gpt pro 5.4 attempt on Math-660

Erdős Problem #62 (Round 3): Common $4$-chromatic subgraphs of $\aleph_1$-chromatic graphs

1) ROUND-3 OBJECTIVE

Path (C): obstruction / prerequisite extraction.

Round 2 reduced Problem #62 to the intersection problem for the sets of finite $4$-vertex-critical subgraphs appearing in two $\aleph_1$-chromatic graphs, and ruled out all proof strategies based on a fixed or finite unavoidable $4$-chromatic template family. The most promising next step is therefore to convert the pairwise-common-subgraph question into a single-graph necessary condition.

The new goal of Round 3 is to prove that if the answer to Problem #62 were positive, then every $\aleph_1$-chromatic graph would have to contain finite $4$-vertex-critical subgraphs with arbitrarily large odd girth and also with arbitrarily large order. This gives a concrete obstruction criterion for any future counterexample search.

2) ROUND-2 FOUNDATION USED

I use the following specific Round 2 results without re-proving them.

• Round 2 Corollary 4: a common $4$-chromatic subgraph exists for $G_1,G_2$ if and only if there exists a common finite $4$-vertex-critical subgraph.
• Round 2 Lemma 5: every fixed non-bipartite graph can be omitted by some $\aleph_1$-chromatic graph.
• Round 2 Corollary 7: every finite family of graphs of chromatic number at least $3$ can be simultaneously omitted by some $\aleph_1$-chromatic graph.
• Round 2 Proposition 8: a counterexample pair would follow from sufficiently flexible omission of prescribed families of finite $4$-vertex-critical graphs.

3) NEW INSIGHT / TOOL (ROUND-3)

The genuinely new ingredient is the following three-step strengthening.

1. Round 2 only formalized omission for fixed graphs and finite families. Round 3 proves an explicit bounded-odd-girth family avoidance theorem: any family of non-bipartite graphs whose odd girths are uniformly bounded can be simultaneously omitted by some $\aleph_1$-chromatic graph.
2. Assuming a positive answer to Problem #62, this immediately forces every $\aleph_1$-chromatic graph $G$ to have finite $4$-vertex-critical subgraphs of unbounded odd girth.
3. Combining the positive answer with Round 2 Corollary 7 also forces every $\aleph_1$-chromatic graph $G$ to have finite $4$-vertex-critical subgraphs of unbounded order.

In contrapositive form, a single $\aleph_1$-chromatic graph with bounded $4$-critical odd-girth spectrum, or bounded $4$-critical order spectrum, would already disprove Problem #62.

4) ATTACK PLAN (ROUND-3)

Gap left after Round 2. Round 2 showed that no fixed graph and no finite family of $4$-chromatic graphs can be unavoidable. But this still left open the possibility that every pair of $\aleph_1$-chromatic graphs might intersect in some infinite, highly nonuniform family of finite $4$-critical graphs.

Precise Round 3 claims. Let $\mathcal C_4$ be the set of isomorphism types of finite $4$-vertex-critical graphs, and let \[ \mathcal C_4(G):=\{F\in \mathcal C_4 : F\text{ embeds into }G\}. \] I will prove:

1. if $S\subseteq \mathcal C_4$ has bounded odd girth, then some $\aleph_1$-chromatic graph omits every member of $S$;
2. if Problem #62 has a positive answer, then for every $\aleph_1$-chromatic graph $G$, the odd girths of graphs in $\mathcal C_4(G)$ are unbounded;
3. if Problem #62 has a positive answer, then for every $\aleph_1$-chromatic graph $G$, the orders of graphs in $\mathcal C_4(G)$ are unbounded.

This plan overcomes the Round 2 obstacle by turning the unresolved pairwise-intersection problem into a concrete single-graph prerequisite. Any future disproof need only construct one $\aleph_1$-chromatic graph violating either prerequisite.

5) WORK (ROUND-3)

5.1. A family-version of Round 2 odd-girth avoidance

Proposition 9 (bounded-odd-girth family avoidance).

Let $\mathcal H$ be any family of non-bipartite graphs such that \[ M:=\sup\{\mathrm{og}(H):H\in \mathcal H\}<\infty. \] Then for every infinite cardinal $\kappa$ there exists a graph $G$ with $|V(G)|=\kappa$ and $\chi(G)=\kappa$ such that no member of $\mathcal H$ embeds into $G$.

Proof. Choose an integer $s>M$. By the external input already used in Round 2 (recorded in Hajnal--Komjáth), there exists a graph $G$ of size and chromatic number $\kappa$ with odd girth $\mathrm{og}(G)>s$. Suppose, toward a contradiction, that some $H\in \mathcal H$ embeds into $G$. Because $H$ is non-bipartite, it contains an odd cycle of length $\mathrm{og}(H)\le M < s < \mathrm{og}(G)$. Its image in $G$ would be an odd cycle of the same length, contradicting the definition of odd girth. Hence no $H\in \mathcal H$ embeds into $G$. $\square$

Corollary 10.

