Erdős Problem #474

Problem 474: an obstruction to splitting a Sierpiński colouring

1) FORMAL RESTATEMENT AND CURRENT STATUS

Attempt dated 2026-09-23. Write $[S]^2$ for the unordered pairs of distinct elements of $S$. In the partition-relation formulation explicitly supplied by the problem page, the question asks for assumptions ensuring a map \[ c:[S]^2\longrightarrow\{0,1,2\},\qquad |S|=2^{\aleph_0}, \] such that $c[[A]^2]=\{0,1,2\}$ for every uncountable $A\subseteq S$. This is $2^{\aleph_0}\not\to[\aleph_1]^2_3$. This formulation removes a dimensional ambiguity in the displayed prose on the source page; only the cardinality of the underlying set matters.

The live editorial page [1], fetched on the attempt date, labels this NOT PROVABLE. It records a positive answer under CH and consistency of a negative answer for a larger continuum; it specifically leaves open consistency of a negative answer when $2^{\aleph_0}=\aleph_2$. Thus an unconditional construction cannot be the objective. Here the objective is to test a concrete family of candidate constructions for that remaining question. No new consistency result is claimed.

2) WORK: THE TWO-COLOUR BASELINE

For completeness, the following is the classical Sierpiński construction, not a new result. Fix a well-order $\prec$ of $\mathbb R$. For $x<y$ in the usual real order define \[ b(\{x,y\})=0\quad\Longleftrightarrow\quad x\prec y. \] Every uncountable subset realizes both colours. Indeed, a set homogeneous for colour 0 is well-ordered in its usual order. Every well-ordered subset $W$ of $\mathbb R$ is countable: for each nonmaximal $x\in W$, let $x^+$ be its successor and choose a rational strictly between $x$ and $x^+$. The intervals are pairwise disjoint, so these choices inject $W$ minus at most one point into $\mathbb Q$. For colour 1 the reversed usual order is a well-order, and the same argument applies after negating the reals.

This suggests refining one of the two colours, but preserving both old colours only guarantees two of the three desired colours.

3) A WHOLE FAMILY OF REFINEMENTS FAILS

Work on $2^{\mathbb N}$, identified with the ternary Cantor subset of $\mathbb R$ by $x\mapsto\sum_{j\ge0}2x(j)3^{-j-1}$. Restrict the preceding $b$ to this subset. For distinct sequences put \[ \Delta(x,y)=\min\{j:x(j)\ne y(j)\}. \] Let $h:\mathbb N\to\{1,2\}$ be arbitrary, and define the proposed refinement \[ c_h(\{x,y\})= \begin{cases} 0,&b(\{x,y\})=0,\\ h(\Delta(x,y)),&b(\{x,y\})=1. \end{cases} \] Proposition. Every $c_h$ fails the required three-colour property, regardless of the chosen well-order.

Proof. One fibre $M=h^{-1}(i)$ is infinite. Let \[ P_M=\{x\in2^{\mathbb N}:x(j)=0\text{ for every }j\notin M\}. \] Restriction to the coordinates in $M$ is a bijection from $P_M$ to $2^M$, so $|P_M|=2^{\aleph_0}$. It is also a perfect set: an infinite set of free coordinates leaves a new free coordinate beyond any prescribed finite prefix. For distinct $x,y\in P_M$, their first differing coordinate lies in $M$, hence $h(\Delta(x,y))=i$. Therefore \[ c_h[[P_M]^2]\subseteq\{0,i\}. \] The remaining colour is absent on this uncountable set. In fact both displayed colours occur, by the baseline argument. This proves failure even on a perfect set.$\square$

The same proof applies if the coordinate statistic is any function of the first differing coordinate followed by a finite-valued rule: compose those rules to obtain $h$. In particular, parity, congruence classes, and predetermined blocks of digit positions cannot repair the construction. More generally, whenever a two-colouring used to split colour 1 is constant on an uncountable set, that set witnesses failure of the refinement.

4) ADVERSARIAL VERIFICATION

The uncountable witness $P_M$ is chosen after $h$, as required to refute a proposed colouring; no single $P_M$ is asserted to refute all $h$ simultaneously. The proof does not assert that all three-colourings, all definable refinements, or all combinations of well-orders have this form. A refinement depending on the actual points and their well-order ranks is not covered. Finite Ramsey experiments cannot verify the uncountable quantifier, so none are offered as evidence for the consistency problem.

5) FINAL

UNRESOLVED. The proved contribution of this attempt is a self-contained obstruction to an explicit family of natural three-colour candidates. The known two-colour baseline is reproduced and identified as known. Neither a colouring under $\mathfrak c=\aleph_2$ nor a model with the opposite partition relation has been obtained. A next construction must evade the uncountable homogeneous sets of its splitting rule; the digit-level rules above cannot do so.

6) SOURCES CHECKED

[1] T. F. Bloom, current statement and editorial status of Erdős problem 474, https://www.erdosproblems.com/474, fetched directly on 2026-09-23. The historical and consistency assertions above are explicitly a report of that page; their forcing proofs are not used here. The two-colour construction and the obstruction are proved in full in this file. No claim of publication novelty is made.

7) COMPLETION ESTIMATE

This subjective estimate measures coverage of the remaining classification problem, not confidence in the proved obstruction. The central consistency question is still untouched. COMPLETION: 5%