sperner’s lemma - university of hawaiiralph/sperner.pdf · sperner’s lemma s is a ﬁnite set...

Sperner’s Lemma

Ralph Freese

Ralph Freese () Sperner’s Lemma Aug 17, 2011 1 / 23

Paul Erdos

Figure: Paul ErdosRalph Freese () Sperner’s Lemma Aug 17, 2011 2 / 23

Pál Erdos

Figure: Pál ErdosRalph Freese () Sperner’s Lemma Aug 17, 2011 3 / 23

Erdos Pál

Figure: Erdos PálRalph Freese () Sperner’s Lemma Aug 17, 2011 4 / 23

Paul Erdos

Figure: Paul ErdosRalph Freese () Sperner’s Lemma Aug 17, 2011 5 / 23

Sperner’s Lemma

S is a finite set with n elements.

P(S) is the set of all subsets of S.An antichain in P(S) is a collections of subsets {S1, . . . ,St} of Ssuch that none is a subset of any of the others.Problem: How big can an antichain in P(S) be?


Sperner’s Lemma

S is a finite set with n elements.P(S) is the set of all subsets of S.

An antichain in P(S) is a collections of subsets {S1, . . . ,St} of Ssuch that none is a subset of any of the others.Problem: How big can an antichain in P(S) be?


Sperner’s Lemma

S is a finite set with n elements.P(S) is the set of all subsets of S.An antichain in P(S) is a collections of subsets {S1, . . . ,St} of Ssuch that none is a subset of any of the others.

Problem: How big can an antichain in P(S) be?


Sperner’s Lemma

S is a finite set with n elements.P(S) is the set of all subsets of S.An antichain in P(S) is a collections of subsets {S1, . . . ,St} of Ssuch that none is a subset of any of the others.Problem: How big can an antichain in P(S) be?


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Sperner’s Lemma

S is a finite set with n elements. That is |S| = n.P(S) is the set of all subsets of S.An antichain in P(S) is a collections of subsets {S1, . . . ,St} of Ssuch that none is a subset of any of the others.Problem: How big can an antichain in P(S) be?

Observations:

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Sperner’s Lemma

S is a finite set with n elements. That is |S| = n.P(S) is the set of all subsets of S.An antichain in P(S) is a collections of subsets {S1, . . . ,St} of Ssuch that none is a subset of any of the others.Problem: How big can an antichain in P(S) be?Observations:


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Sperner’s Lemma



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Sperner’s Lemma



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Sperner’s Lemma



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Theorem (Sperner)No!


Partially Ordered Sets and Lattices

A set P together with a relation ≤ is called a (partially) orderedset if it satisfies:

x ≤ x (reflexive)if x ≤ y and y ≤ x , then x = y (antisymmetric)if x ≤ y and y ≤ z, then x ≤ z (transitive)

Example: P = P(S) with ≤ equal to ⊆.A lattice is an ordered set such that each pair x and y has a leastupper bound, x ∨ y , and greatest lower bound, x ∧ y .x ∨ y is also called the join of x and y .x ∧ y is also called the meet of x and y .Example: P = P(S) with ∨ equal to ∪ and ∧ equal to ∩.





Example: P = P(S) with ≤ equal to ⊆.

A lattice is an ordered set such that each pair x and y has a leastupper bound, x ∨ y , and greatest lower bound, x ∧ y .x ∨ y is also called the join of x and y .x ∧ y is also called the meet of x and y .Example: P = P(S) with ∨ equal to ∪ and ∧ equal to ∩.





Example: P = P(S) with ≤ equal to ⊆.A lattice is an ordered set such that each pair x and y has a leastupper bound, x ∨ y , and greatest lower bound, x ∧ y .

x ∨ y is also called the join of x and y .x ∧ y is also called the meet of x and y .Example: P = P(S) with ∨ equal to ∪ and ∧ equal to ∩.





Example: P = P(S) with ≤ equal to ⊆.A lattice is an ordered set such that each pair x and y has a leastupper bound, x ∨ y , and greatest lower bound, x ∧ y .x ∨ y is also called the join of x and y .x ∧ y is also called the meet of x and y .

