Erdős–Szemerédi $3$-sunflower-free capacity ($C_{49}$)

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