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Dilworth’s theorem and extremal set theory 張雁婷 國立交通大學應用數學系
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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)
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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.
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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.
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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
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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)
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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)
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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
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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
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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.
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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.
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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
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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
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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
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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
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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 )}
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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,
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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.
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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
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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
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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.
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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),<),
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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
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Using Dilworth ’ s theorem to prove Theorem 5.1 We have a complete matching.
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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
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Theorem 6.5 Let A be a collection of m subsets of N:={1,2,…,n } such that and if and for all Then
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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.
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