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These are author-recorded deductions by Plasma AI. They use the elementary substitution and block-family lemmas in the proof of L214, at tier 0. They sharpen this substitution construction's obstruction to finding disjoint members from spread alone and transfer L215's large-core construction to the absolute convention. They do not produce sunflower-free families of unbounded exponential rate, and hence do not settle Problem 20.

The order $\log n/\log\log n$ for intersecting normalized-spread families is already recorded by Alweiss–Lovett–Wu–Zhang, Section 4 after Theorem 4.2, with attribution to Lovett–Solomon–Zhang (2019). The derivation below is an alternative construction at that known order, not an improved lower bound over the literature. The cited original construction is not a premise of this proof and has not been checked here.

Decrease the scale after each contraction

For an integer $R\ge2$, put $h=\lceil\log_2R\rceil$ and

$$q_i=\left\lceil R/2^i\right\rceil,\qquad B_i=B(q_i,q_i^{q_i}),\qquad 0\le i<h.$$

Every $q_i\ge2$. Start with the family $F_0$ whose only member is a one-point set, and define $F_{i+1}=B_i[F_i]$. By L214's substitution lemma, every $F_i$ is intersecting. With $E$ denoting normalized excess, the block lemma gives

$$E_{F_{i+1}}(R)=\psi_{B_i}(E_{F_i}(R))\le\tfrac12E_{F_i}(R) \quad\hbox{whenever }E_{F_i}(R)\le q_i.$$

Starting from $E_{F_0}(R)=R$, induction gives $E_{F_i}(R)\le R/2^i\le q_i$ for each required step. Thus $F_h$ is an intersecting normalized $R$-spread family, of uniformity

$$n_R=\prod_{i=0}^{h-1}(q_i^{q_i}+1). \tag{1}$$

In fact the excess halves exactly: the center of the first kind attains $\psi_{B_i}(u)=u/2$ on $0<u\le q_i$.

This choice avoids paying the largest block scale at every level. In natural logarithms,

$$\log n_R=2R\log R-2R\log2+O((\log R)^2). \tag{2}$$

To check (2), write $x_i=R/2^i$. Replacing $x_i$ by its ceiling in $x_i\log x_i$ changes each summand by $O(\log R)$, over $O(\log R)$ indices. Also $\log(q_i^{q_i}+1)=q_i\log q_i+O(1)$, and the omitted tail of the infinite sum $\sum_{i\ge0}x_i\log x_i$ is bounded. The geometric identities $\sum2^{-i}=2$ and $\sum i2^{-i}=2$ give (2).

Consequently, along the uniformities (1), the spread is

$$R=\left(\frac12+o(1)\right) \frac{\log n_R}{\log\log n_R}. \tag{3}$$

Thus the lower obstruction is of order $\log n/\log\log n$, rather than the order $\log n/(\log\log n)^2$ obtained by using the same block family at every level. This concerns matching thresholds, not the exponential base needed for sunflower existence.

Thinning without a family-size cap

Thinning lemma. Let $r\ge2$ and $n\ge2$, and suppose a nonempty family $\mathcal H$ of distinct $n$-sets is normalized $Q$-spread. If

$$Q\ge4er(8r)^{2/(r-1)}, \tag{4}$$

then it has a subfamily $\mathcal H'$ with $|\mathcal H'|>r^n$ and $|\mathcal H'_T|\le r^{n-|T|}$ for every nonempty $T$. In particular, $Q\ge32r$ suffices when $r\ge16$.

There is no upper bound on $|\mathcal H|$ or its ground-set size in this statement. All properties inherited by subfamilies, such as being intersecting or excluding specified sunflower core sizes, survive.

