### Dilworth`s theorem

```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)
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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}
{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.
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Pf: (by H. Tverberg 1967)
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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
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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.
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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.
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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}
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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)
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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 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 
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Pf: (by Lubell 1966)
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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.
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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.
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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:
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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.
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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)


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
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Pf:
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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
.....
.....

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
.....
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.
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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)

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
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Pf:
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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
   n11   1
i 1  A 1
 i 
▧
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