CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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Infinite Ramsey fails with infinitely many colours: colour $\{i,j\}$ by $\min\{i,j\}$

Statement refuted

The finite-colour hypothesis in Infinite Ramsey theorem on $\mathbb N$ : every finite colouring of $[\mathbb N]^k$ has an infinite monochromatic set, in ZF may be omitted for pair colourings.

Facts & Assumptions

Given: The colouring $c:[\mathbb N]^2\to\mathbb N$ defined by $c(\{i,j\})=\min\{i,j\}$ .

[L1]

Every finite colouring of $[\mathbb N]^k$ has an infinite monochromatic set, in ZF (Infinite Ramsey theorem on $\mathbb N$ : every finite colouring of $[\mathbb N]^k$ has an infinite monochromatic set, in ZF).

Counterexample

technique · constructive

1.1

Every natural $i$ occurs as $c(\{i,i+1\})=i$ , so this colouring genuinely has infinitely many colours.

construct

2.1

If $a<b<c$ , then $c(\{a,b\})=a$ while $c(\{b,c\})=b$ , and these colours differ. Hence no set of size at least three is monochromatic, in particular no infinite set is. This refutes the proposed extension and leaves [L1]'s finite-colour conclusion untouched.

step 1.1L1discharge-construct∎

Depends on

Infinite Ramsey theorem on $\mathbb N$: every finite colouring of $[\mathbb N]^k$ has an infinite monochromatic set, in ZF

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 24 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

I. B. Leader, Ramsey Theory, Theorem 4 and canonical left-dependent colouring (standard reference, not scraped)

Infinite Ramsey fails with infinitely many colours: colour {i,j}\{i,j\}{i,j} by min⁡{i,j}\min\{i,j\}min{i,j}