After we finished proving the EGT last time, to show Corollary 3.2 of Paper S3, it remains to check the inequality (8) for most l-sets \( Y_i \) in the same proof scenario: because of the constraints on \( |{\mathcal M} | \) and \( |{\mathcal D} | \), it’s impossible to have more than \( 2 {n \choose l} \gamma^{-1/3} \) sets \( Y_i \in {X \choose l} \) violating (8). If there were, the total contribution of the wrong \( Y_i \) to \( |{\mathcal D} | \) would be too big (or too small), so no matter how even the marks are distributed in the remaining \( Y_i \), it would violate the conclusion \( |{\mathcal D} | < B \) of the double mark lemma.
Lemma 2.3 and the following paragraphs argue this succinctly. They’re for Theorem 2.1 on p. 5 meeting broader needs with a given sufficiently small \( \epsilon>0 \). For Cor. 3.2, it’s $$ \epsilon = \frac{2}{\sqrt[3] \gamma}. $$ With this considered on p. 8, it says:
Here \( \lfloor x \rfloor \) for \( x \in \mathbb{R} \) is the floor function returning the largest integer that doesn’t exceed x. So \( \lfloor x \rfloor > x- 1 \) holds always.
Now on this setting, it’s enough to check that $$ (11) \qquad |{\mathcal D}|< t'{n \choose l}{l \choose m}^2, \quad \textrm{where} \quad t’ = 1+\frac{u(v-1)^2}{1-u} = 1+\frac{u(v’-1)^2}{1-u}, $$ $$ \qquad \Rightarrow \qquad \qquad v’ {l \choose m} e^{-\kappa({\mathcal F})} < \left| {\mathcal F} \cap {Y \choose m} \right| < v {l \choose m} e^{-\kappa({\mathcal F})}, $$ for some \( (1-2u) {n \choose l} \) sets \( Y \in {X \choose l} \): it suffices because \( |{\mathcal D}| < t < t’ \) from the double mark lemma and the figure above, and also $$ 2u \le \epsilon, \quad v \le 1 + \sqrt{\frac{2}{\epsilon \gamma}} = 1+\frac{1}{\sqrt[3] \gamma},$$ $$ v’ \ge 1 – \sqrt{\frac{2}{\epsilon \gamma}} = 1-\frac{1}{\sqrt[3] \gamma} \quad \textrm{and} \quad e^{-\kappa({\mathcal F})} = \frac{|{\mathcal F}|}{{n \choose m}}. $$ So (11) implies (8) from the last time.
Lemma 2.3 on p.7-8 proves this to confirm the corollary.
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