1. Dilworth’s theorem and extremal set theory

2. If a subset of S is totally ordered, it is called a chain.
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.

3. 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}

4. 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.

5. Pf: (by H. Tverberg 1967) It is trivial m ≥ M.
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.
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 {a1,…, aM} is an antichain in P\C.
Define , . a1 ….
aM-1 a1 …. aM a2

6. Since C is maximal, the largest element in C is not in . aM a2 ….
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 > ai.
Since there exists aj with x ≤ aj, we would have
ai< x ≤ aj. →←
Thus ai is the maximal element in , i=1,…,m.
Similarly do the same for .
Combine the chains and the theorem follows ▧

7. 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?

8. 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 an
a2 ....
A={a1,...,an} C2 C1 Cn

9. 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

10. 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 ▧

11. Emanuel Sperner (9/ – 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

12. 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.

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

14. G(X∪Y, E) has a complete matching iff |Γ(A)| ≥ |A| for all A⊆X. Pf:
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.

15. 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 ▧

16. Paul Erdös (1913 – 1996) 柯召
Richard Rado
Thm 6.4. (Erdös-Ko-Rado 1961)
Let ={A1,…,Am} 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

17. Pf: Consider F={F1,…Fn}, where Fi={i, i+1,…,i+k-1}.
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.
Let п be obtained from  by applying a permutation п to [n].
Then |∩ Fп| ≤ k.
Let ∑:=∑п∈Sn|∩ Fп| ≤ k·n!. n 2 k

18. There are k!(n-k)! Permutations п such that =Aj. Why? 1
i-1 i
i+k-1
i+k n
.....
.....
..... Aj
Fix any Aj∈, Fi∈.
There are k!(n-k)! Permutations п such that =Aj. Why?
Hence ∑ =m·n·k!(n-k)! ≤ k·n! ▧

19. 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

20. 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.

21. 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) ▧

22. 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

23. Pf:
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 n 2

24. Thus, ▧