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These author-recorded arguments are by Plasma AI. The theorems they support are the claims L214, which proves that intersecting families can be arbitrarily spread and refutes Rao's suggested r = O(p) matching lemma, and L215, which shows that spread does not force a sunflower with a bounded core. The proof of L214 holds the lemmas used below. This page explains the mechanism, the further author-recorded observations that are not part of either claim, and what remains open. The attack it serves is recorded in the editable proof sketch for Problem 20.

Intersecting families with unbounded normalized spread are already known: Alweiss–Lovett–Wu–Zhang, Section 4 after Theorem 4.2, records spread $\Omega(\log n/\log\log n)$, citing Lovett–Solomon–Zhang (2019). The explicit constructions here are not a novelty claim for that existence result or its asymptotic order.

The route this closes

Write $n$ for the size of the sets and $k$ for the number of petals, as in Problem 20. A family $\mathcal F$ of $n$-sets is absolutely $R$-spread if every nonempty $T$ lies in at most $R^{n-|T|}$ members, and normalized $R$-spread if every nonempty $T$ lies in at most $R^{-|T|}|\mathcal F|$ members, which forces $|\mathcal F|\ge R^n$.

The matching lemma $M(k)$ asks for a constant $R_k$ such that every absolutely $R_k$-spread family of $n$-sets with more than $R_k^n$ members has $k$ pairwise disjoint members, for every $n$. It would give $f(n,k)\le\lfloor R_k^n\rfloor+1$, and settle the strict form of Problem 20 after enlarging the base, for example to $R_k+2$. Given $|\mathcal F|>R_k^n$, choose an inclusion-maximal $T$ with $|\mathcal F_T|>R_k^{n-|T|}$; the empty set qualifies, and $|T|<n$ since an $n$-set lies in at most one member. The link $\{S\setminus T:S\in\mathcal F_T\}$ is absolutely $R_k$-spread by maximality and has more than $R_k^{n-|T|}$ members, and $k$ disjoint members of it lift to a $k$-sunflower with core $T$. This is the remark accompanying Rao's Lemma 2, whose source says the lemma might hold with $r(p,k)=O(p)$, and Bell–Chueluecha–Warnke's Lemma 2 is $M(k)$ with $R_k$ replaced by $Ck\log n$.

L214 gives intersecting families, hence families without two disjoint members, that are absolutely $r$-spread with more than $r^n$ members for every real $r\ge16$, and normalized $\rho$-spread for every integer $\rho\ge2$. The general thinning lemma extends the absolute examples to every $r\ge2$. Thus $M(k)$ fails for every $k\ge3$: if $r<k-1$, use $k-1$ singletons, and if $r\ge k-1$, use an intersecting example. For $k=2$ the examples refute every $r\ge2$, but not $r=1$, where absolute spread itself forces disjointness. The converse of the link argument needs no lemma: Problem 20 with constant $c_k$ gives a $k$-sunflower in every family with more than $c_k^n$ members, spread or not. So Problem 20 is equivalent to the spread form

SF($k$). For each $k\ge3$ there is $R_k$ such that every absolutely $R_k$-spread family of $n$-sets with more than $R_k^n$ members contains a $k$-sunflower,

and the families of L214 show that the sunflower in SF($k$) cannot always be taken with an empty core.

Substitution and the seed quantity

For an $a$-uniform family $G$ and a uniform family $H$, the substitution $G[H]$ replaces each point $i$ of a member $A\in G$ by a member $S_i$ of a copy of $H$ placed over $i$; its powers are $G^1=G$ and $G^{h+1}=G[G^h]$. Two members of $G[H]$ meet whenever $G$ and $H$ are intersecting, because their root sets share a point $i$ and their fibers over $i$ meet. The key fact is that spread behaves exactly under substitution. With the excess $E_{\mathcal F}(R)=\max_{T\ne\varnothing}\Pr[T\subseteq S]R^{|T|}$ of a uniform random member $S$, and $\psi_G(E)=\max_{I\ne\varnothing}\Pr[I\subseteq A]E^{|I|}$ for a uniform random $A\in G$, Lemma 2 of the L214 proof gives

$$E_{G[H]}(R)=\psi_G\bigl(E_H(R)\bigr),\qquad E_{G^h}(R)=\psi_G^{(h)}(R),$$

and $\mathcal F$ is normalized $R$-spread exactly when $E_{\mathcal F}(R)\le1$.