If $S\subseteq \mathcal C_4$ is any family of finite $4$-vertex-critical graphs with bounded odd girth, then there exists an $\aleph_1$-chromatic graph $G$ such that \[ \mathcal C_4(G)\cap S=\varnothing. \]

Proof. Every member of $S$ is non-bipartite, so Proposition 9 applies with $\kappa=\aleph_1$. $\square$

Round 3 gain over Round 2. Round 2 only made this omission principle explicit for finite families. Proposition 9 shows that the same mechanism works for arbitrary families, provided their odd girths stay uniformly bounded.

5.2. A positive answer would force unbounded odd girth in every $\mathcal C_4(G)$

Theorem 11 (necessary odd-girth richness under a positive answer).

Assume Problem #62 has a positive answer; that is, whenever $G_1,G_2$ have chromatic number $\aleph_1$, they have a common subgraph of chromatic number $4$. Then for every graph $G$ with $\chi(G)=\aleph_1$ and every integer $r\ge 3$, there exists a finite $4$-vertex-critical graph $F\in \mathcal C_4(G)$ with \[ \mathrm{og}(F)>r. \]

Proof. Fix $G$ with $\chi(G)=\aleph_1$ and an integer $r\ge 3$. By the large-odd-girth existence theorem used in Round 2, there exists an $\aleph_1$-chromatic graph $H_r$ with \[ \mathrm{og}(H_r)>r. \] By the assumed positive answer to Problem #62, the graphs $G$ and $H_r$ have a common subgraph of chromatic number $4$. Applying Round 2 Corollary 4, there is a finite $4$-vertex-critical graph $F$ that is a subgraph of both $G$ and $H_r$. Because $F\subseteq H_r$, every odd cycle of $F$ is also an odd cycle of $H_r$. Therefore \[ \mathrm{og}(F)\ge \mathrm{og}(H_r)>r. \] Thus $F\in \mathcal C_4(G)$ and $\mathrm{og}(F)>r$, as required. $\square$

Corollary 12.

Under the positive answer to Problem #62, for every $\aleph_1$-chromatic graph $G$ the set \[ \{\mathrm{og}(F):F\in \mathcal C_4(G)\} \] is unbounded in $\mathbb N$.

Proof. Immediate from Theorem 11. $\square$

Interpretation. A positive solution to the pair problem would already imply a surprisingly strong single-graph consequence: every $\aleph_1$-chromatic graph would contain finite $4$-vertex-critical subgraphs of arbitrarily large odd girth. This is much stronger than anything obtained in Round 2 and sharply narrows the obstruction.

5.3. A positive answer would force unbounded order in every $\mathcal C_4(G)$

Theorem 13 (necessary size richness under a positive answer).

Assume Problem #62 has a positive answer. Then for every graph $G$ with $\chi(G)=\aleph_1$ and every integer $N\ge 1$, there exists a finite $4$-vertex-critical graph $F\in \mathcal C_4(G)$ with \[ |V(F)|>N. \]

Proof. Fix $G$ with $\chi(G)=\aleph_1$ and $N\ge 1$. Let \[ S_N:=\{F\in \mathcal C_4: |V(F)|\le N\}. \] This set is finite, because there are only finitely many isomorphism types of graphs on at most $N$ vertices. By Round 2 Corollary 7, there exists an $\aleph_1$-chromatic graph $H_N$ that contains none of the graphs in $S_N$. By the assumed positive answer to Problem #62, $G$ and $H_N$ have a common $4$-chromatic subgraph; by Round 2 Corollary 4, they therefore have a common finite $4$-vertex-critical subgraph $F$. Now $F\in \mathcal C_4(G)$ and $F\notin S_N$ because $H_N$ omits every member of $S_N$. Hence $|V(F)|>N$. $\square$

Corollary 14.

Under the positive answer to Problem #62, for every $\aleph_1$-chromatic graph $G$ the set \[ \{|V(F)|:F\in \mathcal C_4(G)\} \] is unbounded in $\mathbb N$. In particular, $\mathcal C_4(G)$ is infinite.

Proof. Unboundedness follows from Theorem 13. Infinitude is immediate. $\square$

5.4. A one-graph obstruction criterion for disproving Problem #62

Corollary 15 (single-graph obstruction criterion).

Any one of the following statements would already imply a negative answer to Problem #62.