Example: P = P(S) with ∨ equal to ∪ and ∧ equal to ∩.





Example: P = P(S) with ≤ equal to ⊆.A lattice is an ordered set such that each pair x and y has a leastupper bound, x ∨ y , and greatest lower bound, x ∧ y .x ∨ y is also called the join of x and y .x ∧ y is also called the meet of x and y .Example: P = P(S) with ∨ equal to ∪ and ∧ equal to ∩.



Some ordered sets that aren’t lattices:

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Automorphisms of Ordered Sets

An automorphism of an ordered set P is a one-to-one and ontomap σ : P → P such that x ≤ y in P iff σ(x) ≤ σ(y).

If G is a set of automorphisms closed under function compositionand inverses, then G is called a group.Problem: If P is a finite ordered set and G is a group ofautomorphisms of P, does P have a maximal-sized antichain Ainvariant under G; that is,

σ(A) = A for all σ ∈ G?

(σ(A) = {σ(a) : a ∈ A})

Theorem (Kleitman, Edelberg, Lubell)Yes!



An automorphism of an ordered set P is a one-to-one and ontomap σ : P → P such that x ≤ y in P iff σ(x) ≤ σ(y).If G is a set of automorphisms closed under function compositionand inverses, then G is called a group.

Problem: If P is a finite ordered set and G is a group ofautomorphisms of P, does P have a maximal-sized antichain Ainvariant under G; that is,


(σ(A) = {σ(a) : a ∈ A})




An automorphism of an ordered set P is a one-to-one and ontomap σ : P → P such that x ≤ y in P iff σ(x) ≤ σ(y).If G is a set of automorphisms closed under function compositionand inverses, then G is called a group.Problem: If P is a finite ordered set and G is a group ofautomorphisms of P, does P have a maximal-sized antichain Ainvariant under G; that is,


(σ(A) = {σ(a) : a ∈ A})



Maximal-Sized Antichains

Let P a finite ordered set.

Let L be the collection of all maximal-sized antichains in P.For A, B ∈ L, let A ≤ B if ∀a ∈ A ∃b ∈ B with a ≤ b.Theorem (Dilworth): L is a lattice. For C and D ∈ L

C ∨ D is the maximal elements of C ∪ D.C ∧ D is the miminal elements of C ∪ D.

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Let P a finite ordered set.Let L be the collection of all maximal-sized antichains in P.

For A, B ∈ L, let A ≤ B if ∀a ∈ A ∃b ∈ B with a ≤ b.Theorem (Dilworth): L is a lattice. For C and D ∈ L


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Let P a finite ordered set.Let L be the collection of all maximal-sized antichains in P.For A, B ∈ L, let A ≤ B if ∀a ∈ A ∃b ∈ B with a ≤ b.

Theorem (Dilworth): L is a lattice. For C and D ∈ LC ∨ D is the maximal elements of C ∪ D.C ∧ D is the miminal elements of C ∪ D.

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Let P a finite ordered set.Let L be the collection of all maximal-sized antichains in P.For A, B ∈ L, let A ≤ B if ∀a ∈ A ∃b ∈ B with a ≤ b.Theorem (Dilworth): L is a lattice. For C and D ∈ L


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A Corollary to Dilworth’s Theorem

If A is a maximal-sized antichain of P and σ is an automorphism,the σ(A) is also a maximal-sized antichain.

σ “acts” on L.Since L is a finite lattice, it has a unique maximal (top) element, T .Clearly σ(T ) = T , so the theorem of Kleitman, et al., is true.



If A is a maximal-sized antichain of P and σ is an automorphism,the σ(A) is also a maximal-sized antichain. σ “acts” on L.

Since L is a finite lattice, it has a unique maximal (top) element, T .Clearly σ(T ) = T , so the theorem of Kleitman, et al., is true.



If A is a maximal-sized antichain of P and σ is an automorphism,the σ(A) is also a maximal-sized antichain. σ “acts” on L.Since L is a finite lattice, it has a unique maximal (top) element, T .

Clearly σ(T ) = T , so the theorem of Kleitman, et al., is true.



