Supersaturation in Posets

Jonathan Noel

University of Warwick

Joint work with

Alex Scott and Benny Sudakov

DMO Seminar

McGill University

April 16, 2018

A Classical Result

Defn: Let P(n) := {S : S ⊆ {1, 2, . . . , n}}.

We refer to P(n) as the power set of n.

Sperner’s Theorem (1928):

If A ⊆ P(n) such that there doesnot exist S, T ∈ A with S ( T , then

|A| ≤

(n

bn/2c

).

“⊆” is a partial order on P(n).

Defn: Such a collection A is called an antichain.

A Classical Result

Defn: Let P(n) := {S : S ⊆ {1, 2, . . . , n}}.

We refer to P(n) as the power set of n.

Sperner’s Theorem (1928):

If A ⊆ P(n) such that there doesnot exist S, T ∈ A with S ( T , then

|A| ≤

(n

bn/2c

).

“⊆” is a partial order on P(n).

Defn: Such a collection A is called an antichain.

A Classical Result

Defn: Let P(n) := {S : S ⊆ {1, 2, . . . , n}}.

We refer to P(n) as the power set of n.

Sperner’s Theorem (1928):

If A ⊆ P(n) such that there doesnot exist S, T ∈ A with S ( T , then

|A| ≤

(n

bn/2c

).

“⊆” is a partial order on P(n).

Defn: Such a collection A is called an antichain.

A Classical Result

Defn: Let P(n) := {S : S ⊆ {1, 2, . . . , n}}.

We refer to P(n) as the power set of n.

Sperner’s Theorem (1928): If A ⊆ P(n) such that there doesnot exist S, T ∈ A with S ( T , then

|A| ≤

(n

bn/2c

).

“⊆” is a partial order on P(n).

Defn: Such a collection A is called an antichain.

A Classical Result

Defn: Let P(n) := {S : S ⊆ {1, 2, . . . , n}}.

We refer to P(n) as the power set of n.

Sperner’s Theorem (1928): If A ⊆ P(n) such that there doesnot exist S, T ∈ A with S ( T , then

|A| ≤(

n

bn/2c

).

“⊆” is a partial order on P(n).

Defn: Such a collection A is called an antichain.

A Classical Result

Defn: Let P(n) := {S : S ⊆ {1, 2, . . . , n}}.

We refer to P(n) as the power set of n.

Sperner’s Theorem (1928): If A ⊆ P(n) such that there doesnot exist S, T ∈ A with S ( T , then

|A| ≤(

n

bn/2c

).

“⊆” is a partial order on P(n).

Defn: Such a collection A is called an antichain.

A Classical Result

Defn: Let P(n) := {S : S ⊆ {1, 2, . . . , n}}.

We refer to P(n) as the power set of n.

Sperner’s Theorem (1928): If A ⊆ P(n) such that there doesnot exist S, T ∈ A with S ( T , then

|A| ≤(

n

bn/2c

).

“⊆” is a partial order on P(n).

Defn: Such a collection A is called an antichain.

An Extension of Sperner’s Theorem

Sperner’s Theorem (1928): Every antichain in P(n) hascardinality at most

(nbn/2c

).

Question (Erdos and Katona, 1960s): Given A ⊆ P(n) with|A| =

(nbn/2c

)+ t for t ≥ 1, how many pairs S ( T must be

contained in A?

Kleitman’s Supersaturation Theorem (1968): For|A| =

(nbn/2c

)+ t, the number of comparable pairs is minimised by

a collection of sets of cardinality “as close to n/2 as possible.”

Kleitman’s proof uses induction, Hall’s Theorem and bounds onbinomial coefficients.

Samotij (2017+) recently generalised the theorem to chains oflength k using probabilistic ideas inspired by our approach (solvinga conjecture of Kleitman 1968).

An Extension of Sperner’s Theorem

Sperner’s Theorem (1928): Every antichain in P(n) hascardinality at most

(nbn/2c

).

Question (Erdos and Katona, 1960s): Given A ⊆ P(n) with|A| =

(nbn/2c

)+ t for t ≥ 1, how many pairs S ( T must be

contained in A?

Kleitman’s Supersaturation Theorem (1968): For|A| =

(nbn/2c

)+ t, the number of comparable pairs is minimised by

a collection of sets of cardinality “as close to n/2 as possible.”

