gpt pro 5.2 attempt on Math-5785

Problem 638

1) FORMAL RESTATEMENT

Let $\mathcal S$ be a family of finite (simple, undirected) graphs. Assume that for every integer $n\ge 1$ there exists a graph $G_n\in\mathcal S$ such that \[ G_n \to (K_3)^{e}_n, \] meaning: for every edge-colouring $c:E(G_n)\to\{1,\dots,n\}$ there is a monochromatic triangle.

Question (as written). Must it then be true that for every infinite cardinal $\kappa$ there exists a graph $G$ such that

1. every finite subgraph of $G$ belongs to $\mathcal S$, and
2. $G\to (K_3)^{e}_\kappa$ (every $\kappa$-edge-colouring of $G$ contains a monochromatic triangle)?

2) QUICK LITERATURE/CONTEXT CHECK

The formulation is ambiguous/degenerate unless one assumes additional closure properties of $\mathcal S$ (e.g. being hereditary under taking finite subgraphs). Under the literal reading (no closure assumptions), there are immediate counterexamples (below). A natural "minimal correction" is to add that $\mathcal S$ is closed under taking finite subgraphs (and perhaps has a joint-embedding property so that some infinite $G$ with age contained in $\mathcal S$ even exists); under such strengthened hypotheses the question becomes nontrivial.

3) ATTACK PLAN

Give a concrete family $\mathcal S$ satisfying the hypothesis (Ramsey-for-$K_3$ graphs exist for every $n$), but such that no graph $G$ can have all its finite subgraphs in $\mathcal S$. This disproves the conclusion already at clause (a), hence answers the literal question in the negative.

4) WORK

Step 1: Fix Ramsey witnesses.

For each $n\ge 1$, let $R_3(n)$ denote the (finite) Ramsey number for monochromatic triangles under $n$ edge-colours; i.e. $R_3(n)$ is the least integer $N$ such that \[K_N \to (K_3)^e_n.\] Existence of such $R_3(n)$ follows from the classical finite Ramsey theorem.

Step 2: Define the family $\mathcal S$.

Set \[ \mathcal S := \{\,K_{R_3(n)} : n\in\mathbb N,\ n\ge 1\,\}. \] Then for each $n$ we may take $G_n := K_{R_3(n)}\in\mathcal S$, and by definition $G_n\to (K_3)^e_n$. So the hypothesis of the problem holds.

Step 3: Show the conclusion fails.

Suppose (for contradiction) that there exists some graph $G$ such that every finite subgraph of $G$ belongs to $\mathcal S$. Pick any vertex $v\in V(G)$. The one-vertex graph $K_1$ is a finite subgraph of $G$ (take vertex set $\{v\}$ and no edges), hence $K_1\in\mathcal S$. But by construction $\mathcal S$ consists only of complete graphs $K_{R_3(n)}$ with $R_3(n)\ge 3$, so $K_1\notin\mathcal S$. This contradiction shows that no such $G$ exists.

Therefore the asserted implication is false already at the level of (a), so the answer to the question as written is no.

5) VERIFICATION / SANITY CHECKS

• For each $n$, $G_n:=K_{R_3(n)}\in\mathcal S$ satisfies $G_n\to (K_3)^e_n$ by the definition of $R_3(n)$.
• Any graph $G$ has a finite subgraph on one vertex (and also on two vertices, etc.). Since $K_1\notin\mathcal S$, no graph can have all finite subgraphs in $\mathcal S$.
• Hence the conclusion "for every infinite $\kappa$ there exists $G$ with (a) and (b)" fails.

6) FINAL

LABEL: FULL SOLUTION
SUBLABEL: COUNTEREXAMPLE/DISPROOF
Taking $\mathcal S=\{K_{R_3(n)}:n\ge 1\}$ satisfies the hypothesis, but there is no graph $G$ all of whose finite subgraphs lie in $\mathcal S$ (since $K_1\notin\mathcal S$). Thus the implication is false as stated.

7) COMPLETION ESTIMATE

COMPLETION: 100%

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