Partitioning -fold covers into many subcovers

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Introduction New results Open problems Partitioning κ-fold covers into κ many subcovers Márton Elekes [email protected] www.renyi.hu/˜emarci Rényi Institute, Budapest, Hungary Logic Colloquium 2007 Joint work with Tamás Mátrai and Lajos Soukup. We gratefully acknowledge the support of Öveges Project of and . Márton Elekes [email protected] www.renyi.hu/˜ emarci Partitioning κ-fold covers into κ many subcovers

Transcript of Partitioning -fold covers into many subcovers

Page 1: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Partitioning κ-fold covers into κ many subcovers

Márton [email protected]

www.renyi.hu/˜emarci

Rényi Institute, Budapest, Hungary

Logic Colloquium 2007

Joint work with Tamás Mátrai and Lajos Soukup.

We gratefully acknowledge the support of Öveges Project of and .

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 2: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Outline

1 IntroductionThe problemMotivationTwo easy examples

2 New resultsConvex bodiesClosed setsArbitrary setsGraphs

3 Open problems

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 3: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Definition

Let X be a set and κ be a cardinal (usually infinite). We say that H ⊂ P(X) is a κ-foldcover of X if each x ∈ X is covered at least κ times.

Question

(Main question) Under what assumptions can we decompose a κ-fold cover into κmany disjoint subcovers?

An equivalent formulation:

Definition

Let H ⊂ P(X). We say that c : H → κ is a good colouring with κ colours, (or a goodκ-colouring), if ∀x ∈ X and ∀α < κ ∃H ∈ H such that x ∈ H and c(H) = α.

Fact

H has a good κ-colouring iff it can be decomposed into κ many disjoint subcovers.

Remark

It would also be natural (and useful) to define these notions relative to a set Y ⊂ X , butfor the sake of simplicity we stick to Y = X in this talk.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 4: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Definition

Let X be a set and κ be a cardinal (usually infinite). We say that H ⊂ P(X) is a κ-foldcover of X if each x ∈ X is covered at least κ times.

Question

(Main question) Under what assumptions can we decompose a κ-fold cover into κmany disjoint subcovers?

An equivalent formulation:

Definition

Let H ⊂ P(X). We say that c : H → κ is a good colouring with κ colours, (or a goodκ-colouring), if ∀x ∈ X and ∀α < κ ∃H ∈ H such that x ∈ H and c(H) = α.

Fact

H has a good κ-colouring iff it can be decomposed into κ many disjoint subcovers.

Remark

It would also be natural (and useful) to define these notions relative to a set Y ⊂ X , butfor the sake of simplicity we stick to Y = X in this talk.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 5: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Definition

Let X be a set and κ be a cardinal (usually infinite). We say that H ⊂ P(X) is a κ-foldcover of X if each x ∈ X is covered at least κ times.

Question

(Main question) Under what assumptions can we decompose a κ-fold cover into κmany disjoint subcovers?

An equivalent formulation:

Definition

Let H ⊂ P(X). We say that c : H → κ is a good colouring with κ colours, (or a goodκ-colouring), if ∀x ∈ X and ∀α < κ ∃H ∈ H such that x ∈ H and c(H) = α.

Fact

H has a good κ-colouring iff it can be decomposed into κ many disjoint subcovers.

Remark

It would also be natural (and useful) to define these notions relative to a set Y ⊂ X , butfor the sake of simplicity we stick to Y = X in this talk.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 6: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Definition

Let X be a set and κ be a cardinal (usually infinite). We say that H ⊂ P(X) is a κ-foldcover of X if each x ∈ X is covered at least κ times.

Question

(Main question) Under what assumptions can we decompose a κ-fold cover into κmany disjoint subcovers?

An equivalent formulation:

Definition

Let H ⊂ P(X). We say that c : H → κ is a good colouring with κ colours, (or a goodκ-colouring), if ∀x ∈ X and ∀α < κ ∃H ∈ H such that x ∈ H and c(H) = α.

Fact

H has a good κ-colouring iff it can be decomposed into κ many disjoint subcovers.

Remark

It would also be natural (and useful) to define these notions relative to a set Y ⊂ X , butfor the sake of simplicity we stick to Y = X in this talk.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 7: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Definition

Let X be a set and κ be a cardinal (usually infinite). We say that H ⊂ P(X) is a κ-foldcover of X if each x ∈ X is covered at least κ times.

Question

(Main question) Under what assumptions can we decompose a κ-fold cover into κmany disjoint subcovers?

An equivalent formulation:

Definition

Let H ⊂ P(X). We say that c : H → κ is a good colouring with κ colours, (or a goodκ-colouring), if ∀x ∈ X and ∀α < κ ∃H ∈ H such that x ∈ H and c(H) = α.

Fact

H has a good κ-colouring iff it can be decomposed into κ many disjoint subcovers.

Remark

It would also be natural (and useful) to define these notions relative to a set Y ⊂ X , butfor the sake of simplicity we stick to Y = X in this talk.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 8: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Definition

Let X be a set and κ be a cardinal (usually infinite). We say that H ⊂ P(X) is a κ-foldcover of X if each x ∈ X is covered at least κ times.