Proposition (author-recorded). Let $G$ be an intersecting $a$-uniform family, $a\ge2$, in which no point lies in every member, and put

$$\rho(G)=\min_{|I|\ge2}\Pr[I\subseteq A]^{-1/(|I|-1)},$$

the minimum over sets $I$ contained in a member. Then the largest $R$ for which $G^h$ is normalized $R$-spread tends to $\rho(G)$ as $h\to\infty$, and it is at most $\rho(G)$ for every $h$.

Proof. If $R>\rho(G)$, some $I$ with $|I|\ge2$ has $\Pr[I\subseteq A]R^{|I|-1}>1$, so $\psi_G(E)>E$ for every $E\ge R$ and the iterates from $R$ never fall to $1$. If $1\le R<\rho(G)$, then on $[1,R]$ every term of $\psi_G$ is below the identity: sets with $|I|\ge2$ give $\Pr[I\subseteq A]E^{|I|}\le E(E/\rho(G))^{|I|-1}<E$, and singletons give at most $p_1E<E$, where $p_1<1$ is the largest point probability. By continuity $\psi_G(E)\le E-\delta$ on $[1,R]$ for some $\delta>0$, so the iterates fall below $1$, and they stay there because $\psi_G(E)\le E$ for $E\le1$. $\square$

So building spread intersecting families reduces to one level: a seed must make every set of two or more points rare, with the singletons free except for $p_1<1$. Three seeds show the range.

• The triangle, the three $2$-subsets of a $3$-set, has $\rho=3$. Its powers are the minterm families of recursive majority of three, whose spread rises to $3$; it is $1.5$ at $h=1$ and about $2.12$, $2.52$, $2.75$ and $2.87$ at $h=2,3,4,5$.
• The Hilton–Milner type seed, $\{B\}$ together with all sets $\{x\}\cup B'$ where $B'$ is an $(a-1)$-subset of a set $Y\supseteq B$ meeting the $a$-set $B$, has $\rho$ close to $a$ when $|Y|$ is large compared with $a^2$, so spread is unbounded already. Its point $x$ lies in all members but one, so the leaf reached through $x$ at every level of $G^h$ lies in a fraction $(1-1/|G|)^h$ of the members; spread $R$ needs at least $|G|\ln R$ levels, and the uniformity is doubly exponential in the spread.
• The two-star block seed $B(\rho,\rho^\rho)$ of L214 has $\rho(G)=2^{1/\rho^\rho}\rho$ and $p_1=1/2$: both centers lie in half of the members, the members of the first kind are a product family of spread exactly $\rho$, and a member of the second kind must contain a whole block $Y_i$, which the $\rho^\rho$ blocks make as rare as $\rho$-spread requires. Its excess halves at each level, so $\lceil\log_2\rho\rceil$ levels suffice.

Rate and every large n

The block seed gives normalized spread $\rho$ at uniformity $n=(\rho^\rho+1)^{\lceil\log_2\rho\rceil}$, so $\log_2n=(1+o(1))\rho(\log_2\rho)^2$ and the spread is $(1+o(1))\log_2n/(\log_2\log_2n)^2$. The absolute form loses the factor $16$ of L214's thinning lemma.

