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Erdős Problem 615
A Ramsey–Turán problem of Erdős, Hajnal, Simonovits, Sós, and Szemerédi [EHSSS93]: does there
exist a constant $c > 0$ such that every graph on $n$ vertices with at least $(1/8 - c)n^2$
edges contains either a $K_4$ or an independent set on at least $n/\log n$ vertices? In the
notation of Ramsey–Turán theory this asks whether
$$\mathrm{rt}(n; 4, n/\log n) < (1/8 - c)n^2.$$
This was disproved by Fox, Loh, and Zhao [FLZ15], who showed that
$\mathrm{rt}(n; 4, ne^{-f(n)}) \geq (1/8 - o(1))n^2$ whenever
$f(n) = o(\sqrt{\log n/\log\log n})$. In the other direction Sudakov [Su03] had shown that
$\mathrm{rt}(n; 4, ne^{-f(n)}) = o(n^2)$ whenever $f(n)/\sqrt{\log n} \to \infty$.
[EHSSS93] Erdős, P., Hajnal, A., Simonovits, M., Sós, V. T., and Szemerédi, E.,
Turán-Ramsey theorems and simple asymptotically extremal structures. Combinatorica 13
(1993), 31--56.
[Su03] Sudakov, B., A few remarks on Ramsey-Turán-type problems. J. Combin. Theory Ser. B
88 (2003), 99--106.
[FLZ15] Fox, J., Loh, P.-S., and Zhao, Y., The critical window for the classical
Ramsey-Turán problem. Combinatorica 35 (2015), 435--476.
Does there exist some constant $c > 0$ such that for all sufficiently large $n$, if $G$ is a
graph with $n$ vertices and at least $(1/8 - c)n^2$ edges then $G$ must contain either a $K_4$
or an independent set on at least $n/\log n$ vertices?
The answer is no, as shown by Fox, Loh, and Zhao [FLZ15].
The result of Fox, Loh, and Zhao [FLZ15] disproving the problem: if $f(n) \geq 0$ satisfies
$f(n) = o(\sqrt{\log n/\log\log n})$, then for every $\epsilon > 0$ and all sufficiently
large $n$ there is a $K_4$-free graph on $n$ vertices with independence number at most
$ne^{-f(n)}$ and at least $(1/8 - \epsilon)n^2$ edges; that is,
$\mathrm{rt}(n; 4, ne^{-f(n)}) \geq (1/8 - o(1))n^2$. Applied with
$f(n) = \log\log n$, this disproves the headline problem, since
$ne^{-f(n)} = n/\log n = o(n)$.
The complementary result of Sudakov [Su03]: if $f(n)/\sqrt{\log n} \to \infty$ then
$\mathrm{rt}(n; 4, ne^{-f(n)}) = o(n^2)$; that is, for every $\epsilon > 0$ and all
sufficiently large $n$, every $K_4$-free graph on $n$ vertices with independence number at
most $ne^{-f(n)}$ has at most $\epsilon n^2$ edges.
A sanity check for erdos_615: the empty graph on $n \geq 3$ vertices contains an independent
set on at least $n/\log n$ vertices (namely the whole vertex set), so it satisfies the
conclusion of the implication in the problem statement.