If A is a maximal-sized antichain of P and σ is an automorphism,the σ(A) is also a maximal-sized antichain. σ “acts” on L.Since L is a finite lattice, it has a unique maximal (top) element, T .Clearly σ(T ) = T , so the theorem of Kleitman, et al., is true.


Proof of Dilworth’s Theorem

For A and B ∈ L, let

AB = {a ∈ A : a ≥ b for some b ∈ B}AB = {a ∈ A : a ≤ b for some b ∈ B}

Since every element of P is comparable to something in A

A ∨ B = AB ∪ BA and A ∧ B = AB ∪ BA

AB ∪ AB = A and BA ∪ BA = B

AB ∩ BA = AB ∩ BA = A ∩ B

AB ∩ AB = BA ∩ BA = A ∩ B



Inclusion-exclusion: for sets C and D

|C ∪ D| = |C|+ |D| − |C ∩ D|

|C ∪ D|+ |C ∩ D| = |C|+ |D|



Inclusion-exclusion: for sets C and D

|C ∪ D| = |C|+ |D| − |C ∩ D||C ∪ D|+ |C ∩ D| = |C|+ |D|



The equations from before:





Using inclusion-exclusion:

|A ∨ B|+ |A ∧ B| = |AB ∪ BA|+ |AB ∪ BA|

= |AB|+ |BA| − |AB ∩ BA|+ |AB|+ |BA| − |AB ∩ BA|= |AB|+ |AB|+ |BA|+ |BA| − 2|A ∩ B|= |AB ∪ AB|+ |AB ∩ AB|+ |BA ∪ BA|+ |BA ∩ BA| − 2|A ∩ B|= |A|+ |B|









|A ∨ B|+ |A ∧ B| = |AB ∪ BA|+ |AB ∪ BA|= |AB|+ |BA| − |AB ∩ BA|+ |AB|+ |BA| − |AB ∩ BA|

= |AB|+ |AB|+ |BA|+ |BA| − 2|A ∩ B|= |AB ∪ AB|+ |AB ∩ AB|+ |BA ∪ BA|+ |BA ∩ BA| − 2|A ∩ B|= |A|+ |B|









|A ∨ B|+ |A ∧ B| = |AB ∪ BA|+ |AB ∪ BA|= |AB|+ |BA| − |AB ∩ BA|+ |AB|+ |BA| − |AB ∩ BA|= |AB|+ |AB|+ |BA|+ |BA| − 2|A ∩ B|

= |AB ∪ AB|+ |AB ∩ AB|+ |BA ∪ BA|+ |BA ∩ BA| − 2|A ∩ B|= |A|+ |B|









|A ∨ B|+ |A ∧ B| = |AB ∪ BA|+ |AB ∪ BA|= |AB|+ |BA| − |AB ∩ BA|+ |AB|+ |BA| − |AB ∩ BA|= |AB|+ |AB|+ |BA|+ |BA| − 2|A ∩ B|= |AB ∪ AB|+ |AB ∩ AB|+ |BA ∪ BA|+ |BA ∩ BA| − 2|A ∩ B|

= |A|+ |B|









|A ∨ B|+ |A ∧ B| = |AB ∪ BA|+ |AB ∪ BA|= |AB|+ |BA| − |AB ∩ BA|+ |AB|+ |BA| − |AB ∩ BA|= |AB|+ |AB|+ |BA|+ |BA| − 2|A ∩ B|= |AB ∪ AB|+ |AB ∩ AB|+ |BA ∪ BA|+ |BA ∩ BA| − 2|A ∩ B|= |A|+ |B|



Let k be the number of elements in a maximal-sized antichain.

Since both A ∨ B and A ∧ B are antichains both have at most kelements.But |A∨B|+ |A∧B| = |A|+ |B|. So |A∨B| = |A∧B| = k and soboth are in L.



Let k be the number of elements in a maximal-sized antichain.Since both A ∨ B and A ∧ B are antichains both have at most kelements.

But |A∨B|+ |A∧B| = |A|+ |B|. So |A∨B| = |A∧B| = k and soboth are in L.



Let k be the number of elements in a maximal-sized antichain.Since both A ∨ B and A ∧ B are antichains both have at most kelements.But |A∨B|+ |A∧B| = |A|+ |B|.