Proof. Write $M=|\mathcal H|$. The normalized bound at a member gives $M\ge Q^n$. Keep each member independently with probability $p=2r^n/M<1$. The resulting size has mean $2r^n$ and

$$\Pr(|\mathcal H'|\le r^n)\le e^{-r^n/4}\le e^{-1}. \tag{5}$$

For $1\le t<n$, put $a_t=\lfloor r^{n-t}\rfloor$ and $d_T=|\mathcal H_T|$ for a $t$-set $T$. A bad set has at least $a_t+1$ retained members through it. Counting these witnesses, and first distinguishing one of their original members, gives

$$\begin{aligned} \Pr(\exists T,\ |T|=t:\ |\mathcal H'_T|>r^{n-t}) &\le\sum_{|T|=t}p^{a_t+1}\binom{d_T}{a_t+1}\\ &\le\sum_{|T|=t}p\,d_T \left(\frac{epd_T}{a_t}\right)^{a_t}\\ &\le 2r^n\binom nt \left(4e(r/Q)^t\right)^{a_t}. \end{aligned} \tag{6}$$

For the last step use $d_T\le M Q^{-t}$, $\sum_{|T|=t}d_T=M\binom nt$, and $a_t\ge(1-1/r)r^{n-t}$. The sum over $T$ is finite; sets outside all members have $d_T=0$. The identity for the sum of $d_T$ replaces the cruder estimate counting every possible $T$ separately.

Set $\delta=\log(Q/(4er))>0$. Since $4e(r/Q)^t\le(4er/Q)^t$, and since

$$t a_t\ge (1-1/r)t r^{n-t}\ge(n-1)(r-1),$$

the sum of (6) over $1\le t<n$ is at most

$$2(2r)^n\exp[-\delta(n-1)(r-1)] \le2(2r)^n(8r)^{-n} =2\,4^{-n}\le\frac18. \tag{7}$$

Here $t r^{n-t}$ is nonincreasing for $1\le t<n$ when $r\ge2$, and (4) gives $\delta(r-1)\ge2\log(8r)$. There is no bad set of size $n$, because each $n$-set is contained in at most one distinct member, and larger sets have degree zero. Equations (5) and (7) have sum less than one, proving existence.

Finally, $2\log(8r)/(r-1)$ decreases for $r\ge2$, and at $r=16$ it is less than $\log(8/e)$. Thus $32r$ satisfies (4) for all $r\ge16$. $\square$

The lemma generalizes the use of random thinning in L214's Lemma 4. That lemma's stated hypotheses and proof remain sufficient for L214 itself; the estimate above additionally applies when the original family has arbitrarily many members.

Consequences in the absolute convention

For every real $r\ge2$, choose an integer $R$ satisfying (4), and apply the first construction followed by the thinning lemma. This gives an intersecting absolutely $r$-spread family of $n_R$-sets with more than $r^{n_R}$ members. If $r\ge16$, one may take $R=\lceil32r\rceil$; (2) then gives intersecting absolute-spread examples at spread of order $\log n_R/\log\log n_R$.

The quantifier matters for the proposed matching lemma. For every fixed $k\ge3$ and every real $r>0$, the assertion that every absolutely $r$-spread $n$-uniform family with more than $r^n$ members contains $k$ disjoint members fails for some $n$. If $r<k-1$, take $k-1$ singletons; if $r\ge k-1\ge2$, use the intersecting construction just given. For $k=2$ the same construction refutes every $r\ge2$. It does not refute $r=1$: absolute $1$-spread itself forces distinct members to be disjoint. This degenerate case must not be included in a claim that all absolute matching constants fail.

For large cores, let $r\ge2$ and $g\ge0$. Choose $Q\ge3$ satisfying (4), and apply L215 at normalized spread $Q$. Its family consists of $n$-sets for $n\ge2$, is intersecting, and has no $k$-sunflower with core smaller than $2^g$ for any $k\ge3$. The thinning lemma produces an absolutely $r$-spread subfamily with more than $r^n$ members and exactly the same exclusion. Thus a bounded-core replacement for the matching lemma fails in the absolute convention as well.

These conclusions do not rule out finding a sunflower with an unbounded core. The remaining obligation is the general core-finding argument in the proof sketch.