Having finished our proof of the conversion lemma last time, we are done with the first three tasks of the four listed in the 12/11/23 post. Also recalling our proof scenario of Proposition 2.1 discussed in the 11/27/23 post, it remains to carry out the last task 4 correctly before completing all of the proof: given \( \Gamma(2b) \) of \({\mathcal F} \), we are trying to show that most \( {\mathcal F}_Y \) meet the density requirement, and the exported collective \( \Gamma \)-condition. By this driving force, we can show Proposition 3 given in the 11/13/23 post, which implies \( \Gamma( m^{\frac{1}{2} +\epsilon} ) \) of \( {\mathcal F} \) \( \Rightarrow \) three mutually disjoint sets in it.
The two key properties are shown by the three steps described on p. 8-10 of Paper S4. Step 1 checks the density requirement, and as the 11/27/23 post figured out, it’s our familiar application of the weighted EGT. We just recall here that the EGT is applied to \( {\mathcal H}={X_* \choose m_*} \) for the given \( n_* \)-set \( X_* \), then most \( {\mathcal F}_Y = {\mathcal F} \cap {X – X_* \cup Y \choose m} \) meet the density requirement.
For the second property, we define the following weight with the given sufficiently large constant g:
This is our outcome for the fourth task. In the next two posts, we’ll check the truth of the \( \Gamma_g(b_*, h) \)-condition, and that the \(g^{th} \) mark lemma with it implies the desired collective \( \Gamma\)-condition as in Step 3.
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