So |A∨B| = |A∧B| = k and soboth are in L.



Let k be the number of elements in a maximal-sized antichain.Since both A ∨ B and A ∧ B are antichains both have at most kelements.But |A∨B|+ |A∧B| = |A|+ |B|. So |A∨B| = |A∧B| = k and soboth are in L.


Back to Sperner

Let S be a set with n elements and let P = P(S).Let G be the group of all 1-1 maps from S to S. Of course σ ∈ Gacts on P by X 7→ σ(X ).

If X and Y ⊆ S both have k elements, there is a σ ∈ G withσ(X ) = Y .By Kleitman, Edelberg and Lubell’s Theorem, P(S) has amaximal-sized antichain A invariant under G. (If X ∈ A and σ ∈ Gthen σ(X ) ∈ A.)So if X ∈ A has k elements, then A contains all subsets with kelements. This (easily) implies A must consist of all subsets of Sof size k , for some k , which proves Sperner’s Lemma.


Back to Sperner

Let S be a set with n elements and let P = P(S).Let G be the group of all 1-1 maps from S to S. Of course σ ∈ Gacts on P by X 7→ σ(X ).If X and Y ⊆ S both have k elements, there is a σ ∈ G withσ(X ) = Y .

By Kleitman, Edelberg and Lubell’s Theorem, P(S) has amaximal-sized antichain A invariant under G. (If X ∈ A and σ ∈ Gthen σ(X ) ∈ A.)So if X ∈ A has k elements, then A contains all subsets with kelements. This (easily) implies A must consist of all subsets of Sof size k , for some k , which proves Sperner’s Lemma.


Back to Sperner

Let S be a set with n elements and let P = P(S).Let G be the group of all 1-1 maps from S to S. Of course σ ∈ Gacts on P by X 7→ σ(X ).If X and Y ⊆ S both have k elements, there is a σ ∈ G withσ(X ) = Y .By Kleitman, Edelberg and Lubell’s Theorem, P(S) has amaximal-sized antichain A invariant under G. (If X ∈ A and σ ∈ Gthen σ(X ) ∈ A.)

So if X ∈ A has k elements, then A contains all subsets with kelements. This (easily) implies A must consist of all subsets of Sof size k , for some k , which proves Sperner’s Lemma.


Back to Sperner

Let S be a set with n elements and let P = P(S).Let G be the group of all 1-1 maps from S to S. Of course σ ∈ Gacts on P by X 7→ σ(X ).If X and Y ⊆ S both have k elements, there is a σ ∈ G withσ(X ) = Y .By Kleitman, Edelberg and Lubell’s Theorem, P(S) has amaximal-sized antichain A invariant under G. (If X ∈ A and σ ∈ Gthen σ(X ) ∈ A.)So if X ∈ A has k elements, then A contains all subsets with kelements.

This (easily) implies A must consist of all subsets of Sof size k , for some k , which proves Sperner’s Lemma.


Back to Sperner

Let S be a set with n elements and let P = P(S).Let G be the group of all 1-1 maps from S to S. Of course σ ∈ Gacts on P by X 7→ σ(X ).If X and Y ⊆ S both have k elements, there is a σ ∈ G withσ(X ) = Y .By Kleitman, Edelberg and Lubell’s Theorem, P(S) has amaximal-sized antichain A invariant under G. (If X ∈ A and σ ∈ Gthen σ(X ) ∈ A.)So if X ∈ A has k elements, then A contains all subsets with kelements. This (easily) implies A must consist of all subsets of Sof size k , for some k , which proves Sperner’s Lemma.


Partition Lattices and Rota’s Problem

Figure: Gian-Carlo RotaRalph Freese () Sperner’s Lemma Aug 17, 2011 21 / 23


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Gian-Carlo Rota: Does the largest sized antichain in the lattice ofpartitions of an n element set consist of all partitions with k block,for some k .



Theorem (Canfield)

Rota’s problem has a negative answer—at least if n > 1025.

ProblemWhat is Dilworth’s lattice for the lattice of equivalence relations of an nelement set?


sperner’s lemma - university of hawaiiralph/sperner.pdf · sperner’s lemma s is a ﬁnite set...

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