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 ∈

Dilworthrsquos theorem and extremal set theory

A partial ordered set (poset) is a set S with a binary relation le (or sube) such that (i) a le a for all a isin S (reflexivity)(ii) If a le b and b le c then a le c (transitivity)(iii) If a le b and b le a then a=b (antisymmetry)

If for any a and b in S either a le b or b le 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

Eg sube1 2 3

1 2 2 3 1 3

1 2 3

Dilworth Hartimanis Simon Tsai

Thm 61 (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

Pf (by H Tverberg 1967) It is trivial m ge M Prove M ge m by induction on |P| It is trivial if |P|=0 Let C be a maximal chain in P If every antichain in PC contains at most M-1

elements done Why Assume a1hellip aM is an antichain in PC Define

a1aM-1hellip

i

i

S x P i x a

S x P i a x

m the minimum number of disjoint chains containing all elements of PM the maximum number of elements in an anti-chain of P

a1 a2aM

S

S

hellip

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 gt ai

Since there exists aj with x le aj we would have

ailt x le aj rarrlarr

Thus ai is the maximal element in i=1hellipm Similarly do the same for Combine the chains and the theorem follows

S

1 MS S

i ia S

S

S

ix S

iS

S

a1 a2aM

S

S

hellip

Another proof of Dilworthrsquos theoremSuppose 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 Prsquo = Pa Then Prsquo is the union of n disjoint chains C1Cn

Now every n-element anti-chain in Prsquo consists of one element from each Ci

Let ai be the maximal element in Ci which belongs to some n-element anti-chain in Prsquo Let A=a1anndash an anti-chain

If A a is an anti-chain in P then we are done Otherwise we have ai le a for some i Then K=a x Ci x le ai is a chain in P and

there is no n-element anti-chain in PK Why Thus PK is the union of n-1 chains

Another proof of Dilworthrsquos theoremSuppose the largest anti-chain in the poset P has size r Then P can be partitioned into r chains

aa1

a2

an

A=a1an

C1

C2 Cn

Thm 62 (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 x1ltx2lthellipltxm were a chain in PM

Then it would also be a maximal chain in P and hence xmisinM rarrlarr

Hence PM has no chain of length m By ind hypothesis PM is the union of m-1

antichains This proves the theorem

Thm 63 (Sperner 1928)If A1hellip Am are subsets of N=12hellipn such that Ai is not a subset of Aj if inej then

2

nm n

Emanuel Sperner (912 1905 ndash 311 1980) a German mathematician

Pf (by Lubell 1966) Consider the poset of subsets of N and =A1hellip

Am is an antichain

Oslashsube1 sube1 2 subehellipsube1hellipn ~ 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 Aisin and is

a maximal chain and Aisin Each maximal chain contains at most one

member of an antichain

Let αk denote the number of sets Aisin with |A|=k Then there are maximal chains

passing

since is maximized for and sumαk=m

0( )

n

kkk n k

0

0

( )

1

n

kk

n

nk k

k

k n k n

n

k

2

nk

Prove Thm 51 by Thm 61G(XcupY E) has a complete matching iff |Γ(A)| ge |A| for all AsubeX

Pf Let |X|=n |Y|=nrsquo ge n Introduce a partial order by defining xiltyi iff there

is an edge from xi to yi Suppose the largest antichain contain s

elements say x1hellipxh y1hellip yk where h+k=s

Since Γ(x1hellipxh) subeYy1hellipyk we have h le nrsquo-k

Hence s le nrsquo 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 nrsquo-a elements of Y

Therefore n+nrsquo-a=s le nrsquo size of antichainrArr n le a rArrexista complete matching

Thm 64 (Erdoumls-Ko-Rado 1961)Let =A1hellipAm be a collection of m distinct k-subset of [n] where k le n2 with the property that any two of the subsets have a nonempty intersection Then

1

1

nm

k

柯召 1910--2002 Richard Rado 1906 --1989 Paul Erdoumls (1913 ndash 1996)

Pf Consider F=F1hellipFn

where Fi=i i+1hellipi+k-1 Fi intersects at most one of l l+1 l+k-1 l-khellip

l-1 is in (i lt l lt i+k) Thus |cap F | le k Let

п be obtained from by applying a permutation п to [n]

Then |cap Fп| le k Let sum=sumпisinSn

|cap Fп| le kn

1

k

2n

Fix any Ajisin Fiisin There are k(n-k) Permutations п such that

=Aj Why Hence sum =mnk(n-k) le kn

iF

11

11

nnm

kk n k

1 i-1 i i+k-1 i+k n

Aj

Thm 65Let =A1hellip Am be a collection of m subsets of [n] such that Ai ⊈ Aj and AicapAjneOslash if inej and |Ai| le k le n2 for all i Then

1

1

nm

k

Pf(i) If all |Ai|=k then it has been proved in Thm 64(ii) Let A1hellip As be the subsets with the smallest

cardinality say l le n2 ndash 1 Consider all the (l+1)-subsets Bj of [n] that

contain one or more of the sets Ai 1 le i le s Note that Bj Each Ai is in exactly n-l of the Bjrsquos and each Bj

contains at most l+1 le n-l of the Airsquos By Thm 51 we can match each Ai 1 le i les

with Bi such that AisubeBi

Replace A1hellip As by B1hellip Bs then the new collection rsquo satisfies the conditions of the theorem and the subsets of smallest cardinality have size gt l

By induction we reduce to case (i)

Thm 66 (Bollobagraves 1973)Let =A1hellip Am be a collection of m subsets of [n] where |Ai| le n2 for i=1hellip m with the property that any two of the subsets have a nonempty intersection and Ai

⊈ Aj Then 1

1 1

1 1i

n

ni A

Beacutela Bollobaacutes (born August 3 1943 in Budapest Hungary) a leading Hungarian mathematician

Pf Let п be a permutation of [n] placed

on a circle and say that Aiisinп if the elements of Ai occur consecutively somewhere on that circle

As in Thm 64 if Aiisinп then Ajisinп for at most |Ai| values of j

Define

2n

1 if

0 ow i iA

Af i

Thus

1

n

m

S i

f i n

1 1

1

11 1

1

1

1 1

1

1

n

i

m m

i ii S i i

mi i

i

m

ni A

f i n A n A nA

A n A

n