dilworth’s theorem and extremal set theory. a partial ordered set (poset) is a set s with a binary...
TRANSCRIPT
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-
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
- Dilworthrsquos theorem and extremal set theory
- Slide 2
- Slide 3
- Dilworth Hartimanis Simon Tsai
- Slide 5
- Slide 6
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c
- Another proof of Dilworthrsquos theorem Suppose the largest anti-c (2)
- Slide 9
- Slide 10
- Slide 11
- Slide 12
- Slide 13
- Slide 14
- Slide 15
- Slide 16
- Slide 17
- Slide 18
- Slide 19
- Slide 20
- Slide 21
- Slide 22
- Slide 23
- Slide 24
-