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Source. Published p. 262 and pp. 267–271, Theorem 1.4 (PDF).

Statement. For $0<\eta<1/4$ there is $c=c(\eta)>0$ such that, whenever $\eta n\le l\le(1/2-\eta)n$ is an integer and $|F\cap G|\ne l$ for every $F\in\mathcal F$, $G\in\mathcal G$,

$$|\mathcal F||\mathcal G|\le4^n e^{-cn}.$$

This is the source's $(4-\epsilon_1)^n$ form after changing the constant.

Proof. Use ambient densities as in definitions. For nonempty families in $\mathcal P(X;[a,b])$, choose a coordinate and fix $0<\delta\le1/10$. If the product of the two $1$-slice densities exceeds $(1+\delta)d(\mathcal F)d(\mathcal G)$, take these slices and replace $[a,b]$ by $[a-1,b-1]$.

Otherwise interchange the families if necessary so that $d(\mathcal F_1)/d(\mathcal F)\le\sqrt{1+\delta}$. If $d(\mathcal F_0)d(\mathcal G_0\cup\mathcal G_1)$ exceeds the same threshold, take that pair and leave $[a,b]$ unchanged. In the remaining case take $(\mathcal F_1,\mathcal G_0\cap\mathcal G_1)$ and replace the interval by $[a-1,b]$.

To bound this last product, write

$$\frac{d(\mathcal F_1)}{d(\mathcal F)}=1+y,\quad \frac{d(\mathcal G_0\cup\mathcal G_1)}{d(\mathcal G)}=1+x.$$

Here $x\ge0$, while the complementary ratios are $1-y$ and $1-x$. If $y\ge0$, proposition_2_3 applies. If $y<0$, then $(1+x)(1-y)+(1-x)(1+y)=2-2xy\ge2$, so failure of the preceding growth test gives $(1-x)(1+y)\ge1-\delta$. Thus every step is either a growth step, gaining at least $1+\delta$, or a widening step, retaining at least $b_\delta=1-\delta-2\delta^2>0$. The slice interval identities prove that the forbidden-interval invariant is preserved.

Start with ambient size $n$ and $a=b=l$. Stop when $a=0$ or $b=m$, where $m$ is the remaining ambient size. Before stopping, $0<a\le b<m$, so a coordinate exists and the next step cannot cross an endpoint without hitting it. The ambient size decreases at every step; hence the procedure terminates. Positive density products remain positive. Let $u$ be the number of growth steps and $v$ the number of widening steps. Then $u+v=n-m$ and $b-a=v$. Also $v\le l$ because every widening step decreases $a$.

Suppose for contradiction that the initial density product is at least $e^{-\delta^2 n}$. Since a final density product is at most one,

$$u\log(1+\delta)+v\log b_\delta\le\delta^2n.$$

Uniform Taylor bounds for $0<\delta\le1/10$ imply

$$u-v\le C\delta n \tag{1}$$

for an absolute constant $C$: write $\log(1+\delta)=\delta+O(\delta^2)$ and $\log b_\delta=-\delta+O(\delta^2)$ and use $u,v\le n$. They also give the lower bound $e^{-C_1\delta n}$ for the final density product, for an absolute $C_1$.

If $a=0$, all final cross intersections exceed $b=v$. The number of steps is at least $l$, so (1) gives $v\ge(l-C\delta n)/2\ge\eta n/4$ when $\delta$ is small enough. Theorem 2.1 and the entropy estimate $h(1/2+t)\le\log2-2t^2$ bound the final normalized product by $\exp(-v^2/m)\le\exp(-\eta^2 n/16)$. If $m=0$, positive final families would have intersection zero, already contradicting $a=0$. Choose $\delta$ so that $C_1\delta<\eta^2/16$.

If $b=m$ and $a>0$, all final cross intersections are less than $a=m-v$. Since $m=b\le l\le(1/2-\eta)n$, (1) gives

$$a=m-v\le\frac m2-\eta n+\frac C2\delta n \le\frac m2-\frac\eta2n. \tag{2}$$

The condition $a>0$ implies $m\ge\eta n$, after decreasing $\delta$ if needed; alternatively (1) yields $m\ge n/4$ for small $\delta$. Thus the small-intersection bound of Theorem 2.2, with a fixed positive proportional gap in (2), makes the final product at most $e^{-c_2n}$ for a constant $c_2=c_2(\eta)>0$ and all sufficiently large $n$. Choose $\delta$ still smaller so that $C_1\delta<c_2$. Both stopping cases contradict the lower bound. This proves the result for large $n$.

For the remaining finitely many $n$, the full pair of Boolean cubes realizes every $l\in[0,n]$. An avoiding pair therefore has product strictly less than $4^n$. There are only finitely many possibilities; decreasing $c>0$ covers all of them. $\square$

Source precision. The source's step (f) prints $\mathcal F_2$ where its preceding test and interval invariant require $\mathcal F_0$. The width on p. 269 is $\beta n$, not the printed $\beta/n$. The last line of its Theorem 1.4 proof prints a base $2-\epsilon_1$ for the two-family product; the theorem and density normalization require a base below four. We use a fixed sufficiently small parameter, without assuming that a supremum of admissible parameters is itself admissible. The stopping proof above works on the actual integer ambient set and therefore needs no nonintegral auxiliary padding set.

Dependencies. definitions, proposition_2_3, theorem_2_1, theorem_2_2, entropy_estimates.