Kleitman’s proof uses induction, Hall’s Theorem and bounds onbinomial coefficients.

Samotij (2017+) recently generalised the theorem to chains oflength k using probabilistic ideas inspired by our approach (solvinga conjecture of Kleitman 1968).

An Extension of Sperner’s Theorem

Sperner’s Theorem (1928): Every antichain in P(n) hascardinality at most

(nbn/2c

).

Question (Erdos and Katona, 1960s): Given A ⊆ P(n) with|A| =

(nbn/2c

)+ t for t ≥ 1, how many pairs S ( T must be

contained in A?

Kleitman’s Supersaturation Theorem (1968): For|A| =

(nbn/2c

)+ t, the number of comparable pairs is minimised by

a collection of sets of cardinality “as close to n/2 as possible.”

Kleitman’s proof uses induction, Hall’s Theorem and bounds onbinomial coefficients.

Samotij (2017+) recently generalised the theorem to chains oflength k using probabilistic ideas inspired by our approach (solvinga conjecture of Kleitman 1968).

An Extension of Sperner’s Theorem

Sperner’s Theorem (1928): Every antichain in P(n) hascardinality at most

(nbn/2c

).

Question (Erdos and Katona, 1960s): Given A ⊆ P(n) with|A| =

(nbn/2c

)+ t for t ≥ 1, how many pairs S ( T must be

contained in A?

Kleitman’s Supersaturation Theorem (1968): For|A| =

(nbn/2c

)+ t, the number of comparable pairs is minimised by

a collection of sets of cardinality “as close to n/2 as possible.”

Kleitman’s proof uses induction, Hall’s Theorem and bounds onbinomial coefficients.

Samotij (2017+) recently generalised the theorem to chains oflength k using probabilistic ideas inspired by our approach (solvinga conjecture of Kleitman 1968).

An Extension of Sperner’s Theorem

Sperner’s Theorem (1928): Every antichain in P(n) hascardinality at most

(nbn/2c

).

Question (Erdos and Katona, 1960s): Given A ⊆ P(n) with|A| =

(nbn/2c

)+ t for t ≥ 1, how many pairs S ( T must be

contained in A?

Kleitman’s Supersaturation Theorem (1968): For|A| =

(nbn/2c

)+ t, the number of comparable pairs is minimised by

a collection of sets of cardinality “as close to n/2 as possible.”

Kleitman’s proof uses induction, Hall’s Theorem and bounds onbinomial coefficients.

Samotij (2017+) recently generalised the theorem to chains oflength k using probabilistic ideas inspired by our approach (solvinga conjecture of Kleitman 1968).

Application to Dedekind’s Problem

Dedekind (1897): How many antichains are there in P(n)?

Theorem (Kleitman, 1969): (# antichains) ≤ 2(1+o(1))( n

bn/2c).

Later, Korshunov (1981) obtained precise asymptotics for theproblem

Kleitman’s Supersaturation Theorem implies the following:

“Container” Lemma: For 0 < ε < 1, there exists

g :

{F ⊆ P(n) : |F| ≤ 50 · 2n

εn

}→

{G ⊆ P(n) : |G| ≤

(1 +

ε

10

)( n

bn/2c

)}

such that for every antichain A there is a set F with |F| ≤ 50·2nεn

and A ⊆ F ∪ g(F).

Application to Dedekind’s Problem

Dedekind (1897): How many antichains are there in P(n)?

Theorem (Kleitman, 1969): (# antichains) ≤ 2(1+o(1))( n

bn/2c).

Later, Korshunov (1981) obtained precise asymptotics for theproblem

Kleitman’s Supersaturation Theorem implies the following:

“Container” Lemma: For 0 < ε < 1, there exists

g :

{F ⊆ P(n) : |F| ≤ 50 · 2n

εn

}→

{G ⊆ P(n) : |G| ≤

(1 +

ε

10

)( n

bn/2c

)}

such that for every antichain A there is a set F with |F| ≤ 50·2nεn

and A ⊆ F ∪ g(F).

Application to Dedekind’s Problem

Dedekind (1897): How many antichains are there in P(n)?

Theorem (Kleitman, 1969): (# antichains) ≤ 2(1+o(1))( n

bn/2c).

