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Scope. This elementary auxiliary argument supplies uniformity in the expanded Proposition 7.2 when the size of its containing set approaches the whole ambient set. It is not an extra numbered source theorem.

Statement. Give $\Omega([n];k)$ uniform probability $\nu$, and let $s\ge0$. If every $A\in\mathcal A$, $B\in\mathcal B$ satisfies $|A\setminus B|\ge s$, then

$$\nu(\mathcal A)\nu(\mathcal B)\le e^{-s^2/n}.$$

Proof. Generate a uniform random permutation and let $S$ be its first $k$ entries. A function $f(S)$ which changes by at most one under an exchange of one selected and one unselected element has a reveal martingale with conditional increment range at most one. Indeed, given any revealed prefix, completions after two possible next entries $x,y$ are in bijection by interchanging $x,y$ in the unrevealed positions. The resulting selected sets are equal or differ by one exchange, so conditional expectations differ by at most one. Revealing all $n$ positions and applying the exponential-moment calculation proved in product_measure_separation gives, for every $u\ge0$, the two one-sided bounds

$$\Pr(f-\mathbb Ef\ge u)\le e^{-2u^2/n},\qquad \Pr(f-\mathbb Ef\le-u)\le e^{-2u^2/n}.$$

Take $f(S)=\min_{A\in\mathcal A}|S\setminus A|$, the distance in the exchange graph. It is one-Lipschitz. It vanishes on $\mathcal A$ and is at least $s$ on $\mathcal B$. The same two-tail multiplication as in the product-measure proof gives $e^{-s^2/n}$. Empty families and the one-point layers $k=0,n$ satisfy the statement directly. $\square$

Dependencies. product_measure_separation.