CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck 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 $N$ : every finite colouring of $[N]^{k}$ has an infinite monochromatic set, in ZF may be omitted for pair colourings.

Facts & Assumptions

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

[L1]

Every finite colouring of $[N]^{k}$ has an infinite monochromatic set, in ZF (Infinite Ramsey theorem on $N$ : every finite colouring of $[N]^{k}$ 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 · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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} by min⁡{i,j}min{i,j}