1. There exists an $\aleph_1$-chromatic graph $G$ such that \[ \sup\{\mathrm{og}(F):F\in \mathcal C_4(G)\}<\infty. \]
2. There exists an $\aleph_1$-chromatic graph $G$ such that \[ \sup\{|V(F)|:F\in \mathcal C_4(G)\}<\infty. \]

Proof. If $1$ holds, let \[ R:=\sup\{\mathrm{og}(F):F\in \mathcal C_4(G)\}. \] By Proposition 9, there exists an $\aleph_1$-chromatic graph $H$ with odd girth $>R$. Then no finite $4$-vertex-critical graph embedded in $G$ can embed in $H$. By Round 2 Corollary 4, $G$ and $H$ cannot have a common $4$-chromatic subgraph, so Problem #62 fails.

If $2$ holds, let \[ N:=\sup\{|V(F)|:F\in \mathcal C_4(G)\}. \] By Round 2 Corollary 7, there exists an $\aleph_1$-chromatic graph $H$ omitting every finite $4$-vertex-critical graph on at most $N$ vertices. Again Round 2 Corollary 4 implies that $G$ and $H$ have no common $4$-chromatic subgraph, so Problem #62 fails. $\square$

Remark 16.

Corollary 15 is the sharpest new reduction obtained in Round 3. It shows that the search for a counterexample pair can be replaced by a search for a single $\aleph_1$-chromatic graph whose finite $4$-vertex-critical spectrum is bounded in some globally avoidable invariant. The most natural remaining candidate invariant is odd girth. Hence the new bottleneck is:

Does every $\aleph_1$-chromatic graph contain finite $4$-vertex-critical subgraphs of arbitrarily large odd girth?

A positive answer to Problem #62 would force this statement by Theorem 11. Conversely, a single counterexample to this one-graph statement would disprove Problem #62 by Corollary 15.

6) ADVERSARIAL VERIFICATION

• Does Proposition 9 really extend Round 2? Yes. Round 2 handled fixed graphs and finite families; Proposition 9 handles arbitrary families with uniformly bounded odd girth. No cardinality restriction on $\mathcal H$ is needed.

• Does Theorem 11 improperly assume finiteness of the common subgraph? No. The positive answer only yields a common $4$-chromatic subgraph a priori, but Round 2 Corollary 4 upgrades this to a common finite $4$-vertex-critical subgraph, which is exactly what is used.

• Is odd girth monotone in the right direction? Yes. If $F\subseteq H_r$, then every odd cycle of $F$ is an odd cycle of $H_r$, so $\mathrm{og}(F)\ge \mathrm{og}(H_r)$.

• Is $S_N$ finite in Theorem 13? Yes. There are only finitely many graphs on at most $N$ labeled vertices, hence finitely many isomorphism types, and therefore finitely many finite $4$-vertex-critical graphs of order at most $N$.

• Any hidden converse claim? None is made. Unbounded odd girth or unbounded order in every $\mathcal C_4(G)$ is proved only as a necessary condition for a positive answer, not as a sufficient one.

• Compatibility with Round 2 Proposition 8. There is no conflict. Proposition 8 gave a sufficient condition for a counterexample pair via arbitrary family omission. Round 3 instead gives a one-graph obstruction criterion via boundedness in odd girth or size.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 3 advances beyond Round 2 in a mathematically substantive way.

• It upgrades finite-family omission to bounded-odd-girth family omission (Proposition 9).
• It proves that any positive solution to Problem #62 would force every $\aleph_1$-chromatic graph to contain finite $4$-vertex-critical subgraphs of arbitrarily large odd girth (Theorem 11).
• It proves that any positive solution to Problem #62 would force every $\aleph_1$-chromatic graph to contain finite $4$-vertex-critical subgraphs of arbitrarily large order (Theorem 13).
• It reduces any future disproof to a single-graph construction: it would suffice to build one $\aleph_1$-chromatic graph whose finite $4$-vertex-critical spectrum is bounded in odd girth or in order (Corollary 15).

The main remaining gap is therefore sharply formulated: determine whether every $\aleph_1$-chromatic graph must already have $4$-critical finite subgraphs of unbounded odd girth.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 55%

9) REFERENCES

• T. F. Bloom, Erdős Problem #62, problem page and LaTeX source, accessed 2026-03-07. (Used for the exact current problem statement, the note that Erdős asked the finite-family variant, and the currently listed open status.)
• A. Hajnal, P. Komjáth, What must and what need not be contained in a graph of uncountable chromatic number?, Combinatorica 4$1$ $1984$, 47--52. (Used for the large odd-girth existence input and obligatory-subgraph background.)
• P. Erdős, A. Hajnal, S. Shelah, On some general properties of chromatic numbers, Topics in topology (Proc. Colloq., Keszthely, 1972) $1974$, 243--255. (Used for the common $3$-chromatic / odd-cycle background already invoked in earlier rounds.)