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Dilworth’s theorem and extremal set theory 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. An antichain is a set of elements that are pairwise incomparable. E.g. ⊆: {1, 2, 3} {1, 2} {2, 3} {1} {2} {1, 3} {3} 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. m: the minimum number of disjoint chains containing all elements of P. M: the maximum number of elements in an anti-chain of P. Pf: (by H. Tverberg 1967) It is trivial m ≥ M. …. aM-1 a1 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? S Assume {a1,…, aM} is an antichain in P\C. a1 a2 …. aM Define S x P : i x ai , S S x P : i ai x . S a 1 a2 …. aM S Since C is maximal, the largest element in C is not in S . S By ind. hypothesis, the theorem holds for . Thus, S is the union of M disjoint chains S1 ,..., SM , where ai Si . Suppose x Si and x > ai. Since there exists aj with x ≤ aj, we would have ai< x ≤ aj. →← Thus ai is the maximal element in Si , i=1,…,m. Similarly do the same for S . Combine the chains and the theorem follows. ▧ 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 C1,...,Cn. Now every n-element anti-chain in P’ consists of one element from each Ci. Let ai be the maximal element in Ci which belongs to some n-element anti-chain in P’. Let A={a1,...,an}– an anti-chain? 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. If A {a} is an anti-chain in P, then we are done! Otherwise we have ai ≤ a for some i. Then K={a} {x Ci : x ≤ ai } 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. a a1 a2 .... C1 C2 Cn an A={a1,...,an} 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 Suppose x1<x2<…<xm were a chain in P\M. Then it would also be a maximal chain in P and hence xm∈M. →← Hence P\M has no chain of length m. By ind. hypothesis, P\M is the union of m-1 antichains. This proves the theorem. ▧ Emanuel Sperner (9/12 1905 – 31/1 1980) a German mathematician Thm 6.3. (Sperner 1928) If A1,…, Am are subsets of N={1,2,…,n} such that Ai is not a subset of Aj, if i≠j, then n m n 2 Pf: (by Lubell 1966) Consider the poset of subsets of N and ={A1,…, Am} 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. Let αk denote the number of sets A∈ with |A|=k. n k 0 k k !(n k )! maximal chains passing . Then there are n k k !(n k )! n ! k 0 n k n 1 k 0 k since ▧ n k is maximized for k n2 and ∑αk=m. 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 xi<yi iff there is an edge from xi to yi. Suppose the largest antichain contain s elements, say {x1,…,xh, y1,…, yk}, where h+k=s. Since Γ({x1,…,xh}) ⊆Y\{y1,…,yk}, 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. ▧ Paul Erdös (1913 – 1996) 柯召 1910--2002 Richard Rado 1906 --1989 Thm 6.4. (Erdös-Ko-Rado 1961) Let ={A1,…,Am} be a collection of m distinct ksubset of [n], where k ≤ n/2, with the property that any two of the subsets have a nonempty intersection. Then n 1 m k 1 Pf: n 1 2 Consider F={F1,…Fn}, where Fi={i, i+1,…,i+k-1}. k Fi intersects at most one of {l, l+1, l+k-1}, {l-k,…, l-1} is in , (i < l < i+k) Thus, |∩ F | ≤ k. п Let be obtained from by applying a permutation п to [n]. п Then |∩ F | ≤ k. п Let ∑:=∑п∈Sn|∩ F | ≤ k·n!. i-1 1 i i+k-1 ..... ..... ..... Aj i+k n ........... ........... Fix any Aj∈, Fi∈. There are k!(n-k)! Permutations п such that Fi =Aj. Why? Hence ∑ =m·n·k!(n-k)! ≤ k·n! n 1! n 1 m k 1! n k ! k 1 ▧ Thm 6.5. Let ={A1,…, Am} be a collection of m subsets of [n] such that Ai ⊈ Aj and Ai∩Aj≠Ø if i≠j and |Ai| ≤ k ≤ n/2 for all i. Then n 1 m k 1 Pf: (i) If all |Ai|=k, then it has been proved in Thm 6.4. (ii) Let A1,…, As be the subsets with the smallest cardinality, say l ≤ n/2 – 1. Consider all the (l+1)-subsets Bj of [n] that contain one or more of the sets Ai, 1 ≤ i ≤ s. Note that Bj ∉. Each Ai is in exactly n-l of the Bj’s and each Bj contains at most l+1 ≤ n-l of the Ai’s. By Thm 5.1, we can match each Ai, 1 ≤ i ≤s, with Bi such that Ai⊆Bi. Replace A1,…, As by B1,…, Bs, 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). ▧ Béla Bollobás (born August 3, 1943 in Budapest, Hungary) a leading Hungarian mathematician Thm 6.6. (Bollobàs 1973) Let ={A1,…, Am} be a collection of m subsets of [n], where |Ai| ≤ n/2 for i=1,…, m, with the property that any two of the subsets have a nonempty intersection and Ai ⊈ Aj . Then n 1 n 1 i 1 A 1 i 1 Pf: n 2 Let п be a permutation of [n] placed on a circle and say that Ai∈п if the elements of Ai occur consecutively somewhere on that circle. As in Thm 6.4, if Ai∈п, then Aj∈п for at most |Ai| values of j. Define 1 if Ai Ai f , i o/w 0 m Thus, f , i n ! Sn i 1 m m f , i i 1 m S n i 1 A i 1 n Ai ! n Ai ! n ! Ai 1 ! n Ai ! i 1 n 1! 1 m n11 1 i 1 A 1 i ▧