Later, Korshunov (1981) obtained precise asymptotics for theproblem

Kleitman’s Supersaturation Theorem implies the following:

“Container” Lemma: For 0 < ε < 1, there exists

g :

{F ⊆ P(n) : |F| ≤ 50 · 2n

εn

}→

{G ⊆ P(n) : |G| ≤

(1 +

ε

10

)( n

bn/2c

)}

such that for every antichain A there is a set F with |F| ≤ 50·2nεn

and A ⊆ F ∪ g(F).

Application to Dedekind’s Problem

Dedekind (1897): How many antichains are there in P(n)?

Theorem (Kleitman, 1969): (# antichains) ≤ 2(1+o(1))( n

bn/2c).

Later, Korshunov (1981) obtained precise asymptotics for theproblem

Kleitman’s Supersaturation Theorem implies the following:

“Container” Lemma: For 0 < ε < 1, there exists

g :

{F ⊆ P(n) : |F| ≤ 50 · 2n

εn

}→

{G ⊆ P(n) : |G| ≤

(1 +

ε

10

)( n

bn/2c

)}

such that for every antichain A there is a set F with |F| ≤ 50·2nεn

and A ⊆ F ∪ g(F).

Application to Dedekind’s Problem

Dedekind (1897): How many antichains are there in P(n)?

Theorem (Kleitman, 1969): (# antichains) ≤ 2(1+o(1))( n

bn/2c).

Later, Korshunov (1981) obtained precise asymptotics for theproblem

Kleitman’s Supersaturation Theorem implies the following:

“Container” Lemma: For 0 < ε < 1, there exists

g :

{F ⊆ P(n) : |F| ≤ 50 · 2n

εn

}→

{G ⊆ P(n) : |G| ≤

(1 +

ε

10

)( n

bn/2c

)}

such that for every antichain A there is a set F with |F| ≤ 50·2nεn

and A ⊆ F ∪ g(F).

Application to Dedekind’s Problem

Dedekind (1897): How many antichains are there in P(n)?

Theorem (Kleitman, 1969): (# antichains) ≤ 2(1+o(1))( n

bn/2c).

Later, Korshunov (1981) obtained precise asymptotics for theproblem

Kleitman’s Supersaturation Theorem implies the following:

“Container” Lemma: For 0 < ε < 1, there exists

g :

{F ⊆ P(n) : |F| ≤ 50 · 2n

εn

}→

{G ⊆ P(n) : |G| ≤

(1 +

ε

10

)( n

bn/2c

)}

such that for every antichain A there is a set F with |F| ≤ 50·2nεn

and A ⊆ F ∪ g(F).

Another Application

Kleitman’s Supersaturation Theorem and the container methodcan also be applied to obtain a “sparse random version” ofSperner’s Theorem.

(see Balogh, Mycroft and Treglown (2014) and Collares Netoand Morris (2016))

Another Application

Kleitman’s Supersaturation Theorem and the container methodcan also be applied to obtain a “sparse random version” ofSperner’s Theorem.

(see Balogh, Mycroft and Treglown (2014) and Collares Netoand Morris (2016))

Another Application

Kleitman’s Supersaturation Theorem and the container methodcan also be applied to obtain a “sparse random version” ofSperner’s Theorem.

(see Balogh, Mycroft and Treglown (2014) and Collares Netoand Morris (2016))

A Nastier Poset

The poset P(n) can be equivalently viewed as {0, 1}n where x ≤ yif xi ≤ yi for all 1 ≤ i ≤ n.

Defn: Define a partial order on {0, 1, 2}n where x ≤ y if xi ≤ yifor all 1 ≤ i ≤ n.

Defn: For 0 ≤ k ≤ 2n, let

Lk :=

{x ∈ {0, 1, 2}n :

n∑i=1

xi = k

}.

Theorem (de Bruijn, van Ebbenhorst Tengbergen andKruyswijk, 1951): Every antichain in {0, 1, 2}n has cardinality atmost |Ln|.

A Nastier Poset

The poset P(n) can be equivalently viewed as {0, 1}n where x ≤ yif xi ≤ yi for all 1 ≤ i ≤ n.

Defn: Define a partial order on {0, 1, 2}n where x ≤ y if xi ≤ yifor all 1 ≤ i ≤ n.

Defn: For 0 ≤ k ≤ 2n, let

Lk :=

{x ∈ {0, 1, 2}n :

n∑i=1

xi = k

}.

