Erdős Problem 1128 #
Reference: erdosproblems.com/1128
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 $A_1 \times B_1 \times C_1$.
Equations
- Erdos1128.IsMonochromaticBox f A₁ B₁ C₁ = ∃ (c : Fin 2), ∀ a ∈ A₁, ∀ b ∈ B₁, ∀ c' ∈ C₁, f a b c' = c
Instances For
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This is the unpublished result of Prikry and Mills (1978).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.
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.
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This is the content of the Prikry–Mills theorem (1978, unpublished), stated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 $\omega_1$.
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.