Question

(Main question) Under what assumptions can we decompose a κ-fold cover into κmany disjoint subcovers?

An equivalent formulation:

Definition

Let H ⊂ P(X). We say that c : H → κ is a good colouring with κ colours, (or a goodκ-colouring), if ∀x ∈ X and ∀α < κ ∃H ∈ H such that x ∈ H and c(H) = α.

Fact

H has a good κ-colouring iff it can be decomposed into κ many disjoint subcovers.

Remark

It would also be natural (and useful) to define these notions relative to a set Y ⊂ X , butfor the sake of simplicity we stick to Y = X in this talk.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 9: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Some discrete geometry

Theorem

(Mani-Pach, unpublished, more than 20 years old, ca. 100 pages) Every 33-fold coverof R2 with unit discs has a good 2-colouring.

Theorem

(Tardos-Tóth) Every 43-fold cover of R2 with translates of a triangle has a good2-colouring.

Theorem

(Tóth, ???) For every convex polygon there exists n ∈ N so that every n-fold cover ofR

2 with translates of the polygon has a good 2-colouring.

Conjecture

(Pach) The same holds for every convex set.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 10: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Some discrete geometry

Theorem

(Mani-Pach, unpublished, more than 20 years old, ca. 100 pages) Every 33-fold coverof R2 with unit discs has a good 2-colouring.

Theorem

(Tardos-Tóth) Every 43-fold cover of R2 with translates of a triangle has a good2-colouring.

Theorem

(Tóth, ???) For every convex polygon there exists n ∈ N so that every n-fold cover ofR

2 with translates of the polygon has a good 2-colouring.

Conjecture

(Pach) The same holds for every convex set.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 11: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Some discrete geometry

Theorem

(Mani-Pach, unpublished, more than 20 years old, ca. 100 pages) Every 33-fold coverof R2 with unit discs has a good 2-colouring.

Theorem

(Tardos-Tóth) Every 43-fold cover of R2 with translates of a triangle has a good2-colouring.

Theorem

(Tóth, ???) For every convex polygon there exists n ∈ N so that every n-fold cover ofR

2 with translates of the polygon has a good 2-colouring.

Conjecture

(Pach) The same holds for every convex set.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 12: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Some discrete geometry

Theorem

(Mani-Pach, unpublished, more than 20 years old, ca. 100 pages) Every 33-fold coverof R2 with unit discs has a good 2-colouring.

Theorem

(Tardos-Tóth) Every 43-fold cover of R2 with translates of a triangle has a good2-colouring.

Theorem

(Tóth, ???) For every convex polygon there exists n ∈ N so that every n-fold cover ofR

2 with translates of the polygon has a good 2-colouring.

Conjecture

(Pach) The same holds for every convex set.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 13: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Some discrete geometry

However,

Theorem

(Pach-Tardos-Tóth) For every n ∈ N there is an n-fold cover of R2 with axis-parallelrectangles or with translates of a suitable concave quadrilateral that has no good2-colouring.

Remark

The case of R3 or higher is dramatically different!

Remark

Surprisingly, this theory has applications for sensor networks.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 14: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Some discrete geometry

However,

Theorem

(Pach-Tardos-Tóth) For every n ∈ N there is an n-fold cover of R2 with axis-parallelrectangles or with translates of a suitable concave quadrilateral that has no good2-colouring.

Remark

The case of R3 or higher is dramatically different!

Remark

Surprisingly, this theory has applications for sensor networks.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 15: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Some discrete geometry

However,

Theorem

(Pach-Tardos-Tóth) For every n ∈ N there is an n-fold cover of R2 with axis-parallelrectangles or with translates of a suitable concave quadrilateral that has no good2-colouring.

Remark

The case of R3 or higher is dramatically different!

Remark

Surprisingly, this theory has applications for sensor networks.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 16: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Set theory comes into the picture

J. Pach asked whether such results could be proved for infinite κ.

Theorem

(Aharoni-Hajnal-Milner) Let κ be a cardinal (finite or infinite) and L be a linearly orderedset. Then every κ-fold cover of L consisting of convex sets has a good κ-colouring.

Question

(Pach, Hajnal) How about the higher dimensional versions? E.g. rectangles in R2?

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 17: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Set theory comes into the picture

J. Pach asked whether such results could be proved for infinite κ.

Theorem

(Aharoni-Hajnal-Milner) Let κ be a cardinal (finite or infinite) and L be a linearly orderedset. Then every κ-fold cover of L consisting of convex sets has a good κ-colouring.

Question

(Pach, Hajnal) How about the higher dimensional versions? E.g. rectangles in R2?

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 18: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Set theory comes into the picture

J. Pach asked whether such results could be proved for infinite κ.

Theorem

(Aharoni-Hajnal-Milner) Let κ be a cardinal (finite or infinite) and L be a linearly orderedset. Then every κ-fold cover of L consisting of convex sets has a good κ-colouring.

Question

(Pach, Hajnal) How about the higher dimensional versions? E.g. rectangles in R2?

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 19: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Two easy examples

Statement

Let κ be infinite and X be a set with |X | ≤ κ. Then every κ-fold cover of X has a goodκ-colouring.