Remark (author-recorded): every large $n$. Let $N_0=(\rho^\rho+1)^h$ with $h=\lceil\log_2\rho\rceil$, and let $n\ge(\rho+1)N_0$. Write $n=qN_0+j$ with $q=\lfloor n/N_0\rfloor\ge\rho+1$ and $0\le j<N_0$. The family $B(\rho,q-1)[B(\rho,\rho^\rho)^h]$ is intersecting, $qN_0$-uniform and normalized $\rho$-spread: the inner power is normalized $\rho$-spread by L214, and Lemma 3 of the L214 proof gives $\psi_{B(\rho,q-1)}(E)\le E$ for $E\le\rho$. Add to each member $j$ new points of its own. A set $T$ meeting the new points of a member lies in that member only, so the result is normalized $R$-spread for $R=\rho^{qN_0/n}$, which is at least $\rho^{(\rho+1)/(\rho+2)}=(1-o(1))\rho$. Choosing the largest $\rho$ with $(\rho+1)N_0\le n$, every large $n$ carries an intersecting family of $n$-sets with normalized spread at least $(1-o(1))\log_2n/(\log_2\log_2n)^2$.

In the absolute form, Bell–Chueluecha–Warnke's Lemma 2 caps the spread of an intersecting family at $C\log n$. In the normalized form, a uniformly weighted family is spread in the weighted sense of Alweiss, Lovett, Wu and Zhang exactly when it is normalized spread, and a uniformly random half of the ground set contains a member of an intersecting family with probability at most $1/2$, since the half and its complement cannot both contain one. Their Theorem 2.5 with $\alpha=\beta=1/2$ therefore caps the normalized spread at $O(\log n\log\log n)$, and Rao's improvement, which they record, at $O(\log n)$. So the least spread that forces two disjoint members lies between $c\log n/(\log\log n)^2$ and $C\log n$. Through the link argument a matching lemma at spread $R$ gives $f(n,k)\le\lfloor R^n\rfloor+1$; at the uniformities of L214 no valid matching lemma has $R$ below the spread of these families.

The same step appears outside the sunflower problem. Kupavskii and Zakharov's spread-approximation method (Spread approximations for forbidden intersections problems, arXiv:2203.13379v3, proof of Lemma 9, pp. 7--8) uses that two families of $k$-sets whose measures are $r$-spread with $r>2^{11}\log_2(2k)$ cannot be cross-intersecting, by a random halving. An intersecting family is cross-intersecting with itself, so L214 shows that this logarithm cannot be replaced by a constant, nor by anything below $c\log k/(\log\log k)^2$. This bears only on the generality of that step: their families are spread because they are homogeneous subfamilies of a complete layer, which the families of L214 are not, and no theorem of theirs is contradicted.

Where the sunflowers are

For Problem 20 the relevant question is how spread families with no $k$ disjoint members still contain $k$-sunflowers. The core lemma, Lemma B in the proof of L215, answers this for substitutions. If $G$ is intersecting and $H$ is intersecting and free of $k$-sunflowers, $k\ge3$, then the root sets of a $k$-sunflower of $G[H]$ form a $k$-sunflower of $G$ with a nonempty core $A_*$. Its core is the union of the common fibers over $A_*$, so it consists of whole members of $H$. In particular $G[H]$ is free of $k$-sunflowers when $G$ is.

Three consequences follow.

