---

Source. Published pp. 277–278, Proposition 7.2 (PDF).

Statement used in the full chain. Given $\zeta,\rho,\tau>0$, there is $\epsilon>0$ such that the following holds uniformly for integers $k,h\ge\zeta n$ with $N=k+h\le n$. If $\mathcal A\subseteq\Omega([n];k)$, $\mathcal B\subseteq\Omega([n];h)$ and

$$\frac{|\mathcal A||\mathcal B|}{\binom nk\binom nh} \ge e^{-\epsilon n},$$

there are at least $\binom nN e^{-\rho n}$ sets $C$ of size $N$ for which

$$|\mathcal A\cap2^C|\ge\binom Nk e^{-\tau n},\qquad |\mathcal B\cap2^C|\ge\binom Nh e^{-\tau n}. \tag{1}$$

The input tolerance and the two output tolerances are independent. This is the form needed for every subsequent reduction.

Proof. Choose $0<\sigma<\zeta/4$ small in terms of $\zeta,\rho,\tau$, put $r=\lfloor\sigma n\rfloor$ and $s=N-r$, and initially take $n$ large enough that $r>0$. The containment incidence graph between $k$-sets and $s$-sets is regular; its degree at an $s$-set is $\binom sk$. Lemma 4.1 gives a family $\mathcal F$ of at least $\binom ns e^{-\epsilon n}/2$ sets, each containing at least $\binom sk e^{-\epsilon n}/2$ members of $\mathcal A$. Similarly define $\mathcal G$ using the $h$-sets of $\mathcal B$.

At least half of $\mathcal F$ has a member of $\mathcal G$ whose union with it has size at most $N$. Indeed, if the bad half existed, every such pair of $s$-sets would have exchange distance greater than $r$. The proved slice bound would bound their density product by $e^{-r^2/n}$, whereas the two densities give at least $e^{-2\epsilon n}/8$. Choosing $\epsilon<\sigma^2/16$ contradicts this for large $n$.

For each good $F$, fix a set $C$ of size $N$ containing it and one such $G$. A fixed $C$ receives at most $\binom Nr$ choices of $F$. Thus the number of distinct $C$ is at least

$$\frac{\binom ns}{4\binom Nr}e^{-\epsilon n} \ge\binom nN e^{-o_\sigma(n)-\epsilon n-O(\log(n+1))}. \tag{2}$$

Its two internal family sizes are at least $\binom sk e^{-\epsilon n}/2$ and $\binom sh e^{-\epsilon n}/2$. The uniform entropy estimates compare these with $\binom Nk$ and $\binom Nh$, losing only $e^{-o_\sigma(n)-O(\log(n+1))}$. First choose $\sigma$ so the losses in (1)–(2) fit the prescribed $\rho,\tau$, then $\epsilon$ smaller, and then $n$ large. This proves the conclusion uniformly even as $N/n\to1$. Finitely many smaller $n$ are handled by making the input tolerance force full families, as in Theorem 6.1. $\square$

Source precision. The printed statement uses the same $\epsilon$ in its hypothesis and its final fiber lower bounds. Its proof obtains $\binom{N-r}k(1-\epsilon)^n/2$, not $\binom Nk(1-\epsilon)^n/2$; replacing that binomial coefficient incurs an additional exponential loss. The two-tolerance form above records and absorbs this loss, and suffices for Theorem 1.14. The exact printed same-tolerance assertion is not certified here. For fixed $N/n<1$, the source's Corollary 1.6 supplies the close-pair step. The proved slice bound supplies uniformity at the endpoint $N/n\to1$ without assuming an unproved uniform intersection buffer.

Dependencies. lemma_4_1, slice_separation, entropy_estimates.