Theorem (de Bruijn, van Ebbenhorst Tengbergen andKruyswijk, 1951): Every antichain in {0, 1, 2}n has cardinality atmost |Ln|.

A Nastier Poset

The poset P(n) can be equivalently viewed as {0, 1}n where x ≤ yif xi ≤ yi for all 1 ≤ i ≤ n.

Defn: Define a partial order on {0, 1, 2}n where x ≤ y if xi ≤ yifor all 1 ≤ i ≤ n.

Defn: For 0 ≤ k ≤ 2n, let

Lk :=

{x ∈ {0, 1, 2}n :

n∑i=1

xi = k

}.

Theorem (de Bruijn, van Ebbenhorst Tengbergen andKruyswijk, 1951): Every antichain in {0, 1, 2}n has cardinality atmost |Ln|.

A Nastier Poset

The poset P(n) can be equivalently viewed as {0, 1}n where x ≤ yif xi ≤ yi for all 1 ≤ i ≤ n.

Defn: Define a partial order on {0, 1, 2}n where x ≤ y if xi ≤ yifor all 1 ≤ i ≤ n.

Defn: For 0 ≤ k ≤ 2n, let

Lk :=

{x ∈ {0, 1, 2}n :

n∑i=1

xi = k

}.

Theorem (de Bruijn, van Ebbenhorst Tengbergen andKruyswijk, 1951): Every antichain in {0, 1, 2}n has cardinality atmost |Ln|.

A Nastier Poset

The poset P(n) can be equivalently viewed as {0, 1}n where x ≤ yif xi ≤ yi for all 1 ≤ i ≤ n.

Defn: Define a partial order on {0, 1, 2}n where x ≤ y if xi ≤ yifor all 1 ≤ i ≤ n.

Defn: For 0 ≤ k ≤ 2n, let

Lk :=

{x ∈ {0, 1, 2}n :

n∑i=1

xi = k

}.

Theorem (de Bruijn, van Ebbenhorst Tengbergen andKruyswijk, 1951): Every antichain in {0, 1, 2}n has cardinality atmost |Ln|.

Supersaturation in {0, 1, 2}n

Theorem (N., Scott & Sudakov, 2018): If A ⊆ {0, 1, 2}n suchthat |A| = |Ln|+ t, then A has at least t

(n−12

)comparable pairs.

We also prove stronger bounds for large values of t.

Analogous to P(n), we can deduce counting and probabilisticresults from our theorem using the container method.

Supersaturation in {0, 1, 2}n

Theorem (N., Scott & Sudakov, 2018): If A ⊆ {0, 1, 2}n suchthat |A| = |Ln|+ t, then A has at least t

(n−12

)comparable pairs.

We also prove stronger bounds for large values of t.

Analogous to P(n), we can deduce counting and probabilisticresults from our theorem using the container method.

Supersaturation in {0, 1, 2}n

Theorem (N., Scott & Sudakov, 2018): If A ⊆ {0, 1, 2}n suchthat |A| = |Ln|+ t, then A has at least t

(n−12

)comparable pairs.

We also prove stronger bounds for large values of t.

Analogous to P(n), we can deduce counting and probabilisticresults from our theorem using the container method.

Supersaturation in {0, 1, 2}n

Theorem (N., Scott & Sudakov, 2018): If A ⊆ {0, 1, 2}n suchthat |A| = |Ln|+ t, then A has at least t

(n−12

)comparable pairs.

We also prove stronger bounds for large values of t.

Analogous to P(n), we can deduce counting and probabilisticresults from our theorem using the container method.

Outline of Approach

Theorem (N., Scott & Sudakov, 2018): If A ⊆ {0, 1, 2}n suchthat |A| = |Ln|+ t, then A has at least t

(n−12

)comparable pairs.

Our approach is inspired by the proof of the LYM Inequality.

I Pick a random maximal chain C = (x0, x1, . . . , x2n) in{0, 1, 2}n according to some distribution µ

Some simple calculations show that the theorem holds providedthat µ satisfies:

I P(x ∈ C) ≥ 1|Ln| for all x ∈ {0, 1, 2}n

I P(x ∈ C | y ∈ C) ≤ 2n−1 for all comparable pairs x, y with x at

least as close to the middle level as y.

