Erdős Problem #602

Erdős Problem 602 --- collection record

PROVENANCE AND STATUS

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/602.tex.

REUSED SOURCE 1

Let $(A_i)$ be a family of sets with $\lvert A_i\rvert=\aleph_0$ for all $i$, such that for any $i\neq j$ we have $\lvert A_i\cap A_j\rvert$ finite and $\neq 1$. Is there a $2$-colouring of $\cup A_i$ such that no $A_i$ is monochromatic? A problem of Komjáth. The existence of such a $2$-colouring is sometimes known as Property B.

1) FORMAL RESTATEMENT

Let $I$ be an index set and let $(A_i)_{i\in I}$ be a family of sets with $|A_i|=\aleph_0$ for all $i$. Assume that for all distinct $i,j\in I$, the intersection $A_i\cap A_j$ is finite and has size not equal to $1$.

Question: does there exist a function (2-colouring) \[ c: \bigcup_{i\in I} A_i \to \{0,1\} \] such that for every $i\in I$, the restriction $c|_{A_i}$ is not constant (equivalently, $A_i$ contains at least one point of each colour)?

2) QUICK LITERATURE/CONTEXT CHECK

The statement notes this is known as Property B in this setting. No further results are stated, and I do not import external literature.

3) ATTACK PLAN

• Proof-track idea: attempt a transfinite greedy colouring, using the fact that each $A_i$ is infinite and pairwise intersections are finite. The main difficulty is that at an uncountable stage, $A_i$ might already have many coloured points.
• Disproof-track idea: search for an uncountable family with the given intersection pattern that forces one $A_i$ to become monochromatic under any 2-colouring.
• Sanity check: solve the countable-indexed case explicitly.

4) WORK

Fast reality check (countable family). If $I$ is countable (in particular finite), a direct greedy construction works and does not even need the intersection hypotheses.

Lemma 602.1 (countable family of countably infinite sets has Property B).
Assume $I$ is countable. Then there exists a 2-colouring of $\bigcup_{i\in I}A_i$ such that no $A_i$ is monochromatic.

Proof. Enumerate $I$ as $\{i_0,i_1,i_2,\dots\}$. We build a colouring $c$ in stages. Maintain a finite set $S_n$ of already coloured points after stage $n$, with $S_0=\emptyset$.

At stage $n$, since $A_{i_n}$ is countably infinite and $S_n$ is finite, choose two distinct points \[ x_n, y_n \in A_{i_n}\setminus S_n. \] Define $c(x_n)=0$ and $c(y_n)=1$, and leave all other as-yet-uncoloured points undefined for now. Set $S_{n+1}:=S_n\cup\{x_n,y_n\}$.

After completing all stages, define $c$ on any remaining uncoloured points arbitrarily (e.g., colour them $0$).

Now fix any $i=i_n$. By construction, $x_n,y_n\in A_i$ and $c(x_n)=0$, $c(y_n)=1$. Hence $A_i$ contains both colours and is not monochromatic. This holds for every $i\in I$, so $c$ is a valid 2-colouring. $\square$

Lemma 602.2 (transfinite recursion under "infinitely many new points").
Assume $I$ is well-ordered and that for every $i\in I$, \[ A_i\setminus\bigcup_{j<i}A_j\text{ is infinite.} \] Then there exists a 2-colouring of $\bigcup_{i\in I}A_i$ such that no $A_i$ is monochromatic.

Proof. We define a colouring by transfinite recursion along the well-order of $I$. Suppose colours have been assigned to all points in $\bigcup_{j<i}A_j$. By hypothesis, the set of "new" points \[ B_i:=A_i\setminus\bigcup_{j<i}A_j \] is infinite, hence contains two distinct uncoloured points $x_i,y_i$. Assign $c(x_i)=0$ and $c(y_i)=1$. Leave any other points of $B_i$ either uncoloured until the end or colour them arbitrarily.

At the end, colour any remaining uncoloured points arbitrarily.

