FormalConjectures/ErdosProblems/1128.lean
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9 declarations, 2 contain sorry, 0 declared as axioms
Erdős Problem 1128
Reference: erdosproblems.com/1128
Imports (1)
- def
Erdos1128.IsMonochromaticBoxDocumented declaration: IsMonochromaticBox line 35 sourceSource documentation
A subset $A_1 \times B_1 \times C_1$ of $A \times B \times C$ is monochromatic under a 2-colouring $f : A \to B \to C \to \operatorname{Fin} 2$ if $f$ is constant on $A_1 \times B_1 \times C_1$.
no axioms what this means
Statement uses:
Fin,OfNat.ofNat,Nat,SetProof uses:
Exists,Membership.mem,Eq3 instances, coercions and auxiliary constants left out.
def IsMonochromaticBox {A B C : Type*} (f : A → B → C → Fin 2) (A₁ : Set A) (B₁ : Set B) (C₁ : Set C) : Prop := … - abbrev
Erdos1128.Omega1The set of countable ordinals, expressed using Mathlib'sω_ 1line 45 sourceSource documentation
The set of countable ordinals, expressed using Mathlib's
ω_ 1.standard axioms: Classical.choice, Quot.sound, propext what this means
Proof uses:
Subtype,Ordinal,LT.lt,OrderEmbedding,Ordinal.omega,OfNat.ofNat8 instances, coercions and auxiliary constants left out.
abbrev Omega1 := {o : Ordinal.{0} // o < ω_ 1} … - lemma
Erdos1128.countable_Iio_of_lt_omega1The set of ordinals strictly below any $\gamma < \omega_1$ is countable line 49 sourceSource documentation
The set of ordinals strictly below any $\gamma < \omega_1$ is countable.
private lemma countable_Iio_of_lt_omega1 {γ : Ordinal} (hγ : γ < ω_ 1) : (Set.Iio γ).Countable := by … - lemma
Erdos1128.countable_Iio_omega1Initial segments of ω₁ (as sets ofOmega1elements) are countable line 58 sourceSource documentation
Initial segments of ω₁ (as sets of
Omega1elements) are countable. This is the subtype-order version ofcountable_Iio_of_lt_omega1.private lemma countable_Iio_omega1 (γ : Omega1) : (Set.Iio γ : Set Omega1).Countable := by …
Mentions: Erdos1128.Omega1
- lemma
Erdos1128.countable_subset_bddAny countable subset of $\omega_1$ is bounded strictly below some element of $\omega_1$ line 76 sourceSource documentation
Any countable subset of $\omega_1$ is bounded strictly below some element of $\omega_1$. This uses the key property that $\omega_1$ has uncountable cofinality (it is regular).
private lemma countable_subset_bdd (S : Set Omega1) (hS : S.Countable) : ∃ γ : Omega1, ∀ s ∈ S, s < γ := by …Mentions: Erdos1128.Omega1
- theorem
Erdos1128.erdos_1128.prikryMillsPrikry–Mills counterexample (key lemma): There exists a 2-colouring sorry line 106 sourceSource documentation
Prikry–Mills counterexample (key lemma):
There exists a 2-colouring $f$ of a set of cardinality $\aleph_1$ cubed such that no countable box $A_1 \times B_1 \times C_1$ is monochromatic.
This is the unpublished result of Prikry and Mills (1978). The proof proceeds by transfinite induction along $\omega_1$, which has uncountable cofinality, ensuring every countable box is non-monochromatic.
depends on sorry what this means
Statement uses:
Exists,Eq,Cardinal,Cardinal.mk,OrderEmbedding,Ordinal,Cardinal.aleph,OfNat.ofNat,Fin,Nat,Set,Not, Erdos1128.IsMonochromaticBoxProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous24 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_1128.prikryMills : ∃ (X : Type) (_ : #X = aleph 1) (f : X → X → X → Fin 2), ∀ (A₁ B₁ C₁ : Set X), #A₁ = aleph 0 → #B₁ = aleph 0 → #C₁ = aleph 0 → ¬ IsMonochromaticBox f A₁ B₁ C₁ := by …Mentions: Erdos1128.IsMonochromaticBox
- theorem
Erdos1128.erdos_1128Documented declaration: erdos_1128 line 151 sourceSource documentation
Erdős Problem 1128 (disproved by Prikry–Mills, 1978):
Erdős asked whether every 2-colouring of $A \times B \times C$, where $|A| = |B| = |C| = \aleph_1$, must contain a monochromatic countable box $A_1 \times B_1 \times C_1$ with $|A_1| = |B_1| = |C_1| = \aleph_0$.
The answer is No: Prikry and Mills constructed a 2-colouring of $\omega_1^3$ with no monochromatic countable box.
