Forbidden-intersection density increment ($[a,b]\mapsto [a-1,b]$)

For two set families forbidding all cross-intersections in $[a,b]$, split into absent and present coordinate sections. The section pairs $({\mathcal{F}}_1,{\mathcal{G}}_1)$, $({\mathcal{F}}_0,{\mathcal{G}}_0\cup {\mathcal{G}}_1)$, and $({\mathcal{F}}_1,{\mathcal{G}}_0\cap {\mathcal{G}}_1)$ forbid, respectively, $[a-1,b-1]$, $[a,b]$, and $[a-1,b]$ in one fewer dimension. For $0<\delta \le 1/10$, either one of the first two pairs, allowing exchange of the set family names, increases the density product by at least $1+\delta$, or the third widens the interval and retains at least $1-\delta -2{\delta }^2$ of that product. This converts a single forbidden intersection into a long forbidden interval while recording every density gain and loss.

Let two set families on $n$ coordinates forbid cross-intersection size $\ell$, and let $\rho$ be their density product. For $0<\delta \le 1/10$, put $g=\log (1+\delta )$, $h=-\log (1-\delta -2{\delta }^2)$, and $D=g+h$. Repeated forbidden-intersection density increments terminate when one endpoint of the forbidden interval reaches zero or the remaining dimension. In the first case, if $v$ steps widened the interval, the cross-intersection bound from cube separation gives $\log \rho \le -g\ell +Dv-v^2/n\le -g\ell +D^{2}n/4$. In the second case, at least $n-\ell$ steps occurred and at most $\ell$ widened, giving $\log \rho \le -gn+(2g+h)\ell$. For a single set family take the square of its density. Choosing $\delta =1/50$ proves $|\mathcal{A}|\le 1.999^n$ for forbidden intersections $n/4$ and $\lfloor n/8\rfloor$, with small dimensions handled directly.