For each $i$, we have produced $x_i,y_i\in A_i$ with opposite colours, so $A_i$ is not monochromatic. $\square$

5) VERIFICATION

• Lemma 602.1 does not use the intersection hypotheses; it works for any countable family of infinite sets.
• Lemma 602.2 isolates the obstruction for uncountable families: at stage $i$ we need at least two previously uncoloured points in $A_i$, which is guaranteed if $A_i$ contributes infinitely many new elements beyond earlier sets.
• In the original problem, only pairwise intersections are finite. For uncountable $I$, the union $\bigcup_{j<i}(A_i\cap A_j)$ can be infinite, so the "infinitely many new points" hypothesis of Lemma 602.2 need not follow.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. If the index set $I$ is countable, then such a 2-colouring exists (Lemma 602.1). More generally, if the family can be well-ordered so that each $A_i$ has infinitely many new points outside the union of earlier sets, then a 2-colouring exists (Lemma 602.2).

(ii) First gap (crisp). For an uncountable family $(A_i)_{i\in I}$ with $|A_i|=\aleph_0$ and pairwise intersections finite and $\neq 1$, prove or refute the existence of a 2-colouring of $\cup A_i$ with no monochromatic $A_i$.

(iii) Top 3 next moves.

1. Attempt to show that the condition "$|A_i\cap A_j|\neq 1$" forces the existence of a well-order of $I$ satisfying the "infinitely many new points" hypothesis, or some weaker variant sufficient for a transfinite greedy colouring.
2. Search for a concrete uncountable candidate counterexample where every $A_i$ is "almost covered" by intersections with earlier sets in any well-order, so the greedy method cannot pick fresh points.
3. Finite-model sanity: experiment with finite approximations (finite sets with the same intersection pattern) to see which intersection patterns tend to obstruct 2-colourings, and whether excluding intersection size $1$ removes those obstructions.

(iv) What a minimal counterexample would look like. A minimal counterexample would be an uncountable index set $I$ and countable sets $(A_i)$ with pairwise intersections finite and never of size $1$, such that for every 2-colouring $c$ of $\cup A_i$ there exists some $i\in I$ with $c$ constant on $A_i$. Structurally, such a family would have to defeat every attempt to assign two opposite-coloured "fresh" points per $A_i$.

COMPLETION ESTIMATE

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.

Let $(A_i)$ be a family of sets with $\lvert A_i\rvert=\aleph_0$ for all $i$, such that for any $i\neq j$ we have $\lvert A_i\cap A_j\rvert$ finite and $\neq 1$. Is there a $2$-colouring of $\cup A_i$ such that no $A_i$ is monochromatic? A problem of Komjáth. The existence of such a $2$-colouring is sometimes known as Property B.

1) FORMAL RESTATEMENT

Let $I$ be an index set and let $(A_i)_{i\in I}$ be a family of sets with $|A_i|=\aleph_0$ for all $i$. Assume that for all distinct $i,j\in I$, the intersection $A_i\cap A_j$ is finite and has size not equal to $1$.

Question: does there exist a function (2-colouring) \[ c: \bigcup_{i\in I} A_i \to \{0,1\} \] such that for every $i\in I$, the restriction $c|_{A_i}$ is not constant (equivalently, $A_i$ contains at least one point of each colour)?

2) QUICK LITERATURE/CONTEXT CHECK

The statement notes this is known as Property B in this setting. No further results are stated, and I do not import external literature.

3) ATTACK PLAN

• Proof-track idea: attempt a transfinite greedy colouring, using the fact that each $A_i$ is infinite and pairwise intersections are finite. The main difficulty is that at an uncountable stage, $A_i$ might already have many coloured points.
• Disproof-track idea: search for an uncountable family with the given intersection pattern that forces one $A_i$ to become monochromatic under any 2-colouring.
• Sanity check: solve the countable-indexed case explicitly.

4) WORK

Fast reality check (countable family). If $I$ is countable (in particular finite), a direct greedy construction works and does not even need the intersection hypotheses.

Lemma 602.1 (countable family of countably infinite sets has Property B).
Assume $I$ is countable. Then there exists a 2-colouring of $\bigcup_{i\in I}A_i$ such that no $A_i$ is monochromatic.