Note: The positive statement asserts that every 2-colouring of every $\aleph_1^3$ contains a monochromatic countably infinite box. Since the answer is False, this positive statement fails.
depends on sorry what this means
Statement uses:
Iff,False,Eq,Cardinal,Cardinal.mk,OrderEmbedding,Ordinal,Cardinal.aleph,OfNat.ofNat,Fin,Nat,Exists,Set,And, Erdos1128.IsMonochromaticBoxProof uses:
Iff.intro,False.elim,Not, Erdos1128.erdos_1128.prikryMills26 instances, coercions and auxiliary constants left out.
Tactic trace: 9 steps
Open to load the checked before-and-after goal states.
theorem erdos_1128 : answer(False) ↔ ∀ (A B C : Type) (_ : #A = aleph 1) (_ : #B = aleph 1) (_ : #C = aleph 1) (f : A → B → C → Fin 2), ∃ (A₁ : Set A) (B₁ : Set B) (C₁ : Set C), #A₁ = aleph 0 ∧ #B₁ = aleph 0 ∧ #C₁ = aleph 0 ∧ IsMonochromaticBox f A₁ B₁ C₁ := by …Mentions: Erdos1128.IsMonochromaticBox
- theorem
Erdos1128.erdos_1128.variants.prikryMills_explicitDocumented declaration: prikryMills_explicit sorry line 187 sourceSource documentation
Explicit form of Prikry–Mills:
There exists a 2-colouring of $\omega_1 \times \omega_1 \times \omega_1$ such that for every countably infinite $A_1, B_1, C_1 \subseteq \omega_1$, the box $A_1 \times B_1 \times C_1$ is not monochromatic.
This is the content of the Prikry–Mills theorem (1978, unpublished), stated using Lean's ordinal type
{o : Ordinal // o < ω_ 1}as the representation of $\omega_1$.depends on sorry what this means
Statement uses:
Exists,Subtype,Ordinal,LT.lt,OrderEmbedding,Ordinal.omega,OfNat.ofNat,Fin,Nat,Set,Eq,Cardinal,Cardinal.mk,Cardinal.aleph,Not, Erdos1128.IsMonochromaticBoxProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous26 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_1128.variants.prikryMills_explicit : ∃ (f : {o : Ordinal // o < ω_ 1} → {o : Ordinal // o < ω_ 1} → {o : Ordinal // o < ω_ 1} → Fin 2), ∀ (A₁ B₁ C₁ : Set {o : Ordinal // o < ω_ 1}), #A₁ = aleph 0 → #B₁ = aleph 0 → #C₁ = aleph 0 → ¬ IsMonochromaticBox f A₁ B₁ C₁ := by …Mentions: Erdos1128.IsMonochromaticBox
- theorem
Erdos1128.erdos_1128.variants.two_dimensional_falseThe claim that every 2-colouring of $\omega_1 \times \omega_1$ has an uncountable monochromatic product rectangle is false in ZFC line 236 sourceSource documentation
The claim that every 2-colouring of $\omega_1 \times \omega_1$ has an uncountable monochromatic product rectangle is false in ZFC.
Counterexample: The ordering colouring $f(\alpha, \beta) = 0$ iff $\alpha < \beta$ has no uncountable monochromatic product rectangle $A_1 \times B_1$.
Proof: If $A_1 \times B_1$ were monochromatic with colour 0, then every element of $A_1$ would be strictly less than every element of $B_1$, making $A_1$ bounded above in $\omega_1$; but any bounded subset of $\omega_1$ is countable (since initial segments are countable), contradicting $A_1$ being uncountable. The colour-1 case is symmetric with the roles of $A_1$ and $B_1$ swapped.
Note: The correct classical result for 2-colourings of pairs (not products) is the Erdős–Rado theorem $\omega_1 \to (\omega_1)^2_2$, which concerns unordered pairs.
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Not, Erdos1128.Omega1,Fin,OfNat.ofNat,Nat,Exists,Set,And,Set.Countable,Membership.mem,EqProof uses:
ite,LT.lt,Ordinal,OrderEmbedding,Ordinal.omega,False,LE.le,Classical.byContradiction,Set.Countable.mono,Eq.mp,Eq.trans,binderNameHint,forall_congr,Mathlib.Tactic.Push.not_and_eq,implies_congr,Eq.refl,Set.Nonempty,Eq.rec,EmptyCollection.emptyCollection,Set.countable_emptyand 49 more40 instances, coercions and auxiliary constants left out.
Tactic trace: 53 steps
Open to load the checked before-and-after goal states.
theorem erdos_1128.variants.two_dimensional_false : ¬ ∀ (f : Omega1 → Omega1 → Fin 2), ∃ (A₁ B₁ : Set Omega1), ¬ A₁.Countable ∧ ¬ B₁.Countable ∧ ∃ c : Fin 2, ∀ a ∈ A₁, ∀ b ∈ B₁, f a b = c := by …Mentions: Erdos1128.Omega1, c