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Dilworth’s theorem and extremal set theory

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A partial ordered set (poset) is a set S with a binary relation ≤ (or ⊆ ) such that (i) a ≤ a for all a ∈ S. (reflexivity) (ii) If a ≤ b and b ≤ c then a ≤ c. (transitivity) (iii) If a ≤ b and b ≤ a then a=b. (antisymmetry) If for any a and b in S, either a ≤ b or b ≤ a, then the partial order is called a total order. If a subset of S is totally ordered, it is called a chain.

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An antichain is a set of elements that are pairwise incomparable. E.g. ⊆ : {1, 2, 3} {1, 2} {2, 3} {1, 3} {1}{2} {3}

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Dilworth Hartimanis Simon Tsai Thm 6.1. (Dilworth 1950) 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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Pf: (by H. Tverberg 1967) It is trivial m ≥ M. Prove M ≥ m by induction on |P|. It is trivial if |P|=0. Let C be a maximal chain in P. If every antichain in P\C contains at most M-1 elements, done. Why? Assume {a 1, …, a M } is an antichain in P\C. Define,. a1a1 a M-1 ….…. m: the minimum number of disjoint chains containing all elements of P. M: the maximum number of elements in an anti-chain of P. a1a1 a2a2 aMaM ….….

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Since C is maximal, the largest element in C is not in. By ind. hypothesis, the theorem holds for. Thus, is the union of M disjoint chains, where. Suppose and x > a i. Since there exists a j with x ≤ a j, we would have a i < x ≤ a j. →← Thus a i is the maximal element in, i=1,…,m. Similarly do the same for. Combine the chains and the theorem follows. ▧ a1a1 a2a2 aMaM ….….

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Another proof of Dilworth ’ s theorem: Suppose the largest anti-chain in the poset P has size r. Then P can be partitioned into r chains. Proof: (Fred Galvin 1994 American Math. Monthly) By induction on |P|. Let a be a maximal element of P, and n be the size of largest anti-chain in P ’ = P\{a}. Then P ’ is the union of n disjoint chains C 1,...,C n. Now every n-element anti-chain in P ’ consists of one element from each C i. Let a i be the maximal element in C i which belongs to some n-element anti-chain in P ’. Let A={a 1,...,a n } – an anti-chain?

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If A {a} is an anti-chain in P, then we are done! Otherwise we have a i ≤ a for some i. Then K={a} {x C i : x ≤ a i } is a chain in P and there is no n-element anti-chain in P\K. Why? Thus P\K is the union of n-1 chains. Another proof of Dilworth ’ s theorem: Suppose the largest anti-chain in the poset P has size r. Then P can be partitioned into r chains. a a1a1 a anan A={a 1,...,a n } C1C1 C2C2 CnCn

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Thm 6.2. (Mirsky 1971) If P possesses no chain of m+1 elements, then P is the union of m antichains. Pf: (By ind. on m) It is true for m=1. why? Assume it is true up to m-1. Let P be a poset without chain of m+1 elements. Let M be the set of maximal elements of P. Then M is an antichain. Why? M

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Suppose x 1

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Thm 6.3. (Sperner 1928) If A 1, …, A m are subsets of N={1,2, …,n} such that A i is not a subset of A j, if i≠j, then Emanuel Sperner (9/ – 31/1 1980) a German mathematician

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Pf: (by Lubell 1966) Consider the poset of subsets of N and ={A 1, …, A m } is an antichain. Ø ⊆ {1} ⊆ {1, 2} ⊆ … ⊆ {1, …,n} ~ a maximal chain There are n! different maximal chain. Also, there are exactly k!(n-k)! maximal chains containing a given k-subset of N. Count the ordered pair (A, ), where A ∈ and is a maximal chain and A ∈ . Each maximal chain contains at most one member of an antichain.

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Let α k denote the number of sets A ∈ with |A|=k. Then there are maximal chains passing . since is maximized for and ∑ α k =m. ▧

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Prove Thm 5.1 by Thm 6.1: G(X ∪ Y, E) has a complete matching iff | Γ (A)| ≥ |A| for all A ⊆ X. Pf: Let |X|=n, |Y|=n ’ ≥ n. Introduce a partial order by defining x i

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Since Γ ({x 1,…,x h }) ⊆ Y\{y 1, …,y k }, we have h ≤ n ’ -k. Hence s ≤ n ’. The poset is the union of s disjoint chains. This consists of a matching of size a, the remaining n-a elements of X and n ’ -a elements of Y. Therefore n+n ’ -a=s ≤ n ’, size of antichain. ⇒ n ≤ a ⇒∃ a complete matching. ▧

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Thm 6.4. (Erdös-Ko-Rado 1961) Let ={A 1, …,A m } be a collection of m distinct k- subset of [n], where k ≤ n/2, with the property that any two of the subsets have a nonempty intersection. Then 柯召 Richard Rado Paul Erd ö s (1913 – 1996)

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Pf: Consider F={F 1,…F n }, where F i ={i, i+1,…,i+k-1}. F i intersects at most one of {l, l+1, l+k-1}, {l-k,…, l-1} is in , (i < l < i+k) Thus, | ∩ F | ≤ k. Let п be obtained from by applying a permutation п to [n]. Then | ∩ F п | ≤ k. Let ∑:=∑ п ∈ S n | ∩ F п | ≤ k · n!. 1 k 2 n

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Fix any A j ∈ , F i ∈ . There are k!(n-k)! Permutations п such that =A j. Why? Hence ∑ =m · n · k!(n-k)! ≤ k · n! ▧ 1 i-1ii+k-1i+kn AjAj

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Thm 6.5. Let ={A 1, …, A m } be a collection of m subsets of [n] such that A i ⊈ A j and A i ∩ A j ≠ Ø if i ≠j and |A i | ≤ k ≤ n/2 for all i. Then

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Replace A 1, …, A s by B 1, …, B s, then the new collection ’ satisfies the conditions of the theorem and the subsets of smallest cardinality have size > l. By induction, we reduce to case (i). ▧

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Thm 6.6. (Bollobàs 1973) Let ={A 1, …, A m } be a collection of m subsets of [n], where |A i | ≤ n/2 for i=1, …, m, with the property that any two of the subsets have a nonempty intersection and A i ⊈ A j. Then B é la Bollob á s (born August 3, 1943 in Budapest, Hungary) a leading Hungarian mathematician

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Pf: Let п be a permutation of [n] placed on a circle and say that A i ∈ п if the elements of A i occur consecutively somewhere on that circle. As in Thm 6.4, if A i ∈ п, then A j ∈ п for at most |A i | values of j. Define 2 n

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Thus, ▧

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