Proof. Enumerate $I$ as $\{i_0,i_1,i_2,\dots\}$. We build a colouring $c$ in stages. Maintain a finite set $S_n$ of already coloured points after stage $n$, with $S_0=\emptyset$.

At stage $n$, since $A_{i_n}$ is countably infinite and $S_n$ is finite, choose two distinct points \[ x_n, y_n \in A_{i_n}\setminus S_n. \] Define $c(x_n)=0$ and $c(y_n)=1$, and leave all other as-yet-uncoloured points undefined for now. Set $S_{n+1}:=S_n\cup\{x_n,y_n\}$.

After completing all stages, define $c$ on any remaining uncoloured points arbitrarily (e.g., colour them $0$).

Now fix any $i=i_n$. By construction, $x_n,y_n\in A_i$ and $c(x_n)=0$, $c(y_n)=1$. Hence $A_i$ contains both colours and is not monochromatic. This holds for every $i\in I$, so $c$ is a valid 2-colouring. $\square$

Lemma 602.2 (transfinite recursion under "infinitely many new points").
Assume $I$ is well-ordered and that for every $i\in I$, \[ A_i\setminus\bigcup_{j<i}A_j\text{ is infinite.} \] Then there exists a 2-colouring of $\bigcup_{i\in I}A_i$ such that no $A_i$ is monochromatic.

Proof. We define a colouring by transfinite recursion along the well-order of $I$. Suppose colours have been assigned to all points in $\bigcup_{j<i}A_j$. By hypothesis, the set of "new" points \[ B_i:=A_i\setminus\bigcup_{j<i}A_j \] is infinite, hence contains two distinct uncoloured points $x_i,y_i$. Assign $c(x_i)=0$ and $c(y_i)=1$. Leave any other points of $B_i$ either uncoloured until the end or colour them arbitrarily.

At the end, colour any remaining uncoloured points arbitrarily.

For each $i$, we have produced $x_i,y_i\in A_i$ with opposite colours, so $A_i$ is not monochromatic. $\square$

5) VERIFICATION

• Lemma 602.1 does not use the intersection hypotheses; it works for any countable family of infinite sets.
• Lemma 602.2 isolates the obstruction for uncountable families: at stage $i$ we need at least two previously uncoloured points in $A_i$, which is guaranteed if $A_i$ contributes infinitely many new elements beyond earlier sets.
• In the original problem, only pairwise intersections are finite. For uncountable $I$, the union $\bigcup_{j<i}(A_i\cap A_j)$ can be infinite, so the "infinitely many new points" hypothesis of Lemma 602.2 need not follow.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. If the index set $I$ is countable, then such a 2-colouring exists (Lemma 602.1). More generally, if the family can be well-ordered so that each $A_i$ has infinitely many new points outside the union of earlier sets, then a 2-colouring exists (Lemma 602.2).

(ii) First gap (crisp). For an uncountable family $(A_i)_{i\in I}$ with $|A_i|=\aleph_0$ and pairwise intersections finite and $\neq 1$, prove or refute the existence of a 2-colouring of $\cup A_i$ with no monochromatic $A_i$.

(iii) Top 3 next moves.

1. Attempt to show that the condition "$|A_i\cap A_j|\neq 1$" forces the existence of a well-order of $I$ satisfying the "infinitely many new points" hypothesis, or some weaker variant sufficient for a transfinite greedy colouring.
2. Search for a concrete uncountable candidate counterexample where every $A_i$ is "almost covered" by intersections with earlier sets in any well-order, so the greedy method cannot pick fresh points.
3. Finite-model sanity: experiment with finite approximations (finite sets with the same intersection pattern) to see which intersection patterns tend to obstruct 2-colourings, and whether excluding intersection size $1$ removes those obstructions.

(iv) What a minimal counterexample would look like. A minimal counterexample would be an uncountable index set $I$ and countable sets $(A_i)$ with pairwise intersections finite and never of size $1$, such that for every 2-colouring $c$ of $\cup A_i$ there exists some $i\in I$ with $c$ constant on $A_i$. Structurally, such a family would have to defeat every attempt to assign two opposite-coloured "fresh" points per $A_i$.