GPT 6 Astra Ultra attempt on Math-5479
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Contents
Erdős Problem 180: finite-family extremal compactness
1) Statement and scope
For a finite family $\mathcal F$ of graphs, must there be some $F\in\mathcal F$ for which $\operatorname{ex}(n;\mathcal F)\gg\operatorname{ex}(n;F)$? Here forbidden subgraphs are ordinary, not necessarily induced, subgraphs. In its literal form the answer is no. We give the elementary counterexample recorded on the current problem page, with exact extremal counts. The more restrictive variants excluding these small obstructions are separate questions.2) Two forbidden graphs whose combination is much stronger
Take \[ \mathcal F=\{P_3,2K_2\}, \] where $P_3$ has two adjacent edges and $2K_2$ has two disjoint edges. Any graph with two distinct edges contains one of these: the edges either share a vertex or they do not. Conversely a graph with just one edge avoids both. Therefore, for $n\geq2$, \[ \operatorname{ex}(n;\mathcal F)=1. \tag{1} \] A graph is $P_3$-free exactly when its maximum degree is at most one. It is therefore a union of isolated vertices and disjoint edges, and \[ \operatorname{ex}(n;P_3)=\lfloor n/2\rfloor. \tag{2} \]3) The other individual extremal number
In a $2K_2$-free graph, every pair of edges intersects. If all edges share one vertex, there are at most $n-1$ edges, attained by a star. Otherwise, choose two edges $ab,ac$ and an edge not containing $a$. To meet both chosen edges the latter edge must be $bc$. Any further edge meeting all three of $ab,ac,bc$ must have both endpoints in $\{a,b,c\}$: an edge containing an outside vertex misses the opposite edge of the triangle. Thus in this second case there are at most three edges, attained by a triangle together with isolated vertices. It follows that, for $n\geq3$, \[ \operatorname{ex}(n;2K_2)=\max(n-1,3). \tag{3} \] In particular this equals $n-1$ for $n\geq4$; at $n=4$ the star and triangle constructions tie.Combining $1$--$3$, the ratio of the family extremal number to either individual extremal number tends to zero. Since these are the only members of $\mathcal F$, no member has the asserted comparison with a positive constant independent of $n$.
4) Outcome and source
Outcome: SOLVED by a counterexample to the stated assertion. Every extremal count needed here is proved directly. This is an exposition of the elementary example reported as folklore on the problem page, not a claim to the recent stronger counterexamples for restricted families or a claim of novelty.- T. F. Bloom, Problem 180 and its discussion of the literal formulation,
https://www.erdosproblems.com/180, checked 23 September 2026.