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Source. Theorem 3, p. 2, with its proof in Section 2, pp. 3--4, of Eric Naslund and William F. Sawin, Upper bounds for sunflower-free sets, Forum Math. Sigma 5 (2017), Paper No. e15, doi:10.1017/fms.2017.12. Labels and pages here are those of arXiv:1606.09575v1, the edition named on the source card.

Statement

Definitions (p. 1). Three sets form a 3-sunflower when all three pairwise intersections are equal. A family $\mathcal F$ is sunflower-free when no three of its members form a 3-sunflower. $F_k(n)$ is the largest size of a family of subsets of $\{1,2,\ldots,n\}$ with no $k$ members forming a $k$-sunflower, and the Erdős-Szemerédi $k$-sunflower-free capacity is

$$\mu_k^S=\limsup_{n\to\infty}F_k(n)^{1/n}.$$

Theorem 3 (p. 2). If $\mathcal F$ is a sunflower-free collection of subsets of $\{1,2,\ldots,n\}$, then

$$|\mathcal F|\leq 3(n+1)\sum_{k\leq n/3}\binom{n}{k},$$

and

$$\mu_3^S\leq\frac{3}{2^{2/3}}=1.889881574\ldots$$

The abstract (p. 1) prints the first bound with the factor $3n$ in place of $3(n+1)$, together with the estimate $\le(3/2^{2/3})^{n(1+o(1))}$; the theorem and its proof (p. 4) carry $3(n+1)$. The paper notes (p. 2) that the best known lower bound is $\mu_3^S\ge1.554$, credited to unpublished work of the first author, so a gap remains.

Proof pointer

Section 2, pp. 3--4. Split the family by the number of elements, $l=0,\ldots,n$. Within one layer no member properly contains another, so for $x,y,z$ in the layer the function $\prod_i(2-(x_i+y_i+z_i))$ on $\{0,1\}^n$ vanishes off the diagonal. Lemma 6 (p. 2, the slice-rank lemma of Tao) then bounds the layer by the slice rank of this function, which expanding into monomials and grouping each term by a factor of degree at most $n/3$ bounds by $3\sum_{k\le n/3}\binom nk$. Summing over the $n+1$ layers gives the theorem.

Read depth

Claims checked: the definitions, the statement and the proof outline were read on the print. Nothing here is independently reviewed.

Dependencies

Lemma 6 (p. 2), quoted by the paper from Tao's formulation of the Croot-Lev-Pach and Ellenberg-Gijswijt argument. Nothing in the corpus.

Bears on

• Problem 857: the problem's $m(n,3)$ is $F_3(n)+1$, so the theorem gives $m(n,3)\le3(n+1)\sum_{k\le n/3}\binom nk+1\le(3/2^{2/3})^{(1+o(1))n}$. It is an upper bound for $k=3$ only, with no matching lower bound and no asymptotic formula.