Erdős–Szemerédi $3$-sunflower-free capacity

Description of constant

A family of three distinct sets $A,B,C$ is a $3$-sunflower (or $\Delta$-system) if \(A\cap B \ =\ A\cap C \ =\ B\cap C.\) A family of sets is sunflower-free if it contains no $3$-sunflower (equivalently, no sunflower of any size $\ge 3$).

Let $[n]:={1,2,\dots,n}$ and let $f(n)$ denote the maximum size of a sunflower-free family $\mathcal{F}\subseteq 2^{[n]}$. Define \(C_{49}\ :=\ \mu^{\mathrm S}\_3\ :=\ \limsup\_{n\to\infty} f(n)^{1/n}.\) A standard tensor power argument shows that the limsup is in fact a limit: \(\mu^{\mathrm S}\_3\ =\ \lim\_{n\to\infty} f(n)^{1/n}\) (see e.g. [TZ2025, (1.2)]).

Existence of the limit

The limit exists. One convenient way to see this is to reduce to uniform families and then apply a tensor-power (direct sum) argument (see also [TZ2025]).

For $0\le r\le n$, let $f_{r}(n)$ denote the maximum size of a sunflower-free $r$-uniform family $\mathcal{F}\subseteq \binom{[n]}{r}$, and set \(g(n)\ :=\ \max_{0\le r\le n} f\_{r}(n).\) Then \(g(n)\ \le\ f(n)\ \le\ (n+1)\,g(n),\) since any family $\mathcal{F}\subseteq 2^{[n]}$ decomposes as the disjoint union of its uniform layers $\mathcal{F}\cap\binom{[n]}{r}$ and one layer must have size at least $\lvert\mathcal{F}\rvert/(n+1)$.

Now let $X,Y$ be disjoint sets with $\lvert X\rvert=n$ and $\lvert Y\rvert=m$, and let $\mathcal{F}\subseteq \binom{X}{r}$ and $\mathcal{G}\subseteq \binom{Y}{s}$ be sunflower-free. Define the product family \(\mathcal{F}\otimes\mathcal{G}\ :=\ \{A\cup B:\ A\in\mathcal{F},\ B\in\mathcal{G}\}\ \subseteq\ \binom{X\cup Y}{r+s}.\) Then $\mathcal{F}\otimes\mathcal{G}$ is sunflower-free. Indeed, suppose that three distinct members $A_i\cup B_i$ ($i=1,2,3$) form a $3$-sunflower in $X\cup Y$. Since $X$ and $Y$ are disjoint, the equalities \((A_1\cup B_1)\cap(A_2\cup B_2)\ =\ (A_1\cup B_1)\cap(A_3\cup B_3)\ =\ (A_2\cup B_2)\cap(A_3\cup B_3)\) imply \(A_1\cap A_2\ =\ A_1\cap A_3\ =\ A_2\cap A_3\) and \(B_1\cap B_2\ =\ B_1\cap B_3\ =\ B_2\cap B_3.\) If two of the $A_i$ coincide, say $A_1=A_2$, then $A_1=A_1\cap A_2=A_1\cap A_3\subseteq A_3$, and uniformity forces $A_1=A_3$. Thus either $(A_1,A_2,A_3)$ is a triple of distinct sets forming a $3$-sunflower in $\mathcal{F}$, or all three $A_i$ coincide; similarly for the $B_i$. Since the unions $A_i\cup B_i$ are distinct, at least one of the triples $(A_1,A_2,A_3)$ or $(B_1,B_2,B_3)$ consists of three distinct sets, giving a contradiction.

Consequently $g(n+m)\ge g(n)g(m)$, so $\log g(n)$ is superadditive. By Fekete’s lemma, the limit $\lim_{n\to\infty} g(n)^{1/n}$ exists. Since $f(n)$ and $g(n)$ differ by at most the factor $(n+1)$, it follows that $\lim_{n\to\infty} f(n)^{1/n}$ exists as well.

Trivially $1\le \mu^{\mathrm S}_3 \le 2$.

Known upper bounds

Bound Reference Comments
$2$ Trivial $f(n)\le 2^n$.
$\frac{3}{2^{2/3}} \approx 1.8898815748$ [NS2017] They prove $\lvert\mathcal{F}\rvert \le 3(n+1) \sum_{k=0}^{\lfloor n/3\rfloor} \binom{n}{k} \le \left(\frac{3}{2^{2/3}}\right)^{n(1+o(1))}$ for sunflower-free $\mathcal{F}\subseteq 2^{[n]}$.

Known lower bounds

Bound Reference Comments
$1$ Trivial $f(n)\ge 1$.
$>1.551$ [DEGKM1997] This lower bound is obtained from a construction of Deuber–Erdős–Gunderson–Kostochka–Meyer. The numerical value is stated in [FPP2024], [TZ2025].
$\ge 1.554$ (unpublished) [NS2017] The arXiv preprint version of [NS2017] records $\mu^{\mathrm S}_3\ge 1.554$, citing an unpublished manuscript of the first author.

References

Contribution notes

Prepared with assistance from ChatGPT 5.2.