The short answer
Finite samples cannot establish an eventual property of every later input. Here the formulas, rather than a few observed entries, let us reason about the entire domain.
Original record and worked example
Same homogeneous witness, different row behaviour
All entries follow from the two invented formulas.
| Colouring | Stable on the full domain? | An infinite homogeneous set |
|---|---|---|
| F(i,j)=i mod 2 | Yes; each row is constant | All even integers, colour 0 |
| G(i,j)=j mod 2 | No; each row alternates | All even integers, colour 0 |
Chronology
A timeline of the turning points
Author exposition
Stability and homogeneous sets distinguished.
Original functions
Explicit parity rules, not sampled research data.
Two properties, two quantifiers
Chong and Yang distinguish a stable colouring, whose colour eventually settles for each fixed first argument, from an infinite homogeneous set, on which all pairs share a colour.
The first property concerns rows of a colouring; the second concerns a chosen subset of inputs. The exposition places these ideas within reverse mathematics, where the strength of assumptions matters as well as the truth of an individual statement.
Sources for this section: [1]
Define the original functions
Let the domain be the nonnegative integers. For i<j, set F(i,j)=i mod 2 and G(i,j)=j mod 2, with even parity coloured 0 and odd parity coloured 1. An unordered pair is always written with its smaller member first.
For each function, decide whether every fixed-i row eventually becomes constant on the full domain. Then give one infinite set whose pairs all have the same colour. Explain the infinite claim from the formula; listing the first few values is not enough.
Sources for this section: [1]
Reference answer
For F, fixing i fixes its parity. Every later j gives that same colour, so each row is constant from the outset. F is stable on the full domain.
For G, every fixed-i row alternates as j runs through successive even and odd integers. Neither colour eventually persists, so G is not stable on the full domain.
The even integers H={0,2,4,6,...} are homogeneous with colour 0 for both functions: in every pair from H, both the smaller and larger member are even. Thus an unstable full-domain colouring can nevertheless have an infinite homogeneous subset. Restricting G to H removes the alternating odd inputs; it does not retroactively make G stable on all inputs.
Sources for this section: [1]
What this does not prove
The formulas support these two particular infinite arguments. They do not establish a theorem quantified over every colouring or determine which weak axiom system proves that theorem. The original expositionโs logical programme is larger than this exercise.
A useful error check is to ask which variable was held fixed. Confusing a fixed row with a selected all-even subset turns two different questions into one and gives the wrong answer for G.
Sources for this section: [1]
Continue the chronology
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Biography context
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A career behind the record
Birth year unverified
Chong Chi Tat
Mathematical logicianLogic and computability researcher.Read biographyQuick answers
Common questions
- Does a long constant prefix prove stability?
- No. A finite prefix does not control all later inputs; this exercise uses complete formulas.
- Does an unstable colouring have no homogeneous set?
- No. The second original function provides a counterexample to that inference.
Primary and institutional sources
How this history was checked
Selected original records support the historical discussion. Worked examples are original and explicitly bounded; they are not reconstructed experiments. This article was reviewed on 31 October 2026.