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This author-recorded argument by Plasma AI evaluates the proposed maximal-rate link procedure on the explicit examples. It uses the substitution and block estimates of L214 and the triangle-power calculation of L215, both at tier 0. It proves no bound on the procedure for a general family.

For a family $\mathcal F$ of $n$-sets, $n\ge2$, let

$$R_a(\mathcal F)=\max_{0<|T|<n}|\mathcal F_T|^{1/(n-|T|)}.$$

An extremal set $T$ is a maximizer in this definition. Passing to its link preserves absolute spread at rate $R_a(\mathcal F)$ and makes the link's size exactly that rate to its residual uniformity. The proof sketch asks whether successive such choices can be made without losing too much rate.

An exact formula for a balanced block seed

Fix an integer $\rho\ge2$, and put

$$B=B(\rho,\rho^\rho),\qquad a=\rho^\rho+1, \qquad A=|B|=2\rho^{a-1}.$$

For every $u>0$, L214's block estimate, with equality at a center and at a full member, gives

$$\psi_B(u)=\max\left(\frac u2,\frac{u^a}{A}\right). \tag{1}$$

Indeed the estimate is $\Pr[I\subseteq S]\le\rho^{1-|I|}/2$. When $u\le\rho$ its largest possible contribution is $u/2$, attained by a center. When $u\ge\rho$ the largest possible contribution is $u^a/(2\rho^{a-1})$, attained by a member. The two expressions agree at $u=\rho$.

A link that attains the maximal rate

Let $H$ be a nonempty $b$-uniform family, $b\ge1$, with $M$ members. Suppose that for

$$R=(\rho M)^{1/b} \quad\hbox{one has}\quad E_H(R)=\rho. \tag{2}$$

Then $R_a(B[H])=R$. Moreover, some extremal set has a link containing $\rho$ pairwise disjoint members.

Proof. By (1) and the substitution identity,

$$E_{B[H]}(R)=\psi_B(E_H(R))=\rho/2 =\frac{R^{ab}}{A M^a}.$$

Since $|B[H]|=AM^a$, this is precisely the absolute $R$-spread bound for every nonempty subset, and hence $R_a(B[H])\le R$.

Fix a member $S_0\in H$ and let $T=\{x\}\times S_0$, the whole fiber over the first center of $B$. It is a nonempty proper subset of a member of $B[H]$. Members through $T$ have a first-kind root and the fixed fiber $S_0$ over $x$. Their number is

$$|B[H]_T|=\rho^{a-1}M^{a-1}=R^{b(a-1)}.$$

Thus $T$ attains rate $R$. After removing $T$, each of the $a-1$ blocks independently chooses one of $\rho$ root points and then a member of the corresponding copy of $H$. Choosing a different root point in every block for each of $\rho$ members makes those members pairwise disjoint. No disjointness inside $H$ is required. $\square$

Application to L214

For $F=B^h$, $h\ge1$, take $H=B^{h-1}$, interpreting $B^0$ as the one-point family. Put $d=a^{h-1}$, so $H$ has uniformity $d$ and $M=A^{(d-1)/(a-1)}$ members. Then

$$R_a(F)=(\rho M)^{1/d} =\rho\,2^{(1-1/d)/(a-1)}. \tag{3}$$

To verify (2), set $K=A^{1/(a-1)}$. The number $R$ in (3) lies in $[\rho,K)$, and the first $h-1$ iterates of (1) follow the branch $u\mapsto u^a/A$. Explicitly those iterates are $K(R/K)^{a^i}$, for $0\le i\le h-1$, decreasing to $\rho$ at the final index. They all lie above or at the branch threshold $\rho$. Thus $E_H(R)=\rho$, as needed.

The extremal link constructed above has $\rho$ disjoint members. For each $k\le\rho$, the procedure can therefore find a $k$-sunflower at its first step. This proves existence of a successful maximizer; it does not assert that every tie-breaking rule succeeds.

Application to L215

Now let $F=(B^j)[D^g]$, with the parameters of L215. Write $s=2^g$ and $Q=|D^g|=3^{s-1}$. By the natural identification of nested fibers,

$$F=B[H],\qquad H=(B^{j-1})[D^g].$$

Put $d=a^{j-1}$. Then $H$ has uniformity $sd$ and $M=A^{(d-1)/(a-1)}Q^d$ members. Its relevant rate is

$$R=(\rho M)^{1/(sd)},\qquad y=R^s/Q=\rho^{1/d}A^{(d-1)/((a-1)d)}.$$

As above, $\psi_B^{(j-1)}(y)=\rho$ with all the iterates at least $\rho\ge3$. Since $R^s=Qy\ge3^s$, we have $R\ge3$, and L215's triangle formula applies: $E_{D^g}(R)=R^s/Q=y$. Therefore $E_H(R)=\psi_B^{(j-1)}(y)=\rho$, verifying (2). Hence

$$R_a(F)=(\rho M)^{1/(sd)},$$

and again an extremal link has $\rho$ pairwise disjoint members. Its core has $sd$ points, consistent with L215's lower bound $s$ on core size. The procedure is allowed to choose a large core.

These exact successes do not extend to arbitrary families. The chain obstruction gives families of arbitrarily large exponential rate whose unique maximal-rate choices discard a sunflower and lead to a sunflower-free family. A general argument therefore needs more than the favorable maximizer exhibited here, such as control of the discarded families.