系

Partially ordered set ( poset ) A partially ordered set is a set with a binary relation (sometimes is used) such that : i) for all (reflexivity) ii) if and then (transitivity) iii) if and then (antisymmetry)

Total order If for any and in, either or, then the partial order is called a total order. If a subset of is totally ordered, it is called a chain. An antichain is a set of elements that are pairwise incomparable.

Dilworth ’ s theorem The theorem is due to R. Dilworth(1950) The proof is due to H. Tverberg(1967) 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.

Dilworth ’ s theorem P 是一個 partially ordered set, 若 P 中最大的 anticain size 為 a, 而且最少可用 b 個 disjoint chains 的聯集來涵蓋 P 中所有的元素, 則 a = b

Proof of Dilworth ’ s theorem a b : trivial a b Use strong induction on  Suppose when < n, a (P) b(P)  When = n, let C be a maximal chain in P. < n Hence, a (P\C) b (P\C)

Proof of Dilworth ’ s theorem If every antichain in P\C contains at most a (P)-1 elements, we are done.  a (P)-1 a (P\C) b (P\C) b (P) b (P\C)+1 a (P)-1+1 = a (P)

Proof of Dilworth ’ s theorem Assume that { } is an antichain in P\C.  Define  C is a maximal chain in P the largest element in C is not in < n b( ) a( P ) 而且每個 都是各個 chain 的 maximal element

Proof of Dilworth ’ s theorem  同理， 也可以被分成 a( P ) 個 disjoint chains 的聯集，而且每個 都是各個 chain 的 minimal element  可找到 a( P ) 條 disjoint chains 來涵蓋 P 中所有元 素

A “ dual ” to Dilworth ’ s theorem A “dual” to Dilworth’s theorem was given by Mirsky(1971) Let P be a partially ordered set. If P possesses no chain of m+1 elements, then P is the union of m antichains.

Proof of the “ dual ” to Dilworth ’ s theorem Use induction on m  m =1, the theorem is trivial.  Suppose the theorem is true for m -1  Let P be the poset that has no chain of m + 1elements.  Let M be the set of maximal elements of P, then M is an antichain of P.

Proof of the “ dual ” to Dilworth ’ s theorem ( 反證法 )  Suppose there is a chain { } in P\M and  P has no chain of m+1 elements { } is a maximal chain of P, a contradiction P\M has no chain of m elements P\M is the union of m-1antichains P\M M =P is the union of m antichains

Sperner ’ s theorem The theorem is due to Sperner.(1928) The proof is due to Lubell.(1966) If are subsets of N :={1,2,…,n} such that is not a subset of if, then

Proof of Sperner ’ s theorem Let S(N) be the set of all subsets of N Consider the poset (S(N), ), then A := { } is an antichain in S(N) The maximal chain of S(N) can be expressed as the form { { },{ },…,{ } } there are maximal chains

Proof of Sperner ’ s theorem For a given k-subset A of N, there are maximal chains which contain A Use two way counting to count the number of ordered pairs (A,C), such that A A, C is a maximal chain, and A C

Proof of Sperner ’ s theorem  每個 maximal chain 裡最多只能有一個 member 是來自於同一組 antichain # {(A,C )}

Proof of Sperner ’ s theorem  令 A 中的 k-subsets 的個數為 個, 對每一個 A 中 的 k-subset 而言, 包含此 k-subset 的 maximal chains 會有 個 # {(A,C )} =  Hence, 

Problem 6A Let be a permutation of the integers. Show that Dilworth’s theorem implies that the sequence has a subsequence of length n +1 that is monotone.

Proof of Problem 6A 令 S = Define a partial order of S:  Note that a chain in S is an increasing subsequence of  目的 : 找到一個 chain, 其 size 為 n+1 或找到一個 antichain, 其 size 為 n+1

Proof of Problem 6A If there is no chain in S with size n+1, suppose that the size of the maximal chain in S is n.  S can be partitioned into at least disjoint chains.  By the Dilworth’s theorem, the size of maximal antichain in S is  There is a decreasing subsequence in

Using Dilworth ’ s theorem to prove Theorem 5.1 Theorem 5.1 A necessary and sufficient condition for there to be a complete matching from X to Y in G is that for every  ( ) is trivial.

Using Dilworth ’ s theorem to prove Theorem 5.1 ( )  Let  Define a partial order, in G  Suppose is a maximal antichain of poset (V(G),<), 

Using Dilworth ’ s theorem to prove Theorem 5.1  the maximal antichain size in V(G) is s by Dilworth’s theorem, V(G) can be partitioned into at least s disjoint chains.  This will consist of a matching of size. ＸＹ

Using Dilworth ’ s theorem to prove Theorem 5.1   We have a complete matching.

Theorem 6.4 Let A be a collection of m distinct k-subsets of {1,2,…,n},where with the property that any two of the subsets have a nonempty intersection. Then

Theorem 6.5 Let A be a collection of m subsets of N:={1,2,…,n } such that and if and for all Then

Proof of Theorem 6.5 If all the subsets have size k, then we are done by Theorem 6.4. Let be the subsets with the smallest cardinality, say.  There is a complete matching from to  Replacing  Finally, all the subsets have the same size.