Outline of Approach

Theorem (N., Scott & Sudakov, 2018): If A ⊆ {0, 1, 2}n suchthat |A| = |Ln|+ t, then A has at least t

(n−12

)comparable pairs.

Our approach is inspired by the proof of the LYM Inequality.

I Pick a random maximal chain C = (x0, x1, . . . , x2n) in{0, 1, 2}n according to some distribution µ

Some simple calculations show that the theorem holds providedthat µ satisfies:

I P(x ∈ C) ≥ 1|Ln| for all x ∈ {0, 1, 2}n

I P(x ∈ C | y ∈ C) ≤ 2n−1 for all comparable pairs x, y with x at

least as close to the middle level as y.

Outline of Approach

Theorem (N., Scott & Sudakov, 2018): If A ⊆ {0, 1, 2}n suchthat |A| = |Ln|+ t, then A has at least t

(n−12

)comparable pairs.

Our approach is inspired by the proof of the LYM Inequality.

I Pick a random maximal chain C = (x0, x1, . . . , x2n) in{0, 1, 2}n according to some distribution µ

Some simple calculations show that the theorem holds providedthat µ satisfies:

I P(x ∈ C) ≥ 1|Ln| for all x ∈ {0, 1, 2}n

I P(x ∈ C | y ∈ C) ≤ 2n−1 for all comparable pairs x, y with x at

least as close to the middle level as y.

Outline of Approach

Theorem (N., Scott & Sudakov, 2018): If A ⊆ {0, 1, 2}n suchthat |A| = |Ln|+ t, then A has at least t

(n−12

)comparable pairs.

Our approach is inspired by the proof of the LYM Inequality.

I Pick a random maximal chain C = (x0, x1, . . . , x2n) in{0, 1, 2}n according to some distribution µ

Some simple calculations show that the theorem holds providedthat µ satisfies:

I P(x ∈ C) ≥ 1|Ln| for all x ∈ {0, 1, 2}n

I P(x ∈ C | y ∈ C) ≤ 2n−1 for all comparable pairs x, y with x at

least as close to the middle level as y.

Outline of Approach

Theorem (N., Scott & Sudakov, 2018): If A ⊆ {0, 1, 2}n suchthat |A| = |Ln|+ t, then A has at least t

(n−12

)comparable pairs.

Our approach is inspired by the proof of the LYM Inequality.

I Pick a random maximal chain C = (x0, x1, . . . , x2n) in{0, 1, 2}n according to some distribution µ

Some simple calculations show that the theorem holds providedthat µ satisfies:

I P(x ∈ C) ≥ 1|Ln| for all x ∈ {0, 1, 2}n

I P(x ∈ C | y ∈ C) ≤ 2n−1 for all comparable pairs x, y with x at

least as close to the middle level as y.

Outline of Approach

Theorem (N., Scott & Sudakov, 2018): If A ⊆ {0, 1, 2}n suchthat |A| = |Ln|+ t, then A has at least t

(n−12

)comparable pairs.

Our approach is inspired by the proof of the LYM Inequality.

I Pick a random maximal chain C = (x0, x1, . . . , x2n) in{0, 1, 2}n according to some distribution µ

Some simple calculations show that the theorem holds providedthat µ satisfies:

I P(x ∈ C) ≥ 1|Ln| for all x ∈ {0, 1, 2}n

I P(x ∈ C | y ∈ C) ≤ 2n−1 for all comparable pairs x, y with x at

least as close to the middle level as y.

Constructing the Distribution

Theorem (Anderson, 1968): There is a multiset M of maximalchains in {0, 1, 2}n such that any x, y ∈ Lk are contained in thesame number of chains in M.

This is known as a regular covering by chains.

We choose C uniformly at random from M. Each x ∈ Lk isincluded with probability 1

|Lk| ≥1|Ln| .

The problem reduces to bounding P(y ∈ C | x ∈ C).

This becomes complicated...

Constructing the Distribution

Theorem (Anderson, 1968): There is a multiset M of maximalchains in {0, 1, 2}n such that any x, y ∈ Lk are contained in thesame number of chains in M.

This is known as a regular covering by chains.

We choose C uniformly at random from M. Each x ∈ Lk isincluded with probability 1

|Lk| ≥1|Ln| .

The problem reduces to bounding P(y ∈ C | x ∈ C).

This becomes complicated...

Constructing the Distribution

Theorem (Anderson, 1968): There is a multiset M of maximalchains in {0, 1, 2}n such that any x, y ∈ Lk are contained in thesame number of chains in M.

This is known as a regular covering by chains.

We choose C uniformly at random from M. Each x ∈ Lk isincluded with probability 1

|Lk| ≥1|Ln| .

The problem reduces to bounding P(y ∈ C | x ∈ C).

This becomes complicated...

Constructing the Distribution

Theorem (Anderson, 1968): There is a multiset M of maximalchains in {0, 1, 2}n such that any x, y ∈ Lk are contained in thesame number of chains in M.

This is known as a regular covering by chains.

We choose C uniformly at random from M. Each x ∈ Lk isincluded with probability 1

|Lk| ≥1|Ln| .

The problem reduces to bounding P(y ∈ C | x ∈ C).

This becomes complicated...

Constructing the Distribution

Theorem (Anderson, 1968): There is a multiset M of maximalchains in {0, 1, 2}n such that any x, y ∈ Lk are contained in thesame number of chains in M.

This is known as a regular covering by chains.

We choose C uniformly at random from M. Each x ∈ Lk isincluded with probability 1

|Lk| ≥1|Ln| .

The problem reduces to bounding P(y ∈ C | x ∈ C).

This becomes complicated...

Constructing the Distribution

Theorem (Anderson, 1968): There is a multiset M of maximalchains in {0, 1, 2}n such that any x, y ∈ Lk are contained in thesame number of chains in M.

This is known as a regular covering by chains.

We choose C uniformly at random from M. Each x ∈ Lk isincluded with probability 1

|Lk| ≥1|Ln| .

The problem reduces to bounding P(y ∈ C | x ∈ C).

This becomes complicated...

Conclusion

Naturally, one would guess that the “strong version” of Kleitman’sTheorem holds in {0, 1, 2}n.

We (and, independently, Balogh and Wagner) conjectured that thebest construction is always to take elements “as close to themiddle layer as possible.”

Balogh, Petrıckova and Wagner (2017+) disproved this.

Open Problem: Extend our result to {0, 1, . . . , k}n for fixed kand n ≥ n0(k).

This poset has a regular covering by chains, but analysing arandom chain from the covering becomes even more complicated...

Conclusion

Naturally, one would guess that the “strong version” of Kleitman’sTheorem holds in {0, 1, 2}n.

We (and, independently, Balogh and Wagner) conjectured that thebest construction is always to take elements “as close to themiddle layer as possible.”

Balogh, Petrıckova and Wagner (2017+) disproved this.

Open Problem: Extend our result to {0, 1, . . . , k}n for fixed kand n ≥ n0(k).

This poset has a regular covering by chains, but analysing arandom chain from the covering becomes even more complicated...

Conclusion

Naturally, one would guess that the “strong version” of Kleitman’sTheorem holds in {0, 1, 2}n.

We (and, independently, Balogh and Wagner) conjectured that thebest construction is always to take elements “as close to themiddle layer as possible.”

Balogh, Petrıckova and Wagner (2017+) disproved this.

Open Problem: Extend our result to {0, 1, . . . , k}n for fixed kand n ≥ n0(k).

This poset has a regular covering by chains, but analysing arandom chain from the covering becomes even more complicated...

Conclusion

Naturally, one would guess that the “strong version” of Kleitman’sTheorem holds in {0, 1, 2}n.

We (and, independently, Balogh and Wagner) conjectured that thebest construction is always to take elements “as close to themiddle layer as possible.”

Balogh, Petrıckova and Wagner (2017+) disproved this.

Open Problem: Extend our result to {0, 1, . . . , k}n for fixed kand n ≥ n0(k).

This poset has a regular covering by chains, but analysing arandom chain from the covering becomes even more complicated...

Conclusion

Naturally, one would guess that the “strong version” of Kleitman’sTheorem holds in {0, 1, 2}n.

We (and, independently, Balogh and Wagner) conjectured that thebest construction is always to take elements “as close to themiddle layer as possible.”

Balogh, Petrıckova and Wagner (2017+) disproved this.

Open Problem: Extend our result to {0, 1, . . . , k}n for fixed kand n ≥ n0(k).

This poset has a regular covering by chains, but analysing arandom chain from the covering becomes even more complicated...