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The key is to widen the forbidden intersection while controlling a product of densities. We first derive the two ingredients, with constants explicit enough to yield the numerical base in the question.

For two nonempty set families $\mathcal{F},\mathcal{G}$ on $m$ coordinates, suppose every cross-intersection has size greater than an integer $w\ge 0$. Their complements are separated from $\mathcal{F}$ by Hamming distance greater than $w$. Let $a$ be the least integer with $|\mathcal{F}|\le S_m(a)$, where $S_m(a)={\sum }_{j=0}^a\left(\genfrac{}{}{0ex}{}{m}{j}\right)$. Iterating Harper inequality, and using $|\mathcal{F}|>S_m(a-1)$, shows that its radius-$w$ neighbourhood has at least $S_m(a+w-1)$ vertices. The complementary set family is disjoint from that neighbourhood, soFor $a=0$ the needed weaker neighbourhood bound follows simply by taking the ball around its single vertex; define $S_m(-1)=0$. If a displayed index forces a full neighbourhood, the other set family would be empty, already excluded.

The provided binary entropy function bound givesIndeed, for $t\le m/2$ this follows from $H_2(1/2-v)\le 1-2v^2/\log 2$, obtained by integrating $H_2^{\prime \prime }\le -4/\log 2$; for larger $t$ use $S_m(t)\le 2^m$. The two nonnegative deficits in the preceding product have sum at least $w$. Their squared sum is at least $w^2/2$. Thus the cross-intersection bound from cube separation is

Now suppose a pair of set families forbids every cross-intersection in $[a,b]\subseteq [0,m]$. Put $\rho ={\mu }_m(\mathcal{F}){\mu }_m(\mathcal{G})$, and fix $\delta =1/50$. Split both set families by the last coordinate. The following three replacements preserve the indicated forbidden intervals on $m-1$ coordinates:For example, in the third row the same second set has witnesses in both sections of $\mathcal{G}$, so its intersection with a first-section member must avoid both shifted intervals; their union is $[a-1,b]$.

If the first replacement increases the density product by a factor at least $1+\delta$, use it. Otherwise interchange the names of the two set families if needed so thatIf the second replacement gives a factor at least $1+\delta$, use it. If neither does, use the third. To bound its loss, putWe have $z\ge 1$, $(2-x)z<1+\delta$, and consequently $1-\delta <x\le \sqrt{1+\delta }\le 1+\delta /2$. The third replacement's product factor isSince $h^{\prime \prime }(x)=-4(1+\delta )/(2-x{)}^3<0$, its minimum on $[1-\delta ,1+\delta /2]$ is at an endpoint. ThereThus each step either gains a factor $1+\delta$ without increasing interval width, or widens the interval by one and loses at most a factor $1-\delta -2{\delta }^2$. This is the forbidden-intersection density increment, with its loss proved rather than hidden in an asymptotic error.

Start with $\mathcal{F}=\mathcal{G}=\mathcal{A}$ and $a=b=\ell$. Stop when $a=0$ or $b=m$. Dimension drops at every step. The positive lower product factors ensure that a nonempty starting pair stays nonempty; it therefore cannot reach a forbidden interval containing all $0,\dots ,m$. Write $v$ for the number of widening steps, $u$ for the number of gain steps, and putThe terminal density product ${\rho }_{\ast }$ satisfies ${\rho }_{\ast }\ge {\rho }_0{e}^{gu-hv}$. Always $v\le \ell$, since a widening step lowers $a$ by one.

If the procedure stops at $a=0$, the terminal forbidden interval is $[0,v]$. Every cross-intersection is therefore greater than $v$, and the cross-intersection bound from cube separation gives ${\rho }_{\ast }\le e^{-v^2/m}$. At least $\ell$ steps were needed to lower $a$ to zero, so $u\ge \ell -v$. Since $m\le n$,If instead the procedure stops at $b=m$, the number of steps is at least $n-\ell$, so $u\ge n-\ell -v$. Using ${\rho }_{\ast }\le 1$ and $v\le \ell$ givesThese estimates hold for every nonempty pair forbidding the original intersection, with ${\rho }_0=(|\mathcal{A}|/2^n{)}^2$ in our application.

For $\ell =n/4$, the two positive exponential decay rates are, respectively,Both exceed $c=-2\log (1.999/2)<0.001001$. Hence ${\rho }_0\le e^{-cn}$ andThe empty set family is immediate. This proves the numerical bound directly from Harper inequality and the supplied binary entropy function estimate.

For $\ell =\lfloor n/8\rfloor$, the identical forbidden-intersection density increment applies. For $n\ge 17$, use $\ell \ge n/8-7/8$ in the first stopping estimate. Its decay rate is at leastFor the second stopping estimate, $\ell \le n/8$ gives the still stronger rate $g-(2g+h)/8>0.01222$. At $n=16$ the floor is exact and the first rate is $g/8-D^2/4>0.002058>c$.

It remains to handle $1\le n\le 15$ without silently absorbing a constant into an exponential. Then $\ell$ is zero or one. Fix an $\ell$-set $T$. The $2^{n-\ell }$ sets containing $T$ are paired as $T\cup S$ and $T\cup ([n]\setminus (T\cup S))$, and each pair intersects in exactly $T$. Since $n>\ell$, these are distinct pairs, and at most one member of each belongs to $\mathcal{A}$. ThereforeFor $n=0$, the forbidden self-intersection is zero, so the set family is empty. We have proved the claimed base for all dimensions in the second case as well.

Finally, the same base is not forced by a forbidden intersection near $2n/3$. Take $n=2000$ and every subset of size at most $1000$. Every intersection has size at most $1000$, below $\lfloor 2n/3\rfloor =1333$, including self-intersections. By pairing complementary sets this set family has more than $2^{1999}$ members, whereassince $2000\log (0.9995)<-1<-\log 2$. Thus it is a large family avoiding a high intersection, and it contradicts the proposed bound.

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.