Erdős Problem #1175
Status Open · erdosproblems.comLLM status Open unverified
Erdős problems openaigpt 6 astra ultra 3% cataloguedOpenOpenStatement
For every uncountable cardinal $\kappa$, is there a cardinal $\lambda$ such that each graph of chromatic number exactly $\lambda$ contains a triangle-free subgraph of chromatic number exactly $\kappa$?
Restatement from Math-6229-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-6229
Erdős Problem 1175 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
For every uncountable cardinal $\kappa$, is there a cardinal $\lambda$ such that each graph of chromatic number exactly $\lambda$ contains a triangle-free subgraph of chromatic number exactly $\kappa$?CONTEXT AND SCOPE
This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.Subgraph means an arbitrary subgraph, not necessarily induced. The quantifier allows lambda larger than kappa; a consistency obstruction at lambda=kappa=aleph_1 alone does not settle it.
ATTACK PLAN AND WORK
The usual finite chromatic escalation has a precise ceiling here. Starting with a triangle-free graph $H$ with vertices $v_i$, form a new graph with the original vertices, copies $u_i$, and a new vertex $w$. Retain the original edges, join $u_i$ to every original neighbor of $v_i$, and join $w$ to every $u_i$. This graph remains triangle-free: copies are independent, a triangle through $w$ is impossible, and a triangle through one copy would give a triangle in $H$.If $\chi(H)=k<\infty$, the new graph has chromatic number $k+1$. The upper bound colors $u_i$ like $v_i$ and gives $w$ a new color. For the lower bound, suppose there were $k$ colors and call the color of $w$ color $c$. Recolor each original vertex of color $c$ by its copy's color. No copy has color $c$; and on an original edge only one endpoint could originally have color $c$, while its copy is adjacent to the other endpoint. This would give a proper $(k-1)$-coloring of $H$, a contradiction.
Iterating from $K_2$ and taking the disjoint union produces a countable triangle-free graph of chromatic number exactly $\aleph_0$. Any infinite complete host contains this graph as a subgraph. But the construction stays countable, so it cannot supply the uncountable chromatic target in the problem. Also an arbitrary high-chromatic host need not contain a large complete graph.