• No sunflowers at spread below $\sqrt{10}$. The triangle has no $3$-sunflower, so the recursive majority families are intersecting, $3$-sunflower-free and have normalized spread approaching $3$. A better seed is the family of ten triples of a $6$-set $\{012,013,024,035,045,125,134,145,234,235\}$. It contains one triple from each complementary pair, so it is intersecting, and every pair of points lies in exactly two triples. It is $3$-sunflower-free, since a core of two points would need three triples through one pair, and the link of each point is a five-cycle, which has no three disjoint edges. Its seed quantity is $\rho=\sqrt{10}$, from the triples, with $p_1=1/2$. Its powers are therefore intersecting, $3$-sunflower-free and normalized $R$-spread for every $R<\sqrt{10}$ once the power is large enough. So the threshold $R_3$ of SF($3$) is at least $\sqrt{10}$ in the normalized form. The same powers have $10^{(3^h-1)/2}$ members of size $3^h$; how this compares with the published lower bounds for three-petal sunflowers was not checked here.
• Cores of every size are forced. Substituting the triangle powers $D^g$ into the block-family powers of L214 gives L215: for every $R\ge3$ and $g$ an intersecting family with $R$-spread uniform measure in which every $k$-sunflower, for every $k\ge3$, has a core of at least $2^g$ points. The outer block levels absorb the excess $3(R/3)^{2^g}$ that the inner triangle powers carry, halving it at each level. At fixed $R>3$ these cores have about $\log_2\log_2n/\log_2(R/3)$ points.
• The block powers have one-point cores. In $B(\rho,m)$ the members $\{x\}\cup T_1,\dots,\{x\}\cup T_\rho$ with pairwise disjoint transversals form a $\rho$-sunflower with core $\{x\}$. In $B(\rho,m)^h$, take these members at the root, a $\rho$-sunflower of $B(\rho,m)^{h-1}$ with a one-point core over $x$, and arbitrary members over the other points. The core is the single leaf reached through $x$ at every level.

Remark (author-recorded): the core point can be light. With $B=B(\rho,\rho^\rho)$, the power $B^h$ is normalized $\rho$-spread for every $h\ge\lceil\log_2\rho\rceil$, since the excess only decreases at further levels. Every point of $B$ lies in at most half of its members, so every point of $B^h$ lies in at most a fraction $2^{-h}$ of the members, far below the fraction $1/\rho$ that spread allows. So a spread family without two disjoint members need not have any point of degree at least $|\mathcal F|/(C\rho)$, however large $C$ is. A recursion in the style of Erdős and Rado, which passes to the link of a point of large degree and loses only a constant factor, therefore fails at its first step. The leaf $\ell$ reached through $x$ at every level nevertheless carries a good link: it lies in exactly a fraction $2^{-h}$ of the members, and since $E_{B^h}(\rho)\le\rho2^{-h}$, every $U$ outside $\ell$ satisfies $\Pr[U\cup\{\ell\}\subseteq S]\le2^{-h}\rho^{-|U|}$. So the link at $\ell$, with its own uniform measure, is again normalized $\rho$-spread, and it contains the sunflower petals above. The point that leads to a sunflower is light, but no superset of it is heavier relative to its size.

So a proof of SF($k$) must produce sunflowers with nonempty cores, and in some spread families with cores of unbounded size; the core cannot be bounded in terms of $k$ and the spread alone.

Open questions

• The exact threshold for two disjoint members. Close the gap between $c\log n/\log\log n$ and $C\log n$. The lower order is known from the source comparison above and is also attained by the varying-scale construction. A seed with $\rho(G)\ge\rho$, $p_1\le1/2$ and uniformity $e^{o(\rho\log\rho)}$ would improve the lower bound; the block seed needs $\rho^\rho$ blocks because each member of its second kind contains a whole block. No seed can do much better: if $\rho(G)\ge\rho$ and $p_1\le1/2$, then $\psi_G(E)\le\max(E/2,E^2/\rho)=E/2$ for $E\le\rho/2$, so $G^h$ with $h=\lceil\log_2\rho\rceil$ is normalized $(\rho/2)$-spread, and the normalized cap of order $\log n$ above forces $\log a\ge c\rho/\log\rho$ for the uniformity $a$ of an intersecting seed $G$.
• Many petals. For $k$ disjoint members the products of blocks of size $k-1$ give spread $k-1$, and the intersecting examples give $c\log n/\log\log n$. The combined dependence on $k$ and $n$ is not determined by these constructions.
• A core-finding lemma. SF($k$) is equivalent to Problem 20. A route through spread families must replace $M(k)$ by a statement that produces a sunflower with a nonempty and possibly large core when the family has no $k$ disjoint members. By L215 such a core can need about $\log\log n/\log R$ points; a single choice of a large core or a recursion into links would both be consistent with this obstruction. The proof sketch records the candidate statements and their status.