Proof Trivial transfinite recursion. Let {xα : α < κ} be so that each x ∈ X occurs κtimes. When x shows up for the α’s time, there is an uncoloured H containing x , give itcolour α. �

Statement

Let κ be infinite and X be a set with |X | ≥ 2κ. Then there is a κ-fold cover of X thathas not even a good 2-colouring.

Proof We may assume X = [κ]κ. The cover H will be of the form {Hα : α < κ}. Theidea is that for every A ∈ [κ]κ there will be an x ∈ X so that x ∈ Hα ⇐⇒ α ∈ A. Butthis is easily achieved by choosing x = A, that is, by setting Hα = {A ∈ [κ]κ : α ∈ A}.�

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 20: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Two easy examples

Statement

Let κ be infinite and X be a set with |X | ≤ κ. Then every κ-fold cover of X has a goodκ-colouring.

Proof Trivial transfinite recursion. Let {xα : α < κ} be so that each x ∈ X occurs κtimes. When x shows up for the α’s time, there is an uncoloured H containing x , give itcolour α. �

Statement

Let κ be infinite and X be a set with |X | ≥ 2κ. Then there is a κ-fold cover of X thathas not even a good 2-colouring.

Proof We may assume X = [κ]κ. The cover H will be of the form {Hα : α < κ}. Theidea is that for every A ∈ [κ]κ there will be an x ∈ X so that x ∈ Hα ⇐⇒ α ∈ A. Butthis is easily achieved by choosing x = A, that is, by setting Hα = {A ∈ [κ]κ : α ∈ A}.�

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 21: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Two easy examples

Statement

Let κ be infinite and X be a set with |X | ≤ κ. Then every κ-fold cover of X has a goodκ-colouring.

Proof Trivial transfinite recursion. Let {xα : α < κ} be so that each x ∈ X occurs κtimes. When x shows up for the α’s time, there is an uncoloured H containing x , give itcolour α. �

Statement

Let κ be infinite and X be a set with |X | ≥ 2κ. Then there is a κ-fold cover of X thathas not even a good 2-colouring.

Proof We may assume X = [κ]κ. The cover H will be of the form {Hα : α < κ}. Theidea is that for every A ∈ [κ]κ there will be an x ∈ X so that x ∈ Hα ⇐⇒ α ∈ A. Butthis is easily achieved by choosing x = A, that is, by setting Hα = {A ∈ [κ]κ : α ∈ A}.�

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 22: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Two easy examples

Statement

Let κ be infinite and X be a set with |X | ≤ κ. Then every κ-fold cover of X has a goodκ-colouring.

Proof Trivial transfinite recursion. Let {xα : α < κ} be so that each x ∈ X occurs κtimes. When x shows up for the α’s time, there is an uncoloured H containing x , give itcolour α. �

Statement

Let κ be infinite and X be a set with |X | ≥ 2κ. Then there is a κ-fold cover of X thathas not even a good 2-colouring.

Proof We may assume X = [κ]κ. The cover H will be of the form {Hα : α < κ}. Theidea is that for every A ∈ [κ]κ there will be an x ∈ X so that x ∈ Hα ⇐⇒ α ∈ A. Butthis is easily achieved by choosing x = A, that is, by setting Hα = {A ∈ [κ]κ : α ∈ A}.�

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 23: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Two easy examples

Statement

Let κ be infinite and X be a set with |X | ≤ κ. Then every κ-fold cover of X has a goodκ-colouring.

Proof Trivial transfinite recursion. Let {xα : α < κ} be so that each x ∈ X occurs κtimes. When x shows up for the α’s time, there is an uncoloured H containing x , give itcolour α. �

Statement

Let κ be infinite and X be a set with |X | ≥ 2κ. Then there is a κ-fold cover of X thathas not even a good 2-colouring.

Proof We may assume X = [κ]κ. The cover H will be of the form {Hα : α < κ}. Theidea is that for every A ∈ [κ]κ there will be an x ∈ X so that x ∈ Hα ⇐⇒ α ∈ A. Butthis is easily achieved by choosing x = A, that is, by setting Hα = {A ∈ [κ]κ : α ∈ A}.�

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 24: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Two easy examples

Statement

Let κ be infinite and X be a set with |X | ≤ κ. Then every κ-fold cover of X has a goodκ-colouring.

Proof Trivial transfinite recursion. Let {xα : α < κ} be so that each x ∈ X occurs κtimes. When x shows up for the α’s time, there is an uncoloured H containing x , give itcolour α. �

Statement

Let κ be infinite and X be a set with |X | ≥ 2κ. Then there is a κ-fold cover of X thathas not even a good 2-colouring.

Proof We may assume X = [κ]κ. The cover H will be of the form {Hα : α < κ}. Theidea is that for every A ∈ [κ]κ there will be an x ∈ X so that x ∈ Hα ⇐⇒ α ∈ A. Butthis is easily achieved by choosing x = A, that is, by setting Hα = {A ∈ [κ]κ : α ∈ A}.�

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 25: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Two easy examples

Statement

Let κ be infinite and X be a set with |X | ≤ κ. Then every κ-fold cover of X has a goodκ-colouring.

Proof Trivial transfinite recursion. Let {xα : α < κ} be so that each x ∈ X occurs κtimes. When x shows up for the α’s time, there is an uncoloured H containing x , give itcolour α. �

Statement

Let κ be infinite and X be a set with |X | ≥ 2κ. Then there is a κ-fold cover of X thathas not even a good 2-colouring.

Proof We may assume X = [κ]κ. The cover H will be of the form {Hα : α < κ}. Theidea is that for every A ∈ [κ]κ there will be an x ∈ X so that x ∈ Hα ⇐⇒ α ∈ A. Butthis is easily achieved by choosing x = A, that is, by setting Hα = {A ∈ [κ]κ : α ∈ A}.�

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 26: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Two easy examples

Statement

Let κ be infinite and X be a set with |X | ≤ κ. Then every κ-fold cover of X has a goodκ-colouring.

Proof Trivial transfinite recursion. Let {xα : α < κ} be so that each x ∈ X occurs κtimes. When x shows up for the α’s time, there is an uncoloured H containing x , give itcolour α. �

Statement

Let κ be infinite and X be a set with |X | ≥ 2κ. Then there is a κ-fold cover of X thathas not even a good 2-colouring.

Proof We may assume X = [κ]κ. The cover H will be of the form {Hα : α < κ}. Theidea is that for every A ∈ [κ]κ there will be an x ∈ X so that x ∈ Hα ⇐⇒ α ∈ A. Butthis is easily achieved by choosing x = A, that is, by setting Hα = {A ∈ [κ]κ : α ∈ A}.�

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 27: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

The problemMotivationTwo easy examples

Two easy examples

Statement

Let κ be infinite and X be a set with |X | ≤ κ. Then every κ-fold cover of X has a goodκ-colouring.

Proof Trivial transfinite recursion. Let {xα : α < κ} be so that each x ∈ X occurs κtimes. When x shows up for the α’s time, there is an uncoloured H containing x , give itcolour α. �

Statement

Let κ be infinite and X be a set with |X | ≥ 2κ. Then there is a κ-fold cover of X thathas not even a good 2-colouring.

Proof We may assume X = [κ]κ. The cover H will be of the form {Hα : α < κ}. Theidea is that for every A ∈ [κ]κ there will be an x ∈ X so that x ∈ Hα ⇐⇒ α ∈ A. Butthis is easily achieved by choosing x = A, that is, by setting Hα = {A ∈ [κ]κ : α ∈ A}.�

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 28: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

The case κ < ω is very well studied by geometers.

For κ = ω there are many counterexamples.

Theorem

There is an ω-fold cover of R2 by axis-parallel closed rectangles that has no good2-colouring.

However, we do not know if the cover can consist of translates of a fixed set.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 29: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

The case κ < ω is very well studied by geometers.

For κ = ω there are many counterexamples.

Theorem

There is an ω-fold cover of R2 by axis-parallel closed rectangles that has no good2-colouring.

However, we do not know if the cover can consist of translates of a fixed set.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 30: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

The case κ < ω is very well studied by geometers.

For κ = ω there are many counterexamples.

Theorem

There is an ω-fold cover of R2 by axis-parallel closed rectangles that has no good2-colouring.

However, we do not know if the cover can consist of translates of a fixed set.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 31: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

The case κ < ω is very well studied by geometers.

For κ = ω there are many counterexamples.

Theorem

There is an ω-fold cover of R2 by axis-parallel closed rectangles that has no good2-colouring.

However, we do not know if the cover can consist of translates of a fixed set.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 32: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

Let now κ be uncountable.Recall

Statement

If H is a κ-fold cover of a set X and |X | ≤ κ then H has a good κ-colouring.

Hence κ = 2ω is easy, and so the nontrivial questions are ω1 ≤ κ < 2ω .Hence under CH everything is clear.The next slide summarises what we know if we do not assume CH.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 33: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

Let now κ be uncountable.Recall

Statement

If H is a κ-fold cover of a set X and |X | ≤ κ then H has a good κ-colouring.

Hence κ = 2ω is easy, and so the nontrivial questions are ω1 ≤ κ < 2ω .Hence under CH everything is clear.The next slide summarises what we know if we do not assume CH.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 34: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

Let now κ be uncountable.Recall

Statement

If H is a κ-fold cover of a set X and |X | ≤ κ then H has a good κ-colouring.

Hence κ = 2ω is easy, and so the nontrivial questions are ω1 ≤ κ < 2ω .Hence under CH everything is clear.The next slide summarises what we know if we do not assume CH.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 35: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

Let now κ be uncountable.Recall

Statement

If H is a κ-fold cover of a set X and |X | ≤ κ then H has a good κ-colouring.

Hence κ = 2ω is easy, and so the nontrivial questions are ω1 ≤ κ < 2ω .Hence under CH everything is clear.The next slide summarises what we know if we do not assume CH.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 36: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

Let now κ be uncountable.Recall

Statement

If H is a κ-fold cover of a set X and |X | ≤ κ then H has a good κ-colouring.

Hence κ = 2ω is easy, and so the nontrivial questions are ω1 ≤ κ < 2ω .Hence under CH everything is clear.The next slide summarises what we know if we do not assume CH.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 37: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

Theorem

Let κ ≥ ω1. Then every κ-fold cover of R2 by closed polygons has a good κ-colouring.

Theorem

Let κ ≥ ω1. Then every κ-fold cover of R2 by closed discs has a good κ-colouring.

But!

Theorem

Assume MAκ(σ-centered). Then there exists a κ-fold cover of R2 by isometric copies ofa strictly convex compact set that has no good 2-colouring.

We do not now if the isometries can be replaced by translations.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 38: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

Theorem

Let κ ≥ ω1. Then every κ-fold cover of R2 by closed polygons has a good κ-colouring.

Theorem

Let κ ≥ ω1. Then every κ-fold cover of R2 by closed discs has a good κ-colouring.

But!

Theorem

Assume MAκ(σ-centered). Then there exists a κ-fold cover of R2 by isometric copies ofa strictly convex compact set that has no good 2-colouring.

We do not now if the isometries can be replaced by translations.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 39: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

Theorem

Let κ ≥ ω1. Then every κ-fold cover of R2 by closed polygons has a good κ-colouring.

Theorem

Let κ ≥ ω1. Then every κ-fold cover of R2 by closed discs has a good κ-colouring.

But!

Theorem

Assume MAκ(σ-centered). Then there exists a κ-fold cover of R2 by isometric copies ofa strictly convex compact set that has no good 2-colouring.

We do not now if the isometries can be replaced by translations.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 40: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Convex bodies

Theorem

Let κ ≥ ω1. Then every κ-fold cover of R2 by closed polygons has a good κ-colouring.

Theorem

Let κ ≥ ω1. Then every κ-fold cover of R2 by closed discs has a good κ-colouring.

But!

Theorem

Assume MAκ(σ-centered). Then there exists a κ-fold cover of R2 by isometric copies ofa strictly convex compact set that has no good 2-colouring.

We do not now if the isometries can be replaced by translations.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 41: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Closed sets

Let first κ ≤ ω.

Theorem

There exists an ω-fold cover of R2 with translates of a fixed compact set that has nogood 2-colouring.

Let now κ be uncountable.As mentioned above, if CH holds then all κ-fold covers have good κ-colourings forevery κ ≥ ω1.The next theorem shows that this positive statement is also consistent with anarbitrarily large continuum. More precisely, we can add an arbitrary number of Cohenreals to a suitable model of ZFC.

Theorem

Let λ be a cardinal and V be a model of ZFC satisfying GCH + �µ for everyω = cf (µ) < µ ≤ λ. Denote by VCλ the model obtained by adding λ Cohen reals.Then in VCλ for every κ ≥ ω1 every κ-fold cover of R2 consisting of closed sets has agood κ-colouring.

How about the negative consistency?

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 42: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Closed sets

Let first κ ≤ ω.

Theorem

There exists an ω-fold cover of R2 with translates of a fixed compact set that has nogood 2-colouring.

Let now κ be uncountable.As mentioned above, if CH holds then all κ-fold covers have good κ-colourings forevery κ ≥ ω1.The next theorem shows that this positive statement is also consistent with anarbitrarily large continuum. More precisely, we can add an arbitrary number of Cohenreals to a suitable model of ZFC.

Theorem

Let λ be a cardinal and V be a model of ZFC satisfying GCH + �µ for everyω = cf (µ) < µ ≤ λ. Denote by VCλ the model obtained by adding λ Cohen reals.Then in VCλ for every κ ≥ ω1 every κ-fold cover of R2 consisting of closed sets has agood κ-colouring.

How about the negative consistency?

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 43: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Closed sets

Let first κ ≤ ω.

Theorem

There exists an ω-fold cover of R2 with translates of a fixed compact set that has nogood 2-colouring.

Let now κ be uncountable.As mentioned above, if CH holds then all κ-fold covers have good κ-colourings forevery κ ≥ ω1.The next theorem shows that this positive statement is also consistent with anarbitrarily large continuum. More precisely, we can add an arbitrary number of Cohenreals to a suitable model of ZFC.

Theorem

Let λ be a cardinal and V be a model of ZFC satisfying GCH + �µ for everyω = cf (µ) < µ ≤ λ. Denote by VCλ the model obtained by adding λ Cohen reals.Then in VCλ for every κ ≥ ω1 every κ-fold cover of R2 consisting of closed sets has agood κ-colouring.

How about the negative consistency?

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 44: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Closed sets

Let first κ ≤ ω.

Theorem

There exists an ω-fold cover of R2 with translates of a fixed compact set that has nogood 2-colouring.

Let now κ be uncountable.As mentioned above, if CH holds then all κ-fold covers have good κ-colourings forevery κ ≥ ω1.The next theorem shows that this positive statement is also consistent with anarbitrarily large continuum. More precisely, we can add an arbitrary number of Cohenreals to a suitable model of ZFC.

Theorem

Let λ be a cardinal and V be a model of ZFC satisfying GCH + �µ for everyω = cf (µ) < µ ≤ λ. Denote by VCλ the model obtained by adding λ Cohen reals.Then in VCλ for every κ ≥ ω1 every κ-fold cover of R2 consisting of closed sets has agood κ-colouring.

How about the negative consistency?

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 45: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Closed sets

Let first κ ≤ ω.

Theorem

There exists an ω-fold cover of R2 with translates of a fixed compact set that has nogood 2-colouring.

Let now κ be uncountable.As mentioned above, if CH holds then all κ-fold covers have good κ-colourings forevery κ ≥ ω1.The next theorem shows that this positive statement is also consistent with anarbitrarily large continuum. More precisely, we can add an arbitrary number of Cohenreals to a suitable model of ZFC.

Theorem

Let λ be a cardinal and V be a model of ZFC satisfying GCH + �µ for everyω = cf (µ) < µ ≤ λ. Denote by VCλ the model obtained by adding λ Cohen reals.Then in VCλ for every κ ≥ ω1 every κ-fold cover of R2 consisting of closed sets has agood κ-colouring.

How about the negative consistency?

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 46: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Closed sets

Let first κ ≤ ω.

Theorem

There exists an ω-fold cover of R2 with translates of a fixed compact set that has nogood 2-colouring.

Let now κ be uncountable.As mentioned above, if CH holds then all κ-fold covers have good κ-colourings forevery κ ≥ ω1.The next theorem shows that this positive statement is also consistent with anarbitrarily large continuum. More precisely, we can add an arbitrary number of Cohenreals to a suitable model of ZFC.

Theorem

Let λ be a cardinal and V be a model of ZFC satisfying GCH + �µ for everyω = cf (µ) < µ ≤ λ. Denote by VCλ the model obtained by adding λ Cohen reals.Then in VCλ for every κ ≥ ω1 every κ-fold cover of R2 consisting of closed sets has agood κ-colouring.

How about the negative consistency?

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 47: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Closed sets

Let first κ ≤ ω.

Theorem

There exists an ω-fold cover of R2 with translates of a fixed compact set that has nogood 2-colouring.

Let now κ be uncountable.As mentioned above, if CH holds then all κ-fold covers have good κ-colourings forevery κ ≥ ω1.The next theorem shows that this positive statement is also consistent with anarbitrarily large continuum. More precisely, we can add an arbitrary number of Cohenreals to a suitable model of ZFC.

Theorem

Let λ be a cardinal and V be a model of ZFC satisfying GCH + �µ for everyω = cf (µ) < µ ≤ λ. Denote by VCλ the model obtained by adding λ Cohen reals.Then in VCλ for every κ ≥ ω1 every κ-fold cover of R2 consisting of closed sets has agood κ-colouring.

How about the negative consistency?

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 48: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Closed sets

Theorem

Assume MAκ(σ-centered). Then there exists a κ-fold cover of R2 by translates of acompact set that has a no good 2-colouring.

Remark

Actually, the κ = ω result is a consequence of this one, as MAω(σ-centered) is true.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 49: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Closed sets

Theorem

Assume MAκ(σ-centered). Then there exists a κ-fold cover of R2 by translates of acompact set that has a no good 2-colouring.

Remark

Actually, the κ = ω result is a consequence of this one, as MAω(σ-centered) is true.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 50: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

We look for ‘an optimal bound for the size of elements of the κ-fold cover H’. The rightnotion turns out to be the following.

Definition

Let S(κ) be the minimal cardinal such that for every λ < S(κ) every κ-fold cover Hwith |H| < λ (∀H ∈ H) has a good κ-colouring.

Theorem

κ++ ≤ S(κ) ≤ (2κ)+ for every κ ≥ ω.

Corollary

Assume GCH. Then S(κ) = κ++ = (2κ)+ for every κ ≥ ω.

The next slide shows that neither value is ‘correct’.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 51: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

We look for ‘an optimal bound for the size of elements of the κ-fold cover H’. The rightnotion turns out to be the following.

Definition

Let S(κ) be the minimal cardinal such that for every λ < S(κ) every κ-fold cover Hwith |H| < λ (∀H ∈ H) has a good κ-colouring.

Theorem

κ++ ≤ S(κ) ≤ (2κ)+ for every κ ≥ ω.

Corollary

Assume GCH. Then S(κ) = κ++ = (2κ)+ for every κ ≥ ω.

The next slide shows that neither value is ‘correct’.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 52: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

We look for ‘an optimal bound for the size of elements of the κ-fold cover H’. The rightnotion turns out to be the following.

Definition

Let S(κ) be the minimal cardinal such that for every λ < S(κ) every κ-fold cover Hwith |H| < λ (∀H ∈ H) has a good κ-colouring.

Theorem

κ++ ≤ S(κ) ≤ (2κ)+ for every κ ≥ ω.

Corollary

Assume GCH. Then S(κ) = κ++ = (2κ)+ for every κ ≥ ω.

The next slide shows that neither value is ‘correct’.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 53: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

We look for ‘an optimal bound for the size of elements of the κ-fold cover H’. The rightnotion turns out to be the following.

Definition

Let S(κ) be the minimal cardinal such that for every λ < S(κ) every κ-fold cover Hwith |H| < λ (∀H ∈ H) has a good κ-colouring.

Theorem

κ++ ≤ S(κ) ≤ (2κ)+ for every κ ≥ ω.

Corollary

Assume GCH. Then S(κ) = κ++ = (2κ)+ for every κ ≥ ω.

The next slide shows that neither value is ‘correct’.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 54: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

We look for ‘an optimal bound for the size of elements of the κ-fold cover H’. The rightnotion turns out to be the following.

Definition

Let S(κ) be the minimal cardinal such that for every λ < S(κ) every κ-fold cover Hwith |H| < λ (∀H ∈ H) has a good κ-colouring.

Theorem

κ++ ≤ S(κ) ≤ (2κ)+ for every κ ≥ ω.

Corollary

Assume GCH. Then S(κ) = κ++ = (2κ)+ for every κ ≥ ω.

The next slide shows that neither value is ‘correct’.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 55: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

We look for ‘an optimal bound for the size of elements of the κ-fold cover H’. The rightnotion turns out to be the following.

Definition

Let S(κ) be the minimal cardinal such that for every λ < S(κ) every κ-fold cover Hwith |H| < λ (∀H ∈ H) has a good κ-colouring.

Theorem

κ++ ≤ S(κ) ≤ (2κ)+ for every κ ≥ ω.

Corollary

Assume GCH. Then S(κ) = κ++ = (2κ)+ for every κ ≥ ω.

The next slide shows that neither value is ‘correct’.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 56: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

Theorem

Assume |• κ+ . Then S(κ) = κ++.

Remark

Let κ ≥ ω. Then S(κ) = (2κ)+ can fail, since |• κ+ + 2κ > κ+ is consistent.

Theorem

Assume MA(countable). Then S(ω) = (2ω)+.

Remark

This shows that S(ω) = ω++ can fail, since MA(countable) + ¬CH is consistent.

So far we can only push this one cardinal higher.

Theorem

Assume Baumgartner’s Axiom +CH. Then S(ω1) > ω++1 .

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 57: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

Theorem

Assume |• κ+ . Then S(κ) = κ++.

Remark

Let κ ≥ ω. Then S(κ) = (2κ)+ can fail, since |• κ+ + 2κ > κ+ is consistent.

Theorem

Assume MA(countable). Then S(ω) = (2ω)+.

Remark

This shows that S(ω) = ω++ can fail, since MA(countable) + ¬CH is consistent.

So far we can only push this one cardinal higher.

Theorem

Assume Baumgartner’s Axiom +CH. Then S(ω1) > ω++1 .

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 58: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

Theorem

Assume |• κ+ . Then S(κ) = κ++.

Remark

Let κ ≥ ω. Then S(κ) = (2κ)+ can fail, since |• κ+ + 2κ > κ+ is consistent.

Theorem

Assume MA(countable). Then S(ω) = (2ω)+.

Remark

This shows that S(ω) = ω++ can fail, since MA(countable) + ¬CH is consistent.

So far we can only push this one cardinal higher.

Theorem

Assume Baumgartner’s Axiom +CH. Then S(ω1) > ω++1 .

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 59: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

Theorem

Assume |• κ+ . Then S(κ) = κ++.

Remark

Let κ ≥ ω. Then S(κ) = (2κ)+ can fail, since |• κ+ + 2κ > κ+ is consistent.

Theorem

Assume MA(countable). Then S(ω) = (2ω)+.

Remark

This shows that S(ω) = ω++ can fail, since MA(countable) + ¬CH is consistent.

So far we can only push this one cardinal higher.

Theorem

Assume Baumgartner’s Axiom +CH. Then S(ω1) > ω++1 .

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 60: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

Theorem

Assume |• κ+ . Then S(κ) = κ++.

Remark

Let κ ≥ ω. Then S(κ) = (2κ)+ can fail, since |• κ+ + 2κ > κ+ is consistent.

Theorem

Assume MA(countable). Then S(ω) = (2ω)+.

Remark

This shows that S(ω) = ω++ can fail, since MA(countable) + ¬CH is consistent.

So far we can only push this one cardinal higher.

Theorem

Assume Baumgartner’s Axiom +CH. Then S(ω1) > ω++1 .

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 61: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Arbitrary sets

Remark

By a simple argument all result of this section can be translated to the language ofBernstein property of families of sets.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 62: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Graphs

As this is a very special case, we are more ambitious here. We look for completecharacterisations of good κ-colourable graphs.The case of infinite κ is completely solved.

Theorem

Let κ ≥ ω and G = (V , E) be a graph such that each vertex is of degree at least κ.Then E has a good κ-colouring, that is, the edges can be coloured by κ colours so thatevery vertex is covered by edges of all colours.

κ = 2 is also solved (κ < 2 is trivial).

Theorem

Let G = (V , E) be graph such that each vertex is of degree at least 2. Then E has agood κ-colouring iff no connected component of G is an odd cycle.

Remark

For 3 ≤ κ < ω such a characterisation seems to be difficult. Indeed, even for finite3-regular graphs this is NP-complete.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 63: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Graphs

As this is a very special case, we are more ambitious here. We look for completecharacterisations of good κ-colourable graphs.The case of infinite κ is completely solved.

Theorem

Let κ ≥ ω and G = (V , E) be a graph such that each vertex is of degree at least κ.Then E has a good κ-colouring, that is, the edges can be coloured by κ colours so thatevery vertex is covered by edges of all colours.

κ = 2 is also solved (κ < 2 is trivial).

Theorem

Let G = (V , E) be graph such that each vertex is of degree at least 2. Then E has agood κ-colouring iff no connected component of G is an odd cycle.

Remark

For 3 ≤ κ < ω such a characterisation seems to be difficult. Indeed, even for finite3-regular graphs this is NP-complete.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 64: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Graphs

As this is a very special case, we are more ambitious here. We look for completecharacterisations of good κ-colourable graphs.The case of infinite κ is completely solved.

Theorem

Let κ ≥ ω and G = (V , E) be a graph such that each vertex is of degree at least κ.Then E has a good κ-colouring, that is, the edges can be coloured by κ colours so thatevery vertex is covered by edges of all colours.

κ = 2 is also solved (κ < 2 is trivial).

Theorem

Let G = (V , E) be graph such that each vertex is of degree at least 2. Then E has agood κ-colouring iff no connected component of G is an odd cycle.

Remark

For 3 ≤ κ < ω such a characterisation seems to be difficult. Indeed, even for finite3-regular graphs this is NP-complete.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 65: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Graphs

As this is a very special case, we are more ambitious here. We look for completecharacterisations of good κ-colourable graphs.The case of infinite κ is completely solved.

Theorem

Let κ ≥ ω and G = (V , E) be a graph such that each vertex is of degree at least κ.Then E has a good κ-colouring, that is, the edges can be coloured by κ colours so thatevery vertex is covered by edges of all colours.

κ = 2 is also solved (κ < 2 is trivial).

Theorem

Let G = (V , E) be graph such that each vertex is of degree at least 2. Then E has agood κ-colouring iff no connected component of G is an odd cycle.

Remark

For 3 ≤ κ < ω such a characterisation seems to be difficult. Indeed, even for finite3-regular graphs this is NP-complete.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 66: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Graphs

As this is a very special case, we are more ambitious here. We look for completecharacterisations of good κ-colourable graphs.The case of infinite κ is completely solved.

Theorem

Let κ ≥ ω and G = (V , E) be a graph such that each vertex is of degree at least κ.Then E has a good κ-colouring, that is, the edges can be coloured by κ colours so thatevery vertex is covered by edges of all colours.

κ = 2 is also solved (κ < 2 is trivial).

Theorem

Let G = (V , E) be graph such that each vertex is of degree at least 2. Then E has agood κ-colouring iff no connected component of G is an odd cycle.

Remark

For 3 ≤ κ < ω such a characterisation seems to be difficult. Indeed, even for finite3-regular graphs this is NP-complete.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 67: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Graphs

However, we have the following sufficient condition.

Theorem

Let 1 ≤ κ < ω. Let G = (V , E) be a graph such that every vertex is of degree at leastκ + 1. Then E has a good κ-colouring.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 68: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Convex bodiesClosed setsArbitrary setsGraphs

Graphs

However, we have the following sufficient condition.

Theorem

Let 1 ≤ κ < ω. Let G = (V , E) be a graph such that every vertex is of degree at leastκ + 1. Then E has a good κ-colouring.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 69: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Open problems

Question

Let H be an ω1-fold cover of R2 by closed sets such that |H| = ω1. Does it have agood ω1-colouring?

This follows from CH, but is this true in ZFC?

Question

Is there an ω-fold cover of R2 by translates of a compact convex set that has no a goodω-colouring?

There are so many more! See the preprint that is going to be available soon at

www.renyi.hu/˜emarci.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 70: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Open problems

Question

Let H be an ω1-fold cover of R2 by closed sets such that |H| = ω1. Does it have agood ω1-colouring?

This follows from CH, but is this true in ZFC?

Question

Is there an ω-fold cover of R2 by translates of a compact convex set that has no a goodω-colouring?

There are so many more! See the preprint that is going to be available soon at

www.renyi.hu/˜emarci.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 71: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Open problems

Question

Let H be an ω1-fold cover of R2 by closed sets such that |H| = ω1. Does it have agood ω1-colouring?

This follows from CH, but is this true in ZFC?

Question

Is there an ω-fold cover of R2 by translates of a compact convex set that has no a goodω-colouring?

There are so many more! See the preprint that is going to be available soon at

www.renyi.hu/˜emarci.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 72: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Open problems

Question

Let H be an ω1-fold cover of R2 by closed sets such that |H| = ω1. Does it have agood ω1-colouring?

This follows from CH, but is this true in ZFC?

Question

Is there an ω-fold cover of R2 by translates of a compact convex set that has no a goodω-colouring?

There are so many more! See the preprint that is going to be available soon at

www.renyi.hu/˜emarci.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers

Page 73: Partitioning -fold covers into many subcovers

IntroductionNew results

Open problems

Open problems

Question

Let H be an ω1-fold cover of R2 by closed sets such that |H| = ω1. Does it have agood ω1-colouring?

This follows from CH, but is this true in ZFC?

Question

Is there an ω-fold cover of R2 by translates of a compact convex set that has no a goodω-colouring?

There are so many more! See the preprint that is going to be available soon at

www.renyi.hu/˜emarci.

Márton Elekes [email protected] www.renyi.hu/˜emarci Partitioning κ-